# the weak*-weak* continuous mapping is the conjugate operator of a bounded mapping?

Let $$E,F$$ be Banach spaces, $$u:E^*\rightarrow F^*$$ is a linear mapping. Prove that : the mapping $$u:(E^*,\sigma(E^*,E)\rightarrow (F^*,\sigma(F^*,F))$$ is continuous if and only if there exists $$v\in B(F,E)$$ such that $$u=v^*$$, where $$\sigma(E^*,E)$$ is weak* -topology on $$E^*$$.

I can prove that if there exists $$v\in B(F,E)$$ such that $$u=v^*$$, then $$u$$ is weak*-weak* continuous. but i can not use the weak*-weak* continuous mapping to find a bounded mapping.

The dual of the locally convex topological vector space $$(E^{*},\sigma(E^{*},E))$$ is $$E$$. Fix $$y \in F$$ and consider the map $$x^{*} \to u(x^{*})(y)$$. This is a continuous linear functional on $$(E^{*},\sigma(E^{*},E))$$ and hence it is given by element $$x$$ of $$E$$. This means $$u(x^{*})(y)=x^{*}(x)$$ for all $$x^{*} \in E^{*}$$. You can check that $$x$$ is uniquely determined by $$y$$. Write $$x$$ as $$v(y)$$. This defines your $$v$$. I leave it to you to check that $$v$$ is continuous and $$u=v^{*}$$.