# Thread:

1. ## convergence in Hilbert spaces

I have the following problem:

Let X be a Hilbert space with the inner product <.|.>
Let $C_n \in C(X)$ be a family of compact operators that converges in the norm to operator C.
Let $(x_n)_n$ be a sequence that converges weakly to $x_0$

Does < $C_n x_n | x_n$> converge? If it does, where to?

I'm completely lost, please help.

2. Originally Posted by marianne
Let X be a Hilbert space with the inner product <.|.>
Let $C_n \in C(X)$ be a family of compact operators that converges in the norm to operator C.
Let $(x_n)_n$ be a sequence that converges weakly to $x_0$

Does < $C_n x_n | x_n$> converge? If it does, where to?
Three things you need to know. (1) Weakly convergent sequences are bounded. (This follows from the uniform boundedness principle, because a weakly convergent sequence is obviously weakly bounded.) (2) Compact operators convert weakly convergent sequences into norm-convergent sequences. (3) The set of compact operators is norm-closed, so that if C_n→C_0 in norm and each C_n is compact, then so is C.

We want to show that $\langle C_n x_n | x_n\rangle\to \langle C_0 x_0 | x_0\rangle$. Here's an outline of how to do it. First, for n large enough, Cx_n is close to Cx_0 in norm, by (2) and (3). Next, for n large enough, C_nx_n is close to Cx_n (this follows from (1), together with the fact that C_n→C_0 in norm.

Putting those two statements together, you see that (for n large enough) $\langle C_nx_n|x_n\rangle$ is close to $\langle C_0x_0|x_n\rangle$. But the weak convergence tells you that (again for n large enough) $\langle C_0x_0|x_n\rangle$ is close to $\langle C_0x_0|x_0\rangle$. Therefore $\langle C_n x_n | x_n\rangle\to \langle C_0 x_0 | x_0\rangle$ as n→∞.

3. Thank you so much!

I hope nobody here won't mind if I post another question or two in the next couple of days. :-)

4. THERE ARE LOTS OF PEOPLE WHO MIND THINGS..

just kidding..
post your questions and everyone will do his/her best to help.. Ü

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