Proposition 6.2.1B9A2266E14FE

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 $. $\square$