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This 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.
The requirements for a space to be a normed space are:
The requirements for a space to be a vector space are:
(Mnemonic: ADD1 SIIA)
Answer:
\(l^\infty\) is the space of all bounded sequences, i.e., \(\sum\limits_{i=1}^\infty \vert x_i \vert < \infty\). The norm it is equipped with is \(\|(x_n)\|=\sup \vert x_i\vert\).
The space of sequences with finitely many non-zero elements is an example of a subspace which is not closed.
In particular, the sequence defined as:
\[(x_j^n)=\begin{cases} \displaystyle\frac{1}{2^n} & \text{if } j \leq n \\ 0 & \text{if } j > n \end{cases}\]Assume \(m<n\). Then, we have:
\[\alpha x^{(m)} + \beta x^{(n)} \\ = \alpha (\frac{1}{2} , \frac{1}{2^2} , \cdots + \frac{1}{2^m}, , 0 , 0, \cdots) , \beta (\frac{1}{2} , \frac{1}{2^2} , \cdots , \frac{1}{2^m} , \frac{1}{2^{m+1}} , \cdots , \frac{1}{2^n} , 0 , 0, \cdots) \\ = (\alpha + \beta) \frac{1}{2} , (\alpha + \beta) \frac{1}{2^2} , \cdots , (\alpha + \beta) \frac{1}{2^m} , \frac{\beta}{2^{m+1}} , \cdots , \frac{\beta}{2^n}, 0 , 0, \cdots\]For the \(l^\infty\) case, we have the norm as:
\[{\|x-x^{(n)}\|}_\infty=\sup |x_j-x_j^{(n)}|=\frac{1}{2^{n+1}}\]where \(x=\frac{1}{2} , \frac{1}{2^2} , \cdots\).
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.
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).
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.
Answer:
We have the identity:
\[\|\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_1 e_n\| \geq c(|\alpha_1| + |\alpha_2| + \cdots + |\alpha_n|)\]For \(\mathbb{R}^2\), we have:
\[\|\alpha_1 e_1 + \alpha_2 e_2\| \geq c(|\alpha_1| + |\alpha_2|) \\ {(\alpha_1^2 + \alpha_2^2)}^{1/2} \geq c(|\alpha_1| + |\alpha_2|)\]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:
\[{(\epsilon^2 + \epsilon^2)}^{1/2} \geq c(\epsilon + \epsilon) \\ 2 \epsilon c \leq \sqrt 2 \epsilon \\ c \leq \frac{1}{\sqrt 2}\]For \(\mathbb{R}^3\), we have:
\[\|\alpha_1 e_1 + \alpha_2 e_2 + \alpha_3 e_3\| \geq c(|\alpha_1| + |\alpha_2| + |\alpha_3|) \\ {(\alpha_1^2 + \alpha_2^2 + \alpha_3^2)}^{1/2} \geq c(|\alpha_1| + |\alpha_2| + |\alpha_3|)\]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:
\[{(\epsilon^2 + \epsilon^2 + \epsilon^3)}^{1/2} \geq c(\epsilon + \epsilon + \epsilon) \\ 3 \epsilon c \leq \sqrt 3 \epsilon \\ c \leq \frac{1}{\sqrt 3}\]Proof:
The relation to demonstrate is an equivalence is the following:
\[a{\|x\|}_2 \leq {\|x\|}_1 \leq b{\|x\|}_2\]Reflexive: This is evident since:
\[{\|x\|}_1 \leq {\|x\|}_1 \leq b{\|x\|}_1\]where \(a=b=1\).
Symmetric:
We have:
\[{\|x\|}_1 \leq b{\|x\|}_2 \\ (1/b){\|x\|}_1 \leq {\|x\|}_2 \\ {\|x\|}_2 \geq (1/b){\|x\|}_1\] \[a{\|x\|}_2 \leq {\|x\|}_1 \\ {\|x\|}_2 \leq (1/a){\|x\|}_1\]Thus, we get:
\[\frac{1}{b}{\|x\|}_1 \leq {\|x\|}_2 \leq \frac{1}{a}{\|x\|}_1\]Transitive: Assume that:
\[a_1{\|x\|}_2 \leq {\|x\|}_1 \leq b_1{\|x\|}_2 \\ a_2{\|x\|}_3 \leq {\|x\|}_2 \leq b_2{\|x\|}_3\]We have the following two identities:
\[{\|x\|}_1 \leq b_1{\|x\|}_2 \\ (1/b_1){\|x\|}_1 \leq {\|x\|}_2 \\ (1/b_1){\|x\|}_1 \leq {\|x\|}_2 \leq b_2{\|x\|}_3 \\ {\|x\|}_1 \leq b_1 b_2{\|x\|}_3\] \[a_1{\|x\|}_2 \leq {\|x\|}_1 \\ {\|x\|}_2 \leq (1/a_1){\|x\|}_1 \\ a_2{\|x\|}_3 \leq {\|x\|}_2 \leq (1/a_1){\|x\|}_1 \\ a_1 a_2{\|x\|}_3 \leq {\|x\|}_1 \\\]Putting the two inequalities above together, we get:
\[(a_1 a_2){\|x\|}_3 \leq {\|x\|}_1 \leq (b_1 b_2){\|x\|}_3\] \[\blacksquare\]Proof:
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.
The equivalent norms are related as:
\[a{\|x\|}_2 \leq {\|x\|}_1 \leq b{\|x\|}_2\]From the above relation, we know that \(a \leq b\).
Pick an open ball \(B_1(x_0,r)\). Let \(x \in B_1(x_0,r)\). Then \(d_1(x_0,x) < r\). We then have:
\[a \cdot d_2(x_0,x) \leq d_1(x_0,x) < r \\ d_2(x_0,x) < \frac{r}{a} \\ \Rightarrow x \in B_2(x_0,\frac{r}{a}) \\ \Rightarrow B_1(x_0,r) \in B_2(x_0,\frac{r}{a})\]Conversely, pick an open ball \(B_2(x_0,r)\). Let \(x \in B_2(x_0,r)\). Then \(d_2(x_0,x) < r\). We then have:
\[d_1(x_0,x) \leq b \cdot d_2(x_0,x) < br \\ d_1(x_0,x) < br \\ \Rightarrow x \in B_1(x_0,br) \\ \Rightarrow B_2(x_0,r) \in B_1(x_0,br)\] \[\blacksquare\]Proof:
Assume that the equivalence relation is:
\[a{\|x\|}_0 \leq {\|x\|} \leq b{\|x\|}_0\]Assume a Cauchy sequence \((x_n)\) in \((X,{\|\bullet\|}_0)\), we have that \(\forall \epsilon>0, \exists N_0\) such that \(d_0(x_m,x_n)<\epsilon\) for all \(m,n>N_0\).
Then, we have:
\[d(x_m,x_n) \leq b d_0(x_m,x_n) < b \epsilon \\ d(x_m,x_n) < b \epsilon\]Thus \((x_n)\) is also a Cauchy sequence in \((X,\|\bullet\|)\).
Assume a Cauchy sequence \((x_n)\) in \((X,{\|\bullet\|})\), we have that \(\forall \epsilon>0, \exists N_0\) such that \(d(x_m,x_n)<\epsilon\) for all \(m,n>N_0\).
Then, we have:
\[a d_0(x_m,x_n) \leq b d(x_m,x_n) < \epsilon \\ d_0(x_m,x_n) < \frac{\epsilon}{a}\]Thus \((x_n)\) is also a Cauchy sequence in \((X,{\|\bullet\|}_0)\).
\[\blacksquare\]Proof:
The norms \({\|\bullet\|}_2\) and \({\|\bullet\|}_\infty\) are defined as:
\[{\|x\|}_2={\left(\sum\limits_{i=1}^n {|x_i|}^2\right)}^{1/2} \\ {\|x\|}_\infty=\sup |x_i|\]We have:
\[{|\sup{x_i}|}^2 + {|\sup{x_i}|}^2 + \cdots + {|\sup{x_i}|}^2 \geq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ n{|\sup{x_i}|}^2 \geq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ \sqrt{n}|\sup{x_i}| \geq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2}\]We also have:
\[{|\sup{x_i}|}^2 \leq {|x_1|}^2 + {|x_2|}^2 + \cdots + {|x_n|}^2 \\ {|\sup{x_i}|} \leq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2}\]Thid implies that:
\[{|\sup{x_i}|} \leq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2} \leq \sqrt{n}|\sup{x_i}| \\ {\|\bullet\|}_\infty \leq {\|\bullet\|}_2 \leq \sqrt{n} {\|\bullet\|}_\infty\] \[\blacksquare\]Proof:
We have the vector \(x=\alpha_1 e_1 + \alpha_2 e_2 + \cdots + \alpha_n e_n\). Then we have, by the Triangle Inequality:
\[\|x\|=\|a_1 e_1 + a_2 e_2 + \cdots + a_n e_n\| \leq |a_1| \|e_1\| + |a_2| \|e_2\| + \cdots + |a_n| \|e_n\| \\ \leq \max(\|e_i\|)(|a_1| + |a_2| + \cdots + |a_n|)\]But by the Cauchy-Schwarz Inequality, we have:
\[\sum\limits_{i=1}^n |a_i| \leq {\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} {\left(\sum\limits_{i=1}^n 1\right)}^{1/2} = \sqrt{n} {\|x\|}_2\]This implies that:
\[\|x\| \leq \sqrt{n}\max(\|e_i\|){\|x\|}_2 \\ \|x\| \leq b {\|x\|}_2\]where \(b=\sqrt{n}\max{\|e_i\|}\).
\[\blacksquare\]Proof:
The norms \({\|\bullet\|}_2\) and \({\|\bullet\|}_\infty\) are defined as:
\[{\|x\|}_1=\sum\limits_{i=1}^n {|a_i|} \\ {\|x\|}_2={\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} \\\]By the Cauchy-Schwarz Inequality, we have:
\[{\|x\|}_1 = \sum\limits_{i=1}^n |a_i| \leq {\left(\sum\limits_{i=1}^n {|a_i|}^2\right)}^{1/2} {\left(\sum\limits_{i=1}^n 1\right)}^{1/2} = \sqrt{n} {\|x\|}_2 \\ \Rightarrow {\|x\|}_1 \leq \sqrt{n} {\|x\|}_2 \\ \Rightarrow \displaystyle\frac{1}{\sqrt{}n}{\|x\|}_1 \leq {\|x\|}_2\]By the Cauchy-Schwarz Inequality, we have:
\[{ {\|x\|}_2}^2 = \sum\limits_{i=1}^n {|a_i|}^2 \\ \sum\limits_{i=1}^n |a_i||a_i| \leq \left(\sum\limits_{i=1}^n |a_i|\right)\left(\sum\limits_{i=1}^n |a_i|\right)={\left(\sum\limits_{i=1}^n |a_i|\right)}^2 \\ {\left(\sum\limits_{i=1}^n |a_i||a_i|\right)}^{1/2} \leq {\left(\sum\limits_{i=1}^n |a_i|\right)} \\ \Rightarrow {\|x\|}_2 \leq {\|x\|}_1\]Putting the above inequalities together, we get:
\[\displaystyle\frac{1}{\sqrt{}n}{\|x\|}_1 \leq {\|x\|}_2 \leq {\|x\|}_1\] \[\blacksquare\]Proof:
\[a{\|x\|} \leq {\|x\|}_0 \leq b{\|x\|}\]Let \(\|x_n-x\| \rightarrow 0\). This implies that \(\|x_n-x\| < \epsilon / b\), for some \(\epsilon / b > 0\). By the equivalence relation, we then have:
\[{\|x_n-x\|}_0 \leq b{\|x_n-x\|} < b (\epsilon/b) = \epsilon \\ \Rightarrow {\|x_n-x\|}_0 \rightarrow 0\]Let \({\|x_n-x\|}_0 \rightarrow 0\). This implies that \({\|x_n-x\|}_0 < a \epsilon\), for some \(a \epsilon > 0\). By the equivalence relation, we then have:
\[a {\|x_n-x\|} \leq {\|x_n-x\|}_0 < (1/a) (a \epsilon) = \epsilon \\ \Rightarrow {\|x_n-x\|} \rightarrow 0\] \[\blacksquare\]Proof:
TODO: