# Symmetric bounded linear maps can be approximated by compact symmetric linear maps.

Let $H$ be a separable Hilbert space and let $T:H \rightarrow H$ be a symmetric bound linear map.

a) Show that for every orthogonal projection $P$ on $H$ ($P' = P$, $P^2 = P$) PTP is symmetric.

b) Prove the existens of a sequence $C_n$ of compact symmetric linear maps such that $C_nx\rightarrow Tx$ for $x\in H$.

My try:

a) I don't see that $P^* = P$ for all orthogonal projection, why is it so? but if that is true $(PTP)' = P'T'P' = PTP$

b) I do not really know how to start here, I know that the limit of compact maps are compact if they converge uniformly. Any help or hint would be greatfull

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For (a): Didn't you write in your assumption on $P$ that you want to have $P' = P$? Where is your point? For (b): Let $(e_n)_{n\in\mathbb N}$ an orthogonal basis for $H$. Such a thing exists, as $H$ is seperable. Now let $P_n$ denote the orthogonal projection onto $\operatorname{span} \{e_1, \ldots, e_n\}$ and $C_n := P_nTP_n$. Then $C_n$ is symmetric by (a) and compact by finite-dimensionality. – martini Dec 10 '12 at 14:30

Here's one way to see (a): write $\langle Px, y \rangle = \langle Px, Py \rangle + \langle Px, y-Py \rangle$. Since $P$ is orthogonal projection, $y - Py$ is orthogonal to the range of $P$, so the second term vanishes. Thus $\langle Px, y \rangle = \langle Px, Py \rangle$. By the same argument, $\langle x, Py \rangle = \langle Px, Py \rangle$. Since $\langle Px, y \rangle = \langle x, Py \rangle$, $P$ is symmetric.
For (b): fix an orthonormal basis $\{e_i\}$ for $H$, and let $P_n$ be orthogonal projection onto the span of $e_1, \dots, e_n$. Show that for any $x$, $P_n x \to x$. Now try taking $C_n = P_n T P_n$. (Note that $P_n \to I$ pointwise, but not uniformly, so this doesn't contradict the fact about compact operators that you state.)