# 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 $$S$$.