10.9 Topological vector spaces

中文

A topological vector space is a vector space $V$ that has a Hausdorff topology defined on it, such that the operations of vector addition and scalar multiplication are continuous functions on their respective domains with respect to this topology,

$$\begin{array}{ll}\psi :V\times V\to V& \text{ defined by }\psi (u,v)=u+v,\\ \tau :\mathbb{K}\times V\to V& \text{ defined by }\tau (\lambda ,v)=\lambda v.\end{array}$$

We will always assume that the field ofscalars is either the real or complex numbers, $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C};$ in the latter case the topology is the standard topology in ${\mathbb{R}}^2$

Recall from Section 10.3 that a sequence of vectors $v_n\in V$ is called convergent if there exists a vector $v\in V$ , called its limit, such that for every open neighbourhood $U$ of $v$ there is an integer $N$ such that $v_n\in U$ for all $n\ge N$ . We also say the sequence converges to $v$, denoted

$$v_n\to v\text{ or }\underset{n\to \infty }{\lim }{v}_n=v.$$

The following properties of convergent sequences are easily proved:

$$v_n\to v\text{ and }{v}_n\to v^{\prime }⟹v=v^{\prime },\text{\hspace{1em}(10.3)}$$

$$v_n=v\text{ for all }n⟹v_n\to v,\text{\hspace{1em}(10.4)}$$

$$\text{ if }\{v_n^{\prime }\}\text{ is a subsequence of }{v}_n\to v\text{ then }{v}_n^{\prime }\to v,\text{\hspace{1em}(10.5)}$$

$$u_n\to u,v_n\to v⟹u_n+\lambda v_n\to u+\lambda v,\text{\hspace{1em}(10.6)}$$

where $\lambda \in \mathbb{K}$ is any scalar. Also, if ${\lambda }_n$ is a convergent sequence of scalars in $\mathbb{K}$ then

$${\lambda }_n\to \lambda ⟹{\lambda }_{n}u\to \lambda u.\text{\hspace{1em}(10.7)}$$

Example 10.23

The vector spaces ${\mathbb{R}}^n$ are topological vector spaces with respect to the Euclidean topology. It is worth giving the full proof of this statement, as it sets the pattern for a number of other examples. A set $U\subset {\mathbb{R}}^n$ is open if and only if for every $\mathbf{x}\in U$ there exists $ϵ>0$ such that

$$I_{ϵ}(\mathbf{x})=\{\mathbf{y}||y_i-x_i|<ϵ,\text{ for all }i=1,\dots ,n\}\subseteq U.$$

To show that vector addition $\psi$ is continuous, it is necessary to show that $N={\psi }^{-1}(I_{ϵ}(\mathbf{x}))$ is an open subset of ${\mathbb{R}}^n\times {\mathbb{R}}^n$ for all $\mathbf{x}\in {\mathbb{R}}^n,ϵ>0$ If $\psi (\mathbf{u},\mathbf{v})=\mathbf{u}+\mathbf{v}=\mathbf{x}$ then for any $({\mathbf{u}}^{\prime },{\mathbf{v}}^{\prime })\in I_{ϵ/2}(\mathbf{u})\times I_{ϵ/2}(\mathbf{v})$ , we have for all $i=1,\dots ,n$

$$\begin{array}{c}\left|\right(x_i-(u_i^{\prime }+v_i^{\prime })\left)\right|=\left|\right((u_i-u_i^{\prime })+(v_i-v_i^{\prime })\left)\right|\\ \le \left|\right((u_i-u_i^{\prime })\left)\right|+\left|\right((v_i-v_i^{\prime })\left)\right|\le \frac{ϵ}{2}+\frac{ϵ}{2}=ϵ.\end{array}$$

Hence $I_{ϵ/2}(\mathbf{u})\times I_{ϵ/2}(\mathbf{v})\subset N$ , and continuity of $\psi$ is proved.

For continuity of the scalar multiplication function $\tau$, let $M={\tau }^{-1}(I_{ϵ}(\mathbf{x}))\subset \mathbb{R}\times {\mathbb{R}}^n$ . If $\mathbf{x}=a\mathbf{u}$ , let $\mathbf{v}\in I_{\delta }(\mathbf{u})$ and $b\in I_{{\delta }^{\prime }}(a)$ . Then, setting $A={\max }_i\left|u_i\right|$ , we have

$$\begin{array}{l}|b{v}_i-a{u}_i|=|b{v}_i-b{u}_i+b{u}_i-a{u}_i|\\ \le |b||v_i-u_i|+|b-a||u_i|\\ \le (|a|+{\delta }^{\prime })\delta +{\delta }^{\prime }A\\ \le ϵ\text{ if }{\delta }^{\prime }=\frac{ϵ}{2A}\text{ and }\delta =\frac{ϵ}{2|a|A+ϵ}.\end{array}$$

A similar proof may be used to show that the complex vector space ${\mathbb{C}}^n$ is a topological vector space.

Example 10.24

The vector space ${\mathbb{R}}^{\infty }$ consisting of all infinite sequences $\mathbf{x}=(x_1,x_2,\dots )$ is an infinite dimensional vector space. We give it the product topology as described, whereby a set $U$ is open if for every point $\mathbf{x}\in U$ there is a finite sequence ofintegers $\mathbf{i}=(i_1,i_2,\dots ,i_n)$ such that

$$I_{\mathbf{i},ϵ}(\mathbf{x})=\left\{\mathbf{y}\mid \left|y_{i_k}-x_{i_k}\right|<ϵ\text{ for }k=1,\dots ,n\right\}\subset U.$$

This neighbourhood of $x$ is an infinite product of intervals of which all but a finite number consist of all of $\mathbb{R}$. To prove that $\psi$ and $\tau$ are continuous functions, we again need only show that ${\psi }^{-1}(I_{\mathbf{i},ϵ}(\mathbf{x}))$ and ${\tau }^{-1}(I_{\mathbf{i},ϵ}(\mathbf{x}))$ are open sets. The argument follows along essentially identical lines to that in Example 10.23. To prove continuity of the scalar product $\tau$, we set $A={\max }_{i\in \mathbf{i}}|u_i|$ , where $\mathbf{x}=a\mathbf{u}$ and continue as in the previous example.

Example 10.25

Let $S$ be any set, and set $\mathcal{F}(S)$ to be the set of bounded real-valued functions on $S$. This is obviously a vector space, with vector addition defined by $(f+g)(x)=f(x)+g(x)$ and scalar multiplication by $(af)(x)=af(x)$ . A metric can be defined on this space by setting $d(f,g)$ to be the least upper bound of $|f(x)-g(x)|$ on $S$. Conditions (Met1)–(Met4) are easy to verify. The vector space $\mathcal{F}(S)$ is a topological vector space with respect to the metric topology generated by this distance function. For example, let $f(x)=u(x)+v(x)$ , then if $|u^{\prime }(x)-u(x)|<ϵ/2$ and $|v^{\prime }(x)-v(x)|<ϵ/2$ it follows at once that $|f(x)-(u^{\prime }(x)+v^{\prime }(x))|<ϵ$ . To prove continuity of scalar addition we again proceed as in Example 10.23. If $f(x)=au(x)$ , let $|b-a|<ϵ/2A$ where $A$ is an upper bound of $u(x)$ in $S$ and $|v(x)-u(x)|<ϵ/(2|a|A+ϵ)$ for all $x\in S;$ ; then

$$|bv(x)-f(x)|<ϵ\text{ for all }x\in S.$$

Banach spaces

A norm on a vector space $V$ is a map $\Vert \cdot \Vert :V\to \mathbb{R}$ , associating a real number $\Vert v\Vert$ with every vector $v\in V$ , such that

(Norm1) $\Vert v\Vert \ge 0$ , and $\Vert v\Vert =0$ if and only if $v=0$

(Norm2) $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$

(Norm3) $\Vert u+v\Vert \le \Vert u\Vert +\Vert v\Vert$

In most cases the field of scalars is taken to be the complex numbers, $\mathbb{K}=\mathbb{C}$ , although much of what we say also applies to real normed spaces. We have met this concept earlier, in the context of a complex inner product space (see Section 5.2).

A norm defines a distance function $d:V\times V\to \mathbb{R}$ by

$$d(u,v)=\Vert u-v\Vert .$$

The properties (Met1)–(Met3) are trivial to verify, while the triangle inequality

$$d(u,v)\le d(u,w)+d(w,v)\text{\hspace{1em}(10.8)}$$

is an immediate consequence of (Norm3),

$$\Vert u-v\Vert =\Vert u-w+w-v\Vert \le \Vert u-w\Vert +\Vert w-v\Vert .$$

We give $V$ the standard metric topology generated by open balls $B_a(v)=\{u\mid d(u,v)<a\}$ as in Section 10.3. This makes it into a topological vector space. To show that the function $(u,v)\mapsto u+v$ is continuous with respect to this topology,

$$\Vert u^{\prime }-u\Vert <\frac{ϵ}{2}\text{ and }\Vert v^{\prime }-v\Vert <\frac{ϵ}{2}⟹\Vert u^{\prime }+v^{\prime }-(u+v)\Vert <ϵ$$

on using the triangle inequality. The proof that $(\lambda ,v)\mapsto \lambda v$ is continuous follows the lines of Example 10.23.

Exercise

Show that the norm is a continuous function $\Vert \cdot \Vert :V\to \mathbb{R}$

Example 10.26

The vector space $\mathcal{F}(S)$ ofbounded real-valued functions on a set $S$ defined in Example 10.25 has a norm

$$\Vert f\Vert =\underset{x\in S}{\sup }|f(x)|,$$

giving rise to the distance function $d(f,g)$ of Example 10.25. This is called the supremum norm.

Convergence of sequences is defined on $V$ by

$$u_n\to u\text{ if }d(u_n,u)=\Vert u_n-u\Vert \to 0.$$

As in Section 10.3, every convergent sequence $u_i\to u$ is a Cauchy sequence

$$\Vert u_i-u_j\Vert \le \Vert u-u_i\Vert +\Vert u-u_j\Vert \to 0\text{ as }i,j\to \infty ,$$

but the converse need not always hold. We say a normed vector space $(V,\Vert \cdot \Vert )$ is complete, or is a Banach space, if every Cauchy sequence converges,

$$\Vert u_i-u_j\Vert \to 0\text{ as }i,j\to \infty ⟹u_i\to u\text{ for some }u\in V.$$

Exercise

Give an example of a vector subspace of ${\mathbb{C}}^{\infty }$ that is an incomplete normed vector space.

Example 10.27

On the vector space ${\mathbb{C}}^n$ define the standard norm

$$\Vert \mathbf{x}\Vert =\sqrt{|x_1{|}^2+|x_2{|}^2+\cdots +|x_n{|}^2}.$$

Conditions (Norm1) and (Norm2) are trivial, while (Norm3) follows from Theorem 5.6, since this norm is precisely that defined in Eq. (5.11) from the inner product $⟨x|y⟩={\sum }_{i=1}^n\overline{x_i}{y}_i$ . If $\Vert {\mathbf{x}}_n-{\mathbf{x}}_m\Vert \to 0$ for a sequence of vectors ${\mathbf{x}}_n$ , then each component is a Cauchy sequence $|x_{ni}-x_{mi}|\to 0.$ , and therefore has a limit $x_{ni}\to x_i$ . It is straightforward to show that $\Vert {\mathbf{x}}_n-\mathbf{x}\Vert \to 0$ where $\mathbf{x}=(x_1,x_2,\dots ,x_n)$ . Hence this normed vector space is complete.

Example 10.28

Let $\mathcal{D}([-1,1])$ be the vector space of bounded differentiable complex valued functions on the closed interval $[-1,1]$. As in Example 10.26 we adopt the supremum norm $\Vert f\Vert ={\sup }_{x\in [-1,1]}|f(x)|$ . This normed vector space is not complete, for consider the sequence of functions

$$f_n(x)={\left|x\right|}^{1+1/n}(n=1,2,\dots ).$$

These functions are all differentiable on $[-1,1]$ and have zero derivative at $x=0$ from both the left and the right. Since they approach the bounded function $|x|$ as $n\to \infty$ this is necessarily a Cauchy sequence. However, the limit function $|x|$ is not differentiable at $x=0$ , and the norm is incomplete.

By a linear functional on a Banach space $V$ we always mean a continuous linear map $\phi :V\to \mathbb{C}$ . The vector space of all linear functionals on $V$ is called the dual space $V^{\prime }$ of $V$. If $V$ is finite dimensional then $V^{\prime }$ and $V^{\ast }$ coincide, since all linear functionals are continuous with respect to the norm

$$\Vert \mathbf{u}\Vert =\sqrt{|u_1{|}^2+\cdots +|u_n{|}^2},$$

but for infinite dimensional spaces it is important to stipulate the continuity requirement. A linear map $\phi$ on a Banach space is said to be bounded if there exists $M>0$ such that

$$|\phi (x)|\le M\Vert x\Vert \text{ for all }x\in V.$$

The following theorem shows that the words ‘bounded’ and ‘continuous’ are interchangeable for linear functionals on a Banach space.

Theorem 10.23 · Continuity and boundedness of linear functionals

A linear functional $\phi :V\to \mathbb{R}$ on a Banach space $V$ is continuous if and only if it is bounded

Proof

If $\phi$ is bounded, let $M>0$ be such that $|\phi (x)|\le M\Vert x\Vert$ for all $x\in V$ . Then for any pair of vectors $x$, $y\in V$

$$|\phi (x-y)|\le M\Vert x-y\Vert ,$$

and for any $ϵ>0$ we have

$$\Vert x-y\Vert <\frac{ϵ}{M}⟹|\phi (x)-\phi (y)|=|\phi (x-y)|\le ϵ.$$

Hence $\phi$ is continuous.

Conversely, suppose $\phi$ is continuous. In particular, it is continuous at the origin $x=0$ and there exists $\delta >0$ such that

$$\Vert x\Vert <\delta ⟹|\phi (x)|<1.$$

For any vector $y\in V$ we have

$$\Vert \frac{\delta y}{\Vert y\Vert }\Vert <\delta$$

whence

$$\left|\phi \left(\frac{\delta y}{\Vert y\Vert }\right)\right|<1.$$

Thus

$$|\phi (y)|<\frac{\Vert y\Vert }{\delta },$$

showing that $\phi$ is bounded.

Example 10.29

Let ${\ell }^1$ be the vector space of all complex infinite sequences $\mathbf{X}=(x_1,x_2,\dots )$ that are absolutely convergent,

$$\Vert \mathbf{x}\Vert =\sum _{i=1}^{\infty }|x_i|<\infty .$$

If $\mathbf{c}=(c_1,c_2,\dots )$ is a bounded infinite sequence of complex numbers, $|c_i|\le C$ for all $i=1,2,\dots$ , then

$${\phi }_c(\mathbf{x})=c_1{x}_1+c_2{x}_2+\dots$$

is a continuous linear functional on ${\ell }^1$ . Linearity is obvious as long as the series converges, and convergence and boundedness are proved in one step,

$$\Vert {\phi }_c(\mathbf{x})\Vert =\sum _{i=1}^{\infty }|c_i{x}_i|\le |C|\sum _{i=1}^{\infty }|x_i|=|C|\Vert \mathbf{x}\Vert <\infty .$$

Hence ${\phi }_c$ is a continuous linear operator by Theorem 10.23.

Problems

Problem 10.23

Prove the properties Eq. (10.3)–Eq. (10.7).

Problem 10.24

Show that a linear map $T:V\to W$ between topological vector spaces is continuous everywhere on $V$ if and only if it is continuous at the origin $0\in V$

Problem 10.25

Give an example of a linear map $T:V\to W$ between topological vector spaces $V$ and $W$ that is not continuous.

Problem 10.26

Complete the proof that a normed vector space is a topological vector space with respect to the metric topology induced by the norm.

Problem 10.27

Show that a real vector space $V$ of dimension $\ge 1$ is not a topological vector space with respect to either the discrete or indiscrete topology.

Problem 10.28

Show that the following are all norms in the vector space ${\mathbb{R}}^2$ :

$$\begin{array}{l}\Vert \mathbf{u}{\Vert }_1=\sqrt{(u_1{)}^2+(u_2{)}^2},\\ \Vert \mathbf{u}{\Vert }_2=\max \{|u_1|,|u_2|\},\\ \Vert \mathbf{u}{\Vert }_3=|u_1|+|u_2|.\end{array}$$

What are the shapes of the open balls $B_a(\mathbf{u})$? Show that the topologies generated by these norms are the same.

Problem 10.29

Show that if $x_n\to x$ in a normed vector space then

$$\frac{x_1+x_2+\cdots +x_n}{n}\to x.$$

Problem 10.30

Show that if $x_n$ is a sequence in a normed vector space $V$ such that every subsequence has a subsequence convergent to $x$, then $x_n\to x$

Problem 10.31

Let $V$ be a Banach space and $W$ be a vector subspace of $V.$ Define its closure $\overline{W}$ to be the union of $W$ and all limits of Cauchy sequences of elements of $W$ . Show that $\overline{W}$ is a closed vector subspace of $V$ in the sense that the limit points of all Cauchy sequences in $\overline{W}$ lie in $\overline{W}$ (note that the Cauchy sequences may include the newly added limit points of $W)$

Problem 10.32

Show that every space $\mathcal{F}(S)$ is complete with respect to the supremum norm of Example 10.26. Hence show that the vector space ${\ell }_{\infty }$ of bounded infinite complex sequences is a Banach space with respect to the norm $\Vert \mathbf{x}\Vert =\sup (x_i)$

Problem 10.33

Show that the set $V^{\prime }$ consisting of bounded linear functionals on a Banach space $V$ is a normed vector space with respect to the norm

$$\Vert \phi \Vert =\sup \{M\mid |\phi (x)|\le M\Vert x\Vert \text{ for all }x\in V\}.$$

Show that this norm is complete on $V^{\prime }.$

Problem 10.34

We say two norms $\Vert u{\Vert }_1$ and $\Vert u{\Vert }_2$ on a vector space $V$ are equivalent if there exist constants $A$ and $B$ such that

$$\Vert u{\Vert }_1\le A\Vert u{\Vert }_2\text{ and }\Vert u{\Vert }_2\le B\Vert u{\Vert }_1$$

for all $u\in V$ . If two norms are equivalent then show the following:

(a) If $u_n\to u$ with respect to one norm then this is also true for the other norm.

(b) Every linear functional that is continuous with respect to one norm is continuous with respect to the other norm.

(c) Let $V=\mathcal{C}[0,1]$ be the vector space of continuous complex functions on the interval [0, 1]. By considering the sequence of functions

$$f_n(x)=\frac{n}{\sqrt{\pi }}{\mathrm{e}}^{-n^2{x}^2}$$

show that the norms

$$\Vert f{\Vert }_1=\sqrt{{\int }_0^1|f{|}^2\mathrm{d}x}\text{ and }\Vert f{\Vert }_2=\max \{|f(x)|\mid 0\le x\le 1\}$$

are not equivalent.

(d) Show that the linear functional defined by $F(f)=f(1)$ is continuous with respect to $\Vert \cdot {\Vert }_2$ but not with respect to $\Vert \cdot {\Vert }_1$