6.2 Hilbert spaces and closed operator complexes3F2EA9D1CB5E

Proof. Let $C$ be a nonempty closed convex set and $x\in H$. Replacing $C$ by $C-x$, we may assume that $x=0$. Let $d=\inf _{z\in C}\| z\| $, and let $(x_ j)$ be a sequence in $C$ with $\| x_ j\| \to d$. Since $(x_ j+x_ k)/2\in C$ by convexity, the parallelogram identity gives

The right side tends to zero as $j,k\to \infty $, so $(x_ j)$ is a Cauchy sequence. Its limit lies in $C$, because $C$ is closed, and has norm $d$. If $x,x'\in C$ both have norm $d$, the same inequality with $x_ j=x$ and $x_ k=x'$ gives $\| x-x'\| ^2\le 0$, which proves uniqueness.

Now let $M$ be a closed linear subspace and $x\in H$. Let $p$ be the point of $M$ closest to $x$, and put $r=x-p$. For $v\in M$ and $t\in \mathbb C$ we have $p+tv\in M$, and hence

Taking small real $t$ of both signs gives $\operatorname{Re}\langle r,v\rangle =0$, and taking $t\in \mathrm i\mathbb R$ gives $\operatorname{Im}\langle r,v\rangle =0$. Thus $r\in M^\perp $ and $x=p+r$. Since $M\cap M^\perp =\{ 0\} $, this decomposition is unique, so $H=M\oplus M^\perp $. Every element of $M$ is orthogonal to $M^\perp $, so $M\subset (M^\perp )^\perp $. Conversely, if $x\in (M^\perp )^\perp $ and $x=p+r$ as above, then $r=x-p$ lies in the linear subspace $(M^\perp )^\perp $ and in $M^\perp $, so $\| r\| ^2=\langle r,r\rangle =0$ and $x=p\in M$. We call $p$ the orthogonal projection of $x$ onto $M$; the map $P\colon x\mapsto p$ is linear by the uniqueness of the decomposition, and $\| Px\| \le \| x\| $, because $\| x\| ^2=\| p\| ^2+\| r\| ^2$.

Finally, let $L$ be a bounded linear functional. If $L=0$ we take $y=0$. Otherwise $\ker L$ is a proper closed subspace. Choose $x_0\notin \ker L$, and let $e$ be its component in $(\ker L)^\perp $; then $e\ne 0$ and $L(e)=L(x_0)\ne 0$. For every $x\in H$ the vector $x-(L(x)/L(e))e$ lies in $\ker L$ and is therefore orthogonal to $e$, so that $\langle x,e\rangle =(L(x)/L(e))\| e\| ^2$. With $y=\overline{L(e)}\, e/\| e\| ^2$ this reads $L(x)=\langle x,y\rangle $. The Cauchy–Schwarz inequality shows that $L$ has norm at most $\| y\| $, and $L(y)=\| y\| ^2$ shows equality. If $y'$ also represents $L$, then $\langle x,y-y'\rangle =0$ for all $x$, and $x=y-y'$ gives $y=y'$. If $\ell $ is a bounded conjugate-linear functional, then $\overline\ell \colon x\mapsto \overline{\ell (x)}$ is a bounded linear functional with the same norm, and its representation $\overline{\ell (x)}=\langle x,y\rangle $ reads $\ell (x)=\langle y,x\rangle $. $\square$

Proof. Let $(x_ j)$ be a sequence with $\| x_ j\| \le C$, and let $V$ be the closed linear span of its terms. The finite linear combinations of the $x_ j$ with coefficients in $\mathbb Q+\mathrm i\mathbb Q$ form a countable dense subset of $V$. Applying the Gram–Schmidt process to it, and discarding the vectors that depend linearly on earlier ones, we obtain a finite or countable orthonormal family $(e_ k)$ whose linear span is dense in $V$.

For each $k$ the sequence $(\langle x_ j,e_ k\rangle )_ j$ is bounded by $C$, so by a diagonal argument we may pass to a subsequence, still denoted $(x_ j)$, such that $\langle x_ j,e_ k\rangle $ converges to a number $a_ k$ for every $k$. For every $N$, Bessel's inequality gives $\sum _{k\le N}|\langle x_ j,e_ k\rangle |^2\le C^2$; letting $j\to \infty $ and then $N\to \infty $, we get $\sum _ k|a_ k|^2\le C^2$. Since $\| \sum _{N<k\le N'}a_ ke_ k\| ^2=\sum _{N<k\le N'}|a_ k|^2$ by Pythagoras' theorem, the partial sums of the series $x=\sum _ ka_ ke_ k$ form a Cauchy sequence; the series therefore converges in the closed, hence complete, subspace $V$, and $\| x\| ^2=\sum _ k|a_ k|^2\le C^2$.

By the continuity of the inner product and the orthonormality of the $e_ k$, we have $\langle x,e_ k\rangle =a_ k=\lim _ j\langle x_ j,e_ k\rangle $ for every $k$; hence $\langle x_ j,v\rangle \to \langle x,v\rangle $ for every $v$ in the linear span of the $e_ k$. If $v\in V$ and $v'$ lies in that span with $\| v-v'\| <\delta $, then $|\langle x_ j-x,v\rangle |\le |\langle x_ j-x,v'\rangle |+2C\delta $, so the convergence extends to all $v\in V$. For an arbitrary $v\in H$, write $v=v_1+v_2$ with $v_1\in V$ and $v_2\in V^\perp $ by proposition 6.2.1; since $x_ j$ and $x$ lie in $V$, we get $\langle x_ j-x,v\rangle =\langle x_ j-x,v_1\rangle \to 0$.

For the last assertion we may assume that $x\ne 0$. By the Cauchy–Schwarz inequality, $\| x\| ^2=\lim _ j\langle x_ j,x\rangle \le \varliminf _ j\| x_ j\| \, \| x\| $, and we divide by $\| x\| $. $\square$

Proof. If $y_ j\in \operatorname{Dom}T^*$, $y_ j\to y$ and $T^*y_ j\to w$, then for every $x\in \operatorname{Dom}T$

hence $y\in \operatorname{Dom}T^*$ and $T^*y=w$, so $T^*$ is closed. A vector $y\in H_1$ is orthogonal to $\operatorname {Ran}T$ exactly when the functional $x\mapsto \langle Tx,y\rangle $ vanishes, that is, when $y\in \operatorname{Dom}T^*$ and $T^*y=0$. This proves $\ker T^*=(\operatorname {Ran}T)^\perp $. In the same way, $(a,b)\in H_0\oplus H_1$ is orthogonal to $\operatorname {Gr}T$ if and only if $\langle Tx,b\rangle =\langle x,-a\rangle $ for all $x\in \operatorname{Dom}T$, so

Now let $T$ be closed, and let $v\in H_1$ be orthogonal to $\operatorname{Dom}T^*$. Since $\operatorname {Gr}T$ is closed, proposition 6.2.1 and the description of $(\operatorname {Gr}T)^\perp $ decompose $(0,v)$ as $(x,Tx)+(-T^*y,y)$ with $x\in \operatorname{Dom}T$ and $y\in \operatorname{Dom}T^*$. Thus $x=T^*y$ and $v=Tx+y$. Since $v$ is orthogonal to $y\in \operatorname{Dom}T^*$, we obtain

so $x=y=0$ and $v=0$. Hence $\operatorname{Dom}T^*$ is dense, and $T^{**}$ is defined. Let $J\colon H_1\oplus H_0\to H_0\oplus H_1$ and $J'\colon H_0\oplus H_1\to H_1\oplus H_0$ be the unitary maps $J(b,a)=(-a,b)$ and $J'(a,b)=(-b,a)$. The description of the orthogonal complement of a graph, applied to $T$ and to the densely defined operator $T^*$, reads $(\operatorname {Gr}T)^\perp =J(\operatorname {Gr}T^*)$ and $(\operatorname {Gr}T^*)^\perp =J'(\operatorname {Gr}T^{**})$. For a closed subspace $M$ we have $(M^\perp )^\perp =M$ by proposition 6.2.1, and unitary maps carry orthogonal complements to orthogonal complements. Therefore

because $JJ'=-\mathrm{id}$ and graphs are linear subspaces. Thus $T^{**}=T$. $\square$

Proof. Since $S$ is closed, $\ker S$ is a closed subspace, and $H_1=\ker S\oplus (\ker S)^\perp $ by proposition 6.2.1. Since $\operatorname {Ran}T\subset \ker S$, lemma 6.2.3 gives $(\ker S)^\perp \subset (\operatorname {Ran}T)^\perp =\ker T^*$. Let $f\in \operatorname{Dom}T^*$, and write $f=f_0+f_1$ with $f_0\in \ker S$ and $f_1\in (\ker S)^\perp $. Then $f_1\in \operatorname{Dom}T^*$ and $T^*f_1=0$, so $f_0\in \operatorname{Dom}T^*\cap \operatorname{Dom}S$, $T^*f_0=T^*f$ and $Sf_0=0$. Moreover $\langle g,f\rangle =\langle g,f_0\rangle $, because $g\in \ker S$. The hypothesis applied to $f_0$ therefore gives

Consequently the map $\lambda \colon T^*f\mapsto \langle g,f\rangle $ is well defined on $\operatorname {Ran}T^*$: if $T^*f=T^*f'$, then $|\langle g,f-f'\rangle |\le \sqrt M\, \| T^*(f-f')\| =0$. It is conjugate-linear and bounded by $\sqrt M$, hence uniformly continuous, so it extends uniquely to a conjugate-linear functional $\tilde\lambda $ on the closed subspace $W=\overline{\operatorname {Ran}T^*}$, again bounded by $\sqrt M$. If $P_ W$ is the orthogonal projection onto $W$, then $x\mapsto \tilde\lambda (P_ Wx)$ is a conjugate-linear functional on $H_0$ of norm at most $\sqrt M$, because $\| P_ Wx\| \le \| x\| $. By the representation of conjugate-linear functionals in proposition 6.2.1, there is $u\in H_0$ with $\| u\| \le \sqrt M$ and

Taking complex conjugates, we get $\langle T^*f,u\rangle =\langle f,g\rangle $ for all $f\in \operatorname{Dom}T^*$. Thus the functional $f\mapsto \langle T^*f,u\rangle $ is bounded on $\operatorname{Dom}T^*$, so $u\in \operatorname{Dom}T^{**}$ and $T^{**}u=g$. Since $T^{**}=T$ by lemma 6.2.3, we conclude that $u\in \operatorname{Dom}T$ and $Tu=g$.

The solutions of $Tu=g$ form the affine subspace $u+\ker T$, which is closed because $\ker T$ is closed. If $P$ denotes the orthogonal projection onto $\ker T$, then $\| u+k\| ^2=\| u-Pu\| ^2+\| Pu+k\| ^2$ for $k\in \ker T$. Hence $u-Pu$ is the unique solution of smallest norm; it is orthogonal to $\ker T$, and its norm is at most $\| u\| \le \sqrt M$. $\square$