Hilbert space

We have shown that $L^p (X, \mu)$ is a Banach space - a complete normed space. I shall next discuss the class of Hilbert spaces, a special class of Banach spaces, of which $L^2 (X, \mu )$ is a standard example, in which the norm arises from an inner product, just as it does in Euclidean space.

An inner product on a vector space $V$ over $\mathbb{C}$ (one can do the real case too, not much changes) is a sesquilinear form

$$V \times V \to \mathbb{C}$$

written $(u,v)$, if $u,v \in V$. The `sesqui-' part is just linearity in the first variable

$$(a_1 u_1 + a_2 u_2 , v) = a_1 (u_1 ,v) + a_2 (u_2,v),$$ (50)

anti-linearly in the second

$$(u, a_1 v_1 + a_2 v_2) = \overline{a}_1 (u, v_1) + \overline{a}_2 (u, v_2)$$ (51)

and the conjugacy condition

$$(u,v) = \overline{(v,u)} .$$ (52)

Notice that (4.2) follows from (4.1) and (4.3). If we assume in addition the positivity condition⁸

$$(u, u) \geq 0 , (u, u) =0 \Rightarrow u=0 ,$$ (53)

then

$$\Vert u \Vert = (u, u)^{1/2}$$ (54)

is a norm on $V$, as we shall see.

Suppose that $u,v \in V$ have $\Vert u\Vert=\Vert v\Vert=1$. Then $(u,v)=e^{i\theta } \left\vert(u, v ) \right\vert$ for some $\theta \in \mathbb{R}$. By choice of $\theta$, $e^{-i\theta } (u, v) = \left\vert (u, v) \right\vert$ is real, so expanding out using linearity for $s\in\mathbb{R},$

$$\begin{multline*} 0 \leq (e^{-i\theta } u-sv , e^{-i\theta } u-sv )\\ = ... ...-i\theta } (u, v) + s^2\Vert v \Vert^2 =1-2s\vert(u,v)\vert+s^2. \end{multline*}$$

The minimum of this occurs when $s=\vert(u,v)\vert$ and this is negative unless $\vert(u,v)\vert\le1.$ Using linearity, and checking the trivial cases $u=$ or $v=0$ shows that

$$\left\vert (u, v) \right\vert \leq \Vert u \Vert \Vert v \Vert, \forall u, v \in V .$$ (55)

This is called Schwarz'⁹ inequality.

Using Schwarz' inequality

$$\Vert u+v \Vert^2 = \Vert u \Vert^2 + (u, v) + (v, u) + \Vert v \Vert^2$$

$$\leq \left( \Vert u \Vert + \Vert v \Vert \right)^2$$

$$\Longrightarrow \Vert u+v \Vert \leq \Vert u \Vert + \Vert v \Vert \forall u, v \in V$$

which is the triangle inequality.

Thus we have already shown $L^2 (X, \mu )$ to be a Hilbert space for any positive measure $\mu$. The inner product is

$$(f, g)= \int_X f \overline{g} d \mu ,$$ (56)

since then (4.3) gives $\Vert f \Vert _2$.

Another important identity valid in any inner product spaces is the parallelogram law:

$$\Vert u+v \Vert^2 + \Vert u-v \Vert^2 = 2 \Vert u \Vert^2 + 2 \Vert v \Vert^2 .$$ (57)

This can be used to prove the basic `existence theorem' in Hilbert space theory.

Proof. We can certainly choose a sequence $u_n \in C$ such that

$$\Vert u_n \Vert \to \delta = \inf \left\{ \Vert v \Vert ; v \in C \right\} .$$

By the parallelogram law,

$$\Vert u_n - u_m \Vert^2 = 2 \Vert u_n \Vert^2 + 2 \Vert u_m \Vert^2 - \Vert u_n + u_m \Vert^2$$

$$\leq 2 ( \Vert u_n \Vert^2 + \Vert u_m \Vert^2 ) - 4 \delta^2$$

where we use the fact that $(u_n + u_m)/2 \in C$ so must have norm at least $\delta$. Thus $\left\{ u_n \right\}$ is a Cauchy sequence, hence convergent by the assumed completeness of $H$. Thus $\lim u_n=u \in C$ (since it is assumed closed) and by the triangle inequality

$$\left\vert \Vert u_n \Vert - \Vert u \Vert \right\vert \leq \Vert u_n -u \Vert \to 0$$

So $\Vert u \Vert = \delta$. Uniqueness of $u$ follows again from the parallelogram law which shows that if $\Vert u' \Vert = \delta$ then

$$\Vert u -u' \Vert \leq 2 \delta^2 - 4 \Vert (u+u')/2 \Vert^2 \leq 0 .$$

$\qedsymbol$

The fundamental fact about a Hilbert space is that each element $v\in H$ defines a continuous linear functional by

$$H \ni u \longmapsto (u, v) \in \mathbb{C}$$

and conversely every continuous linear functional arises this way.

Proof. Consider the linear space

$$M= \left\{ u \in H ; Lu=0 \right\}$$

the null space of $L,$ a continuous linear functional on $H$. By the assumed continuity, $M$ is closed. We can suppose that $L$ is not identically zero (since then $v=0$ in (4.9)). Thus there exists $w \notin M$. Consider

$$w+M = \left\{ v \in H ; v=w+u , u \in M \right\} .$$

This is a closed convex subset of $H$. Applying Lemma 4.2 it has a unique smallest element, $v \in w+M$. Since $v$ minimizes the norm on $w+M$,

$$\Vert v+su \Vert^2 = \Vert v \Vert^2 + 2 \Re (su, v) + \Vert s \Vert^2 \Vert u \Vert^2$$

is stationary at $s=0$. Thus $\Re (u,v)=0 \forall u \in M$, and the same argument with $s$ replaced by $is$ shows that $(v, u) =0 \forall u \in M$.

Now $v \in w+M$, so $Lv =Lw \neq 0$. Consider the element $w'=w/Lw \in H$. Since $Lw' =1$, for any $u \in H$

$$L(u-(Lu)w') = Lu - Lu =0 .$$

It follows that $u-(Lu)w' \in M$ so if $w''=w'/ \Vert w' \Vert^2$

$$(u,w'') = \left( (Lu) w' , w'' \right) = Lu \frac{(w' , w')}{\Vert w' \Vert^2} = Lu .$$

The uniqueness of $v$ follows from the positivity of the norm. $\qedsymbol$

Notice the apparent power of `abstract reasoning' here! Although we seem to have constructed $g$ out of nowhere, its existence follows from the completeness of $L^2 (X, \mu )$, but it is very convenient to express the argument abstractly for a general Hilbert space.