I need to prove the following theorem.

Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete. Let $\{T_n\} \subset \mathscr{L}(X,Y)$ with $\Lambda(T_nx)$ converges for all $x \in X$, $\Lambda \in Y^*$. Prove there is a $T \in \mathscr{L}(X,Y)$ such that $T_n \to T$ in the weak operator topology.

Proof so far: $\Lambda(T_nx)$ converges for all $x \in X$, $\Lambda \in Y^*$. Since $Y^*$ is weakly sequentially complete, there is a $y \in Y$ such that $\Lambda(T_nx) \to \Lambda y_x$ for all $\Lambda \in Y^*$. Define $Tx :=y_x$. We need to show $T$ is bounded. Note that because $\Lambda(T_nx)$ converges, it is most certainly true that $\{\Lambda(T_nx) : T_n \in \{T_n\}\} < \infty$. Hence, by a Corollary of the PUB, $\|T_n\|$ is uniformly bounded by say $M > 0$.

So here now I want to prove that $\|T\|$ is bounded. Using reverse triangle inequality, we can get that $|\Lambda(Tx)| \leq M \|\Lambda\| \|x\|$, but the fact that we have $\Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem. I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete. Any hints/help would be much appreciated.


1 Answer 1


As you said, we have $$|\Lambda(Tx)| = \left|\lim_{n\to\infty} \Lambda(T_n x)\right|= \lim_{n\to\infty}|\Lambda(T_n x)| \le \limsup_{n\to\infty} \|\Lambda\| \|T_n\|\|x\| \le M \|\Lambda\| \|x\|$$

so for $\widehat{Tx} \in Y^{**}$, the element of $Y^{**}$ represented by $Tx$ we have

$$\left|\widehat{Tx}(\Lambda)\right| = |\Lambda(Tx)| \le M \|\Lambda\| \|x\|$$

so $\|Tx\| = \left\|\widehat{Tx}\right\| \le M\|x\|$.

Therefore $T$ is bounded and $\|T\| \le M$.

Finally, for any $x \in X$ we have $\Lambda(T_n x) \to \Lambda(Tx), \forall \Lambda\in Y^*$ so $T_nx \xrightarrow{w} Tx$. We conclude $T_n \to T$ in the weak operator topology.

A counterexample when $Y$ is only a Banach space:

Consider $T_n : c_0 \to c_0$ given by $T_n(x_k)_k = (\overbrace{x_1, x_1, \ldots, x_1}^n, 0, 0\ldots)$.

We have $(c_0)^* = \ell^1$ so for any $x \in c_0, \Lambda = (\lambda_n)_n \in \ell^1$ we have

$$\Lambda(T_n x) = \Lambda(\overbrace{x_1, x_1, \ldots, x_1}^n, 0, 0\ldots) = x_1\cdot \sum_{k=1}^n \lambda_n \xrightarrow{n\to\infty} x_1\cdot \sum_{k=1}^\infty \lambda_n$$

so $(\Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, \ldots) \in c_0$ we have $T_nx = (\overbrace{1, \ldots, 1}^n, 0, 0\ldots)$. The coordinate-wise limit of this sequence is $(1, 1, \ldots) \notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.

  • $\begingroup$ Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $\Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $\Lambda$. Is my reasoning incorrect? $\endgroup$ Jul 25, 2018 at 0:49
  • $\begingroup$ @user707959 Yes, $(\Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $\Lambda$. Anyway, I added a possible counterexample, have a look. $\endgroup$ Jul 25, 2018 at 0:55
  • $\begingroup$ @user707959 Wait, actually I agree with you, it should be independent of $\Lambda$. $\endgroup$ Jul 25, 2018 at 0:57
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    $\begingroup$ @user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now. $\endgroup$ Jul 25, 2018 at 10:37
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    $\begingroup$ Technicality for future readers: note that $\widehat{Tx}$ is not yet in $Y^{**}$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^{**}$. $\endgroup$ Jul 25, 2018 at 23:00

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