Functional Analysis Exercises 8 : Normed and Banach Spaces

Avishek Sen Gupta on 11 November 2021

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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:

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

2.2.1 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 \(\theta\).

\[\blacksquare\]

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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)

\[\blacksquare\]

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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:

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

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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 \(\|x\| \geq 0\).

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

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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)

\[\blacksquare\]

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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)

\[\blacksquare\]

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

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

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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.

\[\blacksquare\]

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2.2.9 Verify that (5) defines a norm.

Proof:

(Easy to prove. TODO)

\[\blacksquare\]

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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)

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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.

\[\blacksquare\]

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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.

\[\blacksquare\]

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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.

\[\blacksquare\]

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

\[\blacksquare\]

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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.

\[\blacksquare\]

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