Convergence of dual product of weakly and strongly convergent sequences Say $V$ is a reflexive> Banach space, so that we can identify the dual of the dual $V'$ with $V$, i.e. $(V')'=V$. 
Consider two sequences $\{v_k\} \subset V$ and $\{f_k\} \subset V'$ that converge to $v\in V$ and $f\in V'$ in the strong and in the weak sense, i.e.


*

*$v_k \to v$ in $V$ and

*$f_k \rightharpoonup f$ in V'


as $k \to \infty$. How to show, that 


*

*then $\langle v_k, f_k \rangle \to \langle v, f \rangle$ in $\mathbb R$, 


where $\langle \cdot,\cdot \rangle$ denotes the dual pairing on $V\times V'$?

I see that for the seperate sequences $\langle v_k, f \rangle$ and  $\langle v, f_k \rangle$ both converge (the latter because of reflexivity of $V$) [see comments below] but I haven't come to terms with the mutual convergence.

Write 
$$|\langle f_k,v_k\rangle-\langle f,v\rangle|\leqslant \lVert f_k\rVert\cdot\lVert v_k-v\rVert+|\langle f_k,v\rangle-\langle f,v\rangle|,$$
and notice that the sequence $(\lVert f_k\rVert,k\geqslant 1)$ is bounded.
