This 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.
Notes
The requirements for a space to be a normed space are:
- (N1) Nonnegativity, i.e., \(\|x\| \geq 0, x \in X\)
- (N2) Zero norm implies zero vector and vice versa, i.e., \(\|x\|=0 \Leftrightarrow x=0, x \in X\)
- (N3) Linearity with respect to scalar multiplication, i.e., \(\|\alpha x\|=\vert \alpha \vert \|x\|, x \in X, \alpha \in \mathbb{R}\)
- (N4) Triangle Inequality, i.e., \(\|x+y\| \leq \|x\| + \|y\|, x,y \in X\)
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.
Proof:
\(c \in l^\infty\) is the space of all complex convergent sequences. Let \(x_n \rightarrow x\) and \(y_n \rightarrow y\), such that \(x_n,y_n \in \mathbb{C}\).
Then, \(\forall \epsilon>0, \exists M, N \in \mathbb{N}\), such that \(\|x_m-x\|<\epsilon\) and \(\|y_n-y\|<\epsilon\), for all \(m>M, n>N\). Take \(N_0=\max(M_0,N_0)\), so that \(\|x_n-x\|<\epsilon\) and \(\|y_n-y\|<\epsilon\), for all \(n>N_0\).
\[\|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| = \|\alpha x_n - \alpha x + \beta y_n - \beta y)\| \\ \|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| \leq \|\alpha x_n - \alpha x\| + \|\beta y_n - \beta y\| \\ = |\alpha|\|x_n - x\| + |\beta|\|y_n - y\| < \alpha \epsilon + \beta \epsilon = (\alpha + \beta) \epsilon\]\((\alpha + \beta) \epsilon\) can be made as small as possible. In the limit \(n \rightarrow \infty\), we thus get: \(\|\alpha x_n + \beta y_n - (\alpha x + \beta y)\| \rightarrow \alpha x + \beta y \in c \subset l^\infty\).
Thus, \(c\) is a vector subspace.
\[\blacksquare\]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 \(n \rightarrow \infty\).
Thus, \(c_0\) is a vector subspace.
\[\blacksquare\]2.3.2. Show that \(c_0\) in Prob. 1 is a closed subspace of \(l^\infty\), so that \(c_0\) is complete by 1.5-2 and 1.4-7.
Proof:
Consider a Cauchy sequence \((x_n) \in c_0\). Then, \(\forall \epsilon>0, \exists N \in \mathbb{N}\), such that \(d(x_m-x_n)<\epsilon\), for all \(m,n>N\).
This implies that:
\[\sup d(x_j^m, x_j^n)<\epsilon \\ d(x_j^m, x_j^n)<\epsilon\]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, \(\forall \epsilon>0, \exists M \in \mathbb{N}\) such that \(\|x_j^m-x_j\|<\epsilon\) for all \(m>M\). This holds for every \(j\), yielding a sequence \(x_1, x_2, x_3, \cdots\).
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 \(\forall \epsilon>0, \exists N \in \mathbb{N}\) such that \(\|x_j^m-0\|<\epsilon\) for all \(j>N\) We have:
\[\|x_j-0\| = \|x_j-x_j^m + x_j^m - 0\| \leq \|x_j-x_j^m\| + \|x_j^m - 0\| < \epsilon + \epsilon = 2 \epsilon\]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.
\[\blacksquare\]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.
Proof:
Let \(Y\) be the subset of all sequences with only finitely many nonzero terms.
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 \(m<n\). It is clear that \(\delta(x_n)=\max(x_1, x_2, \cdots, x_m) < \infty\) and \(\delta(y_n)=\max(y_1, y_2, \cdots, y_n) < \infty\), thus \((x_n), (y_n) \in YZ \subset l^\infty\) Then, we have:
\[\alpha(x_n) + \beta(y_n)=\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \cdots, \alpha x_m + \beta y_m, \beta y_{m+1}, \cdots, \beta y_n, 0, 0, \cdots \\ \Rightarrow \delta[\alpha(x_n) + \beta(y_n)]=\max(\alpha x_1 + \beta y_1, \alpha x_2 + \beta y_2, \cdots, \alpha x_m + \beta y_m, \beta y_{m+1}, \cdots, \beta y_n) < \infty \\ \Rightarrow \alpha(x_n) + \beta(y_n) \in Y \subset l^\infty\]Thus \(Y\) is a subspace of \(l^\infty\).
Let there be a Cauchy sequence in \(Y\), where \(y_n=\begin{cases} 1/j & \text{if } j \leq n \\ 0 & \text{if } j > n \end{cases}\)
Assume \(m<n\). 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.
\[\blacksquare\]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.
Proof:
Assume a normed space, with \(\|x-x_0\|<\epsilon\) and \(\|y-y_0\|<\epsilon\)
Vector addition will be continuous if \(\forall \epsilon>0, \exists \delta>0\), such that \(\|x-x_0\| < \delta, \|y-y_0\| < \delta \Rightarrow \|f(x,y)-f(x_0,y_0)\| < \epsilon\).
Then, we have:
\[\|x+y-(x_0+y_0)\|=\|(x-x_0)+(y-y_0)\| \leq \|x-x_0\| + \|y-y_0\| < 2 \epsilon\]\(2 \epsilon\) can be made as small as needed, thus vector addition is continuous in normed space.
\[\blacksquare\]Scalar multiplication will be continuous if \(\forall \epsilon>0, \exists \delta>0\), such that \(\|x-x_0\| < \delta, \|\alpha - \alpha_0\| < \delta \Rightarrow \|\alpha x - \alpha x_0\| < \epsilon\).
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 \((x-x_0)(\alpha-\alpha_0)\).
Then, we have:
\[(\alpha x-\alpha_0)(x-x_0)=\alpha x + \alpha_0 x_0 - \alpha_0 x - \alpha x_0\]Then we can write \(\alpha x - \alpha_0 x_0\) as:
\[\alpha x - \alpha_0 x_0 = \alpha x + \alpha_0 x_0 - \alpha_0 x - \alpha x_0 - \alpha_0 x_0 - \alpha_0 x_0 + \alpha_0 x + \alpha x_0 \\ = (\alpha - \alpha_0)(x-x_0) + \alpha_0 (x-x_0) + x_0(\alpha-\alpha_0)\]Therefore, we can write:
\[\|\alpha x - \alpha_0 x_0\|=\|(\alpha - \alpha_0)(x-x_0) + \alpha_0 (x-x_0) + x_0(\alpha-\alpha_0)\| \\ \|\alpha x - \alpha_0 x_0\| \leq \|\alpha - \alpha_0\|\|x-x_0\| + |\alpha_0| \|x-x_0\| + |x_0|\|\alpha-\alpha_0\| < \epsilon^2 + |\alpha_0| \epsilon + |x_0| \epsilon\]The quantity on the RHS can be made as small as possible, and thus scalar multiplication is continuous in normed space.
\[\blacksquare\]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 \(\alpha_n x_n \rightarrow \alpha x\).
(See Above)
2.3.6. Show that the closure \(\bar{Y}\) of a subspace \(Y\) of a normed space \(X\) is again a vector subspace.
Proof:
Let \(x,y \in \bar{S}\). Thus, \(\forall r>0\), we have \(B(x,r) \cap S \neq \emptyset\) and \(B(y,r) \cap S \neq \emptyset\). Pick \(r<\epsilon\), then \(\|x-x_0\|<\epsilon/2\) and \(\|y-y_0\|<\epsilon/2\).
Then, we have:
\[\|\alpha x + \beta y - (\alpha x_0 + \beta y_0)\|=\|\alpha x - \alpha x_0 + \beta y - \beta y_0\| \\ \leq \|\alpha x - \alpha x_0\| + \|\beta y - \beta y_0\| < \epsilon/2 + \epsilon/2 = \epsilon\]This holds for every \(\epsilon>0\), 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.
\[\blacksquare\]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 \(j \neq n\).
Proof:
We have:
\[y_1=\frac{1}{2},0,0, \cdots \\ y_2=0,\frac{1}{2^2},0, \cdots \\ y_3=0,0,\frac{1}{2^3}, \cdots \\ \vdots \\ y_n=0,0,0, \cdots, \frac{1}{2^n}, \cdots \\ \vdots\]Correspondingly, the norms are:
\[\|y_1\|=\frac{1}{2} \\ \|y_2\|=\frac{1}{2^2} \\ \|y_3\|=\frac{1}{2^3} \\ \|\vdots \\ \|y_n\|=\frac{1}{2^n} \\ \vdots\]Then \(\|y_1\| + \|y_2\| + \|y_3\| + \cdots\) is a convergent series.
The partial sum \(s_n\) is defined as:
\[s_n=\frac{1}{2},\frac{1}{2^2}, \cdots, \frac{1}{2^n}, 0, 0, \cdots\]As \(n \rightarrow \infty\), we get:
\[\lim\limits_{n \rightarrow \infty} s_n=\frac{1}{2},\frac{1}{2^2}, \cdots\]Thus, this is the space of sequences with finite non-zero terms.
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 \(m<n\), then \(\|s_n-s_m\|=\frac{1}{2^{m+1}}\). This can be made as small as possible, and thus \((s_n)\) is Cauchy.
Choose \(s=(\frac{1}{2^n})\), and we have:
\[\|s-s_n\|=\frac{1}{2^{n+1}}\]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.
Thus, \(y_1 +y_2 + y_3 + \cdots\) does not converge even though \(\|y_1\| + \|y_2\| + \|y_3\| + \cdots\) converges.
\[\blacksquare\]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.
Proof:
Take a Cauchy sequence \((x_n)\). Pick \(N_k\) such that \(\|x_m-x_n\|<\frac{1}{2^k}\) for all \(m,n \geq N_k\). Pick the corresponding \(y_k=x_{N_k}\) from \((x_n)\). Then note that \(\|y_k-y_{k+1}\| < \frac{1}{2^k}\).
Then \(\displaystyle \sum\limits_{k=1}^\infty \|y_{k+1}-y_k\| < \sum\limits_{k=1}^\infty \frac{1}{2^k} = 1\). 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 \(x\).
Now, we have:
\[\displaystyle \sum\limits_{k=1}^n y_{k+1}-y_k = \displaystyle \sum\limits_{k=1}^n x_{N_{k+1}}-x_{N_k}\\ =x_{N_{n+1}}-x_{N_1}\]In the limit of \(n \rightarrow \infty\), this expression tends to \(x\), that is:
\[\lim\limits_{n \rightarrow \infty} x_{N_{n+1}}-x_{N_1} = x \\ \lim\limits_{n \rightarrow \infty} x_{N_{n+1}} = x + x_{N_1}\]Thus, this limit exists and since \((x_n)\) was an arbitrary Cauchy sequence, it converges to \(x\). Thus \(X\) is complete.
\[\blacksquare\]2.3.9. Show that in a Banach space, an absolutely convergent series is convergent.
Proof:
Let there be an absolutely convergent series \(\displaystyle \sum\limits_{i=1}^\infty \|x_k\|<\infty\). Since it is convergent, it is also Cauchy, thus we have:
\[\displaystyle \sum\limits_{i=m}^n |x_k|<\epsilon\]By the Triangle Inequality, we have:
\[\displaystyle |\sum\limits_{i=m}^n x_k| \leq \sum\limits_{i=m}^n |x_k| \\ \displaystyle |\sum\limits_{i=m}^n x_k| = s_n - s_{m-1} < \epsilon\]Since the space is Banach, \((s_n)\) is a convergent sequence.
\[\blacksquare\]2.3.10. (Schauder basis) Show that if a normed space has a Schauder basis, it is separable.
Proof:
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 \(n \rightarrow \infty\).
A space is separable if it has a countable subset which is dense in this space.
The partial sum of a Schauder basis is represented as \(s_n=\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3 + \cdots + \alpha_n e_n\).
This implies that \(\forall \epsilon > 0, \exists N\) such that \(\|x-s_n\|<\epsilon\) for all \(n>N\). Thus every neighbourhood of \(x\) has a Schauder representation.
Since \(\alpha_n \in \mathbb{R}\), there exists a \(\beta_n \in \mathbb{Q}\), such that \(\|\alpha_n-\beta_n\|<\epsilon\).
(Prove that \(Y=\sum\limits_{i=1}^n \beta_i e_i\) is countable).
Denote \(s_n'=\beta_1 e_1 + \beta_2 e_2 + \beta_3 e_3 + \cdots + \beta_n e_n\), then we have:
\[\|s_n-s_n'\|=\|(\alpha_1-\beta_1) e_1 + (\alpha_2-\beta_2) e_2 + (\alpha_3-\beta_3) e_3 + \cdots + (\alpha_n-\beta_n) e_n\| \\ \leq \|(\alpha_1-\beta_1) e_1\| + \|(\alpha_2-\beta_2) e_2\| + \|(\alpha_3-\beta_3) e_3\| + \cdots + \|(\alpha_n-\beta_n) e_n\| \\ = |\alpha_1-\beta_1| \|e_1\| + |\alpha_2-\beta_2| \|e_2\| + |\alpha_3-\beta_3| \|e_3\| + \cdots + |\alpha_n-\beta_n| \|e_n\| \\ = \epsilon \|e_1\| + \epsilon \|e_2\| + \epsilon \|e_3\| + \cdots + \epsilon \|e_n\| \\ = \epsilon (\|e_1\| + \|e_2\| + \|e_3\| + \cdots + \|e_n\|)=K \epsilon\](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).
\[\|x-s_n'\| \leq \|x-s_n\| + \|s_n-s_n'\| < \epsilon + K \epsilon = (K+1) \epsilon\]This can be made as small as needed, and thus \(Y\) (countable) is dense in this normed space, and hence the space is separable.
\[\blacksquare\]2.3.11. Show that \((e_n)\), where \(e_n = (\delta_{nj})\), is a Schauder basis for \(l^p\), where \(1 \leq p< +\infty\).
Proof:
\(l^p\) is the space of all bounded sequences. This implies that:
\[\sum\limits_{i=1}^\infty {|x_i|}^p=K<\infty\]Equivalently,
\[\lim\limits_{n \rightarrow \infty} \sum\limits_{i=1}^n {|x_i|}^p=K\]The norm is defined as \({\|x\|}_p={\left(\sum\limits_{i=1}^\infty {|x_i|}^p\right)}^{1/p}\)
Assume the sequence is \(x_1, x_2, x_3, \cdots\). Then, we have:
\[x_1 e_1=x_1, 0, 0, 0, \cdots \\ x_2 e_2=0, x_2, 0, 0, \cdots \\ x_3 e_3=0, x_2, 0, 0, \cdots \\ \vdots \\ x_n e_n=0, 0, 0, 0, \cdots, x_n, 0, 0, \cdots \\\]Then, we get:
\[s_n=\sum\limits_{i=1}^n x_i e_i = x_1, x_2, \cdots, x_n, 0, 0, \cdots \\ x = s_n + \sum\limits_{i=n+1}^\infty x_i e_i \\ x-s_n = \sum\limits_{i=n+1}^\infty x_i e_i \\ \|x-s_n\| = {(\sum\limits_{i=n+1}^\infty {|x_i|}^p)}^{1/p}\]We know that:
\[\sum\limits_{i=1}^\infty {|x_i|}^p=K=\sum\limits_{i=1}^n {|x_i|}^p + \sum\limits_{i=n+1}^\infty {|x_i|}^p\]Taking limits on both sides for \(n \rightarrow \infty\), we get:
\[\lim\limits_{n \rightarrow \infty} \underbrace{\sum\limits_{i=1}^n {|x_i|}^p}_\text{Partial Sum} + \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p=K \\ K + \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p = K \\ \lim\limits_{n \rightarrow \infty} \sum\limits_{i=n+1}^\infty {|x_i|}^p = 0 \\\]Thus \(\|x-s_n\| \rightarrow 0\) as \(n \rightarrow \infty\).
Thus, \(e_n = (\delta_{nj})\) is a Schauder basis for \(l^p\).
\[\blacksquare\]2.3.12. (Seminorm) A seminorm on a vector space \(X\) is a mapping \(p: X \rightarrow \mathbb{R}\) satisfying (N1), (N3), (N4) in Sec. 2.3. (Some authors call this a pseudonorm.) Show that
\[p(0)= 0, \\ |p(y) - p(x)| \leq p(y-x).\](Hence if \(p(x) = 0\) implies \(x = 0\), then \(p\) is a norm.)
Proof:
\[p(\theta)=p(0x)=0 p(x) = 0\]Since the seminorm respects the Triangle Inequality, we have:
\[p(x)=p(x-y+y) \leq p(x-y) + p(y) \\ p(x)-p(y) \leq p(x-y)\]Similarly, we have:
\[p(y)=p(y-x+x) \leq p(y-x) + p(x) \\ p(y)-p(x) \leq p(y-x) = p(x-y) \\\]Therefore:
\[|p(y)-p(x)| \leq p(y-x)\] \[\blacksquare\]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 \(\hat{x} \in X/N\).
Proof:
From the set \(N=\{x \in X : p(x)=0\}\), pick \(x,y \in Y\).
We have:
\[p(\alpha x + \beta y) \leq p(\alpha x) + p(\beta y) = |\alpha| p(x) + |\beta| p(y) = 0\]Thus, \(\alpha x + \beta y \in N\), so \(N\) is a subspace.
The cosets of \(X/N\) are of the form \(x+N\). To prove that \({\|\hat{x}\|}_0=p(x)\) is a norm:
- (N1): 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.
- (N2): 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 \(p(\theta) \leq p(0) + p(y) = 0\).
- (N3): 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 \(\|\alpha \hat{x} \|_0 = p(\alpha x + y)=\vert \alpha \vert p(x+N) = \alpha \|\hat{x}\|_0\).
- (N4): 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 \(\|\hat{x_1} + \hat{x_2}\| \leq \|\hat{x_1}\|_0 + \|\hat{x_2}\|_0\).
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
\[{\|\hat{x}\|}_0 = \inf_{x \in \hat{x}} \|x\|\]where \(\hat{x} \in X/Y\), that is, \(\hat{x}\) is any coset of \(Y\).
Proof:
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:
- (N1): 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.
- (N2): 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.
- (N3): 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 \(\|\alpha \hat{x} \|_0 = \vert \alpha \vert \inf \|x+N\| = \alpha \|\hat{x}\|_0\).
- (N4): 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 \(\|\hat{x_1} + \hat{x_2}\|=\inf \|x_1 + y_1 + x_2 + y_2\| \leq \inf \|x_1 + y_1\| + \|x_2 + y_2\| = \|\hat{x_1}\| + \|\hat{x_2}\|\).
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
\[\|x\|=\max({\|x_1\|}_1, {\|x_2\|}_2)\]where \(x=(x_1,x_2)\).
Proof:
The operations on a product vector space are defined as:
\[(x_1,x_2) + (y_1,y_2) = (x_1+y_1, x_2+y_2) \\ \alpha(x_1, x_2) = (\alpha x_1, \alpha x_2)\]- (N1): Since we take the maximum of nonnegative norms, the norm on the product space is nonnegative.
- (N2): 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., \(x_1=x_2=0\).
- (N3): We have \(\|\alpha x\|=\|(\alpha x_1, \alpha x_2)\| = \max(\|\alpha x_1\|, \|\alpha x_2\|) = \vert\alpha\vert max(\|x_1\|, \|x_2\|) = \alpha \|x\|\).
- (N4): We have: \(\|x+y\|=\|(x_1,x_2) + (y_1,y_2)\| = \|(x_1+y_1, x_2+y_2)\| \\ =\max(\|x_1+y_1\|_1, \|x_2+y_2\|_2) \leq \max(\|x_1\|_1+\|y_1\|_1, \|x_2\|_2+\|y_2\|_2) \\ \leq \max(\|x_1\|_1, \|x_2\|_2) + \max(\|y_1\|_1, \|y_2\|_2) = \|x+y\|\)
Remember that:
\[a \leq max(a,b) \\ b \leq max(a,b) \\ c \leq max(c,d) \\ d \leq max(c,d) \\ \Rightarrow a+c \leq max(a,b) + max(c,d) \\ \Rightarrow b+d \leq max(a,b) + max(c,d) \\ \Rightarrow max(a+c,b+d) \leq max(a,b) + max(c,d)\] \[\blacksquare\]