# Finite codimensional subspaces of L(X,Y)

Let $X$ and $Y$ be separable Banach spaces and $L(X,Y)$ be the Banach space of bounded linear operators from $X$ to $Y$. Suppose $A$ is a norm closed finite codimensional subspace of $L(X,Y)$.

My question is: For which spaces $X$ and $Y$ is $A$ (finite codimensional) Borel in the strong operator topology?

Recall that the strong operator topology is the topology given by the pointwise convergence of nets of operators.

Edit: A previous iteration of this question asked for spaces $X$ and $Y$ so that $A$ closed in the strong operator topology.

• There is something wrong with your final paragraph, Kevin. A normed closed subspace of $X^*$ need not be weak$^*$ closed--consider the kernel of a functional in $X^{**}\sim X$. – Bill Johnson Apr 6 '13 at 1:20
• Bill, that's very sloppy of me. I'm editing my question and removing the final paragraph. – Kevin Beanland Apr 6 '13 at 1:52
• So if $X$ does not have a subspace isomorphic to $\ell_1$ and $Y = \Bbb{R}$, then it is true (because then each functional in $X^{**}$ is a pointwise limit of elements in $X$ by the Odell-Rosenthal theorem). What else do you know about the question, Kevin? – Bill Johnson Apr 6 '13 at 15:12
• And conversely, if $X$ contains $\ell_1$ and $Y=\mathbb R$, then it is not true since there are functionals in $X^{**}$ which are not $w^*$-Borel (the kernel of such a functional is not $w^*$-Borel). – Etienne Apr 6 '13 at 22:18
• Bill and Etienne: Thanks for the answers. What I really want (and didn't ask) is more specific. Suppose that every operator on a Banach space is a weakly compact perturbation of a multiple of identity. Is the space of weakly compact operators Borel in the strong operator topology? There are a few examples of (non-reflexive) spaces having this property. More generally, replace weakly compact with strictly singular or some other operator ideal and ask the same question. – Kevin Beanland Apr 8 '13 at 13:27