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Functional Analysis Exercises 8 : Normed and Banach Spaces

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.

Notes

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

2.2.1 Show that the norm \(\|x\|\) of x is the distance from x to 0.

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 \(\theta\).

\[\blacksquare\]

2.2.2 Verify that the usual length of a vector in the plane or in three dimensional space has the properties (N1) to (N4) of a norm.

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

2.2.3 Prove (2).

Proof:

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

\[|\|y\| - \|x\|| \leq \|y-x\|\]

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:

\[|\|y\| - \|x\|| \leq \|y-x\|\] \[\blacksquare\]

2.2.4 Show that we may replace (N2) by \(\|x\|=0 \Rightarrow x=0\) without altering the concept of a norm. Show that nonnegativity of a norm also follows from (N3) and (N4).

Proof:

We have from (N2) the following:

\[\|\alpha x\|=|\alpha|\|x\|\]

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).

\[\blacksquare\]

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

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

2.2.5 Show that (3) defines a norm.

Proof:

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

(Easy to prove. TODO)

\[\blacksquare\]

2.2.6 Let \(X\) be the vector space of all ordered pairs \(x = (\xi_1, \xi_2), y = (\eta_1, \eta_2), \cdots\) of real numbers. Show that norms on X are defined by

\[{\|x\|}_1=|\eta_1| + |\eta_2| \\ {\|x\|}_2={(\eta_1^2 + \eta_2^2)}^{1/2} \\ {\|x\|}_\infty=\max \{ |\xi_1|, |\xi_2| \}\]

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

2.2.7 Verify that (4) satisfies (N1) to (N4).

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

2.2.8 There are several norms of practical importance on the vector space of ordered n-tuples of numbers (cf. 2.2-2), notably those defined by

\[{\|x\|}_1=|\eta_1| + |\eta_2| + \cdots + |\eta_n| \\ {\|x\|}_p={(|\eta_1|^p + |\eta_2|^p + \cdots + |\eta_n|^p)}^{1/p} \\ {\|x\|}_\infty=\max \{ |\xi_1|, |\xi_2|, \cdots, |\xi_n| \}\]

In each case, verify that (N1) to (N4) are satisfied.

Proof:

(Easy to prove. TODO)

The second result follows from Minkowski’s Inequality.

\[\blacksquare\]

2.2.9 Verify that (5) defines a norm.

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

2.2.10 (Unit sphere) The sphere \(S(0; 1) = \{x \in X : \|x\| = 1\}\) in a normed space \(X\) is called the unit sphere. Show that for the norms in Prob. 6 and for the norm defined by the unit spheres look as shown in Fig. 16.

Answer:

(Check diagram in book after your curve sketching)


2.2.11 (Convex set, segment) A subset \(A\) of a vector space \(X\) is said to be convex if \(x,y \in A\) implies \(M=\{z \in X : z=\alpha x+(1-\alpha)y, 0\leq \alpha \leq 1\} \subset A\). \(M\) is called a closed segment with boundary points \(x\) and \(y\); any other \(z \in M\) is called an interior point of \(M\). Show that the closed unit ball \(B(0; 1) =\{x \in X : \|x\| \leq 1\}\) in a normed space X is convex.

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.

\[\blacksquare\]

2.2.12 Using Prob. 11, show that \(\phi(x)={(\sqrt{\vert\xi_1\vert} + \sqrt{\vert\xi_2\vert})}^2\) does not define a norm on the vector space of all ordered pairs \(x = (\xi_1, \xi_2), \cdots\) of real nwnbers. Sketch the curve \(\phi(x) = 1\) and compare it with Fig. 18.

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.

\[\blacksquare\]

2.2.13 Show that the discrete metric on a vector space \(X \neq \{0\}\) cannot be obtained from a norm. (Cf. 1.1-8.)

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.

\[\blacksquare\]

2.2.14 If \(d\) is a metric on a vector space \(X \neq \{0\}\) which is obtained from a norm, and \(\tilde{d}\) is defined by \(\tilde{d}(x,x) = 0, \tilde{d}(x,y)=d(x,y)+1 (x \neq y)\), show that \(d\) cannot be obtained from a norm.

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 \(\tilde{d}(\alpha x, \alpha y)=d(\alpha x, \alpha y) + 1 = \alpha d(x,y) + 1 \neq \alpha d(x,y) + \alpha = \alpha \tilde{d}(x,y)\).

\[\blacksquare\]

2.2.15 (Bounded set) Show that a subset \(M\) in a normed space \(X\) is bounded if and only if there is a positive number \(c\) such that \(\|x\| \leq c\) for every \(x \in M\). (For the definition, see Prob. 6 in Sec. 1.2.)

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:

\[d(x,y) \leq d(x,\theta) + d(\theta, y) \\ d(x,y) \leq \|x\| + \|y\| \\ d(x,y) \leq c + c \\ d(x,y) \leq 2c \\ \Rightarrow \delta(x,y) = \sup d(x,y) \leq 2c < \infty\]

Thus, \(M\) is bounded.

\[\blacksquare\]
tags: Mathematics - Proof - Functional Analysis - Pure Mathematics - Kreyszig