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Let $X$ and $Y$ be two Banach spaces. Assume that $L(X , Y)$ is equipped with the Weak-operatot topology. Let $\Sigma$ be a set in $L (X , Y)$. Let $T \in \bar\Sigma $.

Can we say that there exists a sequence $T_k \in \Sigma$ such that for all $x \in X$ and for all $y^* \in Y^*$ we have $$ y^* (T_{k} (x) ) \to y^* ( T(x) )$$?

P.S: Definition of Weak-Operator Topology

Red shoes
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  • I don't think so, why would this hold? We merely have a subbase that's given... – Henno Brandsma Oct 28 '21 at 07:30
  • @HennoBrandsma It seems it holds when one uses nets instead of sequences, I was wondering if it still holds for sequences! – Red shoes Oct 28 '21 at 07:33
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    Usually it does not. Such topologies behave much like large products and these are not so-called sequential spaces, except in some special cases (Eberlein-Smulian). – Henno Brandsma Oct 28 '21 at 07:38
  • @HennoBrandsma It makes sense. I was reading a paper where authors defined the weak-operator topology through the convergence of the sequences. I suspected that is the correct way of defining a topology! – Red shoes Oct 28 '21 at 07:43
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    Sequences often don't suffice to completely specify a topology unless you want to only define sequential topologies a priori. Fréchet die research on that way back. – Henno Brandsma Oct 28 '21 at 07:45

1 Answers1

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No. For ease assume that $X=Y=H$, an infinite dimensional separable Hilbert space. Let $\{e_n:n\ge1\}$ be an orthonormal basis for $H$ and let $p_n\in B(H)$ denote the orthogonal projection on the one-dimensional subspace $\text{span}\{e_n\}$. Consider the set $S:=\{\sqrt{n}p_n:n\ge1\}\subset B(H)$. We will show that $0\in\overline{S}^{WOT}$ but there exists no sequence in $S$ that converges to $0$ in the WOT.

Let's see first why $0\in\overline{S}^{WOT}$. It suffices to show that $S$ intersects all WOT-basic neighborhoods of $0$, which are the sets $$W_{\varepsilon,\xi_1,\dots,\xi_n}:=\{x\in B(H): |\langle x\xi_i,\xi_i\rangle|<\varepsilon\text{ for all }i=1,\dots,n\}$$ Assume that this fails, so there exists $\varepsilon>0$ and $\xi_1,\dots,\xi_n\in H$ such that $W_{\varepsilon,\xi_1,\dots,\xi_n}\cap S=\emptyset$. Therefore, if $x\in S$, then there exists a $j\in\{1,\dots,n\}$ such that $|\langle x\xi_j,\xi_j\rangle|\ge\varepsilon$. In particular, for each $k\in\mathbb{N}$ we would have that there exists a $j_k\in\{1,\dots,n\}$ such that $$|\langle\sqrt{k}p_k\xi_{j_k},\xi_{j_k}\rangle|\ge\varepsilon$$ i.e. $$|\langle\xi_{j_k},e_k\rangle|^2\ge\frac{\varepsilon}{k}$$ But now $$\sum_{j=1}^n\|\xi_j\|^2=\sum_{j=1}^n\sum_{k=1}^\infty|\langle\xi_j,e_k\rangle|^2=\sum_{k=1}^\infty\sum_{j=1}^n|\langle\xi_j,e_k\rangle|^2\ge\sum_{k=1}^\infty|\langle\xi_{j_k},e_k\rangle|^2\ge\sum_{k=1}^\infty\frac{\varepsilon}{k}=\infty$$ which is a contradiction.

Let's now see why there is no sequence in $S$ that converges to $0$ in the WOT. A sequence of elements of $S$ is of the form $\sqrt{n_k}p_{n_k}$ for some integers $n_k\ge1$. Assume that this converges to $0$. Note that these terms might not a priori satisfy $n_1<n_2<\dots$, but we can pass to a subsequence by picking the indexes so that each time they get bigger and bigger. In other words, we can assume WLOG that $n_1<n_2<\dots$. But now $\|\sqrt{n_k}p_{n_k}\|=\sqrt{n_k}\to\infty$, and this is a contradiction, because of the following lemma:

Lemma: Let $H$ be a Hilbert space and $(x_n)\subset B(H)$ a sequence. If $x_n\to x\in B(H)$ in WOT, then $(x_n)$ is bounded in norm.

Proof of lemma: See Martin's smart answer in this post.

Just dropped in
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