# Orthogonal Projection on Hilbert Space

Let $A$ be a non-empty subset of a Hilbert space $H$. Suppose that $T$ is a linear operator on $H$ such that $T(H) \subseteq A$ and, for every $x \in H, (x-Tx) \perp A$. Then

1. $T$ is bounded.
2. $A$ is a closed linear subspace.
3. $T$ is the orthogonal projection onto $A$.

Using the facts that $T(H) \subseteq A$ and, for every $x \in H, (x-Tx) \perp A$ I've been able to prove that $T$ is self-adjoint. With that, and the Closed Graph Theorem, I was also able to show that $T$ is bounded.

I'm stuck with (2). For some reason I don't see what characterizes $A$; I don't even see why is it a linear subspace. Having (2), my plan is to prove that

$$Tx = \begin{cases} x & \text{for } x \in A \\ 0 & \text{for } x \notin A \end{cases} \tag{*}$$

since that would imply that, necessarily, $T$ is the orthogonal projection onto $A$.

Any hint, idea, comment, etc. on (2) and (*) will be appreciated.

Let $y \in A$. Then $y-Ty \perp A$. Since $y, Ty \in A$ then we have

$$(y-Ty) \perp (y- Ty)$$

Thus

$$y-Ty =0 \,.$$

In particular, $A=T(A) \subset T(H) \subset A$.

This proves that $T(H)=A$, which shows (2). Moreover, you get $T(y)=y$ for all $y \in A$, from where (3) be easy.

• N.S.: Nice little trick that solves everything. Thank you! – ragrigg Nov 4 '12 at 8:19