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.

Thanks for your help.


1 Answer 1


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 $u=v^{*}$.

  • $\begingroup$ thanks, but i can not prove that v is continuous, may i have a hint? $\endgroup$
    – John
    Commented Apr 19, 2020 at 12:30
  • $\begingroup$ @dapeng It is a simple application of Closed Graph Theorem. $\endgroup$ Commented Apr 19, 2020 at 12:33
  • $\begingroup$ Thank you so much , i got it ! $\endgroup$
    – John
    Commented Apr 19, 2020 at 12:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .