The Projection map on a Hilbert Space

A Hilbert Space is a Vector Space together with an inner product which is complete for the norm given by the inner product.

Parallelogram Law:

\begin{equation} \left|\frac{a+b}{2}\right|^2 + \left|\frac{a-b}{2}\right|^2 = \frac{1}{2} (|a|^2 + |b|^2) \end{equation}

for all $a,b \in H$ .

Theorem Let $K$ be a nonempty convex closed subset of $H$ . For every $f \in H$ there exists unique $u \in K$ such that $|f-u| \le |f-v|$ for all $v \in K$ . We write $u = P_K f$ .

Proof: Let $d = \inf_{v \in K} |f-v|$ and $v_n \in K$ a sequence such that $|f-v_n| \to d$ . Then,

\left|\frac{v_n - v_m}{2}\right|^2 = \frac{1}{2}(|f-v_n|^2 + |f-v_m|^2) - \left| f- \frac{v_n - v_m}{2} \right|^2 \le \frac{1}{2}(|f-v_n|^2 + |f-v_m|^2) - d^2.

Thus $v_n$ is a Cauchy sequence, so $\lim v_n$ exists and is in $K$ since $K$ is closed. Uniqueness follows by the same inequality.

□

An equivalent characterization of $u$ is the following:

Proposition 1: Let $K$ be as in the previous theorem. Then $u = P_K f$ if and only if $u \in K$ and
$(f-u,v-u) \le 0$ for all $v \in K$ .

Proof:

We compute

\begin{equation} |f-u|^2 - |f-v|^2 = 2 (f,v-u) - (v,v) + (u,u) - 2(u, v-u) + 2(u, v-u) \end{equation}

= 2(f-u, v-u) - |v-u|^2.

If $(f-u,v-u) \le 0$ , then $|f-u| \le |f-v|$ for all $v \in K$ so $u=P_k f$ . Conversely if $u=P_K f$ , then for any $w \in K$ , replacing $v=tu + (1-t)w$ in the above inequality yields,

2(f-u, w-u) \le (1-t) |w-u|^2

for all $t \in (0,1)$ . Letting $t \to 1$ implies

(f-u,w-u) \le 0.

□

Corollary 1: Let $K$ be a nonempty closed convex subset of $H$ . Then,

|P_K f_1 - P_K f_2| \le |f_1 - f_2|

for all $f_1, f_2 \in H$ .

Proof. The previous proposition implies

0 \ge (f_1 - P_K f_1, P_K f_2 - P_K f_1) + (f_2 - P_K f_2, P_K f_1 - P_K f_2)

= (P_K f_1 - P_K f_2, P_K f_1 - P_K f_2) - (f_1 - f_2, P_K f_1 - P_K f_2 ).

The result follows since $(f_1 - f_2, P_K f_1 - P_K f_2 ) \le |f_1 - f_2||P_K f_1 - P_K f_2|$ by the Cauchy-Schwartz inequality.

□

Corollary 2 Let $V$ be a closed subspace of $H$ and $f \in H$ . Then $u=P_V f$ if and only if
$(f-u,v) = 0$ for all $v \in V$ .

Proof. Let $\lambda \in \mathbb{R}$ and $v = \lambda u$ , so that by Proposition 1

(\lambda - 1) (f-u, u) \le 0

for all $\lambda \in \mathbb{R}$ . It follows that $(f-u,u) = 0$ .

□

Corollary 3 If $V$ is a closed subspace of $H$ , then $P_V : H \to H$ is a linear operator.

Proof. Let $f,g \in H$ , $\lambda \not = 0$ , $u = P_V f$ and $v=P_V g$ . For any $w \in V$ we have

(\lambda f + g - (\lambda u + v), w - (\lambda u + v)) = \lambda (f - u, \lambda^{-1}(w - v) - u) + (g-v, (w-\lambda u) - v) = 0.

The case $\lambda = 0$ is trivial. Thus, $P_V(\lambda f + g) = \lambda P_V(f) + P_V(g)$ .

□