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Let $X$ be a Banach space and $\mathcal{C}\subset\mathcal{L}(X)$ be a collection of bounded linear operators. A subspace $Y$ is said to be almost invariant under $\mathcal{C}$ if for each $T\in\mathcal{C}$ there is a finite dimensional subspace $F_{T}$ satisfying \begin{equation}TY\subset Y+F_{T}.\end{equation}

Moreover, if $Y$ is almost invariant under $T$, the $\mathit{defect}$ of $Y$ under $T$, $d_{Y,T}$ is defined to be the smallest integer such that there is a subspace $F$ of dimension $d_{Y,T}$ such that the above relation is satisfied.

In Popov's 2009 paper, Almost invariant half-spaces of algebras of operators, he proved one interesting proposition 3.1:

Let $Y$ be a subspace of $X$ and $\mathcal{A}$ an algebra of operators acting on $X$. Let $N\in\mathbb{N}$ be such that $d_{Y,T}\le N$ for all $T\in\mathcal{A}$. Then $d_{Y,T}\le N$ for all $T\in\bar{\mathcal{A}}$, where $\bar{\mathcal{A}}$ is the closure of $\mathcal{A}$ in the weak operator topology.

The author says it is enough to prove the inequality for all $T$ in the closure of $\mathcal{A}$ in the strong operator topology.

So here is something I do not understand. Since in most cases the closure in strong operator topology is strictly smaller than the closure in the weak operator topology, how can an inequality in SOT imply one in WOT?

I wonder whether there is something I do not know about the relation between these two topologies that makes the above implication true. And I guess the solution is actually something not really related to almost invariant subspaces.

Does someone have a hint?

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up vote 1 down vote accepted

The closure of convex sets in SOT and WOT coincides.

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This is true but I cannot really find a proof of it. It would be great if you can give some hint on this. – Hui Yu May 21 '12 at 20:22
The linear functionals on $L(X)$ that are SOT-continues are the same as those that are WOT-continuous. Now use the answer given by Norbert to your other question. – Theo May 22 '12 at 17:06

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