# Total Internal Reflection

Technology and Art

# Functional Analysis Exercises 10 : Finite Dimensional Normed Spaces and Subspaces

Avishek Sen Gupta on 22 November 2021

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.

### 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., $$\|x+y\| \leq \|x\| + \|y\|, x,y \in X$$

The requirements for a space to be a vector space are:

• (V1) Symmetry implies $$x+y=y+x, x,y \in X$$.
• (V2) Identity implies zero vector, i.e., $$x+\theta=x, x,\theta \in X$$.
• (V4) Inverse implies $$x+(-x)=\theta, x,y,\theta \in X$$.
• (V3) Associativity implies $$(x+y)+z=x+(y+z), x,y,z \in X$$.

#### Multiplication

• (V1) Associativity implies $$x(yz)=(xy)z, x,y,z \in X$$.
• (V2) Distributivity with respect to vector addition implies $$\alpha(x+y)=\alpha x + \alpha y, x,y \in X, \alpha \in \mathbb{R}$$.
• (V3) Distributivity with respect to scalar addition implies $$(\alpha + \beta) x = \alpha x + \beta x, x \in X, \alpha, \beta \in \mathbb{R}$$.
• (V4) Identity implies, $$1x=x, x \in X$$.

#### 2.4.1. Give examples of subspaces of $$l^\infty$$ and $$l^2$$ which are not closed.

$$l^\infty$$ 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.

#### 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 $$x_1 = (1,0,0), x_2 = (0,1,0), x_3 = (0,0,1)$$?

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:

${(\epsilon^2 + \epsilon^2 + \epsilon^3)}^{1/2} \geq c(\epsilon + \epsilon + \epsilon) \\ 3 \epsilon c \leq \sqrt 3 \epsilon \\ c \leq \frac{1}{\sqrt 3}$

#### 2.4.3. Show that in Def. 2.4-4 the axioms of an equivalence relation hold (cf. A1.4 in Appendix 1).

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:

$(a_1 a_2){\|x\|}_3 \leq {\|x\|}_1 \leq (b_1 b_2){\|x\|}_3$ $\blacksquare$

#### 2.4.4. Show that equivalent norms on a vector space $$X$$ induce the same topology for $$X$$.

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:

$d_1(x_0,x) \leq b \cdot d_2(x_0,x) < br \\ d_1(x_0,x) < br \\ \Rightarrow x \in B_1(x_0,br) \\ \Rightarrow B_2(x_0,r) \in B_1(x_0,br)$ $\blacksquare$

#### 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.

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 $$(X,{\|\bullet\|}_0)$$.

$\blacksquare$

#### 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.

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:

${|\sup{x_i}|} \leq {\left(\sum_{i=1}^{n}{|x_i|}^2\right)}^{1/2} \leq \sqrt{n}|\sup{x_i}| \\ {\|\bullet\|}_\infty \leq {\|\bullet\|}_2 \leq \sqrt{n} {\|\bullet\|}_\infty$ $\blacksquare$

#### 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 $$b>0$$ such that $$\|x\| \leq b {\|x\|}_2$$ for all $$x$$.

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 $$b=\sqrt{n}\max{\|e_i\|}$$.

$\blacksquare$

#### 2.4.8. Show that the norms $${\|\bullet\|}_1$$ and $${\|\bullet\|}_2$$ in Prob. 8, Sec. 2.2, satisfy $$\frac{1}{\sqrt{n}} {\|x\|}_1 \leq {\|x\|}_2 \leq {\|x\|}_1$$.

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:

$\displaystyle\frac{1}{\sqrt{}n}{\|x\|}_1 \leq {\|x\|}_2 \leq {\|x\|}_1$ $\blacksquare$

#### 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).

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:

$a {\|x_n-x\|} \leq {\|x_n-x\|}_0 < (1/a) (a \epsilon) = \epsilon \\ \Rightarrow {\|x_n-x\|} \rightarrow 0$ $\blacksquare$

#### 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 $$Z$$?

Proof:

TODO:

• Need to prove dimension of space is $$mn$$.
• 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.
$\blacksquare$
tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig