# Total Internal Reflection

Technology and Art

# Functional Analysis Exercises 9 : Further Properties of Normed Spaces

Avishek Sen Gupta on 19 November 2021

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:

$\supd(x_j^m, x_j^n)<\epsilon \\ d(x_j^m, x_j^n)<\epsilon$

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.

$\blacksquare$

(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 $$\|\hat{x_1} + \hat{x_2}\| \leq \|\hat{x_1}\|_0 + \|\hat{x_2}\|_0$$.
$\blacksquare$

#### 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 $$\|\hat{x_1} + \hat{x_2}\|=\inf \|x_1 + y_1 + x_2 + y_2\| \leq \inf \|x_1 + y_1\| + \|x_2 + y_2\| = \|\hat{x_1}\| + \|\hat{x_2}\|$$.
$\blacksquare$

#### 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:

$a \leq max(a,b) \\ b \leq max(a,b) \\ c \leq max(c,d) \\ d \leq max(c,d) \\ \Rightarrow a+c \leq max(a,b) + max(c,d) \\ \Rightarrow b+d \leq max(a,b) + max(c,d) \\ \Rightarrow max(a+c,b+d) \leq max(a,b) + max(c,d)$ $\blacksquare$

tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig