Technology and Art

Avishek Sen Gupta on 11 November 2021

This 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.

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., \(\|x+y\| \leq \|x\| + \|y\|, x,y \in X\)

**Proof:**

We have:

\[\|x\|=\|x + \theta|=\|x + \theta + (-\theta)\|=\|x + (-\theta) + \theta\| \\ \|x\| \leq \|x+(-\theta)\| + \|\theta\|\]We also have:

\[\|x+(-\theta)\| \leq \|x\| + \|-\theta\| \\ \|x+(-\theta)\| \leq \|x\| + |-1|\|\theta\| \\ \|x+(-\theta)\| \leq \|x\| + \|\theta\| = \|x\|\]Thus, \(\|x\| \leq \|x+(-\theta)\|\) and \(\|x\| \geq \|x+(-\theta)\|\). Thus, \(\|x\| = \|x+(-\theta)\|\), which is the distance between \(x\) and \(\theta\).

\[\blacksquare\]**Proof:**

(Easy to prove. TODO)

\[\blacksquare\]**Proof:**

We wish to prove the **Reverse Triangle Inequality**, which is:

We have:

\[\|x\|=\|x-y+y\| \leq \|x-y\| + \|y\| = \|y-x\| + \|y\| \\ \|x\| - \|y\| = \|y-x\| \\\]We also have:

\[\|y\|=\|y-x+x\| \leq \|y-x\| + \|x\| \\ \|y\| - \|x\| \leq \|y-x\|\]Then, we get:

\[|\|y\| - \|x\|| \leq \|y-x\|\] \[\blacksquare\]**Proof:**

We have from **(N2)** the following:

Assuming that \(\alpha=0\), and knowing that \(0x=\theta\), we get:

\[\|0 x\|=|0|\|x\| \\ \|\theta\|=0 \\\]Thus we conclude that \(x=\theta \Rightarrow \|\theta\|=0\) from **(N2)**.

We wish to prove that \(\|x\| \geq 0\).

\[\|x\|=\|x+x-x\| \leq \|x+x\| + \|-x\| = \|2x\| + \|x\| = 2\|x\| + \|x\| \\ 2\|x\| + \|x\| \geq \|x\| \\ 2\|x\| \geq 0 \\ \|x\| \geq 0 \\\] \[\blacksquare\]**Proof:**

(3) defines the norm: \({\|x\|}_2=\sqrt{(|\eta_1|^2 + |\eta_2|^2 + \cdots + |\eta_n|^2)}\)

(Easy to prove. TODO)

\[\blacksquare\]**Proof:**

(Easy to prove. TODO)

\[\blacksquare\]**Proof:**

(Easy to prove. TODO)

\[\blacksquare\]**In each case, verify that (N1) to (N4) are satisfied.**

**Proof:**

(Easy to prove. TODO)

The second result follows from **Minkowski’s Inequality**.

**Proof:**

(Easy to prove. TODO)

\[\blacksquare\]**Answer:**

(Check diagram in book after your curve sketching)

**Proof:**

The norm of the point \(z=\alpha x+(1-\alpha)y\) is:

\[\|z\|=\|\alpha x+(1-\alpha)y\| \leq \|\alpha x\|+\|(1-\alpha)y\| \\ = \alpha \|x\| + (1-\alpha) \|y\|\]Since \(\|x\| \leq 1\) and \(\|y\| \leq 1\), we get:

\[\|z\| \leq \alpha + (1-\alpha) = 1\]Thus \(z \in X\), and thus the closed unit ball is convex.

\[\blacksquare\]**Proof:**

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$.

Set \(\alpha=\frac{1}{2}\), we get \(z=(\frac{1}{2}, \frac{1}{2})\).

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.

\[\blacksquare\]**Proof:**

For any metric derived from a norm, it must be translation invariant, i.e.:

\[d(x+a,y+a)=d(x,y), x,y,a \in X \\ d(\alpha x,\alpha y)=d(x,y), x,y \in X, \alpha \in \mathbb{R}\]The discrete metric is defined as:

\[d(x,y)=\begin{cases} 0 & \text{if } x=y \\ 1 & \text{if } x \neq y \end{cases}\]Assume that \(x \neq y\). Then \(\alpha x \neq \alpha y\). Then \(d(\alpha x, \alpha y)=1 \neq \alpha d(x,y)\).

Thus, the discrete metric cannot be derived from a norm.

\[\blacksquare\]**Proof:**

For any metric derived from a norm, it must be translation invariant, i.e.:

\[d(x+a,y+a)=d(x,y), x,y,a \in X \\ d(\alpha x,\alpha y)=d(x,y), x,y \in X, \alpha \in \mathbb{R}\]Assume that \(x \neq y\). Then \(\alpha x \neq \alpha y\). Then \(\tilde{d}(\alpha x, \alpha y)=d(\alpha x, \alpha y) + 1 = \alpha d(x,y) + 1 \neq \alpha d(x,y) + \alpha = \alpha \tilde{d}(x,y)\).

\[\blacksquare\]**Proof:**

A set is bounded if \(\delta(x,y)<\infty\), where \(\delta(x,y)=\sup d(x,y)\).

\((\Rightarrow)\) Assume that \(M\) is bounded. Then \(\delta(x,y)=\sup d(x,y)<\infty\). 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:

\[d(x,\theta)=\|x\| \leq c\] \[\blacksquare\]\((\Leftarrow)\) Assume that there is a positive number \(c\) such that \(\|x\| \leq c\) for every \(x \in M\).

Then \(\|x\| \leq c\).

Using the **Triangle Inequality**, and noting that \(d(x, \theta)=\|x\|\) and \(d(y, \theta)=\|y\|\), we get:

Thus, \(M\) is bounded.

\[\blacksquare\]tags: