13.1 Definitions and examples

中文

Let $V$ be a complex vector space with an inner product $⟨\cdot |\cdot ⟩:V\times V\to \mathbb{C}$ satisfying (IP1)– (IP3) of Section 5.2. Such a space is sometimes called a pre-Hilbert space. As in Eq. (5.11) define a norm on an inner product space by

$$\Vert u\Vert =\sqrt{⟨u|u⟩}.\text{\hspace{1em}(13.1)}$$

The properties (Norm1)–(Norm3) of Section 10.9 hold for this choice of norm. Condition (Norm1) is equivalent to (IP3), and (Norm2) is an immediate consequence of (IP1) and (IP2), for

$$\Vert \lambda v\Vert =\sqrt{⟨\lambda v|\lambda v⟩}=\sqrt{\lambda \overline{\lambda }⟨v|v⟩}=|\lambda |\Vert v\Vert .$$

The triangle inequality (Norm3) is a consequence of Theorem 5.6. These properties hold equally in finite or infinite dimensional vector spaces. A Hilbert space $(\mathcal{H},⟨\cdot \mid \cdot ⟩)$ is an inner product space that is complete in the induced norm; that is, $(\mathcal{H},\Vert \cdot \Vert )$ is a Banach space. An introduction to Hilbert spaces at the level of this chapter may be found in [1–6], while more advanced topics are dealt with in [7–11].

The parallelogram law

$$\Vert x+y{\Vert }^2+\Vert x-y{\Vert }^2=2\Vert x{\Vert }^2+2\Vert y{\Vert }^2\text{\hspace{1em}(13.2)}$$

holds for all pairs of vectors $x$, $y$ in an inner product space $\mathcal{H}$. The proof is straightforward, by substituting $\Vert x+y{\Vert }^2=⟨x+y|x+y⟩=\Vert x{\Vert }^2+\Vert y{\Vert }^2+2\mathrm{Re}(⟨x|y⟩)$ , etc. It immediately gives rise to the inequality

$$\Vert x+y{\Vert }^2\le 2\Vert x{\Vert }^2+2\Vert y{\Vert }^2.\text{\hspace{1em}(13.3)}$$

For complex numbers Eq. (13.2) and Eq. (13.3) hold with norm replaced by modulus.

Example 13.1

The typical inner product defined on ${\mathbb{C}}^n$ in Example 5.4 by

$$⟨(u_1,\dots ,u_n)|(v_1,\dots ,v_n)⟩=\sum _{i=1}^n\overline{u_i}{v}_i$$

makes it into a Hilbert space. The norm is

$$\Vert \mathbf{v}\Vert =\sqrt{|v_1{|}^2+|v_2{|}^2+\cdots +|v_n{|}^2},$$

which was shown to be complete in Example 10.27. In any finite dimensional inner product space the Schmidt orthonormalization procedure creates an orthonormal basis for which the inner product takes this form (see Section 5.2). Thus every finite dimensional Hilbert space is isomorphic to ${\mathbb{C}}^n$ with the above inner product. The only thing that distinguishes finite dimensional Hilbert spaces is their dimension.

Example 13.2

Let ${\ell }^2$ be the set of all complex sequences $\mathbf{u}=(u_1,u_2,\dots )$ where $u_i\in \mathbb{C}$ such that

$$\sum _{i=1}^{\infty }|u_i{|}^2<\infty .$$

This space is a complex vector space, for if $u$, $v$ are any pair of sequences in ${\ell }^2$ , then $u+v\in {\ell }^2$ For, using the complex number version of the inequality Eq. (13.3),

$$\sum _{i=1}^{\infty }|u_i+v_i{|}^2\le 2\sum _{i=1}^{\infty }|u_i{|}^2+2\sum _{i=1}^{\infty }|v_i{|}^2<\infty .$$

It is trivial that $u\in {\ell }^2$ implies $\lambda u\in {\ell }^2$ for any complex number $\lambda$ .

Let the inner product be defined by

$$⟨u|v⟩=\sum _{i=1}^{\infty }\overline{u_i}{v}_i.$$

This is well-defined for any pair of sequences $u$, $v\in {\ell }^2$ , for

$$\begin{array}{l}\left|{\sum }_{i=1}^{\infty }\overline{u_i}{v}_i\right|\le {\sum }_{i=1}^{\infty }|\overline{u_i}{v}_i|\\ \le \frac{1}{2}{\sum }_{i=1}^{\infty }(|u_i{|}^2+|v_i{|}^2)<\infty .\end{array}$$

The last step follows from

$$2{\left|\bar{a}b\right|}^2=2|a{|}^2|b{|}^2=|a{|}^2+|b{|}^2-(|a|-|b|{)}^2\le |a{|}^2+|b{|}^2.$$

The norm defined by this inner product is

$$\Vert u\Vert =\sqrt{\sum _{i=1}^{\infty }{\left|u_i\right|}^2}\le \infty .$$

Given a Cauchy sequence $u^{(1)},u^{(2)},\dots$ in ${\ell }^2$, for each $i$ the sequence $u_i^{(n)}$ has a limit $u_i$ because $|u_i^{(n)}-u_i^{(m)}|\le \Vert u^{(n)}-u^{(m)}\Vert$. Set $u=(u_1,u_2,\dots )$. For any $ϵ>0$ choose $N$ so that $\Vert u^{(n)}-u^{(m)}\Vert <ϵ$ whenever $n,m>N$. For any integer $M$ and $n,m>N$

$$\sum _{i=1}^M{\left|u_i^{(m)}-u_i^{(n)}\right|}^2\le \sum _{i=1}^{\infty }{\left|u_i^{(m)}-u_i^{(n)}\right|}^2=\Vert u^{(m)}-u^{(n)}{\Vert }^2<{ϵ}^2,$$

and taking the limit $n\to \infty$ we have

$$\sum _{i=1}^M{\left|u_i^{(m)}-u_i\right|}^2\le {ϵ}^2.$$

In the limit $M\to \infty$

$$\sum _{i=1}^{\infty }{\left|u_i^{(m)}-u_i\right|}^2\le {ϵ}^2$$

so that $u^{(m)}-u\in {\ell }^2$ . Hence $u=u^{(m)}-(u^{(m)}-u)$ belongs to ${\ell }^2$ since it is the difference of two vectors from ${\ell }^2$ and it is the limit of the sequence $u^{(m)}$ since $\Vert u^{(m)}-u\Vert <ϵ$ for all $m>N$ . It turns out, as we shall see, that ${\ell }^2$ is isomorphic to most Hilbert spaces of interest – the so-called separable Hilbert spaces.

Example 13.3

On $\mathcal{C}[0,1]$, the continuous complex functions on $[0,1]$, set

$$⟨f|g⟩={\int }_0^1\overline{f}gdx.$$

This is a pre-Hilbert space $\mathcal{H}$, but fails to be a Hilbert space since a sequence of continuous functions may have a discontinuous limit.

Exercise

Find a sequence of functions in $\mathcal{C}[0,1]$ that have a discontinuous step function as thei limit.

Example 13.4

Let $(X,\mathcal{M},\mu )$ be a measure space, and ${\mathcal{L}}^2(X)$ be the set of all square integrable complex-valued functions $f:X\to \mathbb{C},$ , such that

$${\int }_X|f{|}^2\mathrm{d}\mu <\infty .$$

This space is a complex vector space, for if $f$ and $g$ are square integrable then

$${\int }_X|f+\lambda g{|}^{2}d\mu \le 2{\int }_X|f{|}^{2}d\mu +2|\lambda {|}^2{\int }_X|g{|}^{2}d\mu$$

by Eq. (13.3) applied to complex numbers

Write $f\sim f^{\prime }\mathrm{iff}f(x)=f^{\prime }(x)$ almost everywhere on $X;$ ; this is clearly an equivalence relation on $X$. We set $L^2(X)$ to be the factor space ${\mathcal{L}}^2(X)/\sim$ . Its elements are equivalence classes $\tilde{f}$ of functions that differ at most on a set of measure zero. Define the inner product of two classes by

$$⟨\tilde{f}|\tilde{g}⟩={\int }_X\overline{f}g\mathrm{d}\mu ,$$

which is well-defined (see Example 5.6) and independent of the choice of representatives. For, if $f^{\prime }\sim f$ and $g^{\prime }\sim g$ then let $A_f$ and $A_g$ be the sets on which $f(x)\ne f^{\prime }(x)$ and $g^{\prime }(x)\ne g(x)$ , respectively. These sets have measure zero, $\mu (A_f)=\mu (A_g)=0$ . The set on which $\overline{f^{\prime }(x)}{g}^{\prime }(x)\ne \overline{f(x)}g(x)$ is a subset of $A_f\cup A_g$ and therefore must also have measure zero, so that ${\int }_X\overline{f}g\mathrm{d}\mu ={\int }_X\overline{f^{\prime }}{g}^{\prime }\mathrm{d}\mu$

The inner product axioms (IP1) and (IP2) are trivial, and (IP3) follows from

$$\Vert \tilde{f}\Vert =0⟹{\int }_X|f{|}^2\mathrm{d}\mu =0⟹f=0\text{ a.e. }$$

It is common to replace an equivalence class of functions $\tilde{f}\in {\mathcal{L}}^2(X)$ simply by a represen tative function $f$ when there is no danger of confusion.

It turns out that the inner product space ${\mathcal{L}}^2(X)$ is in fact a Hilbert space. The following theorem is needed in order to show completeness.

Theorem 13.1 · Riesz–Fischer theorem

(Riesz–Fischer) If $f_1,f_2,\dots$ is a Cauchy sequence of functions in ${\mathcal{L}}^2(X)$ , there exists afunction $f\in {\mathcal{L}}^2(X)$ such that $\Vert f-f_n\Vert \to 0\text{as}n\to \infty .$

Proof

The Cauchy sequence condition $\Vert f_n-f_m\Vert \to 0$ implies that for any $ϵ>0$ there exists $N$ such that

$${\int }_X{\left|f_n-f_m\right|}^2\mathrm{d}\mu <ϵ\text{ for all }m,n>N.$$

We may, with some relabelling, pick a subsequence such that $f_0=0$ and

$$\Vert f_n-f_{n-1}\Vert <2^{-n}.$$

Setting

$$h(x)=\sum _{n=1}^{\infty }|f_n(x)-f_{n-1}(x)|$$

we have from (Norm3),

$$\Vert h\Vert \le \sum _{n=1}^{\infty }\Vert f_n-f_{n-1}\Vert <\sum _{n=1}^{\infty }{2}^{-n}=1.$$

The function $x\mapsto h^2(x)$ is thus a positive real integrable function on $X,$ and the set of points where its defining sequence diverges, $E=\{x\mid h(x)=\infty \}$ , is a set of measure zero, $\mu (E)=0.\mathrm{Let}{g}_n$ be the sequence of functions

$$g_n(x)=\{\begin{array}{ll}{f}_n-f_{n-1}& \text{ if }x\notin E,\\ 0& \text{ if }x\in E.\end{array}$$

Since $g_n=f_n-f_{n-1}a.e$ . these functions are measurable and $\Vert g_n\Vert =\Vert f_n-f_{n-1}\Vert <2^{-n}$ The function

$$f(x)=\sum _{n=1}^{\infty }{g}_n(x)$$

is defined almost everywhere, since the series is absolutely convergent to $h(x)$ almost everywhere. Furthermore it belongs to ${\mathcal{L}}^2(X)$ , for

$$|f(x){|}^2={\left|\sum g_n(x)\right|}^2\le {\left(\sum |g_n(x)|\right)}^2\le (h(x){)}^2.$$

Since

$$f_n=\sum _{k=1}^n(f_k-f_{k-1})=\sum _{k=1}^n{g}_k\text{ a.e. }$$

it follows that

$$\begin{array}{l}\Vert f-f_n\Vert =\Vert f-{\sum }_{k=1}^n{g}_k\Vert \\ =\Vert {\sum }_{k=n+1}^{\infty }{g}_k\Vert \\ \le {\sum }_{k=n+1}^{\infty }\Vert g_k\Vert \\ <{\sum }_{k=n+1}^{\infty }{2}^{-k}=2^{-n}.\end{array}$$

Hence $\Vert f-f_n\Vert \to 0\mathrm{as}n\to \infty$ and the result is proved.

Problems

Problem 13.1

Let $E$ be a Banach space in which the norm satisfies the parallelogram law Eq. (13.2). Show that it is a Hilbert space with inner product given by

$$⟨x\mid y⟩=\frac{1}{4}(\Vert x+y{\Vert }^2-\Vert x-y{\Vert }^2+i\Vert x-iy{\Vert }^2-i\Vert x+iy{\Vert }^2).$$

Problem 13.2

On the vector space ${\mathcal{F}}^1[a,b]$ of complex continuous differentiable functions on the interval $[a,b]$, set

$$⟨f|g⟩={\int }_a^b\overline{f^{\prime }(x)}{g}^{\prime }(x)dx\text{ where }{f}^{\prime }=\frac{df}{dx},g^{\prime }=\frac{dg}{dx}.$$

Show that this is not an inner product, but becomes one if restricted to the space of functions $f\in$ ${\mathcal{F}}^1[a,b]$ having $f(c)=0$ for some fixed $a\le c\le b$ . Is it a Hilbert space?

Give a similar analysis for the case $a=-\infty ,b=\infty$ , and restricting functions to those ofcompac support.

Problem 13.3

In the space $L^2([0,1])$ which of the following sequences of functions (i) is a Cauchy sequence, (ii) converges to 0, (iii) converges everywhere to 0, (iv) converges almost everywhere to 0, and (v) converges almost nowhere to $0?$

(a) $f_n(x)={\sin }^n(x),n=1,2,\dots$

$$\text{(b)}{f}_n(x)=\{\begin{array}{ll}0& \text{ for }x<1-\frac{1}{n},\\ nx+1-n& \text{ for }1-\frac{1}{n}\le x\le 1.\end{array}$$

(c) $f_n(x)={\sin }^n(nx)$

(d) $f_n(x)={\chi }_{U_n}(x)$ , the characteristic function of the set

$$U_n=\left[\frac{k}{2^m},\frac{k+1}{2^m}\right]\text{ where }n=2^m+k,m=0,1,\dots \text{ and }k=0,\dots ,2^m-1.$$