Orthogonal Projection is Bounded

Theorem
Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.

Let $K$ be a closed linear subspace of $H$.

Let $P_K$ denote the orthogonal projection on $K$.

Then $P_K$ is bounded.

That is:


 * $\norm {\map {P_K} h} \le \norm h$

for each $h \in H$.

Proof
Let $h \in H$.

Note that we can write:


 * $h = \paren {h - \map {P_K} h} + \map {P_K} h$

We have, by the definition of orthogonal projection:


 * $\map {P_K} h \in K$

From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have:


 * $h - \map {P_K} h \in K^\bot$

so that:


 * $\innerprod {\map {P_K} h} {h - \map {P_K} h} = 0$

By Pythagoras's Theorem (Hilbert Space), we therefore have:


 * ${\norm h}^2 = \norm {h - \map {P_K} h}^2 + \norm {\map {P_K} h}^2$

We have that:


 * $\norm {h - \map {P_K} h}^2 \ge 0$

so:


 * ${\norm h}^2 \ge \norm {\map {P_K} h}^2$

giving:


 * $\norm {\map {P_K} h} \le \norm h$