Hilbert spaces. Showing that an a linear operator forms a closed subspace

Question

Let H be a Hilbert space and linear operator $P:H \rightarrow H$ satisfy $\langle Px,y \rangle =\langle x,Py\rangle$ for all $x,y \in H$ and $P^2 = P$. Denote $L=P(H)$. Show that L is a closed linear subspace of H and P is an orthogonal projection on L.

Help! I don't really know where to start!

Answer 1

1. $P$ is symmetric, hence, by Hellinger - Toeplitz, $P$ is bounded.

2. If $y \in L$, then $y=Px$ for some $x \in H$. It follws that $y=Px=P^2x=Py$. This gives:

$$ L=\{y \in H: Py=y\}.$$

3. Let $(y_n)$ be a convergent sequence in $L$ with limit $y$. Since $P$ is bounded, we have: $Py_n \to Py$.

From $Py_n=y_n$ for all $n$ we get: $y_n \to Py$. This gives $Py=y$, hence $y \in L$.

Thus we have shown that $L$ is closed.

Now it is your turn to show that $ker(P)=L^{\perp}$ (which shows that $P$ is an orthogonal projection ).

Answer 2

The orthogonal projection of $x$ onto a subspace $L$ is the unique $l\in L$ such that $(x-l)\perp L$.

$P$ is the orthogonal projection onto $L=P(H)$
You are given that $P$ is symmetric and satisfies $P^2=P$, and $L=P(H)$. For a given $x\in H$, the vector $Px$ is the orthogonal projection of $H$ onto $L$ because the following holds for all $y$: \begin{align} (x-Px,Py) & = (x,Py)-(Px,Py) \\ & =(x,Py)-(x,P^2y) \\ & =(x,Py)-(x,Py)=0. \end{align}

$P$ and $I-P$ are continuous
$P$ and $I-P$ are continuous because $(x-Px)\perp Px$ implies $$ \|x\|^2 = \|x-Px+Px\|^2 = \|x-Px\|^2+\|Px\|^2 \\ \implies \|x-Px\| \le \|x\| \mbox{ and } \|Px\|\le \|x\|,\;\;\; x\in H. $$

$P(H)$ is closed
The range of $P$ is the same as $(I-P)^{-1}\{0\}$, which is the inverse image of a closed set under a continuous function.