# Weak convergence to 0 iff bounded and pointwise convergence to 0

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.

• 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? Mar 14 '19 at 19:38
• 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. Mar 14 '19 at 19:41

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.