Let $A,B \in \mathcal{B}(X)$ where $X$ is a Banach space. Define $T_{A,B}: \mathcal{B}(X) \rightarrow \mathcal{B}(X)$ by $T_{A,B}(S)=ASB$.
Prove that if $A$ and $B$ are compact then $T_{A,B}$ is compact.
I consider a bounded sequence $\{S_n\}$ in $\mathcal{B}(X)$ but I can't find a subsequence of $T_{A,B}(S_n)$ such that is convergent. Any hint?
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In the special case of Hilbert spaces, there is a much simpler proof of this result based on the fact that every compact operator is the limit of finite rank ones (this in fact characterizes the Approximation Property for Banach spaces).
Here is the proof:
If $y\in X$ and $f\in X'$, consider the operator $U_{f,y}\in \mathcal B(X)$ given by $$ U_{f,y}(x) = f(x)y, \quad \forall x\in X. $$ Evidently $U_{f,y}$ has rank-one and it is easy to see that every rank-one operator is of this form.
In case $A=U_{f,y}$ and $B= U_{g,z}$ then $$ T_{AB}(S)(x) = ASBx = $$$$ = f(SBx)y = f(S(g(x)z))y = $$$$ = f(Sz)g(x)y = f(Sz)U_{g, y}(x), $$ which is to say that $T_{AB}(S) = f(Sz)U_{g, y}$. So the range of $T_{A,B}$ is the one-dimensional space spanned by $U_{g, y}$, whence $T_{A,B}$ has rank-one.
If $A$ and $B$ have finite rank, then both $A$ and $B$ may be writen as finite sums of rank-one operators, from where it follows that $T_{A,B}$ has finite rank (actually this proves that $\text{rank}(T_{A, B})\leq \text{rank}(A)\text{rank}(B)$).
Assuming that $X$ has the Approximation Property, such as if $X$ is a Hilbert space, then the given compact operators $A$ and $B$ may be written as $$ A=\lim_{n\to\infty }A_n, \quad \text{and}\quad B=\lim_{n\to\infty }B_n, $$ where $A_n$ and $B_n$ have finite rank, so it easily follows that $T_{A, B}$ is the operator-norm limit of $T_{A_n, B_n}$, hence proving that $T_{A, B}$ is compact.
Since $B$ is compact the image of the unit ball under $B$ is pre-compact, in particular separable. Let $x_m$ be a countable dense subset of this image. Now consider for $m$ fixed the sequence $(S_n x_m)_{n\in\Bbb N}$, since $A$ is compact you have a sub-sequence so that $A S_{n_{k}}x_m$ converges. By the diagonal argument you may assume that $n_k$ is so that $AS_{n_k}x_m$ converges for every $m$.
Denote with $A\!S$ the map $\{x_m\mid m\in\Bbb N\}\to X$ given by $x_m\mapsto \lim_k A S_{n_k}x_m$. Since $S_n$ is bounded you have that $$\|A\!S(x_m) -A\!S(x_l)\| ≤ \sup_n\|AS_n\|\,\|x_m-x_l\|$$ and the map is Lipschitz. You may then extend it to have domain $\overline{\{x_m\mid m\in \Bbb N\}}$, which is the closure of the image of the unit ball under $B$ (denote this map by $\overline{A\!S}$). You can again use the boundedness of $S_n$ to get that $AS_{n_k} x\to \overline{A\!S}x$ for all $x$ in the image of the unit ball under $B$: $$\|\overline{A\!S}x - AS_{n_k}x\| ≤ \|\overline{A\!S}(x-x_m)\|+\|\overline{A\!S}(x_m)- AS_{n_k}(x_m)\|+\|AS_{n_k}(x_m-x)\|$$ you may choose $x_m$ to make the first and the second terms as small as you like, the middle term converges to zero by the arguments before - hence for any $m$ you can find $K$ after which it is as small as you like.
Continue with arguments of this kind to get that $\overline{A\!S}$ is linear and then also itself Lipschitz - you may thus extend it to a continuous linear map $\overline{A\!S}:\overline{B(X)}\to X$. Whats left is to check that $AS_{n_k}B$ converges uniformly on bounded sets to $\overline{A\!S}B$ (ie in operator norm).
Supposing this were not the case you would get a sequence $y_k$ in the unit ball of $X$ with $\|(AS_{n_k}B-\overline{A\!S}B)y_k\|>\epsilon$ for some $\epsilon$. Well you know what to do: $B(y_k)$ has a convergent sub-sequence so just assume that $B(y_k)$ converges with limit $x$. But we have already done this calculation $$\|(\overline{A\!S}-AS_{n_k})(By_k)\| ≤ \|\overline{A\!S}(B(y_k)-x)\|+\|(\overline{A\!S}-AS_{n_k})(x)\| +\|AS_{n_k} (B(y_k)-x)\|$$ Each term converges by the same arguments as before.