Proof. Let $\( \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ be a sequence of elements of $\( E \)$ such that $\( \mathopen{}\left(\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}\right)\mathclose{}_{n=1}^{\infty} \)$ converges to $\( \operatorname{d}\mathopen{}\left( x, E\right)\mathclose{} \)$. We claim that $\( \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ is Cauchy. Indeed, using Theorem I.46, $$\[ 2\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}x-{y}_{m}\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}2\mathopen{}\left(x-\frac{{y}_{m}+{y}_{n}}{2}\right)\mathclose{}\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}+4{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2} \text{.} \]$$ Note that $\( \frac{{y}_{m}+{y}_{n}}{2} \)$ is in $\( E \)$ because $\( E \)$ is convex. So $$\[ {\mathopen{}\left\lVert{}{y}_{m}-{y}_{n}\right\rVert\mathclose{}}^{2}\leq 2\mathopen{}\left(\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{n}\right\rVert\mathclose{}}^{2}-{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2}\right)\mathclose{}+\mathopen{}\left({\mathopen{}\left\lVert{}x-{y}_{m}\right\rVert\mathclose{}}^{2}-{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2}\right)\mathclose{}\right)\mathclose{} \]$$ and it follows that $\( \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ is Cauchy.

As a Hilbert space, $\( H \)$ is complete, so the sequence $\( \mathopen{}\left({y}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ converges to some element $\( y \)$ of $\( H \)$. Because $\( E \)$ is closed, $\( y\in E \)$. Thus $\( \mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}= \operatorname{d}\mathopen{}\left( x, E\right)\mathclose{} \)$, and for any element $\( w \)$ of $\( E \)$ other than $\( y \)$, $$\[ 2\mathopen{}\left({\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}+{\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}}^{2}\right)\mathclose{}= {\mathopen{}\left\lVert{}w-y\right\rVert\mathclose{}}^{2}+4{\mathopen{}\left\lVert{}x-\frac{w+y}{2}\right\rVert\mathclose{}}^{2}\gt 4{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2} \text{.} \]$$ So $\( 2{\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}}^{2}\gt 2{\operatorname{d}\mathopen{}\left( x, E\right)\mathclose{}}^{2} \)$ and $\( \mathopen{}\left\lVert{}x-w\right\rVert\mathclose{}\gt \operatorname{d}\mathopen{}\left( x, E\right)\mathclose{} \)$.

Proof. The subset $\( {S}^{\perp} \)$ is a subspace by Exercise I.54. To show that it is closed, consider a sequence $\( \mathopen{}\left({x}_{n}\right)\mathclose{}_{n=1}^{\infty} \)$ of elements of $\( {S}^{\perp} \)$ that converges in $\( V \)$ to a vector $\( x \)$. Then using Proposition I.3, $$\[ \mathopen{}\left\lvert{}\mathopen{}\left\langle{}x, y\right\rangle\mathclose{}\right\rvert\mathclose{}= \mathopen{}\left\lvert{}\mathopen{}\left\langle{}x-{x}_{n}, y\right\rangle\mathclose{}\right\rvert\mathclose{}\leq \mathopen{}\left\lVert{}y\right\rVert\mathclose{}\mathopen{}\left\lVert{}x-{x}_{n}\right\rVert\mathclose{} \]$$ for all $\( y \)$ in $\( S \)$. Taking limits on each side completes the proof.

Proof. Take an element $\( x \)$ in the Hilbert space $\( H \)$. By Theorem I.49 there exists a point $\( y \)$ in $\( W \)$ nearest to $\( x \)$. In particular, $\( \mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}= \min_{w\in W}{}\mathopen{}\left\lVert{}x-w\right\rVert\mathclose{} \)$. We claim the orthogonal decomposition we seek is $\( x= y+\mathopen{}\left(x-y\right)\mathclose{} \)$.

We must first show that $\( x-y \)$ is in $\( {W}^{\perp} \)$. Indeed, for $\( w\in W \)$ and $\( ε\gt 0 \)$, $$\[ {\mathopen{}\left\lVert{}x-y+εw\right\rVert\mathclose{}}^{2}= {\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2}+2ε\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+{ε}^{2}{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq {\mathopen{}\left\lVert{}x-y\right\rVert\mathclose{}}^{2} \]$$ since $\( y-εw \)$ is in $\( W \)$. So, for $\( ε\gt 0 \)$, $\( 2ε\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+{ε}^{2}{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq 0 \)$. Dividing by $\( ε \)$ gives $\( 2\Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}+ε{\mathopen{}\left\lVert{}w\right\rVert\mathclose{}}^{2}\geq 0 \)$. Thus $\( \Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}\geq 0 \)$ for all $\( w\in W \)$. Repeating the same argument with $\( {-}w \)$ in place of $\( w \)$ leads to the conclusion $\( \Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}\leq 0 \)$ for all $\( w\in W \)$, meaning, in fact, $\( \Re\mathopen{}\left( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}\right)\mathclose{}= 0 \)$ for all $\( w\in W \)$. Repeating the above with $\( \mathrm{i}w \)$ and $\( {-}\mathrm{i}w \)$ shows that $\( \mathopen{}\left\langle{}x-y, w\right\rangle\mathclose{}= 0 \)$ for all $\( w\in W \)$. Thus, $\( x-y \)$ is in $\( {W}^{\perp} \)$.

By Exercise I.56, the decomposition $\( x= y+\mathopen{}\left(x-y\right)\mathclose{} \)$ is unique (since two different decompositions would give you something non-zero in $\( W\cap {W}^{\perp} \)$). Thus $\( H= W\oplus {W}^{\perp} \)$.

Proof. Let $\( W= \overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}} \)$ so that $\( W \)$ is a closed subspace of $\( H \)$. We claim that $\( W= {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp} \)$. Indeed, $\( W\subseteq {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp} \)$ by Exercise I.55. Using Theorem I.59, for $\( x\in {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp} \)$, write $\( x= y+w \)$ where $\( w\in W \)$ and $\( y\in {W}^{\perp} \)$. Then $\( \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= 0 \)$, and so $$\[ \mathopen{}\left\langle{}x, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}w+y, y\right\rangle\mathclose{}= \mathopen{}\left\langle{}w, y\right\rangle\mathclose{}+\mathopen{}\left\langle{}y, y\right\rangle\mathclose{}= 0+\mathopen{}\left\langle{}y, y\right\rangle\mathclose{} \text{.} \]$$ So $\( \mathopen{}\left\langle{}y, y\right\rangle\mathclose{}= 0 \)$ and thus $\( y= 0 \)$. Hence $\( x= w \)$, so $\( x\in W \)$, and thus $\( W= {\mathopen{}\left( {W}^{\perp}\right)\mathclose{}}^{\perp} \)$. Finally, since $\( {W}^{\perp}= {S}^{\perp} \)$ by Exercise I.60, we have $\( {\mathopen{}\left( {S}^{\perp}\right)\mathclose{}}^{\perp}= \overline{\operatorname{span}\mathopen{}\left( S\right)\mathclose{}} \)$.