The image of the weak topology with the canonical injection $J$

Question

Let $E$ be a Banach space. With the weak topology $\sigma(E,E^*)$.

And let $J:E\rightarrow E^{**} $ be the canonical injection.

Can we prove that the image of the weak topology with $J$ is exactly the trace of the topology $\sigma(E^{**},E^*)$ on $J(E)$.

in other words $$J(\sigma(E,E^*))=J(E)\cap \sigma(E^{**},E^*)$$

Answer 1

Yes, it is true that $$ J:\left(E,\sigma(E,E^*)\right)\ni x\mapsto J(x)\in (J(E), \sigma(E^{**},E^*)\big|_{J(E)}) $$ is a homeomorphism where $J(x)\in E^{**}$ is defined as $J(x)(f)=f(x)$ for all $f\in E^*$.

To see that $J$ is continuous, suppose a net $\{x_i\}_{i\in I}$ in $E$ converges to $x$ weakly. Then, for all $f\in E^*$, we have $\lim\limits_{i\in I}J(x_i)(f) = \lim\limits_{i\in I}f(x_i)= f(x)=J(x)(f).$ This says that $J(x_i)$ converges to $J(x)$ in weak-star topology $\sigma(E^{**},E^*)\big|_{J(E)}$ restricted to $J(E)$, and the continuity follows.

Now, note that $J(x)=J(y)$ implies $f(x) = f(y)$ for all $f\in E^*$. Since $E^*$ separates points in $E$, it follows that $x=y$ and hence $J$ is injective. We can also see that $J:E\to J(E)$ is surjective.

It remains to show that $J^{-1}:J(E)\to E$ is continuous. Suppose $\{J(x_i)\}_{i\in I}$ is a net in $J(E)$ such that $J(x_i)$ converges to $J(x)$ in weak-star topology. Then for each $f\in E^*$, we have $$ \lim\limits_{i\in I}f(x_i) = \lim\limits_{i\in I}J(x_i)(f) = J(x)(f)=f(x). $$ This shows that $x_i = J^{-1}J(x_i)$ converges weakly to $x=J^{-1}J(x)$ and the continuity of $J^{-1}$ follows. These arguments show that $J:E\to J(E)$ is a homeomorphism.