I am working on a problem from functional analysis that has me stumped. Let $B$ be a reflexive Banach space on some subset of $\mathbb{R}^n$ s.t. point evaluations are continuous. Show that if $f_n\rightharpoonup0$ then $\sup_n||f_n||<\infty$ and that $f_n$ converges pointwise to 0.

My attempt thus far: If I suppose $f_n\rightharpoonup0$ then I can define the measures $\mu_A(x)=\chi_A(x)$, and then for fixed $x\in X$ we have by weak-convergence to 0 that $f_n(x)=\int_X f_n(t)\mu_{\{x\}}(dt)\rightarrow0$ and so $f_n$ converges pointwise to 0. I can't see why $\sup_n||f_n||<\infty$ or the other direction.

  • $\begingroup$ To clarify, let $B$ be a reflexive Banach space of continuous real- or complex-valued functions on $\Omega\subseteq\mathbb{R}^n$. Could you elaborate on how the Banach-Alaoglu theorem gives me boundedness? $\endgroup$
    – Scott
    Mar 14 '19 at 19:38
  • $\begingroup$ Sorry, I had a brain fart, I meant the Uniform Boundedness Principle. This holds for general normed spaces, i.e. a weakly convergent sequence is necessarily bounded since the topological dual $X^*$ is a Banach space. Can you see why? I can elaborate if needed. $\endgroup$
    – Reveillark
    Mar 14 '19 at 19:41

One direction:

If $f_n$ converges weakly to $0$, since point evaluations are continuous linear functionals, $f_n(x)\to 0$ for every $x$, so $f_n$ converges pointwise to $0$. For boundedness, use the UBP. See, for example, here.

For the converse, if $f_n$ doesn't converge weakly to $0$, there is a functional $\phi$, a subsequence $(f_{n_k})$ and an $\varepsilon>0$ such that $$|\phi(f_{n_k})|\ge \varepsilon$$ for all $k$.

The unit ball of $B$ is isomorphic to the unit ball of $B^{**}$ by reflexiveness and said unit ball is weak$^*$ compact by Banach-Alaoglu. Now, the weak$^*$ topology on the double dual is just the weak topology on the original space. So there is a sub-subsequence which converges weakly to some $f$. Now $f=0$ by pointwise convergence (same argument as before). We now have a contraditction.


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