13.3 Linear functionals

中文

Orthogonal subspaces

Two vectors $u,v\in \mathcal{H}$ are said to be orthogonal if $⟨u|v⟩=0$ , written $u\perp v$ . If $V$ is a subspace of $\mathcal{H}$ we denote its orthogonal complement by

$$V^{\perp }=\{u\mid u\perp v\text{ for all }v\in V\}.$$

Theorem 13.7 · The orthogonal complement is a closed subspace

If $V$ is a subspace of $\mathcal{H}$ then $V^{\perp }$ is also a subspace.

Proof

$V^{\perp }$ is clearly a vector subspace, for $v,v^{\prime }\in V^{\perp }$ since

$$⟨\alpha v+\beta v^{\prime }|u⟩=\overline{\alpha }⟨v|u⟩+\overline{\beta }⟨v^{\prime }|u⟩=0$$

for all $u\in V$ . The space $V^{\perp }$ is closed, for if $v_n\to v$ where $v_n\in V^{\perp }$ , then

$$⟨v\mid u⟩=\underset{n\to \infty }{\lim }⟨v_n\mid u⟩=\underset{n\to \infty }{\lim }0=0$$

for all $u\in V$. Hence $v\in V^{\perp }$

Theorem 13.8 · Orthogonal decomposition and projection theorem

If $V$ is a subspace of a Hilbert space $\mathcal{H}$ then every $u\in \mathcal{H}$ has a unique decomposition

$$u=u^{\prime }+u^{\prime \prime }\text{where}{u}^{\prime }\in V,u^{\prime \prime }\in V^{\perp }.$$

Proof

The idea behind the proof of this theorem is to find the element of $V$ that is ‘nearest to $u$. Just as in Euclidean space, this is the orthogonal projection of the vector $u$ onto the subspace $V$. Let

$$d=\inf \{\Vert u-v\Vert \mid v\in V\}$$

and $v_n\in V$ a sequence of vectors such that $\Vert u-v_n\Vert \to d.$ . The sequence $\{v_n\}$ is Cauchy, for if we set $\begin{array}{r}x=u-\frac{1}{2}(v_n+v_m)\end{array}$ and $\begin{array}{r}y=\frac{1}{2}(v_n-v_m)\end{array}$ in the parallelogram law Eq. (13.2), then

$$\Vert v_n-v_m{\Vert }^2=2\Vert u-v_n{\Vert }^2+2\Vert u-v_m{\Vert }^2-4\Vert u-\frac{1}{2}(v_n+v_m){\Vert }^2.\text{\hspace{1em}(13.8)}$$

For any $ϵ>0$ let $N>0$ be such that for all $\begin{array}{r}k>N,\Vert u-v_k{\Vert }^2\le d^2+\frac{1}{4}ϵ\end{array}$ . Setting $n$, $m$ both > $N$ in Eq. (13.8) we find $\Vert v_n-v_m{\Vert }^2\le ϵ$ . Hence $v_n$ is a Cauchy sequence.

Since $\mathcal{H}$ is complete and $V$ is a closed subspace, there exists a vector $u^{\prime }\in V$ such that $v_n\to u^{\prime }$ . Setting $u^{\prime \prime }=u-u^{\prime }$ , it follows from the exercise after Lemma 13.5 that

$$\Vert u^{\prime \prime }\Vert =\underset{n\to \infty }{\lim }\Vert u-v_n\Vert =d.$$

For any $v\in V$ set $v_0=v/\Vert v\Vert$ , so that $\Vert v_0\Vert =1$ . Then

$$\begin{array}{l}{d}^2\le \Vert u-(u^{\prime }+⟨v_0|u^{\prime \prime }⟩v_0){\Vert }^2\\ =\Vert u^{\prime \prime }-⟨v_0|u^{\prime \prime }⟩v_0{\Vert }^2\\ =⟨u^{\prime \prime }-⟨v_0|u^{\prime \prime }⟩v_0|u^{\prime \prime }-⟨v_0|u^{\prime \prime }⟩v_0⟩\\ =d^2-|⟨v_0|u^{\prime \prime }⟩{|}^2.\end{array}$$

Hence $⟨v_0|u^{\prime \prime }⟩=0$ , so that $⟨v\mid u^{\prime \prime }⟩=0$ . Since $v$ is an arbitrary vector in $V$, we have $u^{\prime \prime }\in V^{\perp }$ A subspace and its orthogonal complement can only have the zero vector in common, $V\cap V^{\perp }=\{0\}$ , for if $w\in V\cap V^{\perp }$ then $⟨w|w⟩=0$ , which implies that $w=0$ . If $u=$ $u^{\prime }+u^{\prime \prime }=v^{\prime }+v^{\prime \prime }$ , with $u^{\prime },v^{\prime }\in V$ and $u^{\prime \prime },v^{\prime \prime }\in V^{\perp }$ , then the vector $u^{\prime }-v^{\prime }\in V$ is equal to $v^{\prime \prime }-u^{\prime \prime }\in V^{\perp }$ . Hence $u^{\prime }=v^{\prime }$ and $u^{\prime \prime }=v^{\prime \prime }$ , the decomposition is unique.

Corollary 13.9 · Double orthogonal complement of a closed subspace

For any subspace $V,V^{\perp \perp }=V.$

Proof

$V\subseteq V^{\perp \perp }$ for if $v\in V$ then $⟨v|u⟩=0$ for all $u\in V^{\perp }$ . Conversely, let $v\in V^{\perp \perp }$ By Theorem 13.8 $v$ has a unique decomposition $\mathbf{v}={\mathbf{v}}^{\prime }+{\mathbf{v}}^{\prime \prime }$ where $v^{\prime }\in V\subseteq V^{\perp \perp }$ and $v^{\prime \prime }\in V^{\perp }$ . Using Theorem 13.8 again but with $V$ replaced by $V^{\perp }$ , it follows that $v^{\prime \prime }=0$ Hence $v=v^{\prime }\in V$

Riesz representation theorem

For every $v\in \mathcal{H}$ the map ${\phi }_v:u\mapsto ⟨v\mid u⟩$ is a linear functional on $\mathcal{H}$. Linearity is obvious and continuity follows from Lemma 13.3. The following theorem shows that all (continuous) linear functionals on a Hilbert space are of this form, a result of considerable significance in quantum mechanics, as it motivates Dirac’s bra-ket notation.

Theorem 13.10 · Riesz representation theorem

(Riesz representation theorem) If $\phi$ is a linear functional on a Hilbert space $\mathcal{H}$, then there is a unique vector $v\in \mathcal{H}$ such that

$$\phi (u)={\phi }_v(u)=⟨v|u⟩\text{ for all }u\in \mathcal{H}.$$

Proof

Since a linear functional $\phi :\mathcal{H}\to \mathbb{C}$ is required to be continuous, we always have

$$|\phi (x_n)-\phi (x)|\to 0\text{ whenever }\Vert x-x_n\Vert \to 0.$$

Let $V$ be the null space of $\phi$ ,

$$V=\{x\mid \phi (x)=0\}.$$

This is a closed subspace, for $\mathrm{if}{x}_n\to x\mathrm{and}\phi (x_n)=0$ for all $n$, then $\phi (x)=0$ by continuity. If $V=\mathcal{H}$ then $\phi$ vanishes on $\mathcal{H}$ and one can set $v=0$ . Assume therefore that $V\ne \mathcal{H}.$ , and let $w$ be a non-zero vector such that $w$ $V$ $w\notin V$ . By Theorem 13.8, there is a unique decomposition

$$w=w^{\prime }+w^{\prime \prime }\text{ where }{w}^{\prime }\in V,w^{\prime \prime }\in V^{\perp }.$$

Then $\phi (w^{\prime \prime })=\phi (w)-\phi (w^{\prime })=\phi (w)\ne 0$ since $w$ $\notin V$ . For any $u\in \mathcal{H}$ we may write

$$u=\left(u-\frac{\phi (u)}{\phi (w^{\prime \prime })}{w}^{\prime \prime }\right)+\frac{\phi (u)}{\phi (w^{\prime \prime })}{w}^{\prime \prime },$$

where the first term on the right-hand side belongs to $V$ since the linear functional $\phi$ gives the value 0 when applied to it, while the second term belongs to $V^{\perp }$ as it is proportional to $w^{\prime \prime }$ . For any $v\in V^{\perp }$ we have then

$$⟨v|u⟩=\frac{\phi (u)}{\phi (w^{\prime \prime })}⟨v|w^{\prime \prime }⟩.$$

In particular, setting

$$v=\frac{\overline{\phi (w^{\prime \prime })}}{\Vert w^{\prime \prime }{\Vert }^2}{w}^{\prime \prime }\in V^{\perp }$$

gives

$$⟨v\mid u⟩=\frac{\phi (u)}{\phi (w^{\prime \prime })}\frac{\phi (w^{\prime \prime })}{\Vert w^{\prime \prime }{\Vert }^2}⟨w^{\prime \prime }\mid w^{\prime \prime }⟩=\phi (u).$$

Hence this $v$ is the vector required for the theorem. It is the unique vector with this property, for if $⟨v-v^{\prime }|u⟩=0$ for all $u\in \mathcal{H}$ then $v=v^{\prime }$ , on setting $u=v-v^{\prime }$ -

Problems

Problem 13.10

If $S$ is any subset of $\mathcal{H}$, and $V$ the closed subspace generated by $S,V=\overline{L(S)}$ , show that $S^{\perp }=\{u\in \mathcal{H}|⟨u|x⟩=0$ for all $x\in S\}=V^{\perp }$

Problem 13.11

Which of the following is a vector subspace of ${\ell }^2$ , and which are closed? In each case find the space of vectors orthogonal to the set.

(a) $V_N=\{(x_1,x_2,\dots )\in {\ell }^2|x_i=0\text{for}i>N\}.$

(b) $V=\bigcup V_N=\{(x_1,x_2,\dots )\in {\ell }^2|x_i=0\text{for}i>\mathrm{some}N\}.$

(c) $U=\{(x_1,x_2,\dots )\in {\ell }^2|x_i=0\text{for}i=2n\}.$

(d) $W=\{(x_1,x_2,\dots )\in {\ell }^2|x_i=0\text{for}\mathrm{some}i\}.$

Problem 13.12

Show that the real Banach space ${\mathbb{R}}^2$ with the norm $\Vert (x,y)\Vert =\max \{|x|,|y|\}$ does not have the closest point property of Theorem 13.8. Namely for a given point $x$ and one-dimensional subspace $L$, there does not in general exist a unique point in $L$ that is closest to $x$.

Problem 13.13

If $A:\mathcal{H}\to \mathcal{H}$ is an operator such that $Au\perp u$ for all $u\in \mathcal{H}.$ , show that $A=0.$