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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 $T$?

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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.

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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 $y$, $n>n_0$.

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

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