3
$\begingroup$

An exam question asks:

Show that if $$x^{(n)} = (x^{(n)}_1,x^{(n)}_2,x^{(n)}_3,...)$$ is a sequence in $\ell^1$ that converges weakly in $\ell^1$ to some $x \in \ell^1$ then $x^{(n)}_j \to x_j$ for all $j$ and $||x^{(n)}||_{\ell^1}$ is bounded but that the converse need not hold. (You need not use the fact that $(\ell^1)^* \simeq \ell^\infty$.)

I can use the linear maps $x \mapsto x_k$ to show $x^{(n)}_k \to x_k$ and I can use the standard basis $e_k$ as the counterexample but for boundedness I don't see an elementary answer.

We have shown (using the Uniform Boundedness Principle) that in fact all weakly convergent sequences on Banach spaces are bounded, and also perhaps the UBP applies to $x^{(n)} \in c_0^*$ as a weak-$*$ sequence.

So I can appeal to a theorem to show that this sequence is bounded, but I believe I am missing a simpler, more elementary, proof.

$\endgroup$
  • 1
    $\begingroup$ Weak convergence in $\ell^1$ coinsides with strong convergence. You may look at Sarason's solution for his class (Problem 51) for two different proofs of the fact, but both are actually similar to two approaches proving UBP (gliding hump argument vs. Baire category theorem). There is a simple proof in the first track here. $\endgroup$ – A.Γ. May 10 '18 at 14:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.