# Bounded operator on Hilbert space is orthogonal projection

*Solution:*

This part wants to show that $S$ is closed.

Indeed, 'operator' here refers to *linear operator*, and the $x_n$'s are taken from $S=\mathrm{im}\, P$, so $P^2 =P$ implies $P(x_n)=x_n$.

Incase someone is interested in a full proof:

Define $S = Im(P)$, first step is to prove $S$ is a closed subspace. Let $x_n \in S$, such that $x_n \to x$.

$P$ is bounded, therefor continuous, so $P(x_n) \to P(x)$. Notice however that since $x_n \in S$ then there is a $y_n \in H$ such that $P(y_n) = x_n$ which implies $P^2(y_n) = P(x_n)$. Since we were given $P^2 = P$, we have $P(y_n) = x_n = P(x_n)$, so overall, $x_n \to P(x)$, so $P(x) = x$ and so $x \in S$ and $S$ is closed.

Now let $w \in S^{\perp}$ and $x \in S$. Notice that $\langle x,Pw\rangle = \langle P^*x, w \rangle = \langle Px, w\rangle = \langle x,w\rangle = 0$

Hence $P(w) \in S^{\perp}$. But by definition $P(w) \in Im(P) = S$ so we must have $P(w) = 0$. for all $w \in S^{\perp}$.

Finally then, let $y = y_{S} + y_{S^\perp}$, then from linearity $P(y) = P(y_S)+P(y_{S^{\perp}}) = P(y_S) = y_S$

So $P$ is the orthogonal projection unto $S$.