We know (and indeed it's trivial to show) that $x_n \to x \implies x_n \rightharpoonup x$. It is a reason for the name "weak-convergence". We have several examples on this site where we have the converse. I just saw a result in our analysis class which wasn't too difficult to prove, namely
Let $X$ be a normed vector space and $x,x_n\in X\ \forall n\in \mathbb{N}$. Similarly let $L,L_n \in X^*$. If $x_n \rightharpoonup x$ and $L_n \to L$, then $L_nx_n\to Lx$ holds.
This got me thinking, does something hold in general here? If $X$ is reflexive, then from this maybe one could do the following with the canonical isometry $\tau$ (here I am quite unsure and I don't really know what I am doing, but bare with me)
$||x-x_n|| = ||\tau^{-1}(\Lambda_x-\Lambda_{x_n})|| \leq ||\tau^{-1}||\ ||\Lambda_x - \Lambda_{x_n}||_{**} = ||\tau^{-1}||\ sup\{|(\Lambda_x - \Lambda_{x_n})L|:\ L\in X^*, ||L||_*\leq1\} = ||\tau^{-1}||\ sup\{|Lx - Lx_n|:\ L\in X^*, ||L||_*\leq1\} \to 0$
where the last argument stems from weak convergence of $x_n$. I wonder if this is correct and if there are other cases where one can deduce strong convergence from the weak in general. Thank you for your time!
EDIT: I will leave the above as is, otherwise the comments below make no sense. I have found a result where it holds and thought it would be appropriate to post it here.
Theorem: Let $X$ be a uniformly convex Banach space. If $x_n \rightharpoonup x$ and $||x||\geq \overline{\lim}||x_n||$, that is $x\mapsto ||x||$ is upper semicontinuous, then $x_n$ converges strongly to $x$.