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 (Inner Product 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:

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

$\blacksquare$

Sources

• 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Theorem $2.7 \text {(b)}$