Let $X$ be a Banach space and $E$ a sublinear subspace . Show there exists a surjective isometry $\phi : E^* \rightarrow \overline{ E}^* $

Question

Let $X$ be a Banach space and $E$ a sublinear subspace of $X$ . Show there exists a surjective isometry $\phi : E^* \rightarrow \overline{ E}^* $

I think it could be a derived fact of the Godlstine lemma or something related to the weak topology but I am really lost at it . Could you please give some guidelines?

Answer 1

Define $$\phi: E ^* \to \overline{E}^*: f \mapsto (\phi_f: \overline{E} \to \Bbb{K}: e \mapsto \lim_n f(e_n))$$ where $e= \lim_n e_n$.

Note first that $\lim_n f(e_n)$ exists because $f$ is continuous. Next, we show that this is well-defined: if $e= \lim_n e_n = \lim_n e_n'$. Then $\lim_n (e_n-e_n') =0$ so $\lim_n (f(e_n)-f(e_n'))=0$. Hence $\lim_n f(e_n) = \lim_n f(e_n')$.

Check that $\phi$ satisfies the required properties.