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?


1 Answer 1


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.

  • $\begingroup$ 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 $\langle Ax,h\rangle=\lim_{n\to\infty}\langle A_n,h\rangle$? $\endgroup$
    – 0xbadf00d
    Dec 12, 2020 at 5:57
  • $\begingroup$ @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. $\endgroup$ Dec 12, 2020 at 7:22
  • $\begingroup$ 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. $\endgroup$
    – 0xbadf00d
    Dec 13, 2020 at 10:01
  • $\begingroup$ 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"? $\endgroup$
    – 0xbadf00d
    Dec 13, 2020 at 10:04
  • 1
    $\begingroup$ @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 $c_0$. $\endgroup$ Dec 13, 2020 at 20:46

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