# Denseness of finite rank operators in $\mathcal{B}(X,Y)$

Let $X$ and $Y$ be Banach spaces and let $\mathcal{B}(X,Y)$ be the space of bounded linear operators from $X$ to $Y$. As noted in the answers to a question on

https://math.stackexchange.com/questions/535645/weak-operator-topology-and-finite-rank-operators

the finite rank operators lie dense in $\mathcal{B}(X,Y)$ with respect to the strong operator topology (SOT-topology). I would like to know under which assumptions on the Banach spaces $X$ and $Y$ one can approximate an arbitrary $T\in\mathcal{B}(X,Y)$ in the SOT-topology by a norm-bounded sequence of finite rank operators.

It seems to me that the proof of denseness of the finite rank operators cited above does not yield approximation by norm-bounded sequences. I can obtain the required approximation under the assumption that $X$ is separable and that either $X$ or $Y$ has the bounded approximation property. However, these assumptions yield a stronger statement, namely that one can approximate a general $T\in\mathcal{B}(X,Y)$ by a norm-bounded sequence of finite-rank operators in the topology of uniform convergence on compact subsets of $X$. This clearly implies SOT-convergence, but it might be a strictly stronger statement than what I am looking for.

Does anyone know whether one can approximate an arbitrary $T\in\mathcal{B}(X,Y)$ by a norm-bounded sequence of finite-rank operators in the SOT-topology for more general Banach spaces than those mentioned above?

• Is it important that you want a sequence, or will a net do? – Yemon Choi Jun 14 '14 at 0:45
• If you can approximate at all, $\|T_n\|\le C$ is automatic from the uniform boundedness principle. – Christian Remling Jun 14 '14 at 0:57
• Sequences will obviously ($T=I$) not suffice on non-separable spaces. – Christian Remling Jun 14 '14 at 0:58
• For bounded nets, strong convergence is the same as uniform convergence on compact sets. – Bill Johnson Jun 14 '14 at 13:21
• @ChristianRemling: Good point, so the question becomes under which assumptions the finite rank operators are sequentially dense in B(X,Y). – Jan Rozendaal Jun 15 '14 at 6:34