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Suppose $(T_n)_{n=1}^\infty$ is a sequence of bounded operators on a Hilbert space $\mathcal{H}$ and that for every $x \in \mathcal{H}$ $$ \langle T_n (x),x \rangle \longrightarrow 0 \quad \text{as } n \to \infty. $$

Is it then true that $T_n (x) \rightarrow 0$ as $n \to \infty$ for every $x \in \mathcal{H}$?


I am not sure whether or not this is true. One the one hand it could be that $T_n (x)$ converges to something in the orthogonal complement of a subsapce containing $x$ but also this seems odd if this holds for any $x \in \mathcal{H}$. We know that the inner-product is continuous in the first entry but I can't see that this is useful here to show the claim.

Any help towards a proof, counter-example or reference is much appreciated!

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  • $\begingroup$ Have you encountered the polar identity? This will show you that the condition $\langle T_nx, x\rangle\to 0$ for all $x\in \mathcal{H}$ is equivalent to $T_n\to 0$ in the weak operator topology. Also, if $T_n(x)\to 0$ for all $x\in\mathcal{H}$ then $T_n$ converges to zero in the strong operator topology. Do you know when these topologies coincide? $\endgroup$ May 1, 2022 at 17:18

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The polar identity states $$ \langle Tx, y\rangle = \frac14\sum_{k=0}^3 i^k \langle T(x + i^ky), x + i^ky\rangle $$ This is a well-known identity, which can be derived from two simpler observations: $$ \begin{aligned} \langle Tx, y\rangle + \langle Ty, x\rangle & = \frac12\left(\langle T(x + y), x + y\rangle - \langle T(x - y), x - y\rangle\right) \\ \langle Tx, y\rangle - \langle Ty, x\rangle & = \frac1{2i}\left(\langle T(x - iy), x - iy\rangle - \langle T(x + iy), x + iy\rangle\right) \end{aligned} $$

Thus, if $\langle T_n x, x\rangle\to 0$ for all $x\in\mathcal{H}$, then $\langle T_n x, y\rangle$, being a sum of four terms of the form $\langle T_n z, z\rangle$ for some $z\in\mathcal{H}$, also tends to zero. So $T_n$ converges to zero in the weak operator topology (or WOT).

On the other hand, $T_nx\to 0$ for all $x\in\mathcal{H}$ is just saying $T_n\to 0$ in the strong operator topology (or SOT).

In general, these topologies are not equivalent; the SOT is strictly finer than the WOT, which is to say that there are more SOT-open sets than WOT-open sets, or equivalently every SOT-convergent net of operators is WOT-convergent (via the Cauchy-Schwarz identity).

Consider the right-shift operator $R \in\mathcal{B}(\ell_2)$ taking $(a_1, a_2, ...)$ to $(0, a_1, a_2, ...)$. It's not too hard to check that $R^n$ converges weakly to zero, but each $R^n$ is an isometry, and so cannot converge strongly to zero.


Clarifying the discussion in the comments, I had mistakenly claimed that the WOT and SOT coincide for certain types of sets. This isn't true, and it's an easy trap to fall into (I've made this mistake before)!

What is true:

  1. If $U\subseteq\mathcal{B}(\mathcal{H})$ is bounded, then the strong and ultra-strong (resp. weak and ultra-weak) topologies on $U$ are identical: they have the same open sets.
  2. If $C\subseteq\mathcal{B}(\mathcal{H})$ is convex, then $\overline{C}^{WOT} = \overline{C}^{SOT}$. This is because the WOT and SOT topologies have precisely the same continuous linear functionals (in fact, any two locally convex topologies on a vector space with the same continuous linear functionals enjoy this property: see this question).

This second fact does not mean that the WOT-open and SOT-open subsets of $C$ are the same.

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  • $\begingroup$ @Zerox Yes, but only under the assumption that $T_n$ is uniformly bounded. Otherwise there are counterexamples. $\endgroup$ May 1, 2022 at 17:31
  • $\begingroup$ I'm curious about the counter-examples, would you include that in your answer? $\endgroup$
    – Zerox
    May 1, 2022 at 17:32
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    $\begingroup$ @Zerox Yeah I'll see if I can find a good example. I'd be surprised if there weren't already examples on this site, all we need is an unbounded net of operators which converges weakly to zero but not strongly to zero. $\endgroup$ May 1, 2022 at 17:34
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    $\begingroup$ @Zerox Give me a sec, I misquoted something. It's not the strong and weak topologies which coincide on bounded sets (though I think they coincide on convex sets), rather it's the strong and ultrastrong (or weak and ultraweak) topologies which coincide. There is an instance when the strong and weak topologies coincide, but I need to check. $\endgroup$ May 1, 2022 at 17:39
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    $\begingroup$ @Zerox Ah I've found my error. So, the strong and ultrastrong (resp. weak and ultraweak) topologies do indeed coincide on bounded sets. But the weak and strong topologies don't coincide on a convex set. Rather, the weak and strong closures of a convex set coincide. $\endgroup$ May 1, 2022 at 17:53
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For seperable real Hilbert spaces there is a counter-example: Let $T_n \equiv T$ such that $Te_{2i}=e_{2i+1} ,Te_{2i+1}=-e_{2i}$ for $i \ge 0$, where $\{ e_i \}_{i \ge 0}$ is an orthonormal base. Then clearly $T_n(x) \equiv T(x) \ne 0$ but $\langle T_n(x),x \rangle \equiv \langle T(x),x \rangle = 0$.
For inseperable real Hilbert spaces similar construction can be made if there is a convolution with no fixed point on the orthonormal base, but I'm afraid some set-theoretic methods will be involved.

Another Construction: Choose a two-dimensional subspace $V$ of the ambient Hilbert space. Suppose $a,b$ is an orthonormal base of $V$. Define the bounded linear map $T$ to be $Ta=b,Tb=-a,T|_{V^{\perp}}=0$ and take $T_n \equiv T$. This works for every real Hilbert space.

For the complex case, as @user3002473 pointed out, a counter-example can be taken to be a WOT-null operator sequence that is not SOT-null. So in the seperable case a counter-example is $T_n=R^n$, where $Re_i=e_{i+1}$ is the right-shift operator. (This apparently also works for the real case.)

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    $\begingroup$ Couldn't we pick two orthonormal vectors, define the action as you did on this subspace and extend the operator by zero on the orthogonal complement of our $2$-dimensional space? This would then work for all real Hilbert spaces without invoking much set theory. $\endgroup$ May 1, 2022 at 16:56
  • $\begingroup$ @SeverinSchraven This is also an example, but I prefer isomorphisms. $\endgroup$
    – Zerox
    May 1, 2022 at 16:57
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    $\begingroup$ That is of course a valid opinion. I tend to prefer low-tech counterexample. $\endgroup$ May 1, 2022 at 16:59
  • $\begingroup$ @SeverinSchraven Fine. I'll include yours. $\endgroup$
    – Zerox
    May 1, 2022 at 17:00
  • $\begingroup$ I am sorry, I did not mean it this way. Math is very much a matter of preferences. Your construction is a bit heavier, but also gives a stronger result. $\endgroup$ May 1, 2022 at 17:23

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