Jekyll2021-12-07T17:01:18+05:30/feed.xmlTotal Internal ReflectionTechnology and ArtFunctional Analysis Exercises 11 : Compactness and Finite Dimension2021-12-02T00:00:00+05:302021-12-02T00:00:00+05:30/2021/12/02/functional-analysis-exercises-11-compactness-and-finite-dimension<p>This post lists solutions to the exercises in the <strong>Compactness and Finite Dimension section 2.5</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h4 id="251-show-that-mathbbrn-and-mathbbcn-are-not-compact">2.5.1. Show that $$\mathbb{R}^n$$ and $$\mathbb{C}^n$$ are not compact.</h4> <p><strong>Proof:</strong></p> <p>$$\mathbb{R}^n$$ has a sequence $$x_n=1,2,3,4, \cdots$$ which does not have a convergent subsequence, since every $$d(x_m,x_n) \geq 1$$ for all $$m,n \in \mathbb{N}$$. The sequence exists in $$\mathbb{C}^n$$, implying that every sequence in that space does not have a convergent subsequence.</p> <p>Thus $$\mathbb{R}^n$$ and $$\mathbb{C}^n$$ are not compact.</p> $\blacksquare$ <hr /> <h4 id="252-show-that-a-discrete-metric-space-x-cf-11-8-consisting-of-infinitely-many-points-is-not-compact">2.5.2. Show that a discrete metric space $$X$$ (cf. 1.1-8) consisting of infinitely many points is not compact.</h4> <p><strong>Proof:</strong></p> <p>The discrete metric space $$X$$ has the distance metric as:</p> $d(x,y)=\begin{cases} 0 &amp; \text{if } x=y \\ 1 &amp; \text{if } x \neq y \end{cases}$ <p>Assume any sequence $$(x_n) \subset X$$ which does not repeat its elements. Then $$d(x_m,x_n)=1$$ for all $$m,n: m \neq n, m,n \in \mathbb{N}$$. Thus, this sequence has no convergent subsequence. Thus the discrete metric space $$X$$ is not compact.</p> $\blacksquare$ <hr /> <h4 id="253-give-examples-of-compact-and-noncompact-curves-in-the-plane-mathbbr2">2.5.3. Give examples of compact and noncompact curves in the plane $$\mathbb{R}^2$$.</h4> <p><strong>Answer:</strong></p> <p>$$y=x$$ is not compact because it is not bounded.<br /> $$y=\sin x, x \in [0, 2\pi]$$ is compact because $$[0,2\pi]$$ is compact and sine is a continuous function, and we know that continuous functions map compact sets to compact sets.</p> <hr /> <h4 id="254--show-that-for-an-infinite-subset-m-in-ihe-space-s-cf-22-8-to-be-compact-it-is-necessary-that-there-arc-numhers-gamma_1-gamma_2-cdots-such-that-for-all-xxi_kx-in-m-we-have-vert-xi_kx-vert-leq-gamma_k-it-can-he-shown-that-the-condition-is-also-sufficient-for-the-compactness-of-m">2.5.4. Show that for an infinite subset $$M$$ in Ihe space $$s$$ (cf. 2.2-8) to be compact, it is necessary that there arc numhers $$\gamma_1, \gamma_2, \cdots$$ such that for all $$x=(\xi_k(x)) \in M$$ we have $$\vert \xi_k(x) \vert \leq \gamma_k$$. (It can he shown that the condition is also sufficient for the compactness of $$M$$.)</h4> <p><strong>Proof:</strong></p> <p>The metric of $$s$$ is defined by:</p> $d(x,y) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \frac{|\xi_j-\eta_j|}{1+|\xi_j-\eta_j|}$ <p>$$\xi_k(x)$$ extracts the $$k$$-th element of the sequence $$x$$.</p> <p>Assume a sequence $$(p_n) \subset M$$, like so:</p> $p_1 = x^1_1, x^1_2, x^1_3, \cdots, x^1_k, \cdots \\ p_2 = x^2_1, x^2_2, x^2_3, \cdots, x^2_k, \cdots \\ \vdots \\ p_m = x^m_1, x^m_2, x^m_3, \cdots, x^m_k, \cdots \\ \vdots$ <p>For fixed $$k$$, we have $$\vert \xi_k(p_m) \vert &lt; \gamma_k$$. Thus $$x^m_k$$ is bounded for fixed $$k$$.</p> <p>Set $$k=1$$, then $$x^1_1, x^2_1, x^3_1, \cdots$$ is bounded. Thus, by the <strong>Bolzano-Weierstrauss Theorem</strong>, this contains a convergent subsequence. Let this convergent subsequence converge to $$x_1$$. Let $$P_1$$ be the set of sequences which contain these subsequence entries. We can repeat the same exercise for $$k=2$$ and apply it to $$P_1$$ to get $$P_2$$, etc.</p> <p>We finally get a subsequence of $$(p_n)$$ (call it $$p_{n_j}$$) where, for any given $$k$$, we have a convergent sequence $$(x^{n_j}_k)$$ converging to $$x_k$$. We construct a sequence out of these limits $$p=x_1, x_2, \cdots$$. For any $$p_m$$, we have:</p> $d(p_m, p) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \frac{|x^m_j-x_j|}{1+|x^m_j-x_j|}=\sum\limits_{j=1}^\infty \frac{1}{2^j} \left( 1 - \frac{1}{1+|x^m_j-x_j|} \right)$ <p>Take $$\epsilon &gt; 0$$. Assume $$N_1 \in \mathbb{N}$$ such that $$\vert x^m_1-x_1 \vert$$, similarly for $$N_2$$, and so on. Now take $$N=\max{(N_1, N_2, \cdots)}$$. Then for $$m&gt;N$$, we have:</p> $d(p_m, p) = \sum\limits_{j=1}^\infty \frac{1}{2^j} \left( 1 - \frac{1}{1+\epsilon} \right) \\ \lim_{\epsilon \rightarrow 0} d(p_m, p) = 0$ <p>Thus, $$p$$ is the limit of the constructed subsequence $$p_{n_j}$$. Thus $$(p_n)$$ has a convergent subsequence. Since $$(p_n)$$ was arbitrary, $$M$$ is compact.</p> $\blacksquare$ <hr /> <h4 id="255--local-compactness-a-metric-space-x-is-said-to-be-locally-compact-if-every-point-of-x-has-a-compact-neighborhood-show-that-mathbbr-and-mathbbc-and-more-generally-mathbbrn-and-mathbbcn-are-locally-compact">2.5.5. (Local compactness) A metric space $$X$$ is said to be locally compact if every point of $$X$$ has a compact neighborhood. Show that $$\mathbb{R}$$ and $$\mathbb{C}$$ and, more generally, $$\mathbb{R}^n$$ and $$\mathbb{C}^n$$ are locally compact.</h4> <p><strong>Proof:</strong></p> <p>Take $$x \in \mathbb{R}$$. Then, for any $$\epsilon&gt;0$$, the set $$[x-\epsilon, x+\epsilon]$$ is closed and bounded and contains the $$\epsilon$$-neighbourhood, and is thus a compact neighbourhood of $$x$$. Thus $$\mathbb{R}$$ is locally compact.</p> <p>Take $$x \in \mathbb{R}^n$$. For any $$\epsilon&gt;0$$, the closed ball $$\bar{B}(x, \epsilon)$$ contains the open $$\epsilon$$-neighbourhood $$B(x,\epsilon)$$ and is thus a compact neighbourhood of $$x$$. Thus $$\mathbb{R}^n$$ is locally compact.</p> $\blacksquare$ <hr /> <h4 id="256-show-that-a-compact-metric-space-x-is-locally-compact">2.5.6. Show that a compact metric space $$X$$ is locally compact.</h4> <p><strong>Proof:</strong></p> <p>The neighbourhood of a point $$x$$ is defined as a set which contains an open ball centered on $$x$$ (or open set containing $$x$$).</p> <p>Since any open ball around $$x \in X$$ is also contained in the set $$X$$, $$X$$ is a neighbourhood of $$x$$. Since $$X$$ is compact, $$x$$ has a compact neighbourhood, implying that $$X$$ is locally compact.</p> $\blacksquare$ <hr /> <h4 id="257-if-dim-y--infty-in-rieszs-lemma-25-4-show-that-one-can-even-choose-theta--1">2.5.7. If $$\dim Y &lt; \infty$$ in Riesz’s lemma 2.5-4, show that one can even choose $$\theta = 1$$.</h4> <p><strong>Proof:</strong></p> <p><strong>(TODO)</strong></p> $\blacksquare$ <hr /> <h4 id="258-in-prob-7-sec-24-show-directly-without-using-24-5-that-there-is-an-a--0-such-that-a-x_2-leq-x-use-25-7">2.5.8. In Prob. 7, Sec. 2.4, show directly (without using 2.4-5) that there is an $$a &gt; 0$$ such that $$a {\|x\|}_2 \leq \|x\|$$. (Use 2.5-7.)</h4> <p><strong>Proof:</strong></p> <p>Norms are continuous. If $$f(x)=\frac{\|x\|}{ {\|x\|}_2}$$ is also continuous, assuming that $${\|x\|}_2 \neq 0$$. Assume a compact subset $$M \subset X$$; then $$f(M)$$ is also compact, and thus contains a maximum and a minimum value. Specifically, it contains a infimum, call it $$a$$. Then we have:</p> $f(x) = \frac{\|x\|}{ {\|x\|}_2} \geq a \\ {\|x\|} \geq a {\|x\|}_2$ <p>If $${\|x\|}_2 =0$$, then obviously $${\|x\|} \geq a {\|x\|}_2 = 0$$.</p> $\blacksquare$ <hr /> <h4 id="259-if-x-is-a-compact-metric-space-and-m-subset-x-is-closed-show-that-m-is-compact">2.5.9. If $$X$$ is a compact metric space and $$M \subset X$$ is closed, show that $$M$$ is compact.</h4> <p><strong>Proof:</strong></p> <p>Every sequence in $$X$$ has a convergent subsequence.</p> <p>Choose an arbitrary sequence $$(x_n) \subset M$$. Since $$(x_n) \subset X$$, it has a convergent subsequence $$(x_{n_k}$$, therefore this subsequence has a limit point $$x$$.</p> <p>We note that $$(x_{n_k}) \subset M$$.</p> <p>Since $$M$$ is closed, it contains all its limit points, thus $$x \in M$$. Thus $$M$$ contains the limit of this subsequence as well. Thus the sequence $$(x_n)$$ has a convergent subsequence in $$M$$. Since $$(x_n)$$ is arbitrary, all sequences in $$M$$ have convergent subsequences. Thus, $$M$$ is compact.</p> $\blacksquare$ <hr /> <h4 id="2510-let-x-and-y-be-metric-spaces-x-compact-and-t-x-rightarrow-y-bijective-and-continuous-show-that-t-is-a-homeomorphism-cf-prob-5-sec-16">2.5.10. Let $$X$$ and $$Y$$ be metric spaces, $$X$$ compact, and $$T: X \rightarrow Y$$ bijective and continuous. Show that $$T$$ is a homeomorphism (cf. Prob. 5, Sec. 1.6).</h4> <p><strong>Proof:</strong></p> <p>For $$T$$ to be a homeomorphism, its inverse should be continuous.</p> <p>$$X$$ is compact and $$T$$ is continuous, thus $$T(X)$$ is compact. Choose any sequence $$(x_n) \subset X$$. Since $$X$$ is compact, $$(x_n)$$ contains a convergent subsequence $$(x_{n_k})$$ which converges to $$x$$. Since continuous functions map convergent sequences to convergent sequences, $$(y_{n_k}) = T((x_{n_k}))$$ is a convergent sequence.</p> <p>Now we define $$T^{-1}$$ as $$T^{-1}: (y_{n_k}) \mapsto (x_{n_k})$$. We know that if a function is continuous if and only if it maps convergent sequences to convergent sequences.</p> <p>Since both $$(x_{n_k})$$ and $$(y_{n_k})$$ are convergent sequences, $$T^{-1}$$ is continuous at $$x$$.</p> <p>Since $$(x_n)$$ was arbitrarily chosen, $$T$$ is a homeomorphism.</p> $\blacksquare$avishekThis post lists solutions to the exercises in the Compactness and Finite Dimension section 2.5 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Functional Analysis Exercises 10 : Finite Dimensional Normed Spaces and Subspaces2021-11-22T00:00:00+05:302021-11-22T00:00:00+05:30/2021/11/22/functional-analysis-exercises-10-finite-dimensional-normed-spaces<p>This post lists solutions to the exercises in the <strong>Finite Dimensional Normed Spaces and Subspaces section 2.4</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h3 id="notes">Notes</h3> <p>The requirements for a space to be a normed space are:</p> <ul> <li><strong>(N1)</strong> <strong>Nonnegativity</strong>, i.e., $$\|x\| \geq 0, x \in X$$</li> <li><strong>(N2)</strong> <strong>Zero norm</strong> implies <strong>zero vector</strong> and vice versa, i.e., $$\|x\|=0 \Leftrightarrow x=0, x \in X$$</li> <li><strong>(N3)</strong> <strong>Linearity</strong> with respect to <strong>scalar multiplication</strong>, i.e., $$\|\alpha x\|=\vert \alpha \vert \|x\|, x \in X, \alpha \in \mathbb{R}$$</li> <li><strong>(N4)</strong> <strong>Triangle Inequality</strong>, i.e., $$\|x+y\| \leq \|x\| + \|y\|, x,y \in X$$</li> </ul> <p>The requirements for a space to be a vector space are:</p> <p>(Mnemonic: <strong>ADD1 SIIA</strong>)</p> <h4 id="addition">Addition</h4> <ul> <li><strong>(V1)</strong> <strong>Symmetry</strong> implies $$x+y=y+x, x,y \in X$$.</li> <li><strong>(V2)</strong> <strong>Identity</strong> implies <strong>zero vector</strong>, i.e., $$x+\theta=x, x,\theta \in X$$.</li> <li><strong>(V4)</strong> <strong>Inverse</strong> implies $$x+(-x)=\theta, x,y,\theta \in X$$.</li> <li><strong>(V3)</strong> <strong>Associativity</strong> implies $$(x+y)+z=x+(y+z), x,y,z \in X$$.</li> </ul> <h4 id="multiplication">Multiplication</h4> <ul> <li><strong>(V1)</strong> <strong>Associativity</strong> implies $$x(yz)=(xy)z, x,y,z \in X$$.</li> <li><strong>(V2)</strong> <strong>Distributivity with respect to vector addition</strong> implies $$\alpha(x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$.</li> <li><strong>(V3)</strong> <strong>Distributivity with respect to scalar addition</strong> implies $$(\alpha + \beta) x = \alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$.</li> <li><strong>(V4)</strong> <strong>Identity</strong> implies, $$1x=x, x \in X$$.</li> </ul> <h4 id="241-give-examples-of-subspaces-of-linfty-and-l2-which-are-not-closed">2.4.1. Give examples of subspaces of $$l^\infty$$ and $$l^2$$ which are not closed.</h4> <p><strong>Answer:</strong></p> <p>$$l^\infty$$ is the space of all bounded sequences, i.e., $$\sum\limits_{i=1}^\infty \vert x_i \vert &lt; \infty$$. The norm it is equipped with is $$\|(x_n)\|=\sup \vert x_i\vert$$.</p> <p>The space of sequences with finitely many non-zero elements is an example of a subspace which is not closed.</p> <p>In particular, the sequence defined as:</p> $(x_j^n)=\begin{cases} \displaystyle\frac{1}{2^n} &amp; \text{if } j \leq n \\ 0 &amp; \text{if } j &gt; n \end{cases}$ <p>Assume $$m&lt;n$$. Then, we have:</p> $\alpha x^{(m)} + \beta x^{(n)} \\ = \alpha (\frac{1}{2} , \frac{1}{2^2} , \cdots + \frac{1}{2^m}, , 0 , 0, \cdots) , \beta (\frac{1}{2} , \frac{1}{2^2} , \cdots , \frac{1}{2^m} , \frac{1}{2^{m+1}} , \cdots , \frac{1}{2^n} , 0 , 0, \cdots) \\ = (\alpha + \beta) \frac{1}{2} , (\alpha + \beta) \frac{1}{2^2} , \cdots , (\alpha + \beta) \frac{1}{2^m} , \frac{\beta}{2^{m+1}} , \cdots , \frac{\beta}{2^n}, 0 , 0, \cdots$ <p>For the $$l^\infty$$ case, we have the norm as:</p> ${\|x-x^{(n)}\|}_\infty=\sup |x_j-x_j^{(n)}|=\frac{1}{2^{n+1}}$ <p>where $$x=\frac{1}{2} , \frac{1}{2^2} , \cdots$$.</p> <p>As $${\|x-x^{(n)}\|}+\infty \rightarrow 0$$ as $$n \rightarrow 0$$, $$x$$ is a limit of $$x^{(n)}$$. Thus, the limit exists, but it is not in this space, since $$x$$ has infinitely many terms.</p> <p>In the case of $$l^2$$, the norm $${\|\bullet\|}_2$$ the partial sum $$s_n=\displaystyle\frac{1}{3}(1-\frac{1}{4^n})$$ (see Rough Work at the end to see how this was calculated).</p> <p>Then $${\|x-x^{(n)}\|}_2=\displaystyle\frac{1}{3\cdot 4^n} \rightarrow 0$$ as $$n \rightarrow \infty$$ where $$x=\frac{1}{4} , \frac{1}{4^2} , \cdots$$. Thus, the limit exists, but it is not in this space, since $$x$$ again has infinitely many terms.</p> <hr /> <h4 id="242-what-is-the-largest-possible-c-in-1-if-x--mathbbr2-and-x_1--10-x_2--01-if-x--mathbbr3-and-x_1--100-x_2--010-x_3--001">2.4.2. What is the largest possible $$c$$ in (1) if $$X = \mathbb{R}^2$$ and $$x_1 = (1,0), x_2 = (0,1)$$? If $$X = \mathbb{R}^3$$ and $$x_1 = (1,0,0), x_2 = (0,1,0), x_3 = (0,0,1)$$?</h4> <p><strong>Answer:</strong></p> <p>We have the identity:</p> $\|\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_1 e_n\| \geq c(|\alpha_1| + |\alpha_2| + \cdots + |\alpha_n|)$ <p>For $$\mathbb{R}^2$$, we have:</p> $\|\alpha_1 e_1 + \alpha_2 e_2\| \geq c(|\alpha_1| + |\alpha_2|) \\ {(\alpha_1^2 + \alpha_2^2)}^{1/2} \geq c(|\alpha_1| + |\alpha_2|)$ <p>If $$\alpha_1=\alpha_2=0$$ gives us an equality above, let us pick an arbitrary small $$\alpha_1=\alpha_2=\epsilon$$, so that we get:</p> ${(\epsilon^2 + \epsilon^2)}^{1/2} \geq c(\epsilon + \epsilon) \\ 2 \epsilon c \leq \sqrt 2 \epsilon \\ c \leq \frac{1}{\sqrt 2}$ <p>For $$\mathbb{R}^3$$, we have:</p> $\|\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3\| \geq c(|\alpha_1| + |\alpha_2| + |\alpha_3|) \\ {(\alpha_1^2 + \alpha_2^2 + \alpha_3^2)}^{1/2} \geq c(|\alpha_1| + |\alpha_2| + |\alpha_3|)$ <p>If $$\alpha_1=\alpha_2=\alpha_3=0$$ gives us an equality above, let us pick an arbitrary small $$\alpha_1=\alpha_2=\alpha_3=\epsilon$$, so that we get:</p> ${(\epsilon^2 + \epsilon^2 + \epsilon^3)}^{1/2} \geq c(\epsilon + \epsilon + \epsilon) \\ 3 \epsilon c \leq \sqrt 3 \epsilon \\ c \leq \frac{1}{\sqrt 3}$ <hr /> <h4 id="243-show-that-in-def-24-4-the-axioms-of-an-equivalence-relation-hold-cf-a14-in-appendix-1">2.4.3. Show that in Def. 2.4-4 the axioms of an equivalence relation hold (cf. A1.4 in Appendix 1).</h4> <p><strong>Proof:</strong></p> <p>The relation to demonstrate is an equivalence is the following:</p> $a{\|x\|}_2 \leq {\|x\|}_1 \leq b{\|x\|}_2$ <p><strong>Reflexive</strong>: This is evident since:</p> ${\|x\|}_1 \leq {\|x\|}_1 \leq b{\|x\|}_1$ <p>where $$a=b=1$$.</p> <p><strong>Symmetric</strong>:</p> <p>We have:</p> ${\|x\|}_1 \leq b{\|x\|}_2 \\ (1/b){\|x\|}_1 \leq {\|x\|}_2 \\ {\|x\|}_2 \geq (1/b){\|x\|}_1$ $a{\|x\|}_2 \leq {\|x\|}_1 \\ {\|x\|}_2 \leq (1/a){\|x\|}_1$ <p>Thus, we get:</p> $\frac{1}{b}{\|x\|}_1 \leq {\|x\|}_2 \leq \frac{1}{a}{\|x\|}_1$ <p><strong>Transitive</strong>: Assume that:</p> $a_1{\|x\|}_2 \leq {\|x\|}_1 \leq b_1{\|x\|}_2 \\ a_2{\|x\|}_3 \leq {\|x\|}_2 \leq b_2{\|x\|}_3$ <p>We have the following two identities:</p> ${\|x\|}_1 \leq b_1{\|x\|}_2 \\ (1/b_1){\|x\|}_1 \leq {\|x\|}_2 \\ (1/b_1){\|x\|}_1 \leq {\|x\|}_2 \leq b_2{\|x\|}_3 \\ {\|x\|}_1 \leq b_1 b_2{\|x\|}_3$ $a_1{\|x\|}_2 \leq {\|x\|}_1 \\ {\|x\|}_2 \leq (1/a_1){\|x\|}_1 \\ a_2{\|x\|}_3 \leq {\|x\|}_2 \leq (1/a_1){\|x\|}_1 \\ a_1 a_2{\|x\|}_3 \leq {\|x\|}_1 \\$ <p>Putting the two inequalities above together, we get:</p> $(a_1 a_2){\|x\|}_3 \leq {\|x\|}_1 \leq (b_1 b_2){\|x\|}_3$ $\blacksquare$ <hr /> <h4 id="244-show-that-equivalent-norms-on-a-vector-space-x-induce-the-same-topology-for-x">2.4.4. Show that equivalent norms on a vector space $$X$$ induce the same topology for $$X$$.</h4> <p><strong>Proof:</strong></p> <p>To prove that $${\|\bullet\|}_1$$ and $${\|\bullet\|}_2$$ are equivalent norms, we need to show that for every open ball $$B_{1}$$, there is an open ball $$B_{2}$$ contained within it, and vice versa.</p> <p>The equivalent norms are related as:</p> $a{\|x\|}_2 \leq {\|x\|}_1 \leq b{\|x\|}_2$ <p>From the above relation, we know that $$a \leq b$$.</p> <p>Pick an open ball $$B_1(x_0,r)$$. Let $$x \in B_1(x_0,r)$$. Then $$d_1(x_0,x) &lt; r$$. We then have:</p> $a \cdot d_2(x_0,x) \leq d_1(x_0,x) &lt; r \\ d_2(x_0,x) &lt; \frac{r}{a} \\ \Rightarrow x \in B_2(x_0,\frac{r}{a}) \\ \Rightarrow B_1(x_0,r) \in B_2(x_0,\frac{r}{a})$ <p>Conversely, pick an open ball $$B_2(x_0,r)$$. Let $$x \in B_2(x_0,r)$$. Then $$d_2(x_0,x) &lt; r$$. We then have:</p> $d_1(x_0,x) \leq b \cdot d_2(x_0,x) &lt; br \\ d_1(x_0,x) &lt; br \\ \Rightarrow x \in B_1(x_0,br) \\ \Rightarrow B_2(x_0,r) \in B_1(x_0,br)$ $\blacksquare$ <hr /> <h4 id="245-if-bullet-and-bullet_0-are-equivalent-norms-on-x-show-that-the-cauchy-sequences-in-x-bullet-and-xbullet_0-are-the-same">2.4.5. If $$\|\bullet\|$$ and $${\|\bullet\|}_0$$ are equivalent norms on X, show that the Cauchy sequences in $$(X, \|\bullet\|)$$ and $$(X,{\|\bullet\|}_0)$$ are the same.</h4> <p><strong>Proof:</strong></p> <p>Assume that the equivalence relation is:</p> $a{\|x\|}_0 \leq {\|x\|} \leq b{\|x\|}_0$ <p>Assume a Cauchy sequence $$(x_n)$$ in $$(X,{\|\bullet\|}_0)$$, we have that $$\forall \epsilon&gt;0, \exists N_0$$ such that $$d_0(x_m,x_n)&lt;\epsilon$$ for all $$m,n&gt;N_0$$.</p> <p>Then, we have:</p> $d(x_m,x_n) \leq b d_0(x_m,x_n) &lt; b \epsilon \\ d(x_m,x_n) &lt; b \epsilon$ <p>Thus $$(x_n)$$ is also a Cauchy sequence in $$(X,\|\bullet\|)$$.</p> <p>Assume a Cauchy sequence $$(x_n)$$ in $$(X,{\|\bullet\|})$$, we have that $$\forall \epsilon&gt;0, \exists N_0$$ such that $$d(x_m,x_n)&lt;\epsilon$$ for all $$m,n&gt;N_0$$.</p> <p>Then, we have:</p> $a d_0(x_m,x_n) \leq b d(x_m,x_n) &lt; \epsilon \\ d_0(x_m,x_n) &lt; \frac{\epsilon}{a}$ <p>Thus $$(x_n)$$ is also a Cauchy sequence in $$(X,{\|\bullet\|}_0)$$.</p> $\blacksquare$ <hr /> <h4 id="246-theorem-24-5-implies-that-bullet_2-and-bullet_infty-in-prob-8-sec-22-are-equivalent-give-a-direct-proof-of-this-fact">2.4.6. Theorem 2.4-5 implies that $${\|\bullet\|}_2$$ and $${\|\bullet\|}_\infty$$ in Prob. 8, Sec. 2.2, are equivalent. Give a direct proof of this fact.</h4> <p><strong>Proof:</strong></p> <p>The norms $${\|\bullet\|}_2$$ and $${\|\bullet\|}_\infty$$ are defined as:</p> ${\|x\|}_2={\left(\sum\limits_{i=1}^n {|x_i|}^2\right)}^{1/2} \\ {\|x\|}_\infty=\sup |x_i|$ <p>We have:</p> ${|\sup{x_i}|}^2 + {|\sup{x_i}|}^2 + \cdots + {|\sup{x_i}|}^2 \geq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ n{|\sup{x_i}|}^2 \geq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ \sqrt{n}|\sup{x_i}| \geq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2}$ <p>We also have:</p> ${|\sup{x_i}|}^2 \leq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ {|\sup{x_i}|} \leq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2}$ <p>Thid implies that:</p> ${|\sup{x_i}|} \leq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2} \leq \sqrt{n}|\sup{x_i}| \\ {\|\bullet\|}_\infty \leq {\|\bullet\|}_2 \leq \sqrt{n} {\|\bullet\|}_\infty$ $\blacksquare$ <hr /> <h4 id="247-let-bullet_2-be-as-in-prob-8-sec-22-and-let-bullet-be-any-norm-on-that-vector-space-call-it-x-show-directly-without-using-24-5-that-there-is-a-b0-such-that-x-leq-b-x_2-for-all-x">2.4.7. Let $${\|\bullet\|}_2$$ be as in Prob. 8, Sec. 2.2, and let $$\|\bullet\|$$ be any norm on that vector space, call it $$X$$. Show directly (without using 2.4-5) that there is a $$b&gt;0$$ such that $$\|x\| \leq b {\|x\|}_2$$ for all $$x$$.</h4> <p><strong>Proof:</strong></p> <p>We have the vector $$x=\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n$$. Then we have, by the <strong>Triangle Inequality</strong>:</p> $\|x\|=\|a_1 e_1 + a_2 e_2 + \cdots + a_n e_n\| \leq |a_1| \|e_1\| + |a_2| \|e_2\| + \cdots + |a_n| \|e_n\| \\ \leq \max(\|e_i\|)(|a_1| + |a_2| + \cdots + |a_n|)$ <p>But by the <strong>Cauchy-Schwarz Inequality</strong>, we have:</p> $\sum\limits_{i=1}^n |a_i| \leq {\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} {\left(\sum\limits_{i=1}^n 1\right)}^{1/2} = \sqrt{n} {\|x\|}_2$ <p>This implies that:</p> $\|x\| \leq \sqrt{n}\max(\|e_i\|){\|x\|}_2 \\ \|x\| \leq b {\|x\|}_2$ <p>where $$b=\sqrt{n}\max{\|e_i\|}$$.</p> $\blacksquare$ <hr /> <h4 id="248-show-that-the-norms-bullet_1-and-bullet_2-in-prob-8-sec-22-satisfy-frac1sqrtn-x_1-leq-x_2-leq-x_1">2.4.8. Show that the norms $${\|\bullet\|}_1$$ and $${\|\bullet\|}_2$$ in Prob. 8, Sec. 2.2, satisfy $$\frac{1}{\sqrt{n}} {\|x\|}_1 \leq {\|x\|}_2 \leq {\|x\|}_1$$.</h4> <p><strong>Proof:</strong></p> <p>The norms $${\|\bullet\|}_2$$ and $${\|\bullet\|}_\infty$$ are defined as:</p> ${\|x\|}_1=\sum\limits_{i=1}^n {|a_i|} \\ {\|x\|}_2={\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} \\$ <p>By the <strong>Cauchy-Schwarz Inequality</strong>, we have:</p> ${\|x\|}_1 = \sum\limits_{i=1}^n |a_i| \leq {\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} {\left(\sum\limits_{i=1}^n 1\right)}^{1/2} = \sqrt{n} {\|x\|}_2 \\ \Rightarrow {\|x\|}_1 \leq \sqrt{n} {\|x\|}_2 \\ \Rightarrow \displaystyle\frac{1}{\sqrt{}n}{\|x\|}_1 \leq {\|x\|}_2$ <p>By the <strong>Cauchy-Schwarz Inequality</strong>, we have:</p> ${ {\|x\|}_2}^2 = \sum\limits_{i=1}^n {|a_i|}^2 \\ \sum\limits_{i=1}^n |a_i||a_i| \leq \left(\sum\limits_{i=1}^n |a_i|\right)\left(\sum\limits_{i=1}^n |a_i|\right)={\left(\sum\limits_{i=1}^n |a_i|\right)}^2 \\ {\left(\sum\limits_{i=1}^n |a_i||a_i|\right)}^{1/2} \leq {\left(\sum\limits_{i=1}^n |a_i|\right)} \\ \Rightarrow {\|x\|}_2 \leq {\|x\|}_1$ <p>Putting the above inequalities together, we get:</p> $\displaystyle\frac{1}{\sqrt{}n}{\|x\|}_1 \leq {\|x\|}_2 \leq {\|x\|}_1$ $\blacksquare$ <hr /> <h4 id="249-if-two-norms-bullet-and-bullet_0-on-a-vector-space-x-are-equivalent-show-that-i-x_n---x-rightarrow-0-implies-ii-x_n---x_0-rightarrow-0-and-vice-versa-of-course">2.4.9. If two norms $$\|\bullet\|$$ and $${\|\bullet\|}_0$$ on a vector space $$X$$ are equivalent, show that (i) $$\|x_n - x\| \rightarrow 0$$ implies (ii) $${\|x_n - x\|}_0 \rightarrow 0$$ (and vice versa, of course).</h4> <p><strong>Proof:</strong></p> $a{\|x\|} \leq {\|x\|}_0 \leq b{\|x\|}$ <p>Let $$\|x_n-x\| \rightarrow 0$$. This implies that $$\|x_n-x\| &lt; \epsilon / b$$, for some $$\epsilon / b &gt; 0$$. By the equivalence relation, we then have:</p> ${\|x_n-x\|}_0 \leq b{\|x_n-x\|} &lt; b (\epsilon/b) = \epsilon \\ \Rightarrow {\|x_n-x\|}_0 \rightarrow 0$ <p>Let $${\|x_n-x\|}_0 \rightarrow 0$$. This implies that $${\|x_n-x\|}_0 &lt; a \epsilon$$, for some $$a \epsilon &gt; 0$$. By the equivalence relation, we then have:</p> $a {\|x_n-x\|} \leq {\|x_n-x\|}_0 &lt; (1/a) (a \epsilon) = \epsilon \\ \Rightarrow {\|x_n-x\|} \rightarrow 0$ $\blacksquare$ <hr /> <h4 id="2410-show-that-all-complex-m-times-n-matrices-a--alpha_jk-with-fixed-m-and-n-constitute-an-mn-dimensional-vector-space-z-show-that-all-norms-on-z-are-equivalent-what-would-be-the-analogues-of-bullet_1-bullet_2-and-bullet_infty-in-prob-8-sec-22-for-the-present-space-z">2.4.10. Show that all complex $$m \times n$$ matrices $$A = (\alpha_{jk})$$ with fixed $$m$$ and $$n$$ constitute an $$mn$$-dimensional vector space $$Z$$. Show that all norms on $$Z$$ are equivalent. What would be the analogues of $${\|\bullet\|}_1$$, $${\|\bullet\|}_2$$ and $${\|\bullet\|}_\infty$$ in Prob. 8, Sec. 2.2, for the present space $$Z$$?</h4> <p><strong>Proof:</strong></p> <p><strong>TODO:</strong></p> <ul> <li>Need to prove dimension of space is $$mn$$.</li> <li>Need to check if matrix norm should be taken as the norm of the vectorised form, if so, then this reduces to $$l^p$$ case.</li> </ul> $\blacksquare$avishekThis post lists solutions to the exercises in the Finite Dimensional Normed Spaces and Subspaces section 2.4 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Functional Analysis Exercises 9 : Further Properties of Normed Spaces2021-11-19T00:00:00+05:302021-11-19T00:00:00+05:30/2021/11/19/functional-analysis-exercises-9-further-properties-of-normed-spaces<p>This post lists solutions to the exercises in the <strong>Further Properties of Normed Spaces section 2.3</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h3 id="notes">Notes</h3> <p>The requirements for a space to be a normed space are:</p> <ul> <li><strong>(N1)</strong> <strong>Nonnegativity</strong>, i.e., $$\|x\| \geq 0, x \in X$$</li> <li><strong>(N2)</strong> <strong>Zero norm</strong> implies <strong>zero vector</strong> and vice versa, i.e., $$\|x\|=0 \Leftrightarrow x=0, x \in X$$</li> <li><strong>(N3)</strong> <strong>Linearity</strong> with respect to <strong>scalar multiplication</strong>, i.e., $$\|\alpha x\|=\vert \alpha \vert \|x\|, x \in X, \alpha \in \mathbb{R}$$</li> <li><strong>(N4)</strong> <strong>Triangle Inequality</strong>, i.e., $$\|x+y\| \leq \|x\| + \|y\|, x,y \in X$$</li> </ul> <h4 id="231-show-that-c-subset-linfty-is-a-vector-subspace-of-linfty-cf-15-3-and-so-is-c_0-the-space-of-all-sequences-of-scalars-converging-to-zero">2.3.1. Show that $$c \subset l^\infty$$ is a vector subspace of $$l^\infty$$ (cf. 1.5-3) and so is $$c_0$$, the space of all sequences of scalars converging to zero.</h4> <p><strong>Proof:</strong></p> <p>$$c \in l^\infty$$ is the space of all complex convergent sequences. Let $$x_n \rightarrow x$$ and $$y_n \rightarrow y$$, such that $$x_n,y_n \in \mathbb{C}$$.</p> <p>Then, $$\forall \epsilon&gt;0, \exists M, N \in \mathbb{N}$$, such that $$\|x_m-x\|&lt;\epsilon$$ and $$\|y_n-y\|&lt;\epsilon$$, for all $$m&gt;M, n&gt;N$$. Take $$N_0=\max(M_0,N_0)$$, so that $$\|x_n-x\|&lt;\epsilon$$ and $$\|y_n-y\|&lt;\epsilon$$, for all $$n&gt;N_0$$.</p> $\|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| = \|\alpha x_n - \alpha x + \beta y_n - \beta y)\| \\ \|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| \leq \|\alpha x_n - \alpha x\| + \|\beta y_n - \beta y\| \\ = |\alpha|\|x_n - x\| + |\beta|\|y_n - y\| &lt; \alpha \epsilon + \beta \epsilon = (\alpha + \beta) \epsilon$ <p>$$(\alpha + \beta) \epsilon$$ can be made as small as possible. In the limit $$n \rightarrow \infty$$, we thus get: $$\|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| \rightarrow \alpha x + \beta y \in c \subset l^\infty$$.</p> <p>Thus, $$c$$ is a vector subspace.</p> $\blacksquare$ <p>For $$c_0$$, we have $$x=y=0$$, thus $$\alpha x + \beta y=0$$, and thus $$\alpha x_n + \beta y_n \rightarrow 0$$ as $$n \rightarrow \infty$$.</p> <p>Thus, $$c_0$$ is a vector subspace.</p> $\blacksquare$ <hr /> <h4 id="232-show-that-c_0-in-prob-1-is-a-closed-subspace-of-linfty-so-that-c_0-is-complete-by-15-2-and-14-7">2.3.2. Show that $$c_0$$ in Prob. 1 is a <em>closed</em> subspace of $$l^\infty$$, so that $$c_0$$ is complete by 1.5-2 and 1.4-7.</h4> <p><strong>Proof:</strong></p> <p>Consider a Cauchy sequence $$(x_n) \in c_0$$. Then, $$\forall \epsilon&gt;0, \exists N \in \mathbb{N}$$, such that $$d(x_m-x_n)&lt;\epsilon$$, for all $$m,n&gt;N$$.</p> <p>This implies that:</p> $\supd(x_j^m, x_j^n)&lt;\epsilon \\ d(x_j^m, x_j^n)&lt;\epsilon$ <p>Thus, for a fixed $$j$$, the sequence of scalars $$x_j^1, x_j^2, \cdots$$ is Cauchy. Since scalars are real numbers, and $$\mathbb{R}$$ is complete, the sequence converges, say to $$x_j$$, that is, $$\forall \epsilon&gt;0, \exists M \in \mathbb{N}$$ such that $$\|x_j^m-x_j\|&lt;\epsilon$$ for all $$m&gt;M$$. This holds for every $$j$$, yielding a sequence $$x_1, x_2, x_3, \cdots$$.</p> <p>Of course, $$x_1^m, x_2^m, x_3^m, \cdots$$ is also Cauchy since it is a convergent sequence (converging to zero), thus we have $$\forall \epsilon&gt;0, \exists N \in \mathbb{N}$$ such that $$\|x_j^m-0\|&lt;\epsilon$$ for all $$j&gt;N$$ We have:</p> $\|x_j-0\| = \|x_j-x_j^m + x_j^m - 0\| \leq \|x_j-x_j^m\| + \|x_j^m - 0\| &lt; \epsilon + \epsilon = 2 \epsilon$ <p>Thus $$(x_j) \rightarrow 0$$, and $$(x_j) \in c_0$$. Since this holds for any arbitrary Cauchy sequence in $$c_0$$, it follows that $$c_0$$ contains all its limits, and is thus a closed subspace.</p> $\blacksquare$ <hr /> <h4 id="233-in-linfty-let-y-be-the-subset-of-all-sequences-with-only-finitely-many-nonzero-terms-show-that-y-is-a-subspace-of-linfty-but-not-a-closed-subspace">2.3.3. In $$l^\infty$$, let $$Y$$ be the subset of all sequences with only finitely many nonzero terms. Show that $$Y$$ is a subspace of $$l^\infty$$ but not a closed subspace.</h4> <p><strong>Proof:</strong></p> <p>Let $$Y$$ be the subset of all sequences with only finitely many nonzero terms.</p> <p>Let $$(x_n)=x_1, x_2, \cdots, x_m, 0, 0, \cdots$$ and let $$(y_n)=y_1, y_2, \cdots, y_n, 0, 0, \cdots$$. Without loss of generality, assume $$m&lt;n$$. It is clear that $$\delta(x_n)=\max(x_1, x_2, \cdots, x_m) &lt; \infty$$ and $$\delta(y_n)=\max(y_1, y_2, \cdots, y_n) &lt; \infty$$, thus $$(x_n), (y_n) \in YZ \subset l^\infty$$ Then, we have:</p> $\alpha(x_n) + \beta(y_n)=\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \cdots, \alpha x_m + \beta y_m, \beta y_{m+1}, \cdots, \beta y_n, 0, 0, \cdots \\ \Rightarrow \delta[\alpha(x_n) + \beta(y_n)]=\max(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \cdots, \alpha x_m + \beta y_m, \beta y_{m+1}, \cdots, \beta y_n) &lt; \infty \\ \Rightarrow \alpha(x_n) + \beta(y_n) \in Y \subset l^\infty$ <p>Thus $$Y$$ is a subspace of $$l^\infty$$.</p> <p>Let there be a Cauchy sequence in $$Y$$, where $$y_n=\begin{cases} 1/j &amp; \text{if } j \leq n \\ 0 &amp; \text{if } j &gt; n \end{cases}$$</p> <p>Assume $$m&lt;n$$. Then $$d(x_m,x_n)=\sup d(x_m^i, x_n^i)=\frac{1}{m+1}$$. Then as $$m \rightarrow \infty$$, $$\text {lim }_{m \rightarrow \infty} d(x_m,x_n) = 0$$, but this limit has more nonzero terms than any sequence in $$Y$$ and is thus not contained in $$Y$$. Thus, $$Y$$ is not complete.</p> $\blacksquare$ <hr /> <h4 id="234-continuity-of-vector-space-operations-show-that-in-a-normed-space-x-vector-addition-and-multiplication-by-scalars-are-continuous-operations-with-respect-to-the-norm-that-is-the-mappings-defined-by-xy-mapsto-xy-and-alpha-x-mapsto-alpha-x-are-continuous">2.3.4. (Continuity of vector space operations) Show that in a normed space $$X$$, vector addition and multiplication by scalars are continuous operations with respect to the norm; that is, the mappings defined by $$(x,y) \mapsto x+y$$ and $$(\alpha, x) \mapsto \alpha x$$ are continuous.</h4> <p><strong>Proof:</strong></p> <p>Assume a normed space, with $$\|x-x_0\|&lt;\epsilon$$ and $$\|y-y_0\|&lt;\epsilon$$</p> <p>Vector addition will be continuous if $$\forall \epsilon&gt;0, \exists \delta&gt;0$$, such that $$\|x-x_0\| &lt; \delta, \|y-y_0\| &lt; \delta \Rightarrow \|f(x,y)-f(x_0,y_0)\| &lt; \epsilon$$.</p> <p>Then, we have:</p> $\|x+y-(x_0+y_0)\|=\|(x-x_0)+(y-y_0)\| \leq \|x-x_0\| + \|y-y_0\| &lt; 2 \epsilon$ <p>$$2 \epsilon$$ can be made as small as needed, thus vector addition is continuous in normed space.</p> $\blacksquare$ <p>Scalar multiplication will be continuous if $$\forall \epsilon&gt;0, \exists \delta&gt;0$$, such that $$\|x-x_0\| &lt; \delta, \|\alpha - \alpha_0\| &lt; \delta \Rightarrow \|\alpha x - \alpha x_0\| &lt; \epsilon$$.</p> <p>We’d like to express $$\alpha x-\alpha_0 x_0$$ using some combination of $$(\alpha-\alpha_0)$$ and $$(y-y_0)$$. As a preliminary test, let’s see what terms fall out of the product $$(x-x_0)(\alpha-\alpha_0)$$.</p> <p>Then, we have:</p> $(\alpha x-\alpha_0)(x-x_0)=\alpha x + \alpha_0 x_0 - \alpha_0 x - \alpha x_0$ <p>Then we can write $$\alpha x - \alpha_0 x_0$$ as:</p> $\alpha x - \alpha_0 x_0 = \alpha x + \alpha_0 x_0 - \alpha_0 x - \alpha x_0 - \alpha_0 x_0 - \alpha_0 x_0 + \alpha_0 x + \alpha x_0 \\ = (\alpha - \alpha_0)(x-x_0) + \alpha_0 (x-x_0) + x_0(\alpha-\alpha_0)$ <p>Therefore, we can write:</p> $\|\alpha x - \alpha_0 x_0\|=\|(\alpha - \alpha_0)(x-x_0) + \alpha_0 (x-x_0) + x_0(\alpha-\alpha_0)\| \\ \|\alpha x - \alpha_0 x_0\| \leq \|\alpha - \alpha_0\|\|x-x_0\| + |\alpha_0| \|x-x_0\| + |x_0|\|\alpha-\alpha_0\| &lt; \epsilon^2 + |\alpha_0| \epsilon + |x_0| \epsilon$ <p>The quantity on the RHS can be made as small as possible, and thus scalar multiplication is continuous in normed space.</p> $\blacksquare$ <hr /> <h4 id="235-show-that-x_n-rightarrow-x-and-y_n-rightarrow-y-implies-x_n--y_n-rightarrow-x--y-show-that-alpha_n-rightarrow-alpha-and-x_n-rightarrow-x-implies-alpha_n-x_n-rightarrow-alpha-x">2.3.5. Show that $$x_n \rightarrow x$$ and $$y_n \rightarrow y$$ implies $$x_n + y_n \rightarrow x + y$$. Show that $$\alpha_n \rightarrow \alpha$$ and $$x_n \rightarrow x$$ implies $$\alpha_n x_n \rightarrow \alpha x$$.</h4> <p>(See Above)</p> <hr /> <h4 id="236-show-that-the-closure-bary-of-a-subspace-y-of-a-normed-space-x-is-again-a-vector-subspace">2.3.6. Show that the closure $$\bar{Y}$$ of a subspace $$Y$$ of a normed space $$X$$ is again a vector subspace.</h4> <p><strong>Proof:</strong></p> <p>Let $$x,y \in \bar{S}$$. Thus, $$\forall r&gt;0$$, we have $$B(x,r) \cap S \neq \emptyset$$ and $$B(y,r) \cap S \neq \emptyset$$. Pick $$r&lt;\epsilon$$, then $$\|x-x_0\|&lt;\epsilon/2$$ and $$\|y-y_0\|&lt;\epsilon/2$$.</p> <p>Then, we have:</p> $\|\alpha x + \beta y - (\alpha x_0 + \beta y_0)\|=\|\alpha x - \alpha x_0 + \beta y - \beta y_0\| \\ \leq \|\alpha x - \alpha x_0\| + \|\beta y - \beta y_0\| &lt; \epsilon/2 + \epsilon/2 = \epsilon$ <p>This holds for every $$\epsilon&gt;0$$, thus $$B(\alpha x + \beta y, r) \cap Y \neq \emptyset$$, since every open ball around it contains $$\alpha x_0 + \beta y_0 \in Y$$. Thus, $$\bar{Y}$$ is an vector subspace.</p> $\blacksquare$ <hr /> <h4 id="237-absolute-convergence-show-that-convergence-of-y_1--y_2--y_3--cdots-may-not-imply-convergence-of-y_1-y_2--y_3--cdots-hint-consider-y-in-prob-3-and-y_n-where-y_n--eta_jn-eta_nn-1n2-eta_jn--0-for-all-j-neq-n">2.3.7. (Absolute convergence) Show that convergence of $$\|y_1\| + \|y_2\| + \|y_3\| + \cdots$$ may not imply convergence of $$y_1 +y_2 + y_3 + \cdots$$. Hint. Consider $$Y$$ in Prob. 3 and $$(y_n)$$, where $$y_n = (\eta_j^{(n)}), \eta_n^{(n)} =1/n^2, \eta_j^{(n)} = 0$$ for all $$j \neq n$$.</h4> <p><strong>Proof:</strong></p> <p>We have:</p> $y_1=\frac{1}{2},0,0, \cdots \\ y_2=0,\frac{1}{2^2},0, \cdots \\ y_3=0,0,\frac{1}{2^3}, \cdots \\ \vdots \\ y_n=0,0,0, \cdots, \frac{1}{2^n}, \cdots \\ \vdots$ <p>Correspondingly, the norms are:</p> $\|y_1\|=\frac{1}{2} \\ \|y_2\|=\frac{1}{2^2} \\ \|y_3\|=\frac{1}{2^3} \\ \|\vdots \\ \|y_n\|=\frac{1}{2^n} \\ \vdots$ <p>Then $$\|y_1\| + \|y_2\| + \|y_3\| + \cdots$$ is a convergent series.</p> <p>The partial sum $$s_n$$ is defined as:</p> $s_n=\frac{1}{2},\frac{1}{2^2}, \cdots, \frac{1}{2^n}, 0, 0, \cdots$ <p>As $$n \rightarrow \infty$$, we get:</p> $\lim\limits_{n \rightarrow \infty} s_n=\frac{1}{2},\frac{1}{2^2}, \cdots$ <p>Thus, this is the space of sequences with finite non-zero terms.</p> <p>We have $$\|s_n-s_{m}\|=\sup\vert s_{n(i)}-s_{n(i)}\vert$$ (note that $$s_n$$ is a sequence, being the sum of sequences). Assume $$m&lt;n$$, then $$\|s_n-s_m\|=\frac{1}{2^{m+1}}$$. This can be made as small as possible, and thus $$(s_n)$$ is Cauchy.</p> <p>Choose $$s=(\frac{1}{2^n})$$, and we have:</p> $\|s-s_n\|=\frac{1}{2^{n+1}}$ <p>which goes to zero in the limit $$n \rightarrow \infty$$, thus $$s$$ is a limit of $$(s_n)$$. However, $$s$$ has infinitely many terms, and thus is not in the space of sequences with finite non-zero terms.</p> <p>Thus, $$y_1 +y_2 + y_3 + \cdots$$ does not converge even though $$\|y_1\| + \|y_2\| + \|y_3\| + \cdots$$ converges.</p> $\blacksquare$ <hr /> <h4 id="238-if-in-a-normed-space-x-absolute-convergence-of-any-series-always-implies-convergence-of-that-series-show-that-x-is-complete">2.3.8. If in a normed space $$X$$, absolute convergence of any series always implies convergence of that series, show that $$X$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>Take a Cauchy sequence $$(x_n)$$. Pick $$N_k$$ such that $$\|x_m-x_n\|&lt;\frac{1}{2^k}$$ for all $$m,n \geq N_k$$. Pick the corresponding $$y_k=x_{N_k}$$ from $$(x_n)$$. Then note that $$\|y_k-y_{k+1}\| &lt; \frac{1}{2^k}$$.</p> <p>Then $$\displaystyle \sum\limits_{k=1}^\infty \|y_{k+1}-y_k\| &lt; \sum\limits_{k=1}^\infty \frac{1}{2^k} = 1$$. Thus, this series is absolutely convergent, and is by assumption, convergent. That is, $$\displaystyle \sum\limits_{k=1}^n y_{k+1}-y_k$$ is convergent, i.e., it converges to some element, say $$x$$.</p> <p>Now, we have:</p> $\displaystyle \sum\limits_{k=1}^n y_{k+1}-y_k = \displaystyle \sum\limits_{k=1}^n x_{N_{k+1}}-x_{N_k}\\ =x_{N_{n+1}}-x_{N_1}$ <p>In the limit of $$n \rightarrow \infty$$, this expression tends to $$x$$, that is:</p> $\lim\limits_{n \rightarrow \infty} x_{N_{n+1}}-x_{N_1} = x \\ \lim\limits_{n \rightarrow \infty} x_{N_{n+1}} = x + x_{N_1}$ <p>Thus, this limit exists and since $$(x_n)$$ was an arbitrary Cauchy sequence, it converges to $$x$$. Thus $$X$$ is complete.</p> $\blacksquare$ <hr /> <h4 id="239-show-that-in-a-banach-space-an-absolutely-convergent-series-is-convergent">2.3.9. Show that in a Banach space, an absolutely convergent series is convergent.</h4> <p><strong>Proof:</strong></p> <p>Let there be an absolutely convergent series $$\displaystyle \sum\limits_{i=1}^\infty \|x_k\|&lt;\infty$$. Since it is convergent, it is also Cauchy, thus we have:</p> $\displaystyle \sum\limits_{i=m}^n |x_k|&lt;\epsilon$ <p>By the <strong>Triangle Inequality</strong>, we have:</p> $\displaystyle |\sum\limits_{i=m}^n x_k| \leq \sum\limits_{i=m}^n |x_k| \\ \displaystyle |\sum\limits_{i=m}^n x_k| = s_n - s_{m-1} &lt; \epsilon$ <p>Since the space is Banach, $$(s_n)$$ is a convergent sequence.</p> $\blacksquare$ <hr /> <h4 id="2310-schauder-basis-show-that-if-a-normed-space-has-a-schauder-basis-it-is-separable">2.3.10. (Schauder basis) Show that if a normed space has a Schauder basis, it is separable.</h4> <p><strong>Proof:</strong></p> <p>A Schauder basis of a space $$X$$ is a sequence $$(e_n)$$ such that $$\|x-(\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \cdots + \alpha_n e_n)\| \rightarrow 0, x \in X$$ as $$n \rightarrow \infty$$.</p> <p>A space is separable if it has a countable subset which is dense in this space.</p> <p>The partial sum of a Schauder basis is represented as $$s_n=\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \cdots + \alpha_n e_n$$.</p> <p>This implies that $$\forall \epsilon &gt; 0, \exists N$$ such that $$\|x-s_n\|&lt;\epsilon$$ for all $$n&gt;N$$. Thus every neighbourhood of $$x$$ has a Schauder representation.</p> <p>Since $$\alpha_n \in \mathbb{R}$$, there exists a $$\beta_n \in \mathbb{Q}$$, such that $$\|\alpha_n-\beta_n\|&lt;\epsilon$$.</p> <p><strong>(Prove that $$Y=\sum\limits_{i=1}^n \beta_i e_i$$ is countable).</strong></p> <p>Denote $$s_n'=\beta_1 e_1 + \beta_2 e_2 + \beta_3 e_3 + \cdots + \beta_n e_n$$, then we have:</p> $\|s_n-s_n'\|=\|(\alpha_1-\beta_1) e_1 + (\alpha_2-\beta_2) e_2 + (\alpha_3-\beta_3) e_3 + \cdots + (\alpha_n-\beta_n) e_n\| \\ \leq \|(\alpha_1-\beta_1) e_1\| + \|(\alpha_2-\beta_2) e_2\| + \|(\alpha_3-\beta_3) e_3\| + \cdots + \|(\alpha_n-\beta_n) e_n\| \\ = |\alpha_1-\beta_1| \|e_1\| + |\alpha_2-\beta_2| \|e_2\| + |\alpha_3-\beta_3| \|e_3\| + \cdots + |\alpha_n-\beta_n| \|e_n\| \\ = \epsilon \|e_1\| + \epsilon \|e_2\| + \epsilon \|e_3\| + \cdots + \epsilon \|e_n\| \\ = \epsilon (\|e_1\| + \|e_2\| + \|e_3\| + \cdots + \|e_n\|)=K \epsilon$ <p>(Note that even though $$K$$ depends upon how far the Schauder basis is expanded, for a fixed Schauder basis, a rational number can be chosen arbitrarily closer to the real number without resorting to going further along the Schauder basis).</p> $\|x-s_n'\| \leq \|x-s_n\| + \|s_n-s_n'\| &lt; \epsilon + K \epsilon = (K+1) \epsilon$ <p>This can be made as small as needed, and thus $$Y$$ (countable) is dense in this normed space, and hence the space is separable.</p> $\blacksquare$ <hr /> <h4 id="2311-show-that-e_n-where-e_n--delta_nj-is-a-schauder-basis-for-lp-where-1-leq-p-infty">2.3.11. Show that $$(e_n)$$, where $$e_n = (\delta_{nj})$$, is a Schauder basis for $$l^p$$, where $$1 \leq p&lt; +\infty$$.</h4> <p><strong>Proof:</strong></p> <p>$$l^p$$ is the space of all bounded sequences. This implies that:</p> $\sum\limits_{i=1}^\infty {|x_i|}^p=K&lt;\infty$ <p>Equivalently,</p> $\lim\limits_{n \rightarrow \infty} \sum\limits_{i=1}^n {|x_i|}^p=K$ <p>The norm is defined as $${\|x\|}_p={\left(\sum\limits_{i=1}^\infty {|x_i|}^p\right)}^{1/p}$$</p> <p>Assume the sequence is $$x_1, x_2, x_3, \cdots$$. Then, we have:</p> $x_1 e_1=x_1, 0, 0, 0, \cdots \\ x_2 e_2=0, x_2, 0, 0, \cdots \\ x_3 e_3=0, x_2, 0, 0, \cdots \\ \vdots \\ x_n e_n=0, 0, 0, 0, \cdots, x_n, 0, 0, \cdots \\$ <p>Then, we get:</p> $s_n=\sum\limits_{i=1}^n x_i e_i = x_1, x_2, \cdots, x_n, 0, 0, \cdots \\ x = s_n + \sum\limits_{i=n+1}^\infty x_i e_i \\ x-s_n = \sum\limits_{i=n+1}^\infty x_i e_i \\ \|x-s_n\| = {(\sum\limits_{i=n+1}^\infty {|x_i|}^p)}^{1/p}$ <p>We know that:</p> $\sum\limits_{i=1}^\infty {|x_i|}^p=K=\sum\limits_{i=1}^n {|x_i|}^p + \sum\limits_{i=n+1}^\infty {|x_i|}^p$ <p>Taking limits on both sides for $$n \rightarrow \infty$$, we get:</p> $\lim\limits_{n \rightarrow \infty} \underbrace{\sum\limits_{i=1}^n {|x_i|}^p}_\text{Partial Sum} + \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p=K \\ K + \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p = K \\ \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p = 0 \\$ <p>Thus $$\|x-s_n\| \rightarrow 0$$ as $$n \rightarrow \infty$$.</p> <p>Thus, $$e_n = (\delta_{nj})$$ is a Schauder basis for $$l^p$$.</p> $\blacksquare$ <hr /> <h4 id="2312-seminorm-a-seminorm-on-a-vector-space-x-is-a-mapping-p-x-rightarrow-mathbbr-satisfying-n1-n3-n4-in-sec-23-some-authors-call-this-a-pseudonorm-show-that">2.3.12. (Seminorm) A seminorm on a vector space $$X$$ is a mapping $$p: X \rightarrow \mathbb{R}$$ satisfying <strong>(N1)</strong>, <strong>(N3)</strong>, <strong>(N4)</strong> in Sec. 2.3. (Some authors call this a pseudonorm.) Show that</h4> $p(0)= 0, \\ |p(y) - p(x)| \leq p(y-x).$ <p><strong>(Hence if $$p(x) = 0$$ implies $$x = 0$$, then $$p$$ is a norm.)</strong></p> <p><strong>Proof:</strong></p> $p(\theta)=p(0x)=0 p(x) = 0$ <p>Since the seminorm respects the <strong>Triangle Inequality</strong>, we have:</p> $p(x)=p(x-y+y) \leq p(x-y) + p(y) \\ p(x)-p(y) \leq p(x-y)$ <p>Similarly, we have:</p> $p(y)=p(y-x+x) \leq p(y-x) + p(x) \\ p(y)-p(x) \leq p(y-x) = p(x-y) \\$ <p>Therefore:</p> $|p(y)-p(x)| \leq p(y-x)$ $\blacksquare$ <hr /> <h4 id="2313-show-that-in-prob-12-the-elements-x-in-x-such-that-px--0-form-a-subspace-n-of-x-and-a-norm-on-xn-cf-prob-14-sec-21-is-defined-by-hatx_0px-where-x-in-hatx-and-hatx-in-xn">2.3.13. Show that in Prob. 12, the elements $$x \in X$$ such that $$p(x) = 0$$ form a subspace $$N$$ of $$X$$ and a norm on $$X/N$$ (cf. Prob. 14, Sec. 2.1) is defined by $${\|\hat{x}\|}_0=p(x)$$, where $$x \in \hat{x}$$ and $$\hat{x} \in X/N$$.</h4> <p><strong>Proof:</strong></p> <p>From the set $$N=\{x \in X : p(x)=0\}$$, pick $$x,y \in Y$$.</p> <p>We have:</p> $p(\alpha x + \beta y) \leq p(\alpha x) + p(\beta y) = |\alpha| p(x) + |\beta| p(y) = 0$ <p>Thus, $$\alpha x + \beta y \in N$$, so $$N$$ is a subspace.</p> <p>The cosets of $$X/N$$ are of the form $$x+N$$. To prove that $${\|\hat{x}\|}_0=p(x)$$ is a norm:</p> <ul> <li><strong>(N1)</strong>: Pick a coset $$x+N \in X/N$$. Now pick any $$y \in N$$. Then the seminorm $$p(x+y) \geq 0$$. Since this holds for any $$y \in N$$, $$\|x+N\|_0 \geq 0$$. Since this is an arbitrary coset, the defined norm is nonnegative.</li> <li><strong>(N2)</strong>: Assume $${\|\hat{x}\|}_0=p(x)=0$$. Then $$\hat{x} \in N$$, since $$p(x)=0, \forall x \in N$$. This is the coset $$(0+N)$$, which is the zero vector. Assume $$\hat{x}=\theta$$. This is the coset $$0+N$$. Pick any element $$y \in N$$. Then $$p(\theta) \leq p(0) + p(y) = 0$$.</li> <li><strong>(N3)</strong>: Assume the coset $$\hat{x} = \alpha(x+N)=\alpha x + N$$. Now pick $$y \in N$$. Then $$\alpha x + y \in \alpha x + N$$. Then $$p(\alpha x + y) = p(\alpha (x+\frac{y}{\alpha})) = \vert \alpha \vert p(x+\frac{y}{\alpha})$$. Note that $$\frac{y}{\alpha} \in N$$, since $$N$$ is a subspace. Thus, we get $$\|\alpha \hat{x} \|_0 = p(\alpha x + y)=\vert \alpha \vert p(x+N) = \alpha \|\hat{x}\|_0$$.</li> <li><strong>(N4)</strong>: Assume the cosets $$\hat{x_1}=x_1 + N, \hat{x_2}=x_2 + N$$. Pick arbitrary elements $$x_1 + y_1 \in \hat{x_1}$$ and $$x_2 + y_2 \in \hat{x_2}$$. Then $$\|\hat{x_1} + \hat{x_2}\|=p(x_1 + y_1 + x_2 + y_2) \leq p(x_1) + p(x_2) + p(y_1) + p(y_2) = p(x_1) + p(x_2)$$. Now, we have: $$p(x_1) \leq p(x_1 + y) + p(y) = p(x_1 + y) = \|\hat{x_1}\|_0$$ and $$p(x_2) \leq p(x_2 + y) + p(y) = p(x_2 + y) = \|\hat{x_2}\|_0$$, for some $$y \in N$$. This gives us $$\|\hat{x_1} + \hat{x_2}\| \leq \|\hat{x_1}\|_0 + \|\hat{x_2}\|_0$$.</li> </ul> $\blacksquare$ <hr /> <h4 id="2314-quotient-space-let-y-be-a-closed-subspace-of-a-normed-space-x-bullet-show-that-a-norm-bullet_0-on-xy-cf-prob-14-sec-21-is-defined-by">2.3.14. (Quotient space) Let Y be a closed subspace of a normed space $$(X, \|\bullet\|)$$. Show that a norm $$\|\bullet\|_0$$ on $$X/Y$$ (cf. Prob. 14, Sec. 2.1) is defined by</h4> ${\|\hat{x}\|}_0 = \inf_{x \in \hat{x}} \|x\|$ <p><strong>where $$\hat{x} \in X/Y$$, that is, $$\hat{x}$$ is any coset of $$Y$$.</strong></p> <p><strong>Proof:</strong></p> <p>The cosets of $$X/N$$ are of the form $$x+N$$. To prove that $${\|\hat{x}\|}_0 = \inf\limits_{x \in \hat{x}} \|x\|$$ is a norm:</p> <ul> <li><strong>(N1)</strong>: Pick a coset $$x+N \in X/N$$. Now pick any $$y \in N$$. Then the $$p(x+y) \geq 0$$, since we take infimum of nonnegative norms. Since this holds for any $$y \in N$$, $$\|x+N\|_0 \geq 0$$. Since this is an arbitrary coset, the defined norm is nonnegative.</li> <li><strong>(N2)</strong>: Assume $${\|\hat{x}\|}_0=p(x)=0$$. Then for this coset we have at least one element $$\|x+y\|=0, y \in N$$. This implies that $$x+y=0 \Rightarrow x=-y=(-1)y$$. This implies that $$x+y=(-1)y+y$$, i.e., it is a linear combination of $$y \in N$$, and thus belongs to $$N$$, which is the zero element.</li> <li><strong>(N3)</strong>: Assume the coset $$\hat{x} = \alpha(x+N)=\alpha x + N$$. Now pick $$y \in N$$. Then $$\alpha x + y \in \alpha x + N$$. Then $${\|\alpha\hat{x}\|}_0 = \inf\|\alpha (x+\frac{y}{\alpha})\| = \vert \alpha \vert \inf \|x+\frac{y}{\vert\alpha\vert}\|$$. Note that $$\frac{y}{\|\alpha\|} \in N$$, since $$N$$ is a subspace. Thus, we get $$\|\alpha \hat{x} \|_0 = \vert \alpha \vert \inf \|x+N\| = \alpha \|\hat{x}\|_0$$.</li> <li><strong>(N4)</strong>: Assume the cosets $$\hat{x_1}=x_1 + N, \hat{x_2}=x_2 + N$$. Pick arbitrary elements $$x_1 + y_1 \in \hat{x_1}$$ and $$x_2 + y_2 \in \hat{x_2}$$. Then $$\|\hat{x_1} + \hat{x_2}\|=\inf \|x_1 + y_1 + x_2 + y_2\| \leq \inf \|x_1 + y_1\| + \|x_2 + y_2\| = \|\hat{x_1}\| + \|\hat{x_2}\|$$.</li> </ul> $\blacksquare$ <hr /> <h4 id="2315-product-of-normed-spaces-if-x_1-bullet_1-and-x_2-bullet_2-are-normed-spaces-show-that-the-product-vector-space-x--x_1-times-x_2-cf-prob-13-sec-21-becomes-a-normed-space-if-we-define">2.3.15. (Product of normed spaces) If $$(X_1, {\|\bullet\|}_1)$$ and $$(X_2, {\|\bullet\|}_2)$$ are normed spaces, show that the product vector space $$X = X_1 \times X_2$$ (cf. Prob. 13, Sec. 2.1) becomes a normed space if we define</h4> $\|x\|=\max({\|x_1\|}_1, {\|x_2\|}_2)$ <p><strong>where $$x=(x_1,x_2)$$</strong>.</p> <p><strong>Proof:</strong></p> <p>The operations on a product vector space are defined as:</p> $(x_1,x_2) + (y_1,y_2) = (x_1+y_1, x_2+y_2) \\ \alpha(x_1, x_2) = (\alpha x_1, \alpha x_2)$ <ul> <li><strong>(N1)</strong>: Since we take the maximum of nonnegative norms, the norm on the product space is nonnegative.</li> <li><strong>(N2)</strong>: Assume that $$(x_1, x_2) = (0,0)$$. Then $$\|x\|=\max(\|x_1\|, \|x_2\|) = 0$$. Conversely, assume that $$\|x\|=\max(\|x_1\|, \|x_2\|) = 0$$. Then, $$\|x_1\|=\|x_2\|=0$$, i.e., $$x_1=x_2=0$$.</li> <li><strong>(N3)</strong>: We have $$\|\alpha x\|=\|(\alpha x_1, \alpha x_2)\| = \max(\|\alpha x_1\|, \|\alpha x_2\|) = \vert\alpha\vert max(\|x_1\|, \|x_2\|) = \alpha \|x\|$$.</li> <li><strong>(N4)</strong>: We have: $$\|x+y\|=\|(x_1,x_2) + (y_1,y_2)\| = \|(x_1+y_1, x_2+y_2)\| \\ =\max(\|x_1+y_1\|_1, \|x_2+y_2\|_2) \leq \max(\|x_1\|_1+\|y_1\|_1, \|x_2\|_2+\|y_2\|_2) \\ \leq \max(\|x_1\|_1, \|x_2\|_2) + \max(\|y_1\|_1, \|y_2\|_2) = \|x+y\|$$</li> </ul> <p>Remember that:</p> $a \leq max(a,b) \\ b \leq max(a,b) \\ c \leq max(c,d) \\ d \leq max(c,d) \\ \Rightarrow a+c \leq max(a,b) + max(c,d) \\ \Rightarrow b+d \leq max(a,b) + max(c,d) \\ \Rightarrow max(a+c,b+d) \leq max(a,b) + max(c,d)$ $\blacksquare$ <hr />avishekThis post lists solutions to the exercises in the Further Properties of Normed Spaces section 2.3 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Two Phase Commit: Indistinguishable Commit Scenario2021-11-18T00:00:00+05:302021-11-18T00:00:00+05:30/2021/11/18/2pc-commit-indistinguishable-commit-scenario<p>We review the most interesting failure scenario for the <strong>Two Phase Commit (2PC) protocol</strong>. There are excellent explanations of 2PC out there, and I won’t bother too much with the basic explanation. The focus of this post is a walkthrough of the <strong>indistinguishable state scenario</strong>, where neither a global commit, nor a global abort command can be issued.</p> <p>The Wikipedia entry on the Two-Phase Commit protocol, under <strong>Disadvantages</strong>, says the following:</p> <blockquote> <p>Since the coordinator can’t distinguish between all cohorts having already voted to commit and the down cohort having committed (in which case it’s invalid to default to abort) and all cohorts but the down cohort having voted to commit and the down cohort having voted to abort (in which case it’s invalid to default to commit).</p> </blockquote> <p>This was somewhat confusing to me at first, but some quick timeline sketches clarified things. We’ll look at the scenarios from the point of view of a participant in the commit protocol. For purposes of this discussion, that participant will be Participant 2. The diagram below shows the fault-free operation of a 2PC-based transaction.</p> <p><img src="/assets/images/two-phase-commit-no-failures.png" alt="Two Phase Commit - No Failures" /></p> <p>Let us assume that Participant 2 receives the Commit-Request message from the Coordinator, to which it replies with a “Vote Commit” message. If there are no failures at all, all the participants voting to commit would perform the operations in the transaction, including acquiring resource locks and so forth, and would wait for either a Commit or an Abort message from the coordinator. In other words, they’d be blocked.</p> <p>Of course, we do not want to hold up other transactions competing for the same resources indefinitely (which will happen if no Commit/Abort messages arrive at all). Thus the (partial) solution is for one of the participants to take over as the coordinator once a timeout has expired.</p> <p>So, this is what our scenario is: Participant 2 has sent a “Vote Commit” message, but is blocked waiting for a global Commit/Abort directive from the coordinator, which does not arrive within Participant 2’s timeout period. It is Participant 2 who decides (in some manner) to take over as the coordinator. Let us see what Participant 2 can deduce about the state of the protocol at this point.</p> <p><img src="/assets/images/two-phase-commit-failure-scenario-1.png" alt="Two Phase Commit - Failure Scenario 1" /></p> <p>It has the following theories about what might have happened.</p> <ul> <li><strong>The coordinator crashed before receipt of all votes</strong> (Participant 2’s vote being one of them).</li> <li><strong>The coordinator received all votes, but crashed before it could send the global Commit/Abort directive to all the participants.</strong> The implication is that some of the participants may have received the global directive, but Participant 2 is not one of them.</li> </ul> <p>Remember that both 2PC and 3PC depend upon the <strong>Fail-Stop model</strong> to satisfy the safety property of the protocol. <strong>If the coordinator were to recover (or maybe it was just slow), and resume operation, these protocols won’t help.</strong> Now, Participant 2 needs to deduce the state of the protocol to complete it. It does this by querying each of the other participants about their knowledge of the protocol (what they voted/whether they received a global Commit/Abort directive). If all the other participants are alive, then the transaction proceeds as usual. Even if some of these participants go down, it is probably possible to deduce what global directive to send. For example:</p> <ul> <li>If even a single participant had voted for an Abort, then it is safe to say that the transaction is definitely not going to complete.</li> <li>If even a single participant had received a global Commit, that implies that the coordinator intended for the transaction to be committed by all participants. The above cases aren’t the ones that are interesting. The edge case is when:</li> </ul> <p><strong>At least one participant has crashed. Assuming this is Participant 1, this means that when Participant 2 asks Participant 1 for its knowledge of the protocol, it receives no reply.</strong></p> <p>The rest of the live participants have voted to <strong>Commit</strong>.</p> <p><img src="/assets/images/two-phase-commit-failure-scenario-2.png" alt="Two Phase Commit - Failure Scenario 2" /></p> <p><strong>Should Participant 2 send a global Commit message, or a global Abort message?</strong> The correct answer is: <strong>it does not know</strong>. Why? Remember the two theories that Participant 2 had about what happened to the coordinator?</p> <p><strong>Scenario 1:</strong> The coordinator crashed before receipt of all votes In this scenario, if Participant 1 had voted to Commit, sending a global Commit might be the right way to go. But if it had voted to Abort, it’d make sense to be pessimistic about it, and just send out a global Abort anyways. This makes sense in this scenario. The situation is depicted below.</p> <p><strong>Scenario 2:</strong> The coordinator received all votes, but crashed before it could send the global Commit/Abort directive to all the participants In this scenario, if Participant 1 had voted to Commit, and if the coordinator had been able to send a global Commit directive to only Participant 1 before the coordinator crashed, then Participant 1 could have committed the transaction before crashing. In this case, it would be a mistake to send out a global Abort directive because, then, effectively, Participant 1 has committed the transaction, while the rest of the participants roll it back. The situation is depicted below.</p> <p>Thus, the <strong>new coordinator simply does not have enough information to unambiguously decide whether to send a global Commit or a global Abort</strong>, at this point.</p> <p><strong>Note:</strong> This is a post rescued from my old blog back from 2013. It is instructive to note that technology has changed much since then, but the ideas behind those technologies haven’t.</p>avishekWe review the most interesting failure scenario for the Two Phase Commit (2PC) protocol. There are excellent explanations of 2PC out there, and I won’t bother too much with the basic explanation. The focus of this post is a walkthrough of the indistinguishable state scenario, where neither a global commit, nor a global abort command can be issued.Functional Analysis Exercises 8 : Normed and Banach Spaces2021-11-11T00:00:00+05:302021-11-11T00:00:00+05:30/2021/11/11/functional-analysis-exercises-8-normed-banach-spaces<p>This post lists solutions to the exercises in the <strong>Normed Space, Banach Space section 2.2</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h3 id="notes">Notes</h3> <p>The requirements for a space to be a normed space are:</p> <ul> <li><strong>(N1)</strong> <strong>Nonnegativity</strong>, i.e., $$\|x\| \geq 0, x \in X$$</li> <li><strong>(N2)</strong> <strong>Zero norm</strong> implies <strong>zero vector</strong> and vice versa, i.e., $$\|x\|=0 \Leftrightarrow x=0, x \in X$$</li> <li><strong>(N3)</strong> <strong>Linearity</strong> with respect to <strong>scalar multiplication</strong>, i.e., $$\|\alpha x\|=\vert \alpha \vert \|x\|, x \in X, \alpha \in \mathbb{R}$$</li> <li><strong>(N4)</strong> <strong>Triangle Inequality</strong>, i.e., $$\|x+y\| \leq \|x\| + \|y\|, x,y \in X$$</li> </ul> <h4 id="221-show-that-the-norm-x-of-x-is-the-distance-from-x-to-0">2.2.1 Show that the norm $$\|x\|$$ of x is the distance from x to 0.</h4> <p><strong>Proof:</strong></p> <p>We have:</p> $\|x\|=\|x + \theta|=\|x + \theta + (-\theta)\|=\|x + (-\theta) + \theta\| \\ \|x\| \leq \|x+(-\theta)\| + \|\theta\|$ <p>We also have:</p> $\|x+(-\theta)\| \leq \|x\| + \|-\theta\| \\ \|x+(-\theta)\| \leq \|x\| + |-1|\|\theta\| \\ \|x+(-\theta)\| \leq \|x\| + \|\theta\| = \|x\|$ <p>Thus, $$\|x\| \leq \|x+(-\theta)\|$$ and $$\|x\| \geq \|x+(-\theta)\|$$. Thus, $$\|x\| = \|x+(-\theta)\|$$, which is the distance between $$x$$ and $$\theta$$.</p> $\blacksquare$ <hr /> <h4 id="222-verify-that-the-usual-length-of-a-vector-in-the-plane-or-in-three-dimensional-space-has-the-properties-n1-to-n4-of-a-norm">2.2.2 Verify that the usual length of a vector in the plane or in three dimensional space has the properties (N1) to (N4) of a norm.</h4> <p><strong>Proof:</strong></p> <p>(Easy to prove. TODO)</p> $\blacksquare$ <hr /> <h4 id="223-prove-2">2.2.3 Prove (2).</h4> <p><strong>Proof:</strong></p> <p>We wish to prove the <strong>Reverse Triangle Inequality</strong>, which is:</p> $|\|y\| - \|x\|| \leq \|y-x\|$ <p>We have:</p> $\|x\|=\|x-y+y\| \leq \|x-y\| + \|y\| = \|y-x\| + \|y\| \\ \|x\| - \|y\| = \|y-x\| \\$ <p>We also have:</p> $\|y\|=\|y-x+x\| \leq \|y-x\| + \|x\| \\ \|y\| - \|x\| \leq \|y-x\|$ <p>Then, we get:</p> $|\|y\| - \|x\|| \leq \|y-x\|$ $\blacksquare$ <hr /> <h4 id="224-show-that-we-may-replace-n2-by-x0-rightarrow-x0-without-altering-the-concept-of-a-norm-show-that-nonnegativity-of-a-norm-also-follows-from-n3-and-n4">2.2.4 Show that we may replace <strong>(N2)</strong> by $$\|x\|=0 \Rightarrow x=0$$ without altering the concept of a norm. Show that nonnegativity of a norm also follows from <strong>(N3)</strong> and <strong>(N4)</strong>.</h4> <p><strong>Proof:</strong></p> <p>We have from <strong>(N2)</strong> the following:</p> $\|\alpha x\|=|\alpha|\|x\|$ <p>Assuming that $$\alpha=0$$, and knowing that $$0x=\theta$$, we get:</p> $\|0 x\|=|0|\|x\| \\ \|\theta\|=0 \\$ <p>Thus we conclude that $$x=\theta \Rightarrow \|\theta\|=0$$ from <strong>(N2)</strong>.</p> $\blacksquare$ <p>We wish to prove that $$\|x\| \geq 0$$.</p> $\|x\|=\|x+x-x\| \leq \|x+x\| + \|-x\| = \|2x\| + \|x\| = 2\|x\| + \|x\| \\ 2\|x\| + \|x\| \geq \|x\| \\ 2\|x\| \geq 0 \\ \|x\| \geq 0 \\$ $\blacksquare$ <hr /> <h4 id="225-show-that-3-defines-a-norm">2.2.5 Show that (3) defines a norm.</h4> <p><strong>Proof:</strong></p> <p>(3) defines the norm: $${\|x\|}_2=\sqrt{(|\eta_1|^2 + |\eta_2|^2 + \cdots + |\eta_n|^2)}$$</p> <p>(Easy to prove. TODO)</p> $\blacksquare$ <hr /> <h4 id="226-let-x-be-the-vector-space-of-all-ordered-pairs-x--xi_1-xi_2-y--eta_1-eta_2-cdots-of-real-numbers-show-that-norms-on-x-are-defined-by">2.2.6 Let $$X$$ be the vector space of all ordered pairs $$x = (\xi_1, \xi_2), y = (\eta_1, \eta_2), \cdots$$ of real numbers. Show that norms on X are defined by</h4> ${\|x\|}_1=|\eta_1| + |\eta_2| \\ {\|x\|}_2={(\eta_1^2 + \eta_2^2)}^{1/2} \\ {\|x\|}_\infty=\max \{ |\xi_1|, |\xi_2| \}$ <p><strong>Proof:</strong></p> <p>(Easy to prove. TODO)</p> $\blacksquare$ <hr /> <h4 id="227-verify-that-4-satisfies-n1-to-n4">2.2.7 Verify that (4) satisfies (N1) to (N4).</h4> <p><strong>Proof:</strong></p> <p>(Easy to prove. TODO)</p> $\blacksquare$ <hr /> <h4 id="228-there-are-several-norms-of-practical-importance-on-the-vector-space-of-ordered-n-tuples-of-numbers-cf-22-2-notably-those-defined-by">2.2.8 There are several norms of practical importance on the vector space of ordered n-tuples of numbers (cf. 2.2-2), notably those defined by</h4> ${\|x\|}_1=|\eta_1| + |\eta_2| + \cdots + |\eta_n| \\ {\|x\|}_p={(|\eta_1|^p + |\eta_2|^p + \cdots + |\eta_n|^p)}^{1/p} \\ {\|x\|}_\infty=\max \{ |\xi_1|, |\xi_2|, \cdots, |\xi_n| \}$ <p><strong>In each case, verify that (N1) to (N4) are satisfied.</strong></p> <p><strong>Proof:</strong></p> <p>(Easy to prove. TODO)</p> <p>The second result follows from <strong>Minkowski’s Inequality</strong>.</p> $\blacksquare$ <hr /> <h4 id="229-verify-that-5-defines-a-norm">2.2.9 Verify that (5) defines a norm.</h4> <p><strong>Proof:</strong></p> <p>(Easy to prove. TODO)</p> $\blacksquare$ <hr /> <h4 id="2210-unit-sphere-the-sphere-s0-1--x-in-x--x--1-in-a-normed-space-x-is-called-the-unit-sphere-show-that-for-the-norms-in-prob-6-and-for-the-norm-defined-by-the-unit-spheres-look-as-shown-in-fig-16">2.2.10 (Unit sphere) The sphere $$S(0; 1) = \{x \in X : \|x\| = 1\}$$ in a normed space $$X$$ is called the unit sphere. Show that for the norms in Prob. 6 and for the norm defined by the unit spheres look as shown in Fig. 16.</h4> <p><strong>Answer:</strong></p> <p>(Check diagram in book after your curve sketching)</p> <hr /> <h4 id="2211-convex-set-segment-a-subset-a-of-a-vector-space-x-is-said-to-be-convex-if-xy-in-a-implies-mz-in-x--zalpha-x1-alphay-0leq-alpha-leq-1-subset-a-m-is-called-a-closed-segment-with-boundary-points-x-and-y-any-other-z-in-m-is-called-an-interior-point-of-m-show-that-the-closed-unit-ball-b0-1-x-in-x--x-leq-1-in-a-normed-space-x-is-convex">2.2.11 (Convex set, segment) A subset $$A$$ of a vector space $$X$$ is said to be convex if $$x,y \in A$$ implies $$M=\{z \in X : z=\alpha x+(1-\alpha)y, 0\leq \alpha \leq 1\} \subset A$$. $$M$$ is called a closed segment with boundary points $$x$$ and $$y$$; any other $$z \in M$$ is called an interior point of $$M$$. Show that the closed unit ball $$B(0; 1) =\{x \in X : \|x\| \leq 1\}$$ in a normed space X is convex.</h4> <p><strong>Proof:</strong></p> <p>The norm of the point $$z=\alpha x+(1-\alpha)y$$ is:</p> $\|z\|=\|\alpha x+(1-\alpha)y\| \leq \|\alpha x\|+\|(1-\alpha)y\| \\ = \alpha \|x\| + (1-\alpha) \|y\|$ <p>Since $$\|x\| \leq 1$$ and $$\|y\| \leq 1$$, we get:</p> $\|z\| \leq \alpha + (1-\alpha) = 1$ <p>Thus $$z \in X$$, and thus the closed unit ball is convex.</p> $\blacksquare$ <hr /> <h4 id="2212-using-prob-11-show-that-phixsqrtvertxi_1vert--sqrtvertxi_2vert2-does-not-define-a-norm-on-the-vector-space-of-all-ordered-pairs-x--xi_1-xi_2-cdots-of-real-nwnbers-sketch-the-curve-phix--1-and-compare-it-with-fig-18">2.2.12 Using Prob. 11, show that $$\phi(x)={(\sqrt{\vert\xi_1\vert} + \sqrt{\vert\xi_2\vert})}^2$$ does not define a norm on the vector space of all ordered pairs $$x = (\xi_1, \xi_2), \cdots$$ of real nwnbers. Sketch the curve $$\phi(x) = 1$$ and compare it with Fig. 18.</h4> <p><strong>Proof:</strong></p> <p>We can see that $$(1,0)$$ and $$(0,1)$$ fall on the unit circle defined by this “norm”. For it to be a valid norm, the unit ball must be convex. Thus all points $$z=\alpha x+(1-\alpha)y$$ must lie in the unit ball, i.e., $$|z| \leq 1.</p> <p>Set $$\alpha=\frac{1}{2}$$, we get $$z=(\frac{1}{2}, \frac{1}{2})$$.</p> <p>However, using this norm gives us $$\|z\|={(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}})}^2=2$$, which implies it does not lie in the unit ball. Thus, this is not a valid norm.</p> $\blacksquare$ <hr /> <h4 id="2213-show-that-the-discrete-metric-on-a-vector-space-x-neq-0-cannot-be-obtained-from-a-norm-cf-11-8">2.2.13 Show that the discrete metric on a vector space $$X \neq \{0\}$$ cannot be obtained from a norm. (Cf. 1.1-8.)</h4> <p><strong>Proof:</strong></p> <p>For any metric derived from a norm, it must be translation invariant, i.e.:</p> $d(x+a,y+a)=d(x,y), x,y,a \in X \\ d(\alpha x,\alpha y)=d(x,y), x,y \in X, \alpha \in \mathbb{R}$ <p>The discrete metric is defined as:</p> $d(x,y)=\begin{cases} 0 &amp; \text{if } x=y \\ 1 &amp; \text{if } x \neq y \end{cases}$ <p>Assume that $$x \neq y$$. Then $$\alpha x \neq \alpha y$$. Then $$d(\alpha x, \alpha y)=1 \neq \alpha d(x,y)$$.</p> <p>Thus, the discrete metric cannot be derived from a norm.</p> $\blacksquare$ <hr /> <h4 id="2214-if-d-is-a-metric-on-a-vector-space-x-neq-0-which-is-obtained-from-a-norm-and-tilded-is-defined-by-tildedxx--0-tildedxydxy1-x-neq-y-show-that-d-cannot-be-obtained-from-a-norm">2.2.14 If $$d$$ is a metric on a vector space $$X \neq \{0\}$$ which is obtained from a norm, and $$\tilde{d}$$ is defined by $$\tilde{d}(x,x) = 0, \tilde{d}(x,y)=d(x,y)+1 (x \neq y)$$, show that $$d$$ cannot be obtained from a norm.</h4> <p><strong>Proof:</strong></p> <p>For any metric derived from a norm, it must be translation invariant, i.e.:</p> $d(x+a,y+a)=d(x,y), x,y,a \in X \\ d(\alpha x,\alpha y)=d(x,y), x,y \in X, \alpha \in \mathbb{R}$ <p>Assume that $$x \neq y$$. Then $$\alpha x \neq \alpha y$$. Then $$\tilde{d}(\alpha x, \alpha y)=d(\alpha x, \alpha y) + 1 = \alpha d(x,y) + 1 \neq \alpha d(x,y) + \alpha = \alpha \tilde{d}(x,y)$$.</p> $\blacksquare$ <hr /> <h4 id="2215-bounded-set-show-that-a-subset-m-in-a-normed-space-x-is-bounded-if-and-only-if-there-is-a-positive-number-c-such-that-x-leq--c-for-every-x-in-m-for-the-definition-see-prob-6-in-sec-12">2.2.15 (Bounded set) Show that a subset $$M$$ in a normed space $$X$$ is bounded if and only if there is a positive number $$c$$ such that $$\|x\| \leq c$$ for every $$x \in M$$. (For the definition, see Prob. 6 in Sec. 1.2.)</h4> <p><strong>Proof:</strong></p> <p>A set is bounded if $$\delta(x,y)&lt;\infty$$, where $$\delta(x,y)=\sup d(x,y)$$.</p> <p>$$(\Rightarrow)$$ Assume that $$M$$ is bounded. Then $$\delta(x,y)=\sup d(x,y)&lt;\infty$$. This implies that $$d(x,y) \leq c, c \in \mathbb{R}$$ for all $$x,y \in M$$. Set $$y=\theta$$ and note that $$d(x,\theta)=\|x\|$$, to get:</p> $d(x,\theta)=\|x\| \leq c$ $\blacksquare$ <p>$$(\Leftarrow)$$ Assume that there is a positive number $$c$$ such that $$\|x\| \leq c$$ for every $$x \in M$$.</p> <p>Then $$\|x\| \leq c$$.</p> <p>Using the <strong>Triangle Inequality</strong>, and noting that $$d(x, \theta)=\|x\|$$ and $$d(y, \theta)=\|y\|$$, we get:</p> $d(x,y) \leq d(x,\theta) + d(\theta, y) \\ d(x,y) \leq \|x\| + \|y\| \\ d(x,y) \leq c + c \\ d(x,y) \leq 2c \\ \Rightarrow \delta(x,y) = \sup d(x,y) \leq 2c &lt; \infty$ <p>Thus, $$M$$ is bounded.</p> $\blacksquare$avishekThis post lists solutions to the exercises in the Normed Space, Banach Space section 2.2 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Resilient Knowledge Bases : Fundamentals, not Buzzwords2021-11-06T00:00:00+05:302021-11-06T00:00:00+05:30/2021/11/06/resilient-knowledge-bases<p>We start this tirade with a memorable quote from <strong>Alice Through the Looking-Glass</strong>:</p> <blockquote> <p>“Well, in our country,” said Alice, still panting a little, you’d generally get to somewhere else — if you ran very fast for a long time, as we’ve been doing.”</p> <p>“A slow sort of country!” said the Queen. “Now, here, you see, it takes all the running you can do, to keep in the same place.</p> <p>If you want to get somewhere else, you must run at least twice as fast as that!” - <em>Lewis Carroll</em>, <strong>Alice Through the Looking-Glass</strong></p> </blockquote> <h2 id="durable-foundations-in-a-shifting-technology-landscape">Durable Foundations in a Shifting Technology Landscape</h2> <p>React. Data Mesh. Keras. By the time, I finish writing this, probably half of these technologies will have been superseded by a new set of challengers. You’d be forgiven for thinking that the race to stay relevant is specifically dictated by the (often breakneck) speed at which you can assimilate the latest framework, or the latest “enterprise innovation”.</p> <p>More importantly, if you are a senior technologist, you are expected to have <strong>fluent opinions</strong> about the latest and greatest “in-thing”; and that’s fine, since that becomes increasingly a major part of how you add value to clients. <strong>It might end up seeming like a race to merely maintain the position you are in.</strong></p> <p>This is obviously encouraged – at times even mandated – by the sometimes-stringent requirements of knowledge in a particular technology or framework. <em>Must know Kafka.</em> <em>Must know Data Engineering</em>. Our way or the highway. It also encourages specialisation: <em>I’m only comfortable doing .NET projects, so that’s the only kind of work I’d like to pick up</em>.</p> <p>Sure, I am not underestimating specialisation: indeed, specialisation results in efficiency of a sort, but only in that particular context. Take someone hyper-specialised in technology X, and drop them into an area where they will spend mostly working on that self-same technology, and you have created a <strong>dependency</strong>, a <strong>bottleneck</strong>, and a recipe for <strong>planned obsolescence</strong>, because sooner or later, that tech is consigned to the dust of technology history. There are hyper-specialisations which are fruitful and absolutely necessary, because the work in those areas have much larger ramifications for technology as a whole, or they serve critical functions in society; <strong>pure mathematics</strong>, for example: we’d be nowhere as advanced without it (and by extension, translating its results via engineering disciplines), or <strong>medical science</strong>: I’d rather hope that a neurosurgeon operating on my brain has studied specifically about the brain extensively (with the accompanying practice).</p> <p>But software technology (at least at the <strong>commercial applications level</strong>) is not that. Sure, you’d need good specialised knowledge if you are designing or programming an embedded system, or if you are writing code to send humans into space. But <em>nooo</em>…we are building a <em>hyper-customisable self-service decentralised platform for accelerating analytics capabilities across the enterprise to provide unparalleled insights into the end-to-end value chain of the organisation</em>. LOL.</p> <p>Now, to be certain, I am not knocking the intent behind the buzzwords which have been strung together to create the description above. Buzzwords take on a life of their own, and become a useful shorthand for conveying a lot of information without too much description. They serve as a <strong>useful layer of abstraction</strong> for all the technology that goes into building a solution.</p> <p>But the technology behind them is always prone to change or fall out of favour, whether it is a product, or a technique. Hanging your star on the <strong>Flavour of the Month</strong> or the <strong>Flavour of the Next Few Months after which Everyone Scrambles for the Next Shiny Thing</strong> is not exactly a sustainable way to prosper in the long term. What I am at pains to point out, is that the future is always ready to eat your hyper-specialty for breakfast. What was amazing technology at the time I began working in 2004, is now commonplace, unremarkable.</p> <h2 id="resilient-knowledge-base">Resilient Knowledge Base</h2> <p>There is no escaping learning new technology. However, I hypothesise that there are certain steps we can take that will contribute to the longevity of our knowledge base without it being buffeted every so often by the winds of change. And that, in my opinion, is a <strong>firm focus on fundamentals</strong>.</p> <p>This is not a new concept, but it is an often deprioritised one. What do I mean by “fundamentals”? Nothing mysterious, just more or less what it says on the tin.</p> <blockquote> <p>fun·da·men·tal /ˌfʌndəˈmentl/ serving as a basis supporting existence or determining essential structure or function - Merriam-Webster</p> </blockquote> <p>The idea is that the <strong>fundamentals constitute a concept (or concepts) which underlie an observation or a practice or another group of concepts</strong>. Usually, the fundamentals end up being a sort of a unifying layer underlying multiple (potentially disparate) ideas.</p> <p>In the spirit of defining new buzzwords, I’d like to propose this term that I like to use: <strong>Resilient Knowledge Bases</strong>.</p> <p><img src="/assets/images/resilient-knowledge-base-structure.png" alt="Structure of a Resilient Knowledge Base" /></p> <p>The structure of a Resilient Knowledge Base might or might not be intuitive, but it is certainly simple. Think of concepts building upon more primitive concepts, using logical reasoning as glue between these stacks. Usually though, the concepts aren’t numerous, and the reasoning can sometimes be loose if lesser rigour suffices; as in mathematics, you start with only a few fundamental concepts, and then derive multiple results from those, and then even more on top of those, like an inverted pyramid. The implicit advantage in understanding the fundamentals is then also that you don’t have a lot to “memorise”; most of what you see at the upper levels of abstraction are built atop the sparser layers below.</p> <h2 id="examples">Examples</h2> <p>Let’s take a look at some examples of what I consider as <strong>‘fundamental’</strong>. Your corresponding definitions and levels for what constitutes “fundamental” for these examples may differ, and that’s quite alright; we’ll address that in a moment.</p> <p>Here’s the lens applied to only a single aspect of the design of a transactional system, maybe a payments integrator.</p> <script src="/assets/javascript/mermaid.min.js"></script> <div class="mermaid"> graph TD; payments[Payments System] system_of_record[System of Record] tradeoffs[Tradeoffs] types_of_consistency_models[Eventual / Causal / etc. Consistency] consistency_models[Consistency Models] properties[Liveness / Safety] model_checking[Model Checking] system_of_record --&gt; payments tradeoffs --&gt; system_of_record types_of_consistency_models --&gt; tradeoffs consistency_models --&gt; types_of_consistency_models consistency_models --&gt; model_checking properties --&gt; consistency_models style payments fill:#006f00,stroke:#000,stroke-width:2px,color:#fff style system_of_record fill:#006fff,stroke:#000,stroke-width:2px,color:#fff style tradeoffs fill:#8f0f00,stroke:#000,stroke-width:2px,color:#fff </div> <p>A few notes before introducing the main lesson. We need to work with the actual system of record (possibly the payment service provider), and thus need to ensure consistency of data between the our system and the service provider. There is not a single way of ensuring consistency, and we would want to make tradeoffs depending upon the operational constraints that are present to us.</p> <p>But given knowledge of those consistency models, we are in a position to precisely state how the system will behave under unexpected conditions. Thus, we are able to enumerate failure scenarios and edge cases in a much more principled and disciplined fashion.</p> <p>Taking it a step further leads us to a somewhat more formal (read mathematical) treatment of the properties of such a system. In a distributed system, those are usually <strong>Safety</strong> and <strong>Liveness</strong>. The next logical step would be an investigation of the techniques which can assure the existence of certain desirable properties (or the absence of undesirable ones). <strong>Model Checking</strong> is one such technique.</p> <p>Of course, this is highly simplified, and the system design of such a component would cover a raft of other considerations. However, the point I’d like to make is this: understanding the technical tradeoffs in this system, adds tools in your kit for designing and reasoning about other types of systems, which are not necessarily related to payments. The snapshot I’ve shown above could form a small facet of this Resilient Knowledge Base, that you can grow and refine over time.</p> <p>Let’s look at another, more detailed map, this time of a certain technique in <strong>Machine Learning</strong>, called <strong>Gaussian Processes</strong>.</p> <script src="/assets/javascript/mermaid.min.js"></script> <div class="mermaid"> graph TD; quad[Quadratic Form of Matrix] chol[Cholesky Factorisation] tri[Triangular Matrices] det[Determinants] cov[Covariance Matrix] mvn[Multivariate Gaussian] mvnlin[MVN as Linearly Transformed Sums of Uncorrelated Random Variables] crv[Change of Random Variable] gp[Gaussian Processes] kernels[Kernels] rkhs[Reproducing Hilbert Kernel Space] mercers_theorem[Mercer's Theorem] quad--&gt;chol; tri--&gt;chol; det--&gt;chol; cov--&gt;mvn chol--&gt;mvn mvn--&gt;mvnlin crv--&gt;mvnlin mvnlin--&gt;Conditioning mvnlin--&gt;Marginalisation Conditioning--&gt;gp Marginalisation--&gt;gp kernels--&gt; gp rkhs--&gt;kernels mercers_theorem--&gt;kernels style chol fill:#006f00,stroke:#000,stroke-width:2px,color:#fff style mvn fill:#006fff,stroke:#000,stroke-width:2px,color:#fff style gp fill:#8f0f00,stroke:#000,stroke-width:2px,color:#fff </div> <p>The exact details of the graph are not very important; what is salient to note is that the nodes closer to the sources are progressively more generally applicable to a problem statement in the domain. As simple examples, understanding <strong>kernels</strong>, and by extension <strong>Mercer’s Theorem</strong> and/or <strong>Reproducing Hilbert Kernel Spaces</strong>, equips us equally well to understand how coefficients are computed in a different ML technique called <strong>Support Vector Machines</strong>. As another, more trivial, example, understanding the concept of a covariance matrix and general matrix theory, equips us very well to understand the tools used in a lot of other ML techniques, ranging from <strong>Linear Regression</strong> to <strong>Principal Components Analysis</strong>.</p> <p>Now, everyone will have a different concept of what “fundamental” is, for a different domain, and this is going to be subjective. For example, you can always go deeper than matrix theory and delve into <strong>Linear Algebra</strong> (and hence to <strong>Real Analysis</strong>, <strong>Abstract Algebra</strong>, etc.), or you might choose to stay in the realm of matrix algebra and be satisfied with its results. That is perfectly fine: everyone has their own comfort level, and it would be unreasonable to ask your vanilla developer to start studying Real Analysis just so they can use TensorFlow.</p> <p>But the point is this: regardless of which level of fundamental detail you are comfortable with, your <strong>Resilient Knowledge Bases</strong> is determinedly more adaptable to change than the latest version of your favourite machine learning library and its APIs, or the latest buzzword that you may have been asked to get behind. Five years ago, it was Data Lake, now people speak of Data Mesh; several years later, there will be a new term for a new something. For sure, you will need to learn that specialised knowledge to some degree to be functional, but the foundations will always equip you well to work in a different environment, e.g., explaining the fundamentals to a curious client, improving existing implementations, making informed technology decisions based on your knowledge of the fundamentals, or even rolling your own implementations, if it comes to that.</p> <p>At its heart, you start to think more critically; you do not see technology as a writhing mass of disparate elements that you are forever racing to catch up to, but more of different aspects of a few underlying principles.</p> <p>I thoroughly and strenuously object to the “Lost without APIs” attitude, which is fine when you have implementations to use in some prescribed fashion, within a limited context. But you do end up doing a disservice to yourself by stunting your growth as a technologist, if this is the stage at which you continue to work in.</p> <h2 id="some-suggestions">Some Suggestions</h2> <p>Far be it from me to claim any authority on how we can get from here to there: I can only offer a few suggestions based on what has worked for me.</p> <h3 id="1-seek-new-ways-of-thinking-about-the-same-thing">1. Seek new ways of thinking about the same thing</h3> <p>Thinking about new ways of dealing with the same problem, can open up many profitable tools at your disposal. As a more mathematical example, take kernels again, like we talked about for Gaussian Processes. Kernels can be understood either through Reproducing Hilbert Kernel Spaces, or through Mercer’s Theorem, which is a statement about eigendecomposition in Functional Analysis. RKHS’s are arguably easier to reason about. Mercer’s Theorem is straightforward, but the fundamentals of it require a more careful study of Functional Analysis to appreciate the results.</p> <p>In the same way, during system design, thinking of tradeoffs forces one to enumerate various “levers” they have at their disposal. This allows you to develop a range of options instead of resigning yourself to the (sometimes unrealistic) requirements set by business. This sets the stage for a more reasoned discussion in otherwise unreasonable scenarios around what is feasible and what isn’t.</p> <p>Taking another example from Machine Learning, even something as “simple” as Linear Regression can be dealt with from a probabilistic perspective, in addition to the normal statistical approach. See <a href="/2021/04/05/linear-regression-maximum-likelihood-estimator.html">Linear Regression: Assumptions and Results using the Maximum Likelihood Estimator</a> for a quick flavour. This opens you up to interpreting Machine Learning from more than one angle, and opens up Bayesian Probability as a more broadly applicable tool for deriving results.</p> <h3 id="2-read-some-theory-and-persevere">2. Read some Theory, and Persevere</h3> <p>Learning on the job is great; it’s usually where a lot of learning usually happens. Unfortunately, in my experience, fundamentals don’t usually present themselves transparently when attempting to solve a problem or write some code or explaining something to stakeholders. Most of that usually can be gained by reading. Here, “reading” is a catch-all term for watching video lectures and discussions with more knowledgeable experts. A lot of theory is not immediately applicable to the situation at hand; instead when you first begin to build your foundations, you will see there’s quite some distance between the result you want to derive (or understand) and where the first building blocks.</p> <p>As another example in Distributed Systems, the FLP Impossibility requires you to understand the system model as well as a working grasp of different types of logical reasoning (proof by contradiction, for example) to understand the proof.</p> <p>When you begin, there will be many hurdles. To take a personal example, I struggled greatly (and still do) while progressing through graduate Functional Analysis, because I’d never taken a formal Real Analysis course in Engineering. However intractable the material though, in time, be confident that it will yield to study, and possibly some careful backtracking (which is what I did, teaching myself most of the Real Analysis material I need to progress through my primary texts).</p> <h3 id="3-dont-mind-the-gap">3. Don’t Mind the Gap</h3> <p>Going back to the fundamentals can be an uneasy experience: suddenly many facts (some disparate, some not) that you took for granted, are either called into question, and need to be proved (as in the case of mathematics), or are shown to be working in your case only because the constraints are relaxed in your situation (for example, you’ve never heard of user complain of data consistency issues in your system because the leader database has never gone down, which could have caused possible desync issues with the replicas).</p> <p>In parallel, there will be many concepts that will seem too abstruse or impenetrable on first reading, and you will be justified in feeling that there are gaps in your understanding. Indeed, in my experience, this feeling of “incompleteness” never goes away. You merely continue to fill the holes in your learning progressively with every new pass of the material. My advice is to not worry too much about getting stuck at the proof / derivation / explanation of a fact / theorem. <strong>Give it a fair shot</strong>, then move on for the moment, accepting the veracity of said fact. You can always give it several more goes, but that shouldn’t stop you from progressing.</p> <h2 id="tldr">TL;DR</h2> <ul> <li><strong>Do not be satisfied</strong> with the minimum needed to get work done.</li> <li>Technology and buzzwords changes; adapt to it, but <strong>strengthen your foundations progressively</strong>.</li> <li><strong>Go a few knowledge levels deeper</strong> than what your current technology competency requires.</li> <li><strong>Persevere</strong> while building your <strong>Resilient Knowledge Base</strong>.</li> <li><strong>Accept that there will be gaps in your understanding</strong>: know that you can always come back and patch them up.</li> </ul> <p>To end this essay, I think I should let <strong>Robert Heinlein</strong> have the last word.</p> <blockquote> <p>“A human being should be able to change a diaper, plan an invasion, butcher a hog, conn a ship, design a building, write a sonnet, balance accounts, build a wall, set a bone, comfort the dying, take orders, give orders, cooperate, act alone, solve equations, analyze a new problem, pitch manure, program a computer, cook a tasty meal, fight efficiently, die gallantly.</p> <p>Specialization is for insects.” - <em>Robert Anson Heinlein</em></p> </blockquote>avishekWe start this tirade with a memorable quote from Alice Through the Looking-Glass:Functional Analysis Exercises 7 : Vector Spaces2021-11-03T00:00:00+05:302021-11-03T00:00:00+05:30/2021/11/03/functional-analysis-exercises-7-vector-spaces<p>This post lists solutions to the exercises in the <strong>Vector Space section 2.1</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h3 id="notes">Notes</h3> <p>The requirements for a space to be a vector space are:</p> <ul> <li><strong>(VA1)</strong> Symmetric with respect to addition, i.e., $$x+y=y+x, x,y \in X$$</li> <li><strong>(VA2)</strong> Existence of identity element, i.e., $$x+\theta=x, x,\theta \in X$$</li> <li><strong>(VA3)</strong> Existence of inverse element, i.e., $$x+(-x)=\theta, x,\theta \in X$$</li> <li> <p><strong>(VA4)</strong> Associative with respect to addition, i.e., $$x+(y+z)=(x+y)+z, x,y,z \in X$$</p> </li> <li><strong>(VM1)</strong> Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM2)</strong> Existence of identity element, i.e., $$\alpha_0 x=x, x \in X, \alpha_0=1 \in \mathbb{R}$$</li> <li><strong>(VM3)</strong> Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM4)</strong> Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$</li> </ul> <h4 id="211-show-that-the-set-of-all-real-numbers-with-the-usual-addition-and-multiplication-constitutes-a-one-dimensional-real-vector-space-and-the-set-of-all-complex-numbers-constitutes-a-one-dimensional-complex-vector-space">2.1.1 Show that the set of all real numbers, with the usual addition and multiplication, constitutes a one-dimensional real vector space, and the set of all complex numbers constitutes a one-dimensional complex vector space.</h4> <p><strong>Proof:</strong></p> <p>Consider $$\mathbb{R}$$.</p> <ul> <li><strong>(VA1)</strong> Symmetric with respect to addition, i.e., $$x+y=y+x, x,y \in \mathbb{R}$$</li> <li><strong>(VA2)</strong> Existence of identity element, i.e., $$x+0=x, x,0 \in \mathbb{R}$$</li> <li><strong>(VA3)</strong> Existence of inverse element, i.e., $$x+(-x)=0, x,0 \in \mathbb{R}$$</li> <li> <p><strong>(VA4)</strong> Associative with respect to addition, i.e., $$x+(y+z)=(x+y)+z, x,y,z \in \mathbb{R}$$</p> </li> <li><strong>(VM1)</strong> Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM2)</strong> Existence of identity element, i.e., $$1x=x, x \in X, 1 \in \mathbb{R}$$</li> <li><strong>(VM3)</strong> Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM4)</strong> Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$</li> </ul> $\blacksquare$ <p>Consider $$\mathbb{C}$$.</p> <ul> <li><strong>(VA1)</strong> Symmetric with respect to addition, i.e., $$(a+ib)+(c+id)=(c+id)+(a+ib)=(a+c) + i(b+d), a+ib,c+id \in \mathbb{C}$$</li> <li><strong>(VA2)</strong> Existence of identity element, i.e., $$(a+ib)+(0_0i)=a+ib, a+ib, c+id \in \mathbb{C}$$</li> <li><strong>(VA3)</strong> Existence of inverse element, i.e., $$(a+ib)+(-a-ib)=0+0i, x,0 \in \mathbb{C}$$</li> <li> <p><strong>(VA4)</strong> Associative with respect to addition, i.e., $$x_1+ix_2+(y_1+iy_2+z_1+iz_2) \\ =(x_1+ix_2+y_1+iy_2)+z_1+iz_2 \\ =(x_1+y_1+z_1)+i(x_2+y_2+z_2), x_1+ix_2,y_1+iy_2,z_1+iz_2 \in \mathbb{C}$$</p> </li> <li><strong>(VM1)</strong> Associative with respect to scalar multiplication, i.e., $$\alpha (\beta x) = (\alpha \beta) x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM2)</strong> Existence of identity element, i.e., $$(1+0i)(a+ib)=a+ib, a+ib \in \mathbb{C}$$</li> <li><strong>(VM3)</strong> Distributive with respect to addition of scalars, i.e., $$(\alpha + \beta) x=\alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$</li> <li><strong>(VM4)</strong> Distributive with respect to addition of vectors, i.e., $$\alpha (x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$</li> </ul> $\blacksquare$ <hr /> <h4 id="212-prove-1-and-2">2.1.2 Prove (1) and (2).</h4> <p><strong>Proof:</strong></p> <p>We have to prove:</p> <p><strong>(1a)</strong> $$0x=\theta$$<br /> <strong>(1b)</strong> $$\alpha \theta=\theta$$<br /> <strong>(1c)</strong> $$(-1) x=-x$$</p> <p>where $$\alpha \in \mathbb{R}, x, \theta \in X$$</p> <p><strong>(1a)</strong> We have:</p> $\begin{array} {lr} (0x) + (0x) = (0+0)x = 0x &amp;&amp; \mathbf{\text{ (by (VM3))}} \\ (0x) + (0x) + (-(0x))= (0x) + (-(0x)) &amp;&amp; \mathbf{\text{ (adding (0x) on both sides)}} \\ (0x) + \theta = \theta &amp;&amp; \mathbf{\text{ (by (VA3))}} \\ 0x = \theta &amp;&amp; \mathbf{\text{ (by (VA2))}} \end{array}$ $\blacksquare$ <p><strong>(1b)</strong> We have:</p> $\begin{array} {lr} \alpha(0x)=(\alpha 0)x &amp;&amp; \mathbf{\text{ (by (VM1))}} \\ (\alpha 0)x=0x=\theta \end{array}$ $\blacksquare$ <p><strong>(1c)</strong> We have:</p> $\begin{array} {lr} 0x = \theta &amp;&amp; \mathbf{\text{ (already proved)}} \\ (1-1)x = \theta \\ 1x + (-1)x = \theta &amp;&amp; \mathbf{\text{ (by (VM3))}} \\ x + (-1)x = \theta &amp;&amp;\mathbf{\text{ (by (VM2))}} \\ x + (-x) + (-1) x = \theta + (-x) &amp;&amp;\mathbf{\text{ (adding (-x) on both sides)}} \\ \theta + (-1) x = \theta + (-x) &amp;&amp; \mathbf{\text{ (by (VA3))}} \\ (-1) x + \theta = (-x) + \theta &amp;&amp; \mathbf{\text{ (by (VA1))}} \\ (-1) x = (-x) &amp;&amp; \mathbf{\text{ (by (VA2))}} \end{array}$ $\blacksquare$ <hr /> <h4 id="213-describe-the-span-of-m--111-002-in-mathbbr1">2.1.3 Describe the span of $$M = {(1,1,1), (0,0,2)}$$ in $$\mathbb{R^1}$$.</h4> <p><strong>Answer:</strong></p> <p>The span of $$M$$ is described by:</p> $\begin{bmatrix} 1 &amp;&amp; 0 \\ 1 &amp;&amp; 0 \\ 1 &amp;&amp; 2 \end{bmatrix} \bullet \begin{bmatrix} x \\ y \end{bmatrix}$ <p>Geometrically, this is a plane whose normal is perpendicular to both $$(1,1,1)$$ and $$(0,0,2)$$. Specifically, from the first perpendicularity with $$(0,0,0)$$:</p> $0x+0y+2z=0 \\ z=0$ <p>From the second perpendicularity with $$(1,1,1)$$:</p> $x+y+z=0 \\ x+y=0 \text{ (since z=0)} \\ x=-y$ <p>Choose $$x=1$$, then $$y=-1$$. Then, one choice of the normal vector is $$(1,-1,0)$$. The equation of the plane then becomes:</p> $x-y=0$ <hr /> <h4 id="214-which-of-the-following-subsets-of-mathbbr3-constitute-a-subspace-of-mathbbr3-here-x--xi_1-xi_2-xi_3">2.1.4 Which of the following subsets of $$\mathbb{R}^3$$ constitute a subspace of $$\mathbb{R}^3$$? [Here, $$x = (\xi_1, \xi_2, \xi_3)$$.]</h4> <p><strong>(a) All $$x$$ with $$\xi_1=\xi_2$$ and $$\xi_3=0$$.</strong><br /> <strong>(b) All $$x$$ with $$\xi_1=\xi_2+1$$.</strong><br /> <strong>(c) All $$x$$ with positive $$\xi_1$$, $$\xi_2$$, $$\xi_3$$.</strong><br /> <strong>(d) All $$x$$ with $$\xi_1-\xi_2+\xi_3=k=\text{const}$$.</strong></p> <p><strong>Answer:</strong></p> <p>For a subset to be a subspace, it needs to satisfy the following criterion:</p> $\alpha x + \beta y \in X, \alpha, \beta \in \mathbb{R}$ <p><strong>(a)</strong> Consider two arbitrary members $$x=(\xi, \xi, 0)$$ and $$y=(\eta, \eta, 0)$$ of the given subset (call it $$X$$).</p> <p>Then, we have:</p> $\alpha x + \beta y=\alpha (\xi, \xi, 0) + \beta (\eta, \eta, 0) \\ = (\alpha\xi, \alpha\xi, \alpha 0) + (\beta\eta, \beta\eta, \beta 0) \\ = (\alpha\xi + \beta\eta, \alpha\xi + \beta\eta, \alpha 0 + \beta 0) \\ = (\alpha\xi + \beta\eta, \alpha\xi + \beta\eta, 0) \in X$ <p>Thus, $$X$$ is a subspace of $$\mathbb{R}^3$$.</p> <p><strong>(b)</strong> Consider two arbitrary members $$x=(\xi+1, \xi, 0)$$ and $$y=(\eta+1, \eta, 0)$$ of the given subset (call it $$X$$).</p> <p>Then, we have:</p> $\require{cancel} \alpha x + \beta y=\alpha (\xi+1, \xi, 0) + \beta (\eta+1, \eta, 0) \\ = (\alpha\xi + \alpha, \alpha\xi, \alpha 0) + (\beta\eta + \beta, \beta\eta, \beta 0) \\ = [(\alpha\xi + \beta\eta) + (\alpha + \beta), (\alpha\xi + \beta\eta), \alpha 0 + \beta 0] \\ = [(\alpha\xi + \beta\eta) + (\alpha + \beta), (\alpha\xi + \beta\eta), 0] \cancel{\in} X\\$ <p>Thus, $$X$$ is not a subspace of $$\mathbb{R}^3$$.</p> <p><strong>(c)</strong> Consider two arbitrary members $$x=(\xi_1, \xi_2, \xi_3)$$ and $$y=(\eta_1, \eta_2, \eta_3)$$ of the given subset (call it $$X$$), with $$\xi_i,\eta_i \geq 0$$.</p> <p>Then, we have:</p> $\alpha x + \beta y=\alpha (\xi_1, \xi_2, \xi_3) + \beta (\eta_1, \eta_2, \eta_3) \\ = (\alpha\xi_1, \alpha\xi_2, \alpha\xi_3) + (\beta\eta_1, \beta\eta_2, \beta\eta_3) \\ = (\alpha\xi_1 + \beta\eta_1, \alpha\xi_2 + \beta\eta_2, \alpha\xi_3 + \beta\eta_3)$ <p>Choose any $$\xi_1&gt;\eta_1$$, and $$\alpha=-1$$, $$\beta=1$$. Then, we have:</p> $\require{cancel} \alpha\xi_1 + \beta\eta_1 = -\xi_1 + \eta_1 &lt; 0 \notin X$ <p>Thus, $$X$$ is not a subspace of $$\mathbb{R}^3$$.</p> <p><strong>(d)</strong> Consider two arbitrary members $$x=(\xi_1, \xi_2, k-\xi_1+\xi_2)$$ and $$y=(\eta_1, \eta_2, k-\eta_1+\eta_2)$$ of the given subset (call it $$X$$).</p> <p>Then we have:</p> $\require{cancel} \alpha x + \beta y=\alpha (\xi_1, \xi_2, k-\xi_1+\xi_2) + \beta (\eta_1, \eta_2, k-\eta_1+\eta_2) \\ = [\alpha\xi_1, \alpha\xi_2, \alpha(k-\xi_1+\xi_2)] + [\beta\eta_1, \beta\eta_2, \beta(k-\eta_1+\eta_2)] \\ = [\alpha\xi_1 + \beta\eta_1, \alpha\xi_2 + \beta\eta_2, (\alpha + \beta) k - (\alpha\xi_1 + \beta\eta_1) + (\alpha\xi_2 + \beta\eta_2)] \notin X$ <p>Thus, $$X$$ is not a subspace of $$\mathbb{R}^3$$.</p> $\blacksquare$ <hr /> <h4 id="215-s-show-that-x_1-cdots-x_n-where-x_jt--tj-is-a-linearly-independent-set-in-the-space-cab">2.1.5 s. Show that $${x_1, \cdots, x_n}$$, where $$x_j(t) = t^j$$, is a linearly independent set in the space $$C[a,b]$$.</h4> <p><strong>Proof:</strong></p> <p>A set $$\{x_1, x_2, \cdots, x_n\}$$ is linearly independent if:</p> $\alpha_1 x_1 + \alpha_2 x_2 + \cdots \alpha_n x_n = x_0$ <p>only for all $$\alpha_i=0$$.</p> <p>Any linear combination of the given set, call it $$X$$, is given by:</p> $f(t)=\alpha_1 t^1 + \alpha_2 t^2 + \cdots + \alpha_n t^n$ <p>To prove that $$X$$ is linearly independent, we need to prove that that the zero vector $$f(t)=0=f_0(t)$$ is not possible for any combination of $$\alpha_i$$.</p> <p>This is a polynomial of degree $$n$$. Fix all $$\alpha_i$$, with not all of them zero.<br /> By the <strong>Fundamental Theorem of Algebra</strong>, we know that it can have at most $$n$$ roots of this polynomial. Thus, there are at most $$n$$ values of $$t$$ for which $$f(t)=0$$. Thus, it is not zero for the remaining uncountable values of $$t \in [a,b]$$, therefore for an arbitrary combination of $$\alpha_i$$, $$f(t) \neq f_0(t)$$.</p> <p>Thus, $$X$$ is a linearly independent set.</p> $\blacksquare$ <hr /> <h4 id="216-show-that-in-an-n-dimensional-vector-space-x-the-representation-of-any-x-as-a-linear-combination-of-given-basis-vectors-e_1-cdots-e_n-is-unique">2.1.6 Show that in an $$n$$-dimensional vector space $$X$$, the representation of any $$x$$ as a linear combination of given basis vectors $$e_1, \cdots, e_n$$ is unique.</h4> <p><strong>Proof:</strong></p> <p>Assume that $$x=\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n$$. Assume that $$x$$ can also be represented by a different linear combination $$\beta_i$$, such that:</p> $x=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n$ <p>Then we have:</p> $\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n \\ (\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n)+(-\beta_1 e_1) +(-\beta_2 e_2) + \cdots + (-\beta_n e_n)=\beta_1 e_1 + \beta_2 e_2 + \cdots + \beta_n e_n + (-\beta_1 e_1) +(-\beta_2 e_2) + \cdots + (-\beta_n e_n) \\ (\alpha_1 - \beta_1) e_1 + (\alpha_2 - \beta_2) e_2 + \cdots + (\alpha_n -\beta_n) e_n =(\beta_1 - \beta_1) e_1 + (\beta_2 - \beta_2) e_2 + \cdots + (\beta_n - \beta_n) e_n \\ (\alpha_1 - \beta_1) e_1 + (\alpha_2 - \beta_2) e_2 + \cdots + (\alpha_n -\beta_n) e_n =0 e_1 + 0 e_2 + \cdots + 0 e_n \\$ <p>Then equating the coefficients on both sides, we get:</p> $\alpha_i-\beta_i=0 \\ \alpha_i=\beta_i$ <p>Thus, the representation of any $$x$$ as a linear combination of given basis vectors $$e_1, \cdots, e_n$$ is unique.</p> $\blacksquare$ <hr /> <h4 id="217-let-e_1-cdots-e_n-be-a-basis-for-a-complex-vector-space-x-find-a-basis-for-x-regarded-as-a-real-vector-space-what-is-the-dimension-of-x-in-either-case">2.1.7 Let $$\{e_1, \cdots, e_n\}$$ be a basis for a complex vector space $$X$$. Find a basis for $$X$$ regarded as a real vector space. What is the dimension of $$X$$ in either case?</h4> <p><strong>Answer:</strong></p> <p>A complex vector space is a vector space whose field of scalars is the complex numbers. Then any vector $$x$$ in this complex vector space is representable as:</p> $x=(a_1 + ib_1) e_1 + (a_2 + ib_2) e_2 + \cdots + (a_n + ib_n) e_n \\ =(a_1 e_1 + a_2 e_2 + \cdots + a_n e_n) + (b_1 ie_1 + b_2 ie_2 + \cdots + b_n ie_n)$ <p>The basis for $$X$$ regarded as a real vector space are:</p> $(e_1, e_2, \cdots, e_n, ie_1, ie_2, \cdots, ie_n)$ <p>The dimension of the complex space is $$n$$.<br /> The dimension of the complex space regarded as a real vector space is $$2n$$.</p> <hr /> <h4 id="218-if-m-is-a-linearly-dependent-set-in-a-complex-vector-space-x-is-m-linearly-dependent-in-x-regarded-as-a-real-vector-space">2.1.8 If $$M$$ is a linearly dependent set in a complex vector space $$X$$, is $$M$$ linearly dependent in $$X$$, regarded as a real vector space?</h4> <p><strong>Proof:</strong></p> <p>Consider the set $$\{u=i,v=-1\}$$. This is a linearly dependent set because $$iv=u$$, since $$i.i=-1$$. But there is no $$\alpha \in \mathbb{C}$$ which gives $$\alpha u=v$$, i.e., $$\alpha i=-1$$.</p> <p>Thus, $$X$$, regarded as a real vector space, is not necessarily dependent.</p> $\blacksquare$ <hr /> <h4 id="219-on-a-fixed-interval-ab-subset-mathbbr-consider-the-set-x-consisting-of-all-polynomials-with-real-coefficients-and-of-degree-not-exceeding-a-given-n-and-the-polynomial-x0-for-which-a-degree-is-not-defined-in-the-usual-discussion-of-degree-show-that-x-with-the-usual-addition-and-the-usual-multiplication-by-real-numbers-is-a-real-vector-space-of-dimension-n1-find-a-basis-for-x-show-that-we-can-obtain-a-complex-vector-space-tildex-in-a-similar-fashion-if-we-let-those-coefficients-be-complex-is-x-a-subspace-of-tildex">2.1.9 On a fixed interval $$[a,b] \subset \mathbb{R}$$, consider the set $$X$$ consisting of all polynomials with real coefficients and of degree not exceeding a given $$n$$, and the polynomial $$x=0$$ (for which a degree is not defined in the usual discussion of degree). Show that $$X$$, with the usual addition and the usual multiplication by real numbers, is a real vector space of dimension $$n+1$$. Find a basis for $$X$$. Show that we can obtain a complex vector space $$\tilde{X}$$ in a similar fashion if we let those coefficients be complex. Is $$X$$ a subspace of $$\tilde{X}$$?</h4> <p><strong>Proof:</strong></p> <p>A polynomial not exceeding degree $$n$$ is given as: $$f(x)=\sum\limits_{i=0}^n \alpha_i x^i$$.</p> $\blacksquare$ <hr /> <h4 id="2110-if-y-and-z-are-subspaces-of-a-vector-space-x-show-that-ycap-z-is-a-subspace-of-x-but-ycup-z-need-not-be-one-give-examples">2.1.10 If $$Y$$ and $$Z$$ are subspaces of a vector space $$X$$, show that $$Y\cap Z$$ is a subspace of $$X$$, but $$Y\cup Z$$ need not be one. Give examples.</h4> <p><strong>Proof:</strong></p> <p>We have, for all $$\alpha, \beta \in \mathbb{R}$$:</p> $\alpha y_1 + \beta y_2 \in Y \\ \alpha z_1 + \beta z_2 \in Z$ <p>Assume $$x \in Y \cap Z$$.<br /> Then $$x \in Y$$ and $$x \in Z$$.<br /> Then $$\alpha x_1 + \beta x_2 \in Y$$ and $$\alpha x_1 + \beta x_2 \in Z$$.<br /> Then $$\alpha x_1 + \beta x_2 \in Y \cap Z$$</p> <p>Thus, $$Y \cap Z$$ is a subspace of $$X$$.</p> $\blacksquare$ <p>In $$\mathbb{R}^2$$, the vector space with the basis vector $$u=(1,0)$$ and the vector space with the basis vector $$v=(0,1)$$ gives two vector spaces $$Y$$ and $$Z$$. Choose $$\alpha=1$$, $$\beta=1$$, then $$\alpha u + \beta v = (1,1)$$ does not belong to $$Y \cup Z$$.</p> <p>Thus, $$Y \cap Z$$ is not necessarily a subspace of $$X$$.</p> $\blacksquare$ <hr /> <h4 id="2111-if-m-neq-emptyset-is-any-subset-of-a-vector-space-x-show-that-span-m-is-a-subspace-of-x">2.1.11 If $$M \neq \emptyset$$ is any subset of a vector space $$X$$, show that span $$M$$ is a subspace of $$X$$.</h4> <p><strong>Proof:</strong></p> <p>Let $$M \subset X$$. Let $$e_1, e_2, \cdots, e_n \in M$$. Then, the span of $$M$$ is $$\sum\limits_{i=1}^n\alpha_i e_n$$.</p> <p>Then, every $$x=\sum\limits_{i=1}^n\alpha_i e_n, x \in M$$. Pick any two arbitrary points in $$M$$, so that:</p> $x_1=\sum\limits_{i=1}^n\alpha_i e_i \\ x_2=\sum\limits_{i=1}^n\beta_i e_i \\$ <p>Then, we get:</p> $ax_1+bx_2=\sum\limits_{i=1}^n a\alpha_i e_i + \sum\limits_{i=1}^n b\beta_i e_i \\ =\sum\limits_{i=1}^n (a\alpha_i+b\beta_i) e_i = \sum\limits_{i=1}^n k_i e_i \in M$ <p>Thus, span $$M$$ is a subspace of $$X$$.</p> $\blacksquare$ <hr /> <h4 id="2112-show-that-the-set-of-all-real-two-rowed-square-matrices-forms-a-vector-space-x-what-is-the-zero-vector-in-x-determine-dim-x-find-a-basis-for-x-give-examples-of-subspaces-of-x-do-the-symmetric-matrices-x-in-x-form-a-subspace-the-singular-matrices">2.1.12 Show that the set of all real two-rowed square matrices forms a vector space $$X$$. What is the zero vector in $$X$$? Determine dim $$X$$. Find a basis for $$X$$. Give examples of subspaces of X. Do the symmetric matrices $$x \in X$$ form a subspace? The singular matrices?</h4> <p><strong>Proof:</strong></p> <p>A real-valued $$2 \times 2$$ matrix is of the form:</p> $\begin{bmatrix} a &amp;&amp; b \\ c &amp;&amp; d \end{bmatrix}$ <p>Assume $$x_1$$, $$x_2$$ as below:</p> $x_1=\begin{bmatrix} a &amp;&amp; b \\ c &amp;&amp; d \end{bmatrix}\\ x_2=\begin{bmatrix} p &amp;&amp; q \\ q &amp;&amp; s \end{bmatrix}$ <p>Then, we have:</p> $\alpha x_1 + \beta x_2 = \alpha \begin{bmatrix} a &amp;&amp; b \\ c &amp;&amp; d \end{bmatrix} + \beta \begin{bmatrix} p &amp;&amp; q \\ r &amp;&amp; s \end{bmatrix} \\ = \begin{bmatrix} \alpha a &amp;&amp; \alpha b \\ \alpha c &amp;&amp; \alpha d \end{bmatrix} + \begin{bmatrix} \beta p &amp;&amp; \beta q \\ \beta r &amp;&amp; \beta s \end{bmatrix} \\ = \begin{bmatrix} \alpha a + \beta p &amp;&amp; \alpha b + \beta q \\ \alpha c + \beta r &amp;&amp; \alpha d + \beta s \end{bmatrix}$ <p>This is also a $$2 \times 2$$ matrix, thus the set of all real two-rowed square matrices forms a vector space.</p> <p>The zero vector of $$X$$ is $$\begin{bmatrix} 0 &amp;&amp; 0 \\ 0 &amp;&amp; 0 \end{bmatrix}$$</p> <p>The dimension of $$X$$ is 4. A possible basis for $$X$$ is:</p> $\begin{bmatrix} 1 &amp;&amp; 0 \\ 0 &amp;&amp; 0 \end{bmatrix}, \begin{bmatrix} 0 &amp;&amp; 1 \\ 0 &amp;&amp; 0 \end{bmatrix}, \begin{bmatrix} 0 &amp;&amp; 0 \\ 1 &amp;&amp; 0 \end{bmatrix}, \begin{bmatrix} 0 &amp;&amp; 0 \\ 0 &amp;&amp; 1 \end{bmatrix}$ <p>An example of a subspace of $$X$$ is $$\begin{bmatrix} k &amp;&amp; 0 \\ 0 &amp;&amp; 0 \end{bmatrix}, k \in \mathbb{R}$$.</p> <p>Yes, the symmetric matrices form a subspace. No, the singular matrices do not form a subspace. Here is a counter-example. Take $$x$$ and $$y$$ to be as follows:</p> $x=\begin{bmatrix} 2 &amp;&amp; 6 \\ 4 &amp;&amp; 12 \end{bmatrix}, y=\begin{bmatrix} 1 &amp;&amp; 1 \\ 1 &amp;&amp; 1 \end{bmatrix}$ <p>The determinants for both $$x$$ and $$y$$ are zero, thus, they are singular.</p> <p>Then, setting $$\alpha=1$$, $$\beta=1$$, we get:</p> $\alpha x + \beta = \begin{bmatrix} 2 &amp;&amp; 6 \\ 4 &amp;&amp; 12 \end{bmatrix} + \begin{bmatrix} 1 &amp;&amp; 1 \\ 1 &amp;&amp; 1 \end{bmatrix}= \begin{bmatrix} 3 &amp;&amp; 7 \\ 5 &amp;&amp; 13 \end{bmatrix}$ <p>The determinant of the result is $$39-35=4 \neq 0$$, thus it is not a singular matrix.</p> $\blacksquare$ <hr /> <h4 id="2113-product-show-that-the-cartesian-product-x--x_1-times-x_2-of-two-vector-spaces-over-the-same-field-becomes-a-vector-space-if-we-define-the-two-algebraic-operations-by">2.1.13 (Product) Show that the Cartesian product $$X = X_1 \times X_2$$ of two vector spaces over the same field becomes a vector space if. we define the two algebraic operations by</h4> <p>$$(x_1. x_2) + (y_1, y_2) = (x_1 +y_1. x_2 + y_2)$$,<br /> $$\alpha(x_1, x_2) = (\alpha x_1, \alpha x_2)$$.</p> <p><strong>Proof:</strong></p> <p>[Easy. TODO]</p> $\blacksquare$ <hr /> <h4 id="2114-quotient-space-codimension-let-y-be-a-subspace-of-a-vector-space-x-the-coset-of-an-element-x-in-x-with-respect-to-y-is-denoted-by-x--y-and-is-defined-to-be-the-set-see-fig-12">2.1.14 (Quotient space, codimension) Let $$Y$$ be a subspace of a vector space $$X$$. The coset of an element $$x \in X$$ with respect to $$Y$$ is denoted by $$x + Y$$ and is defined to be the set (see Fig. 12)</h4> <p>$$x+Y={v\vert v=x+y, \in Y}$$.</p> <p><strong>Show that the distinct cosets form a partition of X. Show that under algebraic operations defined by (see Figs. 13, 14)</strong></p> $(w+Y)+(x+Y)=(w+x)+Y \alpha(x+Y)=\alpha x+Y$ <p><strong>these cosets constitute the elements of a vector space. This space is called the quotient space (or sometimes factor space) of $$X$$ by $$Y$$ (or modulo $$Y$$) and is denoted by $$X/Y$$. Its dimension is called the codimension of $$Y$$ and is denoted by codim $$Y$$, that is, $$\text{codim } Y=\text{dim }(X/Y)$$.</strong></p> <p><strong>Proof:</strong></p> <p>To prove that the cosets form a partition of $$X$$, we need to prove that an arbitrary element $$x \in X$$ belongs to one and only one coset.</p> <p>Suppose $$x \in X$$ belongs to two cosets $$u+Y$$ and $$v+Y$$. Then, we have from the definition:</p> $x=v|v=u+y_1,y \in Y \\ x=v|v=v+y_2,y \in Y$ <p>Then, we have:</p> $u+y_1=v+y_2 \\ u-v=y_2-y_1 \in Y\\$ <p>Then $$u-v \in Y$$. Then $$u-v=y_0, y_0 \in Y$$.<br /> Then $$u=v+y_0$$, where $$y_0 \in Y$$. Then $$u \in v+Y$$. Since $$u \in u+Y$$, $$u+Y$$ and $$v+Y$$ are the same coset.</p> $\blacksquare$ <p>[Easy to prove that it’s a vector space. TODO]</p> <hr /> <h4 id="2115-let-xmathbbr3-and-yxi_100-vert-xi-in-mathbbr-find-xy-xx-x0">2.1.15 Let $$X=\mathbb{R}^3$$ and $$Y=\{(\xi_1,0,0) \vert \xi \in \mathbb{R}\}$$. Find $$X/Y$$, $$X/X$$, $$X/{0}$$.</h4> <p><strong>Answer:</strong></p> <p>Loosely, the quotient space is the set of points which can translate cosets to cover the entire vector space.</p> <p>For $$X/Y$$, we get the coset as the set of parallel subspaces along the vector $$(1,0,0)$$.</p> <p>For $$X/X$$, any translation of the coset $$X$$ covers the entire vector space $$X$$. No translation also covers the entire space. This implies that $$x=\{0\}$$.</p> <p>For $$X/\{0\}$$, we have $$x+\{0\}=\{v:v=x+0\}$$, i.e., each coset is the point itself. Thus the set of points required to partition $$X$$ into cosets is $$X$$ itself.</p>avishekThis post lists solutions to the exercises in the Vector Space section 2.1 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Functional Analysis Exercises 6 : Completion of Metric Spaces2021-10-25T00:00:00+05:302021-10-25T00:00:00+05:30/2021/10/25/functional-analysis-exercises-6-completion-metric-spaces<p>This post lists solutions to the exercises in the <strong>Completion of Metric Spaces section 1.6</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h4 id="161-show-that-if-a-subspace-y-of-a-metric-space-consists-of-finitely-manypoints-then-y-is-complete">1.6.1 Show that if a subspace $$Y$$ of a metric space consists of finitely manypoints, then $$Y$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>By definition, a limit point $$L$$ of the set $$Y$$ has at least one point $$x \neq L$$ within every neightbourhood $$\epsilon$$. Since $$Y$$ has a finite number of points, then it has no limit points, and thus (vacuously) contains all its limit points.</p> <p>Thus, $$Y$$ is a complete metric subspace.</p> $\blacksquare$ <hr /> <h4 id="162-what-is-the-completion-of-xd-where-x-is-the-set-of-all-rational-numbers-and-dxyvert-x-y-vert">1.6.2 What is the completion of $$(X,d)$$, where $$X$$ is the set of all rational numbers and $$d(x,y)=\vert x-y \vert$$?</h4> <p><strong>Proof:</strong> The completion of $$(X,d)$$, where $$X$$ is the set of all rational numbers and $$d(x,y)=\vert x-y \vert$$, is $$\mathbb{R}$$, since every real number is the limit of a sequence of rational numbers.</p> $\blacksquare$ <hr /> <h4 id="163-what-is-the-completion-of-a-discrete-metric-space-x-cf-11-8">1.6.3. What is the completion of a discrete metric space $$X$$? (Cf. 1.1-8.).</h4> <p><strong>Proof:</strong> A discrete metric space $$X$$ has no limit points, since no point in it has at least one point in every neighbourhood $$\epsilon$$. Thus, it vacuously contains all its limit points. Thus, the completion of the discrete metric space $$X$$ is itself.</p> $\blacksquare$ <hr /> <h4 id="164-if-x_1-and-x_2-are-isometric-and-x_1-is-complete-show-that-x_2-is-complete">1.6.4 If $$X_1$$ and $$X_2$$ are isometric and $$X_1$$ is complete, show that $$X_2$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>Assume $$x,y \in X_1$$, and let $$T:X_1 \rightarrow X_2$$. Since $$X_1$$ and $$X_2$$ are isometric, we have:</p> $d(x,y)=\bar{d}(Tx,Ty)$ <p>We know that $$X_1$$ is complete: let us assume, for an arbitrary $$\epsilon$$, a point $$x_1 \in X_1$$ lying in the $$\epsilon$$-neighbourhood of a limit point $$x \in X_1$$. Then, we have:</p> $d(x,x_1)=d(Tx,Tx_1) &lt; \epsilon$ <p>Thus, for an arbitrary $$\epsilon$$, there is a point $$Tx \in X_2$$ which has a point $$Tx_1$$ in its $$\epsilon$$-neighbourhood as well. Thus, $$Tx$$ is a limit point of $$X_2$$ as well. Since $$Tx \in X_2$$ for all $$x \in X_1$$, $$X_2$$ contains all its limit points as well.</p> <p>Thus, $$X_2$$ is complete.</p> $\blacksquare$ <hr /> <h4 id="165-homeomorphism-a-homeomorphism-is-a-continuous-bijective-mapping-t-x-rightarrow-y-whose-inverse-is-continuous-the-metric-spaces-x-and-y-are-then-said-to-be-homeomorphic-a-show-that-if-x-and-y-are-isometric-they-are-homeomorphic-b-illustrate-with-an-example-that-a-complete-and-an-incomplete-metric-space-may-be-homeomorphic">1.6.5 (Homeomorphism) A homeomorphism is a continuous bijective mapping $$T: X \rightarrow Y$$ whose inverse is continuous; the metric spaces $$X$$ and $$Y$$ are then said to be homeomorphic. (a) Show that if $$X$$ and $$Y$$ are isometric, they are homeomorphic. (b) Illustrate with an example that a complete and an incomplete metric space may be homeomorphic.</h4> <p><strong>Proof:</strong></p> <p>Consider a Cauchy sequence $$(x_n)$$ in $$X$$. Then, we have, $$\forall \epsilon&gt;0$$, $$\exists N$$, such that $$d(x_m,x_n) &lt; \epsilon$$ for all $$m,n&gt;N$$.</p> <p>Let $$T:X \rightarrow Y$$. Since $$X$$ is isometric to $$Y$$, we have:</p> $d(x_m,x_n)=d(Tx_m,Tx_n) &lt; \epsilon$ <p>This implies that for every $$\epsilon&gt;0$$, there exists a $$\delta&gt;0$$, such that $$d(x_m,x_n) &lt; \delta \Rightarrow d(Tx_m,Tx_n) &lt; \epsilon$$. In this case $$\delta=\epsilon$$. Thus, $$T$$ is continuous at $$x_n$$.</p> <p>The above argument can be used for $$T^{-1}$$ to prove that it is also continuous.</p> <p>To prove injectivity, we note that $$x \neq y \Rightarrow d(x,y) \neq 0 \Rightarrow \Rightarrow d(Tx,Ty) \neq 0 \Rightarrow Tx \neq Ty$$.</p> <p>To prove surjectivity, we pick a point $$y \in Y$$. Assume $$x_1 \in X$$. Then, by isometry we must have: d(y,Tx_1)=d(x, x_1), where $$x \in X$$. Thus, there is a corresponding preimage for every $$y \in Y$$.</p> $\blacksquare$ <p>Consider the $$f:(0,1) \rightarrow \mathbb{R}$$ defined as $$f(x)=x$$. Then $$f(x)$$ and its inverse are continuous and bijective. $$(0,1)$$ is an incomplete metric space and $$\mathbb{R}$$ is complete.</p> <hr /> <h4 id="166-show-that-c01-and-cab-are-isometric">1.6.6 Show that $$C[0,1]$$ and $$C[a,b]$$ are isometric.</h4> <p><strong>Proof:</strong></p> <p>We note that $$f(t)=\displaystyle\frac{t-a}{b-a}, a \neq b$$ is a mapping $$f: [a,b] \rightarrow [0,1]$$, and that $$f^{-1}(t)=a+(b-a)t$$ is a mapping $$f^{-1}: [0,1] \rightarrow [a,b]$$.</p> <p>We note that $$f$$ and $$f^{-1}$$ are bijections. The distance metric in $$C$$ is defined as $$d(x,y)=\sup\vert x(t) - y(t) \vert$$.</p> <p>Define a mapping $$T:C_{t \in [0,1]}(t) \rightarrow C_{t \in [a,b]}(f^{-1}(t))$$</p> <p>Think of $$C(f(t)$$ as the original function applied to $$[0,1]$$ even though the input $$t \in [a,b]$$. Then, practically we have $$C_{t \in [0,1]}(t)=C_{t \in [a,b]}(f(t))$$.</p> <p>Then:</p> $d(Tx,Ty)=\sup_{[a,b]} |x(f(t)) - y(f(t))|=\sup_{[0,1]} |x(t) - y(t)|=d(x,y)$ <p>Thus, $$T$$ preserves distances.</p> <p>To prove injectivity, suppose $$Tx=Ty$$, then we have:</p> $d(Tx,Ty)=\sup_{[a,b]} |x(f(t)) - y(f(t))|=0 \\ \Rightarrow \sup_{[0,1]} |x(t) - y(t)|=d(x,y)=0 \\ \Rightarrow x=y$ <p>For surjectivity, we note that for an arbitrary function $$y(f(t)) \in C[a,b]$$, we always have $$x(t) \in C[0,1]$$, since $$T^{-1}x=x(f^{-1}(f(t)))=x(t)$$.</p> $\blacksquare$ <hr /> <h4 id="167-if-xd-is-complete-show-that-xtilded-where-tilded--dl--d-is-complete">1.6.7 If $$(X,d)$$ is complete, show that $$(X,\tilde{d})$$, where $$\tilde{d} = d/(l + d)$$, is complete.</h4> <p><strong>Proof:</strong></p> <p>Let $$(x_n)$$ be a Cauchy sequence in $$(X,\bar{d})$$, so that we have, $$\forall \epsilon&gt;0, \exists N$$, such that $$d(x_m,x_n) &lt; \epsilon$$ for all $$m,n&gt;N$$.</p> <p>Then, we have:</p> $\frac{d(x_m,x_n)}{1+d(x_m,x_n)} &lt; \epsilon \\ d(x_m,x_n) &lt; \epsilon + \epsilon d(x_m,x_n) \\ d(x_m,x_n)(1-\epsilon) &lt; \epsilon \\ d(x_m,x_n) &lt; \frac{\epsilon}{1-\epsilon}$ <p>Set $$\epsilon=\frac{1}{k}$$, so that we get:</p> ${d}(x_m,x_n) &lt; \frac{1}{k-1} \\$ <p>$$k$$ can be made as large as needed to make $$\epsilon$$ as small as needed. Thus, the sequence $$(x_n)$$ is Cauchy in $$(X,d)$$, and thus has a limit $$x$$, i.e., $$x_n \rightarrow x$$.</p> <p>Then, $$d(x_n,x)&lt;\epsilon$$.</p> $\bar{d}(x_n,x)&lt;d(x_n,x)&lt;\epsilon$ $\blacksquare$ <hr /> <h4 id="168-show-that-in-prob-7-completeness-of-xtilded-implies-completeness-of-xd">1.6.8 Show that in Prob. 7, completeness of $$(X,\tilde{d})$$ implies completeness of $$(X,d)$$.</h4> <p><strong>Proof:</strong> Suppose $$(X,\tilde{d})$$ is complete. Then, we have:</p> $\blacksquare$ <hr /> <h4 id="169--if-x_n-and-x_n-in-xd-are-such-that-1-holds-and-x_n-rightarrow-l-show-that-x_n-converges-and-has-the-limit-l">1.6.9 If $$(x_n)$$ and $$(x_n')$$ in $$(X,d)$$ are such that (1) holds and $$x_n \rightarrow l$$, show that $$(x_n')$$ converges and has the limit $$l$$.</h4> <p><strong>Proof:</strong></p> $\blacksquare$ <hr /> <h4 id="1610--if-x_n-and-x_n-are-convergent-sequences-in-a-metric-space-xd-and-have-the-same-limit-l-show-that-they-satisfy-1">1.6.10 If $$(x_n)$$ and $$(x_n')$$ are convergent sequences in a metric space $$(X,d)$$ and have the same limit $$l$$, show that they satisfy (1).</h4> <p>(1) defines equivalence of two sequences as $$(x_n)\tilde(x_n') \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0$$.</p> <p><strong>Proof:</strong></p> <p>Since $$(x_n)$$ and $$(x_n')$$ are convergent, we have, $$\forall \epsilon&gt;0, \exists N_1, N_2$$ such that $$d(x_m,l)&lt;\epsilon$$ and $$d(x_n',l)&lt;\epsilon$$, for $$m&gt;N_1, n&gt;N_2$$. Choose $$N=\max(N_1,N_2)$$, so that we have $$d(x_n,l)&lt;\epsilon$$ and $$d(x_n',l)&lt;\epsilon$$ for all $$n&gt;N$$.</p> $d(x_n,x_n') \leq d(x_n,l) + d(l,x_n') &lt; \epsilon+\epsilon=2 \epsilon \\ \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n') = 0$ $\blacksquare$ <hr /> <h4 id="1611---show-that-1-defines-an-equivalence-relation-on-the-set-of-all-cauchy-sequences-of-elements-of-x">1.6.11 Show that (1) defines an equivalence relation on the set of all Cauchy sequences of elements of $$X$$.</h4> <p>(1) defines equivalence of two sequences as $$(x_n)\tilde{}(x_n') \Rightarrow \lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0$$.</p> <p><strong>Proof:</strong></p> <p>We will check for the following properties:</p> <ul> <li>Reflexive</li> <li>Symmetric</li> <li>Transitive</li> </ul> <p>We know that $$d(x_n,x_n)=0$$ always because of the <strong>Principle of Indiscernibles</strong>. Thus, we get:</p> $\lim\limits_{n \rightarrow \infty} d(x_n,x_n)=0$ <p>By the <strong>Symmetry Property</strong> of a distance metric, we know that $$d(x_n,x_n')=d(x_n',x_n)$$. Thus if we have $$\lim\limits_{n \rightarrow \infty} d(x_n,x_n')=0$$, then we also have:</p> $\lim\limits_{n \rightarrow \infty} d(x_n',x_n)=0$ <p>By the <strong>Triangle Inequality</strong>, we have:</p> $d(x_n,z_n)&lt;d(x_n,y_n)+d(y_n,z_n)$ <p>Taking limits, we get:</p> $\lim\limits_{n \rightarrow \infty} d(x_n,z_n) \leq \lim\limits_{n \rightarrow \infty} d(x_n,y_n) + \lim\limits_{n \rightarrow \infty} d(y_n,z_n)$ <p>If we have $$\lim\limits_{n \rightarrow \infty} d(x_n,y_n)=0$$ and $$\lim\limits_{n \rightarrow \infty} d(y_n,z_n)=0$$, we get:</p> $\lim\limits_{n \rightarrow \infty} d(x_n,z_n) \leq 0$ <p>Since distances are always nonnegative, we have: $$\lim\limits_{n \rightarrow \infty} d(x_n,z_n) = 0$$.</p> $\blacksquare$ <hr /> <h4 id="1612---if-x_n-is-cauchy-in-xd-and-x_n-in-x-satisfies-1-show-that-x_n-is-cauchy-in-x">1.6.12 If $$(x_n)$$ is Cauchy in $$(X,d)$$ and $$(x_n')$$ in $$X$$ satisfies (1), show that $$(x_n')$$ is Cauchy in $$X$$.</h4> <p><strong>Proof:</strong></p> <p>Since $$(x_n)$$ is Cauchy, we have, $$\forall \epsilon&gt;0, \exists N$$ such that $$d(x_m,x_n)&lt;\epsilon$$ for $$m,n&gt;N$$.</p> $d(x_m,x_n) &lt; \epsilon$ <p>We also have $$(x_n)$$ and $$(x_n')$$ being equivalent, so we can write:</p> $\lim\limits_{n \rightarrow \infty} d(x_n,x_n') = 0$ <p>By the <strong>Triangle Inequality</strong>, we have:</p> $d(x_m',x_n') \leq d(x_m',x_m) + d(x_m,x_n) + d(x_n,x_n')$ <p>Taking limits on both sides, we get:</p> $\lim\limits_{n \rightarrow \infty} d(x_m',x_n') \leq \underbrace{\lim\limits_{n \rightarrow \infty} d(x_m',x_m)}_\text{0 because equivalent} + \underbrace{\lim\limits_{n \rightarrow \infty} d(x_m,x_n)}_\text{0 because Cauchy} + \underbrace{\lim\limits_{n \rightarrow \infty} d(x_n,x_n')}_\text{0 because equivalent} \\$ <p>Since distance metric has to be nonnegative, we conclude that:</p> $\lim\limits_{n \rightarrow \infty} d(x_m',x_n')=0$ $\blacksquare$ <hr /> <h4 id="1613-pseudometric-a-finite-pseudometric-on-a-set-x-is-a-function-d-x-times-x-rightarrow-r-satisfying-m1-m3-m4-sec-11-and-m2-dxx0-what-is-the-difference-between-a-metric-and-a-pseudometric-show-that-dxyvert-xi_1---eta_1vert-defines-a-pseudometric-on-the-set-of-all-ordered-pairs-of-real-numbers-where-x--xi_1xi_2-y--eta_1eta_2-we-mention-that-some-authors-use-the-term-semimetric-instead-of-pseudometric">1.6.13 (Pseudometric) A finite pseudometric on a set $$X$$ is a function $$d: X \times X \rightarrow R$$ satisfying (M1), (M3), (M4), Sec. 1.1, and (M2*) $$d(x,x)=0$$. What is the difference between a metric and a pseudometric? Show that $$d(x,y)=\vert \xi_1 - \eta_1\vert$$ defines a pseudometric on the set of all ordered pairs of real numbers, where $$x = (\xi_1,\xi_2), y = (\eta_1,\eta_2)$$. (We mention that some authors use the term semimetric instead of pseudometric.)</h4> <p><strong>Proof:</strong></p> <p><strong>(M1)</strong> We know that $$d(x,y)=\vert \xi_1 - \eta_1\vert$$ is always nonnegative, real-valued, and finite.<br /> <strong>(M3)</strong> Because $$d(x,y)=\vert \xi_1 - \eta_1\vert$$ has a modulus sign, we always have: $$\vert \xi_1 - \eta_1\vert = \vert \eta_1 - \xi_1\vert$$, and thus we have symmetry.<br /> <strong>(M4)</strong> We have: $$d(x,y)=\vert \xi_1 - \eta_1\vert = \vert \xi_1 - \kappa_1 + \kappa_1 - \eta_1\vert \leq \vert \xi_1 - \kappa_1 \vert + \vert \kappa_1 - \eta_1\vert$$. Thus, the <strong>Triangle Inequaity</strong> is shown.</p> <p><strong>(Modified M2)</strong> An example pair which satisfies this condition is $$(1,2)$$ and $$(1,3)$$. We see that any pair $$(\kappa,\xi)$$ and $$(\kappa,\eta)$$ will satisfy <strong>(Modified M2)</strong>. We see that if $$x_1=(\kappa,\xi)$$ and $$x_2=(\kappa,\eta)$$, then $$d(x,y)=\vert \kappa - \kappa\vert=0$$.</p> $\blacksquare$ <hr /> <h4 id="1614-does-dxyintlimits_abvert-xt-ytvert-dt-define-a-metric-or-pseudometric-on-x-if-x-is-i-the-set-of-all-real-valued-continuous-functions-on-ab-ii-the-set-of-all-real-value-riemann-integrable-functions-on-ab">1.6.14 Does $$d(x,y)=\int\limits_a^b\vert x(t)-y(t)\vert dt$$ define a metric or pseudometric on $$X$$ if $$X$$ is (i) the set of all real-valued continuous functions on $$[a,b]$$, (ii) the set of all real-value Riemann integrable functions on $$[a,b]$$?</h4> <p><strong>Proof:</strong></p> <p><strong>[TODO]</strong></p> $\blacksquare$ <hr /> <h4 id="1615-if-xd-is-a-pseudometric-space-we-call-a-set-bx_0-r--x-in-x--dxx_0--r-ro-an-open-ball-in-x-with-center-x_0-and-radius-r-note-that-this-is-analogous-to-13-1-what-are-open-balls-of-radius-1-in-prob-13">1.6.15 If $$(X,d)$$ is a pseudometric space, we call a set $$B(x_0; r) = {x \in X : d(x,x_0) &lt; r} (r&gt;O)$$ an open ball in $$X$$ with center $$x_0$$ and radius $$r$$. (Note that this is analogous to 1.3-1.) What are open balls of radius $$1$$ in Prob. 13?</h4> <p><strong>Answer:</strong></p> <p>The open ball in this case is a vertical rectangles with open width 2 centered at $$x_0$$.</p>avishekThis post lists solutions to the exercises in the Completion of Metric Spaces section 1.6 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Functional Analysis Exercises 5 : Completeness Proofs2021-10-16T00:00:00+05:302021-10-16T00:00:00+05:30/2021/10/16/functional-analysis-exercises-5-completeness-proofs<p>This post lists solutions to the exercises in the <strong>Completeness Proofs section 1.5</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h4 id="151-let-ab-in-mathbbr-and-ab-show-that-the-open-interval-ab-is-an-incomplete-subspace-of-mathbbr-whereas-the-closed-interval-a-b-is-complete">1.5.1 Let $$a,b \in \mathbb{R}$$ and $$a&lt;b$$. Show that the open interval $$(a,b)$$ is an incomplete subspace of $$\mathbb{R}$$, whereas the closed interval $$[a, b]$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>$$(a,b)$$ is not a closed set, since the limits of Cauchy sequences which converge to $$a$$ and $$b$$ are not contained in $$(a,b)$$. Thus $$(a,b)$$ is not complete.</p> <p>$$[a,b]$$ is a closed set since it contains the limits of all the Cauchy sequences which converge, including $$a$$ and $$b$$. Thus, $$[a.b]$$ is a complete metric space.</p> $\blacksquare$ <hr /> <h4 id="152-let-x-be-the-space-of-all-ordered-n-tuples-x--xi_1-cdots-xi_n-of-real-numbers-and-dxymax_j-vert-xi_j-eta_jvert-where-yeta_j-show-that-xd-is-complete">1.5.2 Let $$X$$ be the space of all ordered n-tuples $$x = (\xi_1, \cdots, \xi_n)$$ of real numbers and $$d(x,y)=\max_j \vert \xi_j-\eta_j\vert$$ where $$y=(\eta_j)$$. Show that $$(X,d)$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>Consider a Cauchy sequence of ordered n-tuples $$(\xi^m) \in X$$. By the Cauchy criterion, we have:</p> $d(\xi^m, \xi^n)=\max|\xi^m_j - \xi^n_j|&lt;\epsilon$ <p>This implies that $$\vert\xi^m_j - \xi^n_j\vert &lt; \epsilon$$.<br /> For a fixed $$j$$, we have a sequence of reals $$\xi^1_j, \xi^2_j, \cdots$$ which is then a Cauchy sequence, and because of the completeness of $$\mathbb{R}$$, this sequence converges to $$L_j \in \mathbb{R}$$.</p> <p>Then, we have for any $$m,j$$, $$\vert\xi^m_j - L_j\vert &lt; \epsilon$$. It follows then that $$\max\vert\xi^m_j - L_j\vert &lt; \epsilon$$. This implies that the n-tuple formed by $$(L)=L_1,L_2,\cdots,L_n$$ is the limit of the Cauchy sequence $$(\xi^m)$$. Since $$(\xi^m)$$ was arbitrary, every Cauchy sequence in this space converges to a limit. Also, $$L \in X$$, hence the limit is contained within this metric space.</p> <p>Hence, this is a complete metric space.</p> $\blacksquare$ <hr /> <h4 id="153-let-m-subset-linfty-be-the-subspace-consisting-of-all-sequences-x--xi_j-with-at-most-finitely-many-nonzero-terms-find-a-cauchy-sequence-in-m-which-does-not-converge-in-m-so-that-m-is-not-complete">1.5.3 Let $$M \subset l^\infty$$ be the subspace consisting of all sequences $$x = (\xi_j)$$ with at most finitely many nonzero terms. Find a Cauchy sequence in $$M$$ which does not converge in $$M$$, so that $$M$$ is not complete.</h4> <p><strong>Proof:</strong></p> <p>$$l^\infty$$ is the space of all bounded sequences. Let there be a Cauchy sequence $$(x_n)$$, where the $$n$$th sequence has $$n$$ terms $$1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\cdots,\frac{1}{n}$$.</p> <p>This is a Cauchy sequence because for any $$m,n&gt;N$$, we have $$d(x^m,x^n)=\sup\vert x^m_j - x^n_j\vert=\displaystyle\frac{1}{\min(m,n)+1}$$ and $$\min(m,n)$$ can be made as large as possible to make $$\epsilon$$ as small as possible (by the Archimedean Principle).</p> <p>The limit of the Cauchy sequence is the infinite sequence $$1, \frac{1}{2}, \frac{1}{3}, \cdots$$. Call it $$x$$.</p> <p>The distance between any $$(x^n)$$ and $$x$$ will be $$\frac{1}{n+1}$$. However, $$x$$ is not contained in $$M$$, since $$x$$ does not have finitely many nonzero terms.</p> <p>Thus, $$M \subset l^\infty$$ is not a complete subset.</p> $\blacksquare$ <hr /> <h4 id="154-show-that-m-in-prob-3-is-not-complete-by-applying-theorem-14-7">1.5.4 Show that $$M$$ in Prob. 3 is not complete by applying Theorem 1.4-7.</h4> <p><strong>Proof:</strong></p> <p>The limit of the Cauchy sequence in $$M$$ described in the previous problem, does not belong to $$M$$, thus $$M$$ is not a closed subset, and is thus not complete.</p> $\blacksquare$ <hr /> <h4 id="155-show-that-the-set-x-of-all-integers-with-metric-d-defined-by-dmn--vert-m-nvert-is-a-complete-metric-space">1.5.5 Show that the set $$X$$ of all integers with metric $$d$$ defined by $$d(m,n) = \vert m-n\vert$$ is a complete metric space.</h4> <p><strong>Proof:</strong></p> <p>The only possible Cauchy sequences in this set are the ones ultimately yielding to the subsequence $$a,a,a,\cdots, a \in \mathbb{Z}$$, because only then will the Cauchy criterion of $$d(x_m,x_n)&lt;\epsilon, m,n&gt;N$$ hold.</p> <p>Thus, every Cauchy sequence in this set has as its limit $$x \in \mathbb{Z}$$. Thus, the set of all integers contains the limits of all its Cauchy sequences, and is thus a complete metric space.</p> $\blacksquare$ <hr /> <h4 id="156-show-that-the-set-of-all-real-numbers-constitutes-an-incomplete-metric-space-if-we-choose-dxy--vert-textarc-tan--x---textarc-tan--y-vert">1.5.6 Show that the set of all real numbers constitutes an incomplete metric space if we choose $$d(x,y) = \vert \text{arc tan } x - \text{arc tan } y \vert$$.</h4> <p><strong>Proof:</strong></p> <p>Note that $$\text{arc tan }_{n\rightarrow\infty} n=\frac{\pi}{2}$$.</p> <p>We need to find a Cauchy sequence which has a limit not contained in $$\mathbb{R}$$.</p> <p>Assume $$(x_n)=n$$ We note that $$\text{arc tan } x \rightarrow \frac{\pi}{2}$$ as $$x \rightarrow \infty$$. Then $$\vert \text{arc tan } x - \frac{\pi}{2} \vert &lt; \frac{\epsilon}{2}$$. Then, we use this to prove that $$(x_n)$$ is Cauchy using the <strong>Triangle Inequality</strong>. That is:</p> $d(x_m,x_n) \leq d(x_m,x) + d(x,x_n) = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ <p>Then $$(x_n)$$ has a limit at $$\infty$$. However, $$\mathbb{R}$$ does not contain $$\infty$$. Thus, this set if an incomplete metric space.</p> $\blacksquare$ <hr /> <h4 id="157-let-x-be-the-set-of-all-positive-integers-and-dmnvert-m-1-n-1vert-show-that-xd-is-not-complete">1.5.7 Let $$X$$ be the set of all positive integers and $$d(m,n)=\vert m^{-1}-n^{-1}\vert$$. Show that $$(X,d)$$ is not complete.</h4> <p><strong>Proof:</strong></p> <p>Let there be a sequence $$(x_n)=n$$. The distance $$d(x_n,x)=\vert \frac{1}{x_n} - \frac{1}{x} \vert$$ as $$x \rightarrow \infty$$ approaches $$\frac{1}{n}$$ which can be made as small as needed by choosing a large enough $$n$$. Thus we have:</p> $d(x_n,x)=\vert \frac{1}{x_n} - \frac{1}{x} \vert &lt; \epsilon$ <p>Then, by the <strong>Triangle Inequality</strong>, we have:</p> $d(x_m,x_n) \leq d(x_m,x) + d(x,x_n) &lt; \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ <p>Hence $$(x_n)$$ is Cauchy. However, this Cauchy does not converge in $$\mathbb{Z}_+$$, because there is no element in the set where $$\frac{1}{x_m}=0$$.</p> $\blacksquare$ <hr /> <h4 id="158-space-ca-b-show-that-the-subspace-y-subset-cab-consisting-of-all-x-in-ca-b-such-that-xa--xb-is-complete">1.5.8 (Space $$C[a, b]$$) Show that the subspace $$Y \subset C[a,b]$$ consisting of all $$x \in C[a, b]$$ such that $$x(a) = x(b)$$ is complete.</h4> <p><strong>Proof:</strong></p> <p>The distance metric on $$C[a.b]$$ is defined as: $$d(f_1,f_2)=\sup\vert f_1(x), f_2(x)\vert$$. We assume a Cauchy sequence of functions $$(f_n)=f_1,f_2,f_3,\cdots$$.</p> <p>We have, by the Cauchy criterion:</p> $d(f_m(x),f_n(x))&lt;\epsilon \\ \Rightarrow \sup |f_m(x) - f_n(x)| &lt; \epsilon \\ \Rightarrow |f_m(x) - f_n(x)| &lt; \epsilon$ <p>Fix $$x=t$$ such that $$f_1(t), f_2(t), \cdots$$ forms a Cauchy sequence in $$\mathbb{R}$$ since $$\vert f_m(t) - f_n(t) \vert &lt; \epsilon$$. Since $$\mathbb{R}$$ is complete, this sequence also converges to a real number $$f_L(t)$$.</p> <p>Since $$t$$ is arbitrary, all the limits of all $$t \in [a,b]$$ form the values of a limit function; call it $$f_L(x)$$.</p> <p>Since this limit function exists, we have:</p> $d(f_n(x),d_L(x))&lt;\epsilon$ <p>Since we know that $$d(f_n(a), f_n(b))=0$$, the above reduces to:</p> $d(f_L(a), f_L(b)) \leq d(f_L(a), f_n(a)) + d(f_n(a), f_n(b)) + d(f_n(b),f_L(b)) \\ \Rightarrow d(f_L(a), f_L(b)) \leq \epsilon + 0 + \epsilon = 2 \epsilon$ <p>Since $$d(f_L(a), f_L(b))&lt;2 \epsilon$$ for all $$\epsilon&gt;0$$, it follows that $$d(f_L(a), f_L(b))=0$$. Thus $$f_L$$ is contained in the set of all $$x \in C[a, b]$$ such that $$x(a) = x(b)$$ is complete.</p> <p>Hence this subset is complete.</p> $\blacksquare$ <hr /> <h4 id="159-in-15-5-we-referred-to-the-following-theorem-of-calculus-if-a-sequence-x_m-of-continuous-functions-on-ab-converges-on-ab-and-the-convergence-is-uniform-on-ab-then-the-limit-function-x-is-continuous-on-ab-prove-this-theorem">1.5.9 In 1.5-5 we referred to the following theorem of calculus. If a sequence $$(x_m)$$ of continuous functions on $$[a,b]$$ converges on $$[a,b]$$ and the convergence is uniform on $$[a,b]$$, then the limit function $$x$$ is continuous on $$[a,b]$$. Prove this theorem.</h4> <p><strong>Proof:</strong></p> <p>A sequence of functions $$(f_n)=f_1,f_2,\cdots$$ is uniformly convergent onto a limit function $$f$$ if $$\forall \epsilon&gt;0, \exists N$$, such that $$\vert f_n(x)-f(x) \vert &lt; \epsilon$$ for all $$x$$ and $$n&gt;N$$.</p> <p>A function is continuous at a point $$x_0$$ if $$\forall \epsilon&gt;0, \exists \delta&gt;0$$ such that $$\vert x-x_0 \vert &lt;\delta \Rightarrow \vert f(x)- f(x_0)\vert &lt; \epsilon$$.</p> <p>Choose a function $$f_n$$ where $$n&gt;N$$. Fix a point $$x_0$$. Then, because of uniform convergence, we have:</p> $d[f(x_0),f_n(x_0)]&lt;\epsilon \\ d[f_n(x),f(x)]&lt;\epsilon$ <p>Since $$f_n$$ is continuous, it is also continuous at $$x_0$$, pick an $$\epsilon&gt;0$$ such that $$d[f_n(x), f_n(x_0)]&lt;\epsilon$$.</p> <p>Then, by the <strong>Triangle Inequality</strong>, we have:</p> $d[f(x),f(x_0)] \leq d[f(x),f_n(x)] + d[f_n(x),f_n(x_0)] + d[f_n(x_0),f(x_0)] \\ \Rightarrow d[f(x),f(x_0)] &lt; \epsilon + \epsilon + \epsilon = 3 \epsilon$ <p>This shows that the limit function $$f$$ is also continuous.</p> $\blacksquare$ <hr /> <h4 id="1510--discrete-metric-show-that-a-discrete-metric-space-cf-11-8-is-complete">1.5.10 (Discrete metric) Show that a discrete metric space (cf. 1.1-8) is complete.</h4> <p><strong>Proof:</strong> The discrete metric is defined as:</p> $d(x,y)=\begin{cases} 0 &amp; \text{if } x=y \\ 1 &amp; \text{if } x \neq y \end{cases}$ <p>The only Cauchy sequences in $$X$$ with this metric are those which yield to subsequences $$a,a,a,\cdots$$. The limit of such a Cauchy sequence is $$a \in X$$. Thus the limit exists and is contained by the set $$X$$.</p> $\blacksquare$ <hr /> <h4 id="1511--space-s-show-that-in-the-space-s-cf-12-1-we-have-x_n-rightarrow-x-if-and-only-if-xin_j-rightarrow-xi_j-for-all-j--1-2-cdots--where-x_nxin_j-and-xxi_j">1.5.11 (Space s) Show that in the space $$s$$ (cf. 1.2-1) we have $$x_n \rightarrow x$$ if and only if $$\xi^{(n)}_j \rightarrow \xi_j$$ for all $$j = 1, 2, \cdots$$ , where $$x_n=(\xi^{(n)}_j)$$ and $$x=(\xi_j)$$.</h4> <p><strong>Proof:</strong></p> <p>The space $$s$$ consists of all bounded and unbounded sequences with the metric:</p> $d(x,y)=\displaystyle\sum\frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|}$ <p>Assume $$x_n \rightarrow x$$, so that $$d(\xi^n,\xi)&lt;\epsilon/2$$. Then, by the <strong>Triangle Inequality</strong>, we have:</p> $d(\xi^m,\xi^n) \leq d(\xi^m,\xi) + d(\xi, \xi^n) &lt; \epsilon/2 + \epsilon/2 = \epsilon$ <p>Thus, the Cauchy criterion also holds, so that: $$d(\xi^m,\xi^n)=\displaystyle\sum\frac{1}{2^j} \frac{|\xi^m_j-\xi^n_j|}{1+|\xi^m_j-\xi^n_j|} &lt; \epsilon$$</p> <p>Denote $$D_j=\vert\xi^m_j-\xi_j\vert$$ for simplicity, so that we get:</p> <p>We also note that $$\displaystyle\sum\frac{1}{2^j}=1$$, so that we can write the following: $$\displaystyle\sum\frac{1}{2^j} \frac{D_j}{1+D_j} &lt; \epsilon\sum\frac{1}{2^j} \\ \Rightarrow \displaystyle\sum\frac{1}{2^j} \frac{D_j}{1+D_j} &lt; \sum\frac{\epsilon}{2^j}$$</p> <p>Comparing term by term, we can say that:</p> $\frac{D_j}{1+D_j}&lt;\epsilon \\ \Rightarrow D_j&lt;\epsilon(1+D_j) \\ \Rightarrow D_j(1-\epsilon)&lt;\epsilon \\ \Rightarrow D_j&lt;\frac{\epsilon}{1-\epsilon}$ <p>Note that $$\epsilon&lt;1$$ simply because $$\frac{D_j}{1+D_j}&lt;1$$ and $$\frac{1}{2^j}&lt;1$$. Setting $$\epsilon=\frac{1}{k}$$, we get:</p> $\frac{\epsilon}{1-\epsilon}=\frac{\frac{1}{n}}{1-\frac{1}{n}} = \frac{1}{n-1}$ <p>Thus, $$D_j$$ can be made as small as needed by choosing a large $$n$$, i.e., small $$\epsilon$$, i.e.:</p> $\vert\xi^m_j-\xi\vert &lt; \epsilon_0$ <p>This implies that, for an arbitrary $$j$$:</p> $\xi^{(n)}_j \rightarrow \xi_j$ $\blacksquare$ <p>Conversely, assume that $$\xi^{(n)}_j \rightarrow \xi_j$$ for $$j=1,2,\cdots$$.</p> <p>This implies that:</p> $\vert\xi^m_j-\xi_j\vert &lt; \epsilon_0$ <p>Assume there is a sequence $$\xi_1, \xi_j, \cdots$$.</p> <p>We can see that:</p> $\frac{|\xi^m_j-\xi_j|}{1+|\xi^m_j-\xi_j|}&lt;|\xi^m_j-\xi_j|&lt;\epsilon&lt;1$ <p>Multiplying both sides by $$\frac{1}{2^n}$$, we get:</p> $\frac{1}{2^j}\frac{|\xi^m_j-\xi_j|}{1+|\xi^m_j-\xi_j|}&lt;\frac{1}{2^j} \epsilon$ <p>Summing to $$\infty$$, we get:</p> $\sum\frac{1}{2^j}\frac{|\xi^m_j-\xi_j|}{1+|\xi^m_j-\xi_j|}&lt;\sum\frac{1}{2^j} \epsilon \\ \Rightarrow d(\xi^m,\xi) &lt; \epsilon\sum\frac{1}{2^j} \\ \Rightarrow d(\xi^m,\xi) &lt; \epsilon$ <p>This implies that:</p> $\xi^m \rightarrow \xi$ $\blacksquare$ <hr /> <h4 id="1512--using-prob-11-show-that-the-sequence-space-s-in-12-1-is-complete">1.5.12 Using Prob. 11, show that the sequence space $$s$$ in 1.2-1 is complete.</h4> <p><strong>Proof:</strong></p> <p>Assume a Cauchy sequence in the space $$s$$. Since it is a Cauchy sequence, we have:</p> $d(\xi^m,\xi_n) &lt; \epsilon$ <p>By the previous problem, this would imply that</p> $\vert\xi^m_j-\xi^n_j\vert &lt; \epsilon$ <p>For a fixed $$j$$, we thus have a Cauchy sequence of reals. Since $$\mathbb{R}$$ is complete, this sequence converges so that $$\xi^n_j \rightarrow \xi_j$$.</p> <p>By the previous problem, this implies that the Cauchy sequence in the space $$s$$ converges to a limit $$\xi$$. Moreover, $$\xi \in s$$. Hence $$s$$ is a complete metric space.</p> $\blacksquare$ <hr /> <h4 id="1513--show-that-in-15-9-another-cauchy-sequence-is-x_n-where-x_ntn-text-if--0-leq-t-leq-n-2-and-x_ntt-frac12-text-if--n-2-leq-t-leq-1">1.5.13 Show that in 1.5-9, another Cauchy sequence is $$(x_n)$$, where $$x_n(t)=n \text{ if } 0 \leq t \leq n^{-2}$$ and $$x_n(t)=t^{-\frac{1}{2}} \text{ if } n^{-2} \leq t \leq 1$$.</h4> <p><strong>Proof:</strong></p> <p><img src="/assets/images/cauchy-function-convergence.png" alt="Diagram for this Problem" /></p> <p>We prove that this sequence is Cauchy. The distance metric is defined as:</p> $d(x,y)=\int\limits_0^1 |x(t) - y(t)| dt$ $d(x_m,x_n)=\int\limits_0^1 |x_m(t) - x_n(t)| dt \\ = \int\limits_0^{\frac{1}{n^2}} |x_m(t) - x_n(t)| dt + \int\limits_{\frac{1}{n^2}}^{\frac{1}{m^2}} |x_m(t) - x_n(t)| dt + \int\limits_{\frac{1}{m^2}}^1 |x_m(t) - x_n(t)| dt \\ = \int\limits_0^{\frac{1}{n^2}} |n-m| dt + \int\limits_{\frac{1}{n^2}}^{\frac{1}{m^2}} |m-\frac{1}{\sqrt t}| dt + \int\limits_{\frac{1}{m^2}}^1 |\frac{1}{\sqrt t}-\frac{1}{\sqrt t}| dt \\ = \int\limits_0^{\frac{1}{n^2}} (n-m) dt + \int\limits_{\frac{1}{n^2}}^{\frac{1}{m^2}} (\frac{1}{\sqrt t}-m) dt + \int\limits_{\frac{1}{m^2}}^1 |\frac{1}{\sqrt t}-\frac{1}{\sqrt t}| dt \\$ <p>Note above that for the middle term $$\frac{1}{\sqrt t}&gt;m$$, thus the terms are flipped.</p> <p>Assuming $$m&lt;n$$, we get:</p> $\require{cancel} (n-m)(\frac{1}{n^2} - 0) + m(\frac{1}{n^2} - \frac{1}{m^2}) - (\frac{2}{n} - \frac{2}{m}) + 0 \\ = \frac{1}{n} - \cancel{\frac{m}{n^2}} + \cancel{\frac{m}{n^2}} -\frac{1}{m} - \frac{2}{n} + \frac{2}{m} \\ = \frac{1}{m} - \frac{1}{n}=\frac{n-m}{mn}$ <p>Assume you keep $$n-m=K$$, then you can make $$d(x_m, x_n)&lt;\epsilon$$ by choosing suitable $$m,n$$ so that the product $$mn$$ is as large as needed.</p> <p>Thus $$(x_n)$$ is a Cauchy sequence.</p> $\blacksquare$ <hr /> <h4 id="1514--show-that-the-cauchy-sequence-in-prob-13-does-not-converge">1.5.14 Show that the Cauchy sequence in Prob. 13 does not converge.</h4> <p><strong>Proof:</strong></p> <p>Assume that the limit of $$x_n$$ exists, and is $$x(t)$$. Thus, as $$n \rightarrow \infty$$, $$d(x_m(t),x(t)) \rightarrow 0$$, implying that the integral above approaches zero. Since, all the integrands are positive, each of them should individually approach zero.</p> <p>Then, we get, in the limit of $$n \rightarrow \infty$$:</p> $\int\limits_0^0 |n-x(t)| dt + \int\limits_0^{\frac{1}{m^2}} |m-x(t)| dt + \int\limits_{\frac{1}{m^2}}^1 |\frac{1}{\sqrt t}-x(t)| dt \\$ <p>The three terms above should individually go to zero. Thus, the third term implies that $$x(t)=\frac{1}{\sqrt t}$$.</p> <p>$$x(t)$$ should be in the space of continuous functions. However $$x(t)=\frac{1}{\sqrt t}$$ is not defined for $$t=0$$. Thus, $$x(t)$$ is not continuous, and this Cauchy sequence in $$C[0,1]$$ does not converge.</p> $\blacksquare$ <hr /> <h4 id="1515--let-x-be-the-metric-space-of-all-real-sequences-xxi_j-each-of-which-has-only-finitely-many-nonzero-terms-and-dxydisplaystylesum-vert-xi_j---eta_j-vert-where-y--eta_j-note-that-this-is-a-finite-sum-but-the-number-of-terms-depends-on-x-and-y-show-that-x_n-with-x_n--xin_j">1.5.15 Let $$X$$ be the metric space of all real sequences $$x=(\xi_j)$$ each of which has only finitely many nonzero terms, and $$d(x,y)=\displaystyle\sum \vert \xi_j - \eta_j \vert$$, where $$y = (\eta_j)$$. Note that this is a finite sum but the number of terms depends on $$x$$ and $$y$$. Show that $$(x_n)$$ with $$x_n = (\xi^{(n)}_j)$$,</h4> <p>$$\xi^{(n)}_j=j^{-2}$$ for $$j=1,\cdots,n$$ and $$\xi^{(n)}_j=0$$ for $$j&gt;n$$</p> <p><strong>is Cauchy but does not converge.</strong></p> <p><strong>Proof:</strong></p> <p>Consider the space of sequences $$(x_n)=\frac{1}{1},\frac{1}{2^2},\frac{1}{3^2},\cdots,\frac{1}{n^2},0,0,\cdots$$.</p> <p>We have $$d(x_m,x_n)=\displaystyle\sum\limits_{j=m+1}^n \frac{1}{j^2}$$. Since $$\displaystyle\sum\limits_{j=1}^\infty \frac{1}{j^2}$$ is bounded, $$\displaystyle\sum\limits_{j=n}^\infty \frac{1}{j^2}$$ can be made as small as needed, by choosing a large $$m$$, thus we have:</p> $\displaystyle\sum\limits_{j=m+1}^n \frac{1}{j^2}&lt;\displaystyle\sum\limits_{j=m+1}^\infty \frac{1}{j^2}&lt;\epsilon.$ <p>Thus, this sequence is Cauchy.</p> <p>Assume a limit exists, then $$d(x_n,x) \rightarrow 0$$ as $$n \rightarrow \infty$$.</p> <p>Then $$\require{cancel} d(x_n,x)=\displaystyle\sum\limits_{j=n+1}^\infty \frac{1}{j^2} \cancel\rightarrow 0$$</p> <p>which is a contradiction. Thus, this limit does not exist in $$X$$, and thus $$X$$ is not a complete metric space.</p> $\blacksquare$avishekThis post lists solutions to the exercises in the Completeness Proofs section 1.5 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.Functional Analysis Exercises 4 : Convergence, Cauchy Sequences, and Completeness2021-10-12T00:00:00+05:302021-10-12T00:00:00+05:30/2021/10/12/functional-analysis-exercises-4-convergence-completeness-cauchy-sequences<p>This post lists solutions to the exercises in the <strong>Convergence, Cauchy Sequences, and Completeness section 1.4</strong> of <em>Erwin Kreyszig’s</em> <strong>Introductory Functional Analysis with Applications</strong>. This is a work in progress, and proofs may be refined over time.</p> <h4 id="141-subsequence-if-a-sequence-x-in-a-metric-space-x-is-convergent-and-has-limit-x-show-that-every-subsequence-x_n_k-of-x_n-is-convergent-and-has-the-same-limit-x">1.4.1. (Subsequence) If a sequence $$(x)$$ in a metric space $$X$$ is convergent and has limit $$x$$, show that every subsequence $$(x_{n_k})$$ of $$(x_n)$$ is convergent and has the same limit $$x$$.</h4> <p><strong>Proof:</strong></p> <p>Suppose $$(x_n)$$ is convergent and $$x_n \rightarrow x$$. Let $$(x_{n_k})$$ be a subsequence. Let $$x_{n_m}$$ correspond to $$x_i$$.</p> <p>Since $$(x_n)$$ is convergent, $$\forall \epsilon&gt;0, \exists N$$, such that $$d(x_n,x)&lt;\epsilon$$ for $$n&gt;N$$. Pick $$p \geq N$$, such that $$x_p$$ exists in $$(x_{n_k})$$ and is identified as $$x_{n_m}$$.</p> <p>This implies that $$\forall \epsilon&gt;0, \exists M$$, such that $$d(x_{n_m},x)&lt;\epsilon$$.</p> <p>The above is the definition of the convergence of a sequence to a limit. Thus, $$(x_{n_k})$$ converges to $$x$$.</p> <p>Alternatively, you can prove that the limit of $$(x_{n_k})$$ is $$x$$ in the following manner.<br /> Suppose $$x_{n_k} \rightarrow y$$. Then, by the <strong>Triangle Inequality</strong>, we have:</p> $d(x,y) \leq d(x,x_p) + d(x_p,y) \\ \Rightarrow d(x,y) \leq d(x,x_p) + d(x_{n_m},y)$ <p>Both $$d(x,x_p)$$ and $$d(x_{n_m},y)$$ can be made as small as possible, since $$(x_n)$$ and $$(x_{n_k})$$ converge, implying that $$d(x,y)$$ is smaller than any positive value. Thus:</p> $d(x,y) \leq d(x,x_p) + d(x_{n_m},y) \\ \Rightarrow d(x,y) &lt; \epsilon_1 + \epsilon_2, \epsilon_1, \epsilon_2 &gt; 0 \\ \Rightarrow d(x,y)=0$ <p>Thus, $$x=y$$, i.e., $$x_{n_k}$$ has the same limit as $$(x_n)$$.</p> $\blacksquare$ <hr /> <h4 id="142-if-x_n-is-cauchy-and-has-a-convergent-subsequence-say-x_n-rightarrow-x-show-that-x_n-is-convergent-with-the-limit-x">1.4.2. If $$(x_n)$$ is Cauchy and has a convergent subsequence, say, $$x_n \rightarrow x$$, show that $$(x_n)$$ is convergent with the limit $$x$$.</h4> <p><strong>Proof:</strong></p> <p>Suppose $$(x_n)$$ is Cauchy. Let $$(x_{n_k})$$ be a subsequence. Let $$x_{n_m}$$ correspond to $$x_i$$.</p> <ul> <li>Since $$(x_{n_k})$$ is convergent, $$\forall \epsilon&gt;0, \exists M$$, such that $$d(x_{n_k},x)&lt;\epsilon$$ for $$n&gt;M$$.</li> <li>Since $$(x_n)$$ is convergent, $$\forall \epsilon&gt;0, \exists N$$, such that $$d(x_i,x_j)&lt;\epsilon$$ for $$i,j&gt;N$$.</li> </ul> <p>Pick $$N_0=\max(M,N)$$. Then the above two statements become:</p> <ul> <li>Since $$(x_{n_k})$$ is convergent, $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_{n_i},x)&lt;\epsilon$$ for $$i&gt;N_0$$.</li> <li>Since $$(x_n)$$ is convergent, $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_i,x_j)&lt;\epsilon$$ for $$i,j&gt;N_0$$.</li> </ul> <p>(Note that we picked the same $$i$$ for both $$(x_n)$$ and $$(x_{n_k})$$ because any value greater than $$N_0$$ will fulfil the above conditions, so we might as well pick the same index. For any index $$i$$, we can use $$x_i$$ and $$x_{n_i}$$ interchangeably, since they index the same element in both the sequence and the subsequence.)</p> <p>By the <strong>Triangle Inequality</strong>, we have:</p> $d(x_j,x) \leq d(x_j, x_i) + d(x_i,x) \\ \Rightarrow d(x_j,x) \leq d(x_j, x_i) + d(x_{n_i},x) \\ \Rightarrow d(x_j,x) \leq \epsilon + \epsilon \\ \Rightarrow d(x_j,x) &lt; 2\epsilon$ <p>$$\epsilon$$ can be made as small as possible, implying that $$d(x_j,x)$$ is smaller than any positive value. Thus:</p> $d(x_j,x)=0$ <p>Hence, $$(x_n)$$ is convergent with the limit $$x$$.</p> $\blacksquare$ <hr /> <h4 id="143-show-that-x_n-rightarrow-x-if-and-only-if-for-every-neighborhood-v-of-x-there-is-an-integer-n_0-such-that-x_n-in-v-for-all-n--n_0">1.4.3. Show that $$x_n \rightarrow x$$ if and only if for every neighborhood $$V$$ of $$x$$ there is an integer $$n_0$$ such that $$x_n \in V$$ for all $$n &gt; n_0$$.</h4> <p><strong>Proof:</strong></p> <p>Suppose that $$x_n \rightarrow x$$. Then $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_n,x)&lt;\epsilon$$ for all $$n&gt;N_0$$. This implies that a neighbourhood $$V_\epsilon$$ of $$x$$ exists, which contains all $$x_{n&gt;N_0}$$. Since there are an infinite number of values for $$\epsilon$$, it follows that this applies to every neighbourhood of $$x$$.</p> <p>Conversely, suppose that for every neighborhood $$V$$ of $$x$$ there is an integer $$n_0$$ such that $$x_n \in V$$ for all $$n &gt; n_0$$. Assume each neighbourhood has a size of $$\epsilon$$. Thus, $$x_n \in V$$ implies that $$d(x_n,x)&lt;\epsilon$$. Then, we can restate this as the following: $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_n,x)&lt;\epsilon$$ for all $$n&gt;N_0$$.</p> $\blacksquare$ <hr /> <h4 id="144-boundedness-show-that-a-cauchy-sequence-is-bounded">1.4.4. (Boundedness) Show that a Cauchy sequence is bounded.</h4> <p><strong>Proof:</strong></p> <p>By definition, for a Cauchy sequence, we have: $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_m,x_n)&lt;\epsilon$$ for all m,n&gt;N_0$$.</p> <p>Choose $$\epsilon=1$$. Then, assume the value of $$N_0$$ to be $$N_1$$. For any $$d(x_a,x_b)$$, we have:</p> <ul> <li>$$a&lt;b \leq N_0$$: Then $$d(x_a,x_b) \leq a = max[d(x_a,x_0), d(x_a,x_1), \cdots, d(x_a,x_{N_1})]$$</li> <li>$$N_0&lt;a&lt;b$$: Then $$d(x_a,x_b) &lt; \epsilon = 1$$</li> <li>$$a \leq N_0&lt;b$$: By the <strong>Triangle Inequality</strong>, we have: $$d(x_a,x_b) \leq d(x_a,x_{N_1}) + d(x_{N_1}, x_b) &lt; a + 1$$</li> </ul> <p>Combining these upper bounds, we get: $$\sup d(x_a,x_b) &lt; a+1$$</p> $\blacksquare$ <hr /> <h4 id="145-is-boundedness-of-a-sequence-in-a-metric-space-sufficient-for-the-sequence-to-be-cauchy-convergent">1.4.5. Is boundedness of a sequence in a metric space sufficient for the sequence to be Cauchy? Convergent?</h4> <p><strong>Answer:</strong></p> <p>Consider the discrete metric on $$\mathbb{R}$$. If we have a sequence $$(x_n)=0,1,0,1,\cdots$$, then the series is bounded because $$\sup d(x_m, x_n)=1$$, but for $$\epsilon=\frac{1}{2}$$, there is no $$N$$ for which $$d(x_m,x_n)&lt;\epsilon$$ for $$m,n&gt;N$$. Thus, the sequence is not Cauchy, though it is bounded.</p> <p>Convergence is sufficient for a sequence to be Cauchy. For convergence, we have the condition: if $$x_n \rightarrow x$$, $$\forall \epsilon&gt;0, \exists N_0$$, such that $$d(x_n,x)&lt;\epsilon$$ for all $$n&gt;N_0$$.</p> <p>Consider $$m,n&gt;N_0$$. Then, by the <strong>Triangle Inequality</strong>, we have:</p> $d(x_m,x_n) \leq d(x_m,x) + d(x,x_n) &lt; \epsilon + \epsilon = 2 \epsilon$ <p>thus proving the Cauchy criterion.</p> <hr /> <h4 id="146-if-x_n-and-y_n-are-cauchy-sequences-in-a-metric-space-x-d-show-that-a_n-where-a_n--dx_n-y_n-converges-give-illustrative-examples">1.4.6. If $$(x_n)$$ and $$(y_n)$$ are Cauchy sequences in a metric space $$(X, d)$$, show that $$(a_n)$$, where $$a_n = d(x_n, y_n)$$, converges. Give illustrative examples.</h4> <p><strong>Proof:</strong></p> $d(x_m,x_n)&lt;\epsilon \\ d(y_m,y_n)&lt;\epsilon$ <p>Then, we have:</p> $d(x_m,y_m) \leq d(x_m,x_n) + d(x_n,y_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) \leq d(x_m,x_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) &lt; 2 \epsilon$ <p>Similarly, we have:</p> $d(x_n,y_n) \leq d(x_n,x_m) + d(x_m,y_m) + d(y_m,y_n) \\ d(x_n,y_n) - d(x_m,y_m) \leq d(x_n,x_m) + d(y_m,y_n) \\ \Rightarrow d(x_n,y_n) - d(x_m,y_m) &lt; 2 \epsilon$ <p>The above inequalities imply that: $$\vert d(x_m,y_m) - d(x_n,y_n) \vert &lt; 2 \epsilon \\ d[d(x_m,y_m) - d(x_n,y_n)] &lt; 2 \epsilon$$</p> <p>This implies that $$a_n=d(x_n,y_n)$$ is Cauchy, and thus converges.</p> $\blacksquare$ <hr /> <h4 id="147-give-an-indirect-proof-of-lemma-14-2b">1.4.7. Give an indirect proof of Lemma 1.4-2(b).</h4> <p><strong>Lemma 1.4-2(b)</strong> is: Let $$X=(X,d)$$ be a metric space. Then, if $$x_n \rightarrow x$$ and $$y_n \rightarrow y$$, then $$d(x_n,y_n) \rightarrow d(x,y)$$.</p> <p><strong>Proof:</strong></p> <p>We have $$x_n \rightarrow x$$ and $$y_n \rightarrow y$$. Then $$(x_n)$$ and $$(y_n)$$ are Cauchy. Thus the following two statements hold true:</p> <ul> <li>$$\forall \epsilon/2&gt;0, \exists M$$ such that $$d(x_m,x_n)&lt;\epsilon/2$$ for $$m,n&gt;M$$</li> <li>$$\forall \epsilon/2&gt;0, \exists N$$ such that $$d(y_m,y_n)&lt;\epsilon/2$$ for $$m,n&gt;N$$</li> </ul> <p>Taking $$N_0=\max(M,N)$$, the above statements become:</p> <p>$$\forall \epsilon/2&gt;0, \exists N_0$$ such that $$d(x_m,x_n)&lt;\epsilon/2$$ and $$d(y_m,y_n)&lt;\epsilon/2$$ for $$m,n&gt;N_0$$, i.e., $$d(x_m,x_n)+d(y_m,y_n)&lt;\epsilon/2+\epsilon/2=\epsilon$$</p> <p>We will prove the result using proof by contradiction.</p> <p>Suppose $$(a_n)=(d(x_n,y_n))$$ Suppose the claim is not true. Then, $$\require{cancel} a_n \cancel\rightarrow a$$, thus $$(a_n)$$ is not Cauchy. This implies that: $$\exists \epsilon$$ such that $$\forall N$$, we have $$d(a_m, a_n)&gt;\epsilon$$ for all $$m,n&gt;N$$.</p> <p>By the <strong>Triangle Inequality</strong>, we have:</p> $d(x_m,y_m) \leq d(x_m,x_n) + d(x_n,y_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,x_n) + d(y_n,y_m) \geq d(x_m,y_m) - d(x_n,y_n) \\ \Rightarrow d(x_m,x_n) + d(y_n,y_m) &gt; \epsilon \\$ <p>This is then true for arbitrary $$\epsilon$$. But, this implies that for all $$N$$, we cannot make $$d(x_m,x_n)+d(y_m,y_n)&lt;\epsilon$$. This is a contradiction, since by assumption, we have</p> $d(x_m,x_n)+d(y_m,y_n)&lt;\epsilon$ <p>Thus $$(a_n)$$ is Cauchy, and is thus a convergent sequence.</p> $\blacksquare$ <hr /> <h4 id="148-if-d_1-and-d_2-are-metrics-on-the-same-set-x-and-there-are-positive-numbers-a-and-b-such-that-for-all-x-y-in-x-ad_1xy-leq-d_2xy-leq-bd_1xy-show-that-the-cauchy-sequences-in-x-d_1-and-x-d_2-are-the-same">1.4.8. If $$d_1$$ and $$d_2$$ are metrics on the same set $$X$$ and there are positive numbers $$a$$ and $$b$$ such that for all $$x, y \in X$$, $$a.d_1(x,y) \leq d_2(x,y) \leq b.d_1(x,y)$$, show that the Cauchy sequences in $$(X, d_1)$$ and $$(X, d_2)$$ are the same.</h4> <p><strong>Proof:</strong></p> <p>We wish to show that if $$L_1$$ is the limit of a Cauchy sequence in $$(X,d_1)$$ (call it $$x_n(d_1)$$)and $$L_2$$ is the limit of a Cauchy sequence in $$(X,d_2)$$ (call it $$x_n(d_2)$$), then $$L_1=L_2$$.</p> <p>We have $$\forall x, y \in X$$, $$a.d_1(x,y) \leq d_2(x,y) \leq b.d_1(x,y)$$.</p> <p>Then, the <strong>Triangle Inequality</strong> gives us:</p> $d_1(L_1,L_2) \leq d_1(L_1,x) + d_1(x,L_2)$ <p>Applying the given metric constraints:</p> $d_1(L_1,L_2) \leq d_1(L_1,x) + \frac{1}{a} d_2(x,L_2)$ <p>We know that $$x_n(d_1) \rightarrow L_1$$ and $$x_n(d_2) \rightarrow L_2$$, therefore. If we have the following:</p> <ul> <li>$$\forall \epsilon&gt;0, \exists M$$ such that $$d_1(x_m(d_1),L_1)&lt;\epsilon$$ for $$m&gt;M$$</li> <li>$$\forall \epsilon&gt;0, \exists N$$ such that $$d_1(x_m(d_2),L_2)&lt;\epsilon$$ for $$m&gt;N$$</li> </ul> <p>Pick $$N_0=\max(M,N)$$, so that the above holds true for $$N_0$$.</p> <p>Then $$d_1(L_1,x_{N_0})&lt;\epsilon$$ and $$d_2(L_2,x_{N_0})&lt;\epsilon$$, so that we get:</p> $d_1(L_1,L_2) &lt; \epsilon \left(1+\frac{1}{a} \right)$ <p>Since the above is true for all $$\epsilon&gt;0$$, we can conclude that $$d_1(L_1,L_2)=0$$. Hence $$L_1=L_2$$.</p> <p>The same procedure can also be showing using $$d_2$$.</p> $\blacksquare$ <hr /> <h4 id="149-using-prob-8-show-that-the-metric-spaces-in-probs-13-to-15-sec-12-have-the-same-cauchy-sequences">1.4.9. Using Prob. 8, show that the metric spaces in Probs. 13 to 15, Sec. 1.2, have the same Cauchy sequences.</h4> <p>The three distance metrics mentioned are:</p> <ul> <li> $d_1(x,y)=d(x_1,x_2)+d(y_1,y_2)$ </li> <li> $d_2(x,y)=\sqrt{ {d(x_1,x_2)}^2+{d(y_1,y_2)}^2}$ </li> <li> $d_{max}(x,y)=\max [d(x_1,x_2),d(y_1,y_2)]$ </li> </ul> <p><strong>Proof:</strong></p> <p>For $$d_1$$ and $$d_2$$, let’s determine the conditions.</p> $x+y \leq \sqrt{x^2+y^2} \\ x^2+y^2+2xy \leq x^2+y^2$ <p>This gives us $$2xy \leq 0$$, so that’s invalid; if we however we introduce a $$\sqrt{2}$$ on the right hand side, we get:</p> $x+y \leq \sqrt{2(x^2+y^2)} \\ x^2+y^2+2xy \leq 2x^2+2y^2 \\ x^2+y^2-2xy \geq 0$ <p>which works. For the reverse inequality $$x+y \geq \sqrt{x^2+y^2}$$, note that we immediately get $$2xy&gt;0$$, which works, so we can write the combined inequalities as:</p> $\sqrt{x^2+y^2} \leq x+y \leq \sqrt{2} \sqrt{x^2+y^2} \\ \Rightarrow d_2(x,y) \leq d_1(x,y) \leq \sqrt{2} d_2(x,y)$ <p>For $$d_1$$ and $$d_max$$, note that $$x+y&gt; \geq \max(x,y)$$ and $$2 \max(x,y) \geq x+y$$</p> <p>Then, we get:</p> $\max(x,y) \leq x+y \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_1(x,y) \leq 2 d_{ max}(x,y)$ <p>For $$d_2$$ and $$d_max$$, note that $$x^2+y^2&gt; \geq {\max(x,y)}^2$$ and $$2 {\text{ max }(x,y)}^2 \geq x^2+y^2$$</p> <p>Then, we get:</p> $\max(x,y) \leq \sqrt{x^2+y^2} \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_2(x,y) \leq 2 d_{max}(x,y)$ $\blacksquare$ <hr /> <h4 id="1410-using-the-completeness-of-mathbbr-prove-completeness-of-mathbbc">1.4.10. Using the completeness of $$\mathbb{R}$$, prove completeness of $$\mathbb{C}$$.</h4> <p><strong>Proof:</strong></p> <p>Assume $$\mathbb{R}$$ is complete.</p> <p>Assume two Cauchy Sequences in $$\mathbb{R}$$:</p> <ul> <li>$$(x_n)=x_1,x_2,\cdots$$ converges to $$x$$.</li> <li>$$(y_n)=y_1,y_2,\cdots$$ converges to $$y$$.</li> </ul> <p>Construct a sequence in $$\mathbb{C}$$, like so:</p> $(z_n)=x_1+iy_1,x_2+iy_2,\cdots$ <p>Assume the distance metric for $$\mathbb{Z}$$ is $$d(z_1,z_2)=\sqrt{ {(x_1-x_2)}^2 + {(y_1-y_2)}^2}$$.</p> <ul> <li>$$\forall \epsilon&gt;0, \exists M$$ such that $$x_m-x&lt;\frac{\epsilon}{\sqrt{2}}$$ for $$m&gt;M$$</li> <li>$$\forall \epsilon&gt;0, \exists N$$ such that $$x_m-x&lt;\frac{\epsilon}{\sqrt{2}}$$ for $$m&gt;N$$</li> </ul> <p>Pick $$N_0=\max(M,N)$$, so that the above holds true for $$N_0$$.</p> <p>Pick $$z_i$$ so that $$i&gt;N_0$$. Assume $$z=x+iy$$. Then, we have:</p> $d(z_i,z)=\sqrt{ {(x_i-x)}^2 + {(y_i-y)}^2}&lt;\epsilon$ <p>Then, for an arbitrary $$\epsilon$$, there exists $$N_0$$, such that $$d(z_i,z)&lt;\epsilon$$. Furthermore $$z \in \mathbb{C}$$. Thus, $$\mathbb{C}$$ contains the limits of all its Cauchy sequences. Thus, it is a closed set; hence it is a complete metric space.</p> $\blacksquare$avishekThis post lists solutions to the exercises in the Convergence, Cauchy Sequences, and Completeness section 1.4 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.