2
$\begingroup$

Suppose $T: H \rightarrow H$ is a compact operator, $H$ is a Hilbert space, and let $(A_n)$ be a sequence of bounded linear operators on $H$ converging strongly to $A$. Show that $A_nT$ converges in operator norm to $AT$.

I'm attempting to prove this by contradiction: That is, there exists some $\epsilon > 0$ such that $ \|A_nT-AT\| = \sup_{\|f\|=1}\|(A_nT-AT)f\| > \epsilon$ for all $n$. How can I exploit the compactness of $T$?

$\endgroup$
2
$\begingroup$

You can use the property of the compact operators: that they map bounded sequences in X to sequences in Y with convergent subsequences. So you can choose your sequence to be an orthonormal basis in your Hilbert space and then you can advance with your proof.

$\endgroup$
2
$\begingroup$

Notice that $\sup_{x\in B} \lVert (A_n-A)Tx\lVert=\sup_{y\in {T[B]}}\lVert (A_n-A)x\rVert=\sup_{y\in \overline{T[B]}}\lVert (A_n-A)x\rVert$ (where $B$ is the closed unit ball), and $\overline{T[B]}$ is compact. If you need more hints (almost a solution), it is hidden below.

If fix any $\varepsilon>0$ you pick any $y$, and $n_y$ large enough so that the norm $\lVert (A_n-A)y\rVert<\varepsilon$ for each $n>n_y$, then the norm is smaller than $2\varepsilon$ in its neighbourhood $U_y$. By compactness, we can pick just finitely many of those $U_y$ before we cover the entire compact set, which gives as an $n_0$ such that $\lVert (A_n-A)y\rVert<2\varepsilon$ for all $y$, $n>n_0$.

$\endgroup$
1
$\begingroup$

Hint: The image of the unit ball is totally bounded.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.