# Is the limit of a sequence of continuous linear operators in the weak operator topology again a continuous linear operator?

From the Banach-Steinhaus theorem we know that if $$(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$$, where $$X$$ is a Banach and $$Y$$ a normed space, converges in the strong operator topology, then its limit in the strong operator topology is again a bounded linear operator from $$X$$ to $$Y$$.

Now I've read that in a Hilbert space $$H$$ the following stronger result holds: If $$(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(H)$$ converges in the weak operator topology, then its limit in the weak operator topology is again a bounded linear operator on $$H$$.

Why is it important that $$H$$ is a Hilbert space? Doesn't the claim remain true in the previous considered case $$(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$$, where $$X$$ is a Banach and $$Y$$ a normed space?

If $$E$$ is a normed space, we know that $$B\subseteq E$$ is bounded if and only if it is weakly bounded. Thus, a weakly convergent sequence is norm bounded.

Shouldn't it immediately follow that if $$(A_n)_{n\in\mathbb N}\subseteq\mathfrak L(X,Y)$$ is weakly convergent, it is bounded in the strong operator topology and hence bounded in the uniform operator topology by the Banach-Steinhaus theorem?

I think what you are saying is true. Never thought about it since i've always pre-assumed that the weakly-operator limit $$A$$ of the $$A_n's$$ was always in $$A\in \mathfrak L(X,Y)$$. Am writing the argument just to convience ourselfs. Indeed, we only need to assume that $$Y$$ has a norm, not necessarily a complete one.

So, lets suppose that $$A_n\overset{\text{wo}}{\to}A$$ in the weak operator topology where $$A:X\to Y$$ is a linear operator, not necessarily bounded. Convergence in the weak operator topology is described by $$h(A_n x)\to h(A x)$$ for every $$x\in X$$ and $$h\in Y^*$$. This implies that the set $$\{A_n x: n\in \mathbb{N}\}$$ is weakly bounded in $$Y$$, hence it is also bounded in $$Y$$. By the Banach-Steinhaus it follows that $$\sup_{n}||A_n||=M<\infty$$. Now, for $$x\in X$$ with $$||x||=1$$ we have $$||Ax||=\max_{h\in Y^*,\, ||h||=1}|h(Ax)|$$ So, there is some $$||h||=1$$ in $$Y^*$$ such that $$||Ax||=|h(Ax)|$$. Using the weak convergence for $$A_nx$$ we end up with \begin{align} ||Ax||&=|h(Ax)|\\ &=\lim_{n\to \infty}|h(A_nx)|\\ &\leq \underbrace{||h||}_{=1}\liminf_{n\to \infty}||A_n||\cdot \underbrace{||x||}_{=1} \end{align} Hence, $$||Ax||\leq M$$ for every $$||x||=1$$ and therefore, $$||A||\leq M<\infty$$.

Edit: (Responding to the comment)

The existence of such $$A$$ is trickier. To ensure such existence we need another assumption for $$Y$$, since there is a counter example in here where $$X=Y=c_0$$. The only natural that i could think while i was trying to prove it is that $$Y$$ has to be reflexive (from not being a Banach space we went straight out to reflexivity :P). In the case where $$X=Y=H$$ is a Hilbert space things were slightly more easier since we can identify $$H^*$$ with $$H$$ and dont need to mess with the second duals.

The argument in the case where $$Y$$ is reflexive is the following:

Suppose that $$\lim_{n}\langle A_n x, h \rangle$$ exists for every $$x\in X$$ and $$h\in Y^*$$. For fixed $$x\in X$$ let $$f_x:Y^*\to \mathbb{R}$$ defined by $$\langle h, f_x\rangle =\lim_{n\to \infty}\langle A_n x, h\rangle$$ Its easy to check that $$f_x$$ is a linear functional and by the previous discussion it is also bounded. Meaning, $$f_x \in Y^{**}$$. By reflexivity, there is some $$y_x\in Y$$ such that $$\langle h, f_x\rangle =\langle y_x, h\rangle$$ for all $$h\in Y^*$$. Now, let $$x\overset{A}{\longmapsto} y_x$$. Now, its easy to check that $$A:X\to Y$$ is a linear operator. By the previous discussion it is also bounded.

• Thank you for the validation. There is one (rather elementary) issue left: You've started with the assumption that there is a linear operator $A:X\to Y$ s.t. $(A_n)_{n\in\mathbb N}\to A$ in the WOT. But how do we need to argue formally that there is such a linear operator $A$ with $\langle Ax,h\rangle=\lim_{n\to\infty}\langle A_n,h\rangle$? Dec 12, 2020 at 5:57
• @0xbadf00d Oh, you are right. Again i pre-assumed that $A$ is a given linear operator. Although, if we dont assume the existence of such an $A$ we have to put an extra (strong) hypothesis for $Y$. I edited the answer above, you can check it. Dec 12, 2020 at 7:22
• Well, I think I was a bit sloppy in the terminology I've used. Saying that $(A_n)_{n\in\mathbb N}$ converges in the WOT doesn't merely mean that $(A_nx)_{n\in\mathbb N}$ is weakly convergent for all $x\in X$ or that $$\langle Ax,h\rangle=\lim_{n\to\infty}\langle A_nx,h\rangle\;\;\;\text{for all }x\in X\text{ and }h\in Y'$$ for some linear $A:X\to Y$, but also that $A$ is bounded; which is precisely what we are aiming to show. So that didn't make sense. Dec 13, 2020 at 10:01
• Regarding the counterexample in the link you've provided. I understand that the operators $T_n:c_0\to c_0$ satisfies that $(T_n)_{n\in\mathbb N}$ is weakly convergent for all $x\in c_0$, but I don't understand the argument why they don't converge in the WOT. What's exactly meant by the "componentwise limits"? Dec 13, 2020 at 10:04
• @0xbadf00d Yes, exactly. This the argument, if we assume that there exists $T:c_0\to c_0$ such that $T_n$ converges WOT to $T$, then with this argument we end up to a contradiction, since $Tx$ must be an element of $c_0$. Dec 13, 2020 at 20:46