Normed and Banach Spaces, Examples and Applications

Abstract

In the book’s first proper chapter, we will discuss the fundamental notions and theorems about normed and Banach spaces. We will introduce certain algebraic structures modelled on natural algebras of operators on Banach spaces.

I’m convinced mathematics is the most important investigating tool of the legacy of the human enterprise, it being the source of everything.

René Descartes

Exercises

2.1

Prove that any seminorm p satisfies $p(x)=p(-x)$.

2.2

Let K be a compact set, ${\mathsf X}$ a normed space and $f : K \rightarrow {\mathsf X}$ a continuous map. Show f is bounded, i.e. there exists $M\ge 0$ such that $||f(k)||\le M$ for any $k\in K$.

Hint. Adapt the proof of Proposition 1.21.

2.3

Prove that if ${\mathsf S}$ denotes a vector space of bounded maps from ${\mathsf X}$ to ${\mathbb C}$ (or to ${\mathbb R}$), then

$${\mathsf S}\ni f \mapsto ||f||_\infty := \sup _{x\in {\mathsf X}} |f(x)|$$

defines a norm on ${\mathsf S}$.

2.4

Let ${\mathsf X}$ be a topological space. Prove that the spaces of bounded complex functions $L({\mathsf X})$, and of measurable and bounded complex functions $M_b({\mathsf X})$ (cf. Examples 2.29), are Banach spaces for the norm $||\,\,||_\infty$.

Solution. We shall prove the claim for $M_b({\mathsf X})$, the other one being exactly the same. The claim is that an arbitrary Cauchy sequence $\{f_n\}_{n\in {\mathbb N}} \subset M_b({\mathsf X})$ converges uniformly to some $f\in M_b({\mathsf X})$. By assumption the numerical sequence $\{f_n(x)\}_{n\in {\mathbb N}}$ is Cauchy, for any $x\in {\mathsf X}$. Therefore there exists $f: {\mathsf X}\rightarrow {\mathbb C}$ such that $f_n(x) \rightarrow f(x)$, as $n\rightarrow + \infty$, for any $x\in {\mathsf X}$. This function will be measurable because it arises as limit of measurable maps. We are left to prove that f is bounded and $f_n \rightarrow f$ uniformly. Start from the latter. Since

$$|f(x)-f_m(x)| = \lim _{n\rightarrow +\infty } |f_n(x)-f_m(x)| \le \lim _{n\rightarrow +\infty } ||f_n-f_m||_\infty \,,$$

and using the fact that the initial sequence is Cauchy for $||\,\,||_\infty$, we have that for any $\varepsilon >0$ there is $N_\varepsilon$ such that:

$$\lim _{n\rightarrow +\infty } ||f_n-f_m||_\infty < \varepsilon \quad \text{ for } m> N_\varepsilon .$$

Hence:

$$|f(x)-f_m(x)| < \varepsilon \quad \text{ for }\, m> N_\varepsilon \,\text {and any}\, x\in {\mathsf X}.$$

In other words $||f-f_m||_\infty \rightarrow 0$ as $m\rightarrow +\infty$, as required. Now the boundedness is obvious:

$$\sup _{x\in {\mathsf X}} |f(x)| \le \sup _{x\in {\mathsf X}} |f(x) -f_m(x)| + \sup _{x\in {\mathsf X}} |f_m(x)|< \varepsilon + ||f_m||_\infty < + \infty \,.$$

2.5

Show that the Banach spaces $(L({\mathsf X}), ||\,\,||_\infty )$ and $(M_b({\mathsf X}), ||\,\,||_\infty )$ (cf. Examples 2.29) are Banach algebras with unit.

Sketch. The unit is clearly the constant map 1. The property $||f\cdot g||_\infty \le ||f||_\infty ||g||_\infty$ follows from the definition of $||\,\,||_\infty$, and the remaining conditions are easy.

2.6

Prove that the space $C_0({\mathsf X})$ of continuous, complex functions on ${\mathsf X}$ that vanish at infinity (cf. Examples 2.29) is a Banach algebra for $||\,\,||_\infty$. Explain in which circumstances the algebra has a unit.

Solution. We take a Cauchy sequence $\{f_n\}_{n\in {\mathbb N}} \subset C_0({\mathsf X})$ and prove it converges uniformly to $f\in C_0({\mathsf X})$. By hypothesis the numerical sequence $\{f_n(x)\}_{n\in {\mathbb N}}$ is Cauchy for any $x\in {\mathsf X}$. Therefore there exists a function $f: {\mathsf X}\rightarrow {\mathbb C}$ such that $f_n(x) \rightarrow f(x)$ for any $x\in {\mathsf X}$, as $n\rightarrow + \infty$. The proof that $||f-f_n||_\infty \rightarrow 0$, $n\rightarrow +\infty$, goes exactly as in Exercise 2.4. Since continuity is preserved by uniform limits, there remains to show $f\in C_0({\mathsf X})$. Given $\varepsilon >0$, pick n such that $||f-f_n|| < \varepsilon /2$, and choose a compact set $K_\varepsilon \subset {\mathsf X}$ so that $|f_n(x)| < \varepsilon /2$ for $x\in {\mathsf X}\setminus K_\varepsilon$. By construction

$$|f(x)| \le |f(x)-f_n(x)| + |f_n(x)| < \varepsilon , \quad {x \in {\mathsf X}\setminus K_\varepsilon .}$$

The Banach space thus found is a Banach algebra for the familiar operations, as one proves without difficulty.

If the unit is present, it must be the constant map 1. If ${\mathsf X}$ is compact, the function 1 belongs to the space. But if ${\mathsf X}$ is not compact, then 1 cannot be in ${\mathsf X}$, because the elements of $C_0({\mathsf X})$ can be shrunk arbitrarily outside compact subsets, and no constant map does that.

2.7

Prove the space $C_b({\mathsf X})$ of continuous and bounded complex functions on ${\mathsf X}$ (see Examples 2.29) is a Banach space for $||\,\,||_\infty$ and a Banach algebra with unit.

2.8

Prove that in Proposition 2.17 the converse implication holds as well. In other words, the proposition may be rephrased like this:

Proposition. Let $({\mathsf X}, ||\,\,||)$ be a normed space. Every absolutely convergent series $\sum _{n=0}^{+\infty } x_n$ (i.e. $\sum _{n=0}^{+\infty } ||x_n||< +\infty$) converges in ${\mathsf X}$ iff $({\mathsf X}, ||\,\,||)$ is a Banach space.

Solution. Take an absolutely convergent series $\sum _{n=0}^{+\infty } x_n$ in ${\mathsf X}$. The partial sums of the norms have to be a Cauchy sequence, i.e. for any $\varepsilon >0$ there is $M_\varepsilon >0$ with

$$\left| \sum ^n_{j=0} ||x_j|| - \sum ^m_{j=0} ||x_j|| \right| < \varepsilon \,, \quad \text{ for }\, n, m> M_\varepsilon .$$

Supposing $n\ge m$:

$$\left| \sum ^n_{j=m+1} ||x_j|| \right| < \varepsilon \,, \quad {n, m> M_\varepsilon .}$$

Therefore:

$$\left| \left| \sum ^n_{j=0} x_j - \sum ^m_{j=0} x_j \right| \right| = \left| \left| \sum ^n_{j=m+1} x_j \right| \right| \le \sum ^n_{j=m+1} ||x_j|| < \varepsilon , \quad {n, m> M_\varepsilon .}$$

We proved the sequence of partial sums $\sum _{j=0}^n x_n$ is Cauchy. But the space is complete, so the series converges to a point in ${\mathsf X}$.

2.9

Show that the space $C_c({\mathsf X})$ of complex functions with compact support on ${\mathsf X}$ (cf. Examples 2.29) is not, in general, a Banach space for $||\,\,||_\infty$, nor is it dense in $C_b({\mathsf X})$ if ${\mathsf X}$ is not compact.

Outline of proof. For the first statement we need to exhibit a counterexample for ${\mathsf X}= {\mathbb R}$. Consider for instance the continuous maps $f_n: {\mathbb R}\rightarrow {\mathbb C}$, $n=1,2,\ldots ,$: $f_n(x) := \frac{\sin x}{x}$ for $0<|x|<2n\pi$, $f_n(0)=1$ and $f_n(x)=0$ at other points of ${\mathbb R}$. The sequence evidently converges pointwise to the continuous map defined as $\frac{\sin x}{x}$ on ${\mathbb R}\setminus \{0\}$ and set to 1 at the origin. It is easy to convince ourselves that the convergence is uniform. But the limit function does not have compact support. As for the second part, note that any constant map $c\ne 0$ belongs in $C_c({\mathsf X})$. But if ${\mathsf X}$ is not compact, then $||f-c||_\infty \ge |c|>0$ for any function $f\in C_c({\mathsf X})$ because of the values attained outside the support of f.

2.10

Given a compact space K and a Banach space ${\mathsf B}$, let $C(K; {\mathsf B})$ be the space of continuous maps $f: K \rightarrow {\mathsf B}$ in the norm topologies of domain and codomain. Define

$$||f||_\infty := \sup _{x\in K} ||f(x)||\quad f\in C(K; {\mathsf B})\,,$$

where the norm on the right is the one on ${\mathsf B}$. Show this norm is well defined, and that it turns $C(K; {\mathsf B})$ into a Banach space.

Hint. Keep in mind Exercise 2.2 and adjust the proof of Proposition 2.18.

2.11

Let $({\mathfrak A}, \circ )$ be a Banach algebra without unit. Consider the direct sum ${\mathfrak A}\oplus {\mathbb C}$ and define the product:

$$(x, c)\cdot (y,c') := (x\circ y + cy + c'x, cc'),\quad {(x',c'), (x, c)\in {\mathfrak A}\oplus {\mathbb C}}$$

and the norm:

$$||(x,c)|| := ||x|| + |c|,\quad {(x, c)\in {\mathfrak A}\oplus {\mathbb C}.}$$

Show that the vector space ${\mathfrak A}\oplus {\mathbb C}$ with this product and norm becomes a Banach algebra with unit.

2.12

Take a Banach algebra ${\mathfrak A}$ with unit ${\mathbb I}$ and an element $a\in {\mathfrak A}$ with $||a||<1$. Prove that the series $\sum _{n=0}^{+\infty } (-1)^n a^{2n}$, $a^0 := {\mathbb I}$, converges in the topology of ${\mathfrak A}$. What is the sum?

Hint. Show the series of partial sums is a Cauchy series. The sum is $({\mathbb I}+ a^2)^{-1}$.

2.13

(Hard.) Prove Hölder’s inequality:

$$\int _X |f(x)g(x)| d\mu (x) \le \left( \int _{\mathsf X}|f(x)|^p d\mu (x)\right) ^{1/p} \left( \int _{\mathsf X}|g(x)|^q d\mu (x)\right) ^{1/q}$$

where $p, q>0$ satisfy $1= \frac{1}{p}+\frac{1}{q}$, f and g are measurable and $\mu$ is a positive measure on ${\mathsf X}$.

Solution. Define $I:=\int _X |f(x)|\, |g(x)| d\mu (x)$, $A:= \left( \int _{\mathsf X}|f(x)|^p d\mu (x) \right) ^{1/p}$ and $B:= \left( \int _{\mathsf X}|g(x)|^q d\mu (x)\right) ^{1/q}$. If either A or B is zero or infinite (conventionally, $\infty \cdot 0 = 0 \cdot \infty =0$), the inequality is trivial. So let us assume $0<A, B < +\infty$ and define $F(x):= |f(x)|/A$, $G(x):= |g(x)|/B$. Thus

$$\ln (F(x)G(x)) = \frac{1}{p} \ln (F(x)^p) + \frac{1}{q} \ln (G(x)^q) \le \ln \left( \frac{1}{p}F(x)^p + \frac{1}{q} G(x)^q \right) \,,$$

because the logarithm is a convex function. Exponentiating gives

$$F(x)G(x) \le \frac{1}{p}F(x)^p + \frac{1}{q} G(x)^q\,.$$

Integrating the above, and noting that the right-hand-side integral is $1/p + 1/q =1$ we recover Hölder’s inequality in the form:

$$\frac{\int _X |f(x)g(x)| d\mu (x)}{\left( \int _{\mathsf X}|f(x)|^p d\mu (x)\right) ^{1/p} \left( \int _{\mathsf X}|g(x)|^q d\mu (x)\right) ^{1/q}} \le 1\,.$$

2.14

(Hard.) Making use of Hölder’s inequality, prove Minkowski’s inequality:

$$\left( \int _X |f(x)+g(x)|^p d\mu (x)\right) ^{1/p} \le \left( \int _X |f(x)|^p d\mu (x)\right) ^{1/p} + \left( \int _X |g(x)|^p d\mu (x)\right) ^{1/p} $$

where $p\ge 1$, f and g are measurable and $\mu$ a positive measure on ${\mathsf X}$.

Solution. Define $I:=\int _X |f(x)+ g(x)|^p d\mu (x)$, $A:= \left( \int _{\mathsf X}|f(x)|^p d\mu (x) \right) ^{1/p}$ and $B:= \left( \int _{\mathsf X}|g(x)|^p d\mu (x)\right) ^{1/p}$. The inequality is trivial in case $p=1$ or if one of A, B is infinite. So we assume $p>1$, $A, B < +\infty$. Then I must be finite too, because $(a+b)^p \le 2^p(a^p+ b^p)$ for any $a, b\ge 0$ and $p\ge 1$.³ Minkowski’s inequality is trivial also when $I=0$, so we consider only $p>1$, $A, B < +\infty$, $0<I <+\infty$. Note $|f+g|^p = |f|\, |f+g|^{p-1} + |g|\, |f+g|^{p-1}$. Using Hölder’s inequality on each summand on the right we have:

$$\int _{\mathsf X}|f(x)+g(x)|^p d\mu (x) \le \left( \left( |f(x)+g(x)|^{(p-1)q} d\mu (x)\right) ^{1/q} \right) $$

$$\times \left( \left( \int _{\mathsf X}|f(x)|^p d\mu (x)\right) ^{1/p} + \left( \int _{\mathsf X}|g(x)|^p d\mu (x)\right) ^{1/p}\right) \,,$$

where $1= \frac{1}{p}+\frac{1}{q}$. This last inequality can be written as $I \le I^{1/q}(A+B)$, dividing which by $I^{1/q}$ produces $I^{1/p} \le A+B$, i.e. Minkowski’s inequality.

2.15

Take two finite-dimensional normed spaces ${\mathsf X},{\mathsf Y}$ and consider $T\in {\mathfrak L}({\mathsf X},{\mathsf Y})= {\mathfrak B}({\mathsf X},{\mathsf Y})$. Fix bases in ${\mathsf X}$ and ${\mathsf Y}$ and represent T by the matrix M(T). Show that one can choose bases for the dual spaces ${\mathsf X}'$, ${\mathsf Y}'$ so that the operator $T'$ is determined by the transpose matrix $M(T)^t$.

2.16

Prove Proposition 2.70.

2.17

Consider the space ${\mathfrak B}({\mathsf X})$ for some normed space ${\mathsf X}$. Prove the strong topology is finer than the weak topology (put loosely: weakly open sets are strongly open), and the uniform topology is finer than the strong one.

2.18

Prove Propositions 2.74–2.77.

2.19

In a normed space ${\mathsf X}$ prove that if $\{x_n\}_{n\in {\mathbb N}}\subset {\mathsf X}$ tends to $x\in {\mathsf X}$ weakly (cf. Proposition 2.74), the set $\{x_n\}_{n\in {\mathbb N}}$ is bounded.

Hint. Use Corollary 2.64.

2.20

If ${\mathsf B}$ is a Banach space and $T, S\in {\mathfrak B}({\mathsf B})$, show:

(i) $(TS)' = S'T'$,

(ii) $(T')^{-1}= (T^{-1})'$, provided T is bijective.

2.21

Prove that if ${\mathsf X}$ and ${\mathsf Y}$ are reflexive Banach spaces, and $T\in {\mathfrak B}({\mathsf X},{\mathsf Y})$, then $(T')' = T$.

2.22

If ${\mathsf X}$ is normed, the function that maps $(T, S) \in {\mathfrak B}({\mathsf X})\times {\mathfrak B}({\mathsf X})$ to $TS \in {\mathfrak B}({\mathsf X})$ is continuous in the uniform topology. What can be said regarding the strong and weak topologies?

Solution. For both topologies the map is separately continuous in either argument, but not continuous as a function of two variables, in general.

2.23

If we define an isomorphism of normed spaces as a continuous linear map with continuous inverse, does an isomorphism preserve completeness?

Hint. Extend Proposition 2.105 to the case of a continuous linear map between normed spaces with continuous inverse.

2.24

Using weak equi-boundedness, prove this variant of the Banach–Steinhaus Theorem 2.62.

Proposition. Let ${\mathsf X}$ be a Banach space and ${\mathsf Y}$ a normed space over the same field ${\mathbb C}$, or ${\mathbb R}$. Suppose the family of operators $\{T_\alpha \}_{\alpha \in A}\subset {\mathfrak B}({\mathsf X},{\mathsf Y})$ satisfies:

$$\sup _{\alpha \in A} |f(T_\alpha x)| <+\infty \quad \text{ for } \text{ any }\, x\in {\mathsf X}, f\in {\mathsf Y}'\,.$$

Then there exists a uniform bound $K\ge 0$:

$$||T_\alpha || \le K \,\,\,\,\text{ for } \text{ any }\, \alpha \in A\,.$$

Solution. Referring to Corollary 2.59, for any given $x\in {\mathsf X}$ define $F_{\alpha , x} := {\mathfrak I}(T_\alpha x) \in ({\mathsf Y}')'$. Then

$$\sup _{\alpha \in A} |F_{\alpha , x}(f)| <+\infty \quad \text{ for } \text{ any }\, f\in {\mathsf Y}'\,.$$

As ${\mathsf Y}'$ is complete, we can use Theorem 2.62 to infer the existence, for any $x\in {\mathsf X}$, of $K_x\ge 0$ that bounds uniformly the family $F_{\alpha , x}: {\mathsf Y}' \rightarrow {\mathbb C}$:

$$||F_{\alpha , x}|| \le K_x \,\,\,\,\text{ for } \text{ any }\, \alpha \in A\,.$$

But ${\mathfrak I}$ is isometric, so:

$$||T_{\alpha }(x)|| \le K_x \,\,\,\,\text{ for } \text{ any }\, \alpha \in A$$

and hence

$$\sup _{\alpha \in A} ||T_\alpha x|| <+\infty \quad \text{ for } \text{ any }\, x\in {\mathsf X}\,.$$

The Banach–Steinhaus Theorem 2.62 ends the proof.

2.25

Let K be compact, ${\mathsf X}$ a Banach space, and equip ${\mathfrak B}({\mathsf X})$ with the strong topology. Prove that any continuous map $f: K \rightarrow {\mathfrak B}({\mathsf X})$ belongs to $C(K;{\mathfrak B}({\mathsf X}))$. (The latter is a Banach space, defined in Exercise 2.10, if we view ${\mathfrak B}({\mathsf X})$ as a Banach space.)

Solution. We must prove

$$\sup _{k\in K} ||f(k)|| <+\infty \,$$

where on the left we used the operator norm of ${\mathfrak B}({\mathsf X})$. As f is continuous in the strong topology, for any given $x\in {\mathsf X}$ the map $K \ni k \mapsto f(k)x \in {\mathsf X}$ is continuous. If we fix $x\in {\mathsf X}$, by Exercise 2.2 there exists $M_x\ge 0$ such that:

$$\sup _{k\in K}||f(k)x||_{\mathsf X}< M_x\,.$$

The Banach–Steinhaus Theorem 2.62 ends the proof.