In our lecture on partial differential equations we proved the following statement
Let $V$ be reflexive and $A: V \to V^*$ a weak-strong continuous operator ($u_n \rightharpoonup u$ in $V$ $\implies A u_n \to A u$ in $V^*$). Then $A$ is compact.
In the proof the reflexivity of $V$ is essential.
Therefore, I was looking for a counterexample for this statement if we drop reflexivity, i.e. I want to find an operator $A: V \to V^*$, where $V$ is a non-reflexive Banach space which is weak-strong continuous but not compact.
The simplest non-reflexive Banach space, whose dual is not too complicated I could think of is $V = c_0$, the space of zero sequences and therefore $V^* = \ell_1$.
Any hints are welcome.
Update 1:
As noted in the comments by @DanielFisher we can rule out linear maps $A: c_0 \to \ell_1$ as Pitt's theorem states all bounded linear operators $c_0 \to \ell_{p}$, $p < \infty$ are compact.
Best Answer
What about $V=l^1$, $V^*=l^\infty$, $A=id$ the continuous embedding $l^1\hookrightarrow l^\infty$? Due to the Schur-property, it is trivially weak-strong continuous, but not compact.