Convergence of $A_nT$ to $AT$ in operator norm for compact $T$ $A_n:Y\rightarrow Z$ are operators that strongly converge to $A$. Also, $\|A_n\|_\text {op}\le c$ for $c>0$. Given a compact operator $T:X\rightarrow Y$, I need to show that $A_nT$ converges to $AT$ in operator norm (all spaces in this question are Banach spaces).   
I was unable to prove this and I also do not understand why we need to assume that $T$ is compact. Any ideas?
 A: You need $T$ to be compact because otherwise, by taking $T=\Bbb1$, the statement would imply that strong convergence is equivalent to norm convergence, this is not true.
Suppose $\|AT-A_nT\|\not\to0$, this means that there exists a sequence of unit vectors $x_n$ so that $\|(AT-A_nT)x_n\|>\epsilon$ for some $\epsilon$ and for all $n$. Now $T$ is compact so the image of the unit ball under it is pre-compact. This means that $Tx_n$ has a convergent subsequence, so lets actually just assume $Tx_n$ to be convergent with limit $x$.
From strong convergence of the $A_n$ you see that $\|(A-A_n)x\|\to0$. Now lets combine our information:
$$\|(AT-A_nT)x_n\|=\|(A-A_n)(Tx_n-x+x)\|≤\|(A-A_n)x\|+\|A_n-A\|\,\|T x_n-x\|.$$
Now $\|A_n-A\|≤(\|A_n\|+\|A\|)$ is bounded by some constant by assumption. Every other term on the right converges to zero. This is a contradiction to $\|(AT-A_nT)x_n\|>\epsilon$ for all $n$.
A: In order to prove that a sequence $(S_n)_n$ of operators converges to $S$ in operator norm, we have to prove that 
$$\lim_{n\to\infty}\sup_{x:\lVert x\rVert\leqslant 1}\left\lVert S_nx-Sx\right\rVert =0.$$
The supremum is taken on the closed unit ball $B$, which can be "big", hence the convergence may fail. Here, we have to prove that 
$$\lim_{n\to\infty}\sup_{y\in T(B)}\left\lVert A_ny-Ay\right\rVert =0.$$
This can be done by using precompactness of $T(B)$: for any fixed $\varepsilon$, there exists a finite set $F\subset Y$ such that for any $y\in Y$, there exists $y'\in F$ such that $\lVert y-y'\rVert\lt \varepsilon$.
A: Take a sequence $\{x_n\}$ with $\|x_n\|=1$ such that $\|A_nTx_n-ATx_n\|\ge\|A_nT-AT\|-\frac1n$. Let $y_n=Tx_n$. It suffices to show that $s_n:=\|A_ny_n-Ay_n\|\to0$. Let $\{s_{n_k}\}$ be any subsequence. Since $T$ is compact, the sequence $\{y_{n_k}\}$ is relatively compact, so there is a further subsequence $\{y_{n_{k_\ell}}\}$ such that $y_{n_{k_\ell}}\to y$. Then
$$s_{n_{k_\ell}}\le\|A_{n_{k_\ell}}(y_{n_{k_\ell}}-y)\|+\|(A_{n_{k_\ell}}-A)y\|+\|A(y_{n_{k_\ell}}-y)\|\to0$$
where the first term can be seen to converge to zero since $\|A_{n_{k_\ell}}\|\le c$. Thus for any subsequence $\{s_{n_k}\}$, there is a further subsequence converging to zero, so in particular $\{s_n\}$ cannot have  a subsequence bounded away from zero. This implies $s_n\to0$. (Note: the subsequences of subsequences argument is a fairly standard compactness argument - if you are familiar with it, you can streamline the proof somewhat.)
