# 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.

• I don't think that you need reflexivity. This looks like a general fact of Banach spaces. Jul 11 '13 at 13:25
• Maybe. You need to get from $\langle f_k,v''\rangle \to \langle f,v''\rangle$ for any $v'' \in V''$ to $\langle f_k,v\rangle \to \langle f,v\rangle$ for any $v \in V$.
– Jan
Jul 11 '13 at 13:34
• Which you always can do in Banach spaces, since there is a norm-preserving injection $J\colon V \to V''$
– Jan
Jul 11 '13 at 13:40
• Oh well, I had missed that point. Since you spoke of "weak convergence in $V'$" I had automatically assumed that it was what is usually called weak-$\star$ convergence, that is $$\lvert\langle f_k - f, v\rangle\rvert\to 0, \quad \forall v \in V.$$ With this notion of convergence everything goes on fine as in Davide's proof. You need Banach-Steinhaus's theorem to ensure that the sequence $f_j$ is norm bounded. Jul 11 '13 at 13:57
• If your sequence converges in weak sense, meaning that $$\lvert \langle f_k-f, v''\rangle\rvert \to 0, \quad \forall v''\in V'',$$ then better still: using the norm-preserving injection that you have rightfully recalled we can prove that the sequence converges in weak-$\star$ sense too, and we are finished. Jul 11 '13 at 14:00

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.