Proof of Orthogonal Projection onto a Closed Linear Subspace of a Real Hilbert Space

We use the notation and claims of Elementary Identities in a Real Inner Product Space, The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity and Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space, the latter applied to $K=M$. For a subset $A\subseteq H$, $A^{\perp}$ is its orthogonal complement.

Preliminaries. Every closed linear subspace $N$ of $H$ is nonempty since $0_{H}\in N$ (condition 1 of Linear Subspace), and convex, since for $x,y\in N$ and every $t\in\mathbb{R}$ with $0\le t$ and $t\le 1$ the vector $tx+(1-t)y$ lies in $N$ by conditions 2 and 3 of Linear Subspace. So Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space applies to every closed linear subspace, in particular to $M$. We record two facts used repeatedly.

(i) For every subset $A\subseteq H$, $A^{\perp}$ is a closed linear subspace of $H$. By Elementary Identities in a Real Inner Product Space §zero, $0_{H}\in A^{\perp}$, and by conditions (b) and (c) of Real Inner Product Space §inner-product, $A^{\perp}$ is closed under sums and scalar multiples; so it is a linear subspace. If $(x_{m})$ is a sequence in $A^{\perp}$ converging to $x\in H$, then for each $a\in A$ the real sequence $(\langle x_{m},a\rangle)$, which is constantly $0$, converges to $\langle x,a\rangle$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §continuity; by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences (the constant sequence converges to $0$), $\langle x,a\rangle=0$. Hence $x\in A^{\perp}$, and $A^{\perp}$ is closed by Sequential Characterization of Closed Subsets of a Metric Space.

(ii) $A\subseteq(A^{\perp})^{\perp}$ and $A\cap A^{\perp}\subseteq\{0_{H}\}$. For $a\in A$ and $w\in A^{\perp}$, $\langle a,w\rangle=\langle w,a\rangle=0$ by symmetry; and if $a\in A\cap A^{\perp}$ then $|a|^{2}=\langle a,a\rangle=0$, so $a=0_{H}$ by Elementary Identities in a Real Inner Product Space §vanishing.

Claim 1. Let $x\in H$ and $z\in M$. By Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space §characterisation, $z=P_{M}x$ if and only if $\langle x-z,y-z\rangle\le 0$ for every $y\in M$. If $x-z\in M^{\perp}$, then for $y\in M$ we have $y-z\in M$, so $\langle x-z,y-z\rangle=0\le 0$. Conversely, suppose $\langle x-z,y-z\rangle\le 0$ for every $y\in M$, and let $m\in M$. For $t\in\{1,-1\}$ the point $y=z+tm$ lies in $M$ and $y-z=tm$, so $t\langle x-z,m\rangle\le 0$ by Elementary Identities in a Real Inner Product Space §bilinear; $t=1$ gives $\langle x-z,m\rangle\le 0$ and $t=-1$ gives $0\le\langle x-z,m\rangle$, hence $\langle x-z,m\rangle=0$. Thus $x-z\in M^{\perp}$.

Claim 2. Let $x,x'\in H$ and $\lambda\in\mathbb{R}$. The vector $P_{M}x+P_{M}x'$ lies in $M$, and $(x+x')-(P_{M}x+P_{M}x')=(x-P_{M}x)+(x'-P_{M}x')$ lies in $M^{\perp}$ by claim 1 and (i). By claim 1, $P_{M}(x+x')=P_{M}x+P_{M}x'$. Likewise $\lambda P_{M}x\in M$ and $\lambda x-\lambda P_{M}x=\lambda(x-P_{M}x)\in M^{\perp}$, so $P_{M}(\lambda x)=\lambda P_{M}x$; thus $P_{M}$ is linear. For $x\in M$, $P_{M}x=x$ by Nearest-Point Projection onto a Nonempty Closed Convex Subset of a Real Hilbert Space §fixed; applying this to $P_{M}x\in M$ gives $P_{M}(P_{M}x)=P_{M}x$.

Claim 3. Let $x\in H$. Since $P_{M}x\in M$ and $x-P_{M}x\in M^{\perp}$ (claim 1), $\langle P_{M}x,x-P_{M}x\rangle=0$, so Elementary Identities in a Real Inner Product Space §expansion applied to $x=P_{M}x+(x-P_{M}x)$ gives $|x|^{2}=|P_{M}x|^{2}+|x-P_{M}x|^{2}$. Since squares are nonnegative, the translation rule (claim 3 of Elementary Arithmetic in an Ordered Field) gives $|P_{M}x|^{2}\le|x|^{2}$ and $|x-P_{M}x|^{2}\le|x|^{2}$, and claim 2 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field gives the two inequalities.

Claim 4. $M^{\perp}$ is a closed linear subspace by (i), and $M\cap M^{\perp}=\{0_{H}\}$ by (ii) and $0_{H}\in M\cap M^{\perp}$. Let $x\in H$. The vector $x-P_{M}x$ lies in $M^{\perp}$ (claim 1), and $x-(x-P_{M}x)=P_{M}x\in M\subseteq(M^{\perp})^{\perp}$ by (ii); so by claim 1 applied to the closed linear subspace $M^{\perp}$, $P_{M^{\perp}}x=x-P_{M}x$. Finally, $M\subseteq(M^{\perp})^{\perp}$ by (ii); conversely let $x\in(M^{\perp})^{\perp}$. Then $x-P_{M}x\in M^{\perp}$, so $\langle x,x-P_{M}x\rangle=0$, and also $\langle P_{M}x,x-P_{M}x\rangle=0$ since $P_{M}x\in M$; subtracting (Elementary Identities in a Real Inner Product Space §bilinear) gives $|x-P_{M}x|^{2}=\langle x-P_{M}x,x-P_{M}x\rangle=0$, so $x=P_{M}x\in M$ by Elementary Identities in a Real Inner Product Space §vanishing.

Claim 5. Existence: $x=P_{M}x+(x-P_{M}x)$ with $P_{M}x\in M$ and $x-P_{M}x\in M^{\perp}$ by claim 1. Uniqueness: if $m+m'=n+n'$ with $m,n\in M$ and $m',n'\in M^{\perp}$, then $m-n=n'-m'$ lies in $M\cap M^{\perp}=\{0_{H}\}$ by claim 4, so $m=n$ and $m'=n'$.

Claim 6. Let $L$ be a linear subspace of $H$ and $\bar{L}$ its closure, which is closed by claim 2 of The Closure is the Smallest Closed Superset and contains $L$ by claim 1 there. We first show that $\bar{L}$ is a linear subspace and that $\bar{L}^{\perp}=L^{\perp}$. By Sequential Characterization of the Closure in a Metric Space, a point lies in $\bar{L}$ if and only if it is the limit of a sequence in $L$. If $x,y\in\bar{L}$ are limits of sequences $(x_{m})$, $(y_{m})$ in $L$, then $x+y$ and $\lambda x$ are limits of the sequences $(x_{m}+y_{m})$ and $(\lambda x_{m})$ in $L$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §linear-limits, so lie in $\bar{L}$; and $0_{H}\in L\subseteq\bar{L}$. Since $L\subseteq\bar{L}$, every element of $\bar{L}^{\perp}$ lies in $L^{\perp}$. Conversely let $w\in L^{\perp}$ and $x\in\bar{L}$ with $x_{m}\to x$, $x_{m}\in L$; then $\langle x,w\rangle$ is the limit of the constant sequence $(\langle x_{m},w\rangle)=(0)$ by The Norm Metric of a Real Inner Product Space: Triangle Inequalities, Limits and Continuity §continuity, so $\langle x,w\rangle=0$ by claim 1 of Uniqueness of Limits and Boundedness of Convergent Real Sequences, and $w\in\bar{L}^{\perp}$.

Now $L$ is dense in $H$ if and only if $\bar{L}=H$ (Dense Subset of a Topological Space). If $\bar{L}=H$, then $L^{\perp}=H^{\perp}\subseteq H\cap H^{\perp}=\{0_{H}\}$ by (ii), and $0_{H}\in L^{\perp}$, so $L^{\perp}=\{0_{H}\}$. Conversely, if $L^{\perp}=\{0_{H}\}$, then $\bar{L}^{\perp}=\{0_{H}\}$, and applying claim 4 to the closed linear subspace $\bar{L}$ gives $\bar{L}=(\bar{L}^{\perp})^{\perp}=\{0_{H}\}^{\perp}=H$, the last equality by Elementary Identities in a Real Inner Product Space §zero.