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.
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.