Why is the canonical injection always continuous from $\sigma(E,E^*)$ into $\sigma(E^{**},E^*)$?

Question

I am told that this is because for a fixed $f \in E^*$, $x \mapsto \langle Jx, f \rangle = \langle f, x \rangle$ is continuous wrt $\sigma(E,E^*)$.

Now, I understand this is the case. But why does this imply that the canonical injection $J$ is continuous wrt the topologies $\sigma(E,E^*)$ and $\sigma(E^{**},E^*)$?

Recall: $J: E \to E^{**}$ and $\langle Jx,f\rangle =\langle f,x\rangle$.

Thank you!

Answer 1

$\sigma(E^{**},E^*)$ is the initial topology on $E^{**}$ with respect to evaluations on elements of $E^*$. ("Evaluations" means maps of the form $\kappa \in E^{**}\mapsto \kappa(f),$ for fixed $f\in E^*$.) This means that a linear map from any topological vector space into $(E^{**},\sigma(E^{**},E^*))$ is continuous if and only if its compositions with all evaluations are continuous. In particular $J:(E,\sigma(E,E^*))\to (E^{**},\sigma(E^{**},E^*))$ is continuous iff for every $f\in E^*,$ $x\in E\mapsto J(x)(f)=f(x)$ is continuous from $(E,\sigma(E,E^*))$ to the base field. But it is continuous, because functionals in $E^*$ are $\sigma(E,E^*)$-continuous.

Answer 2

1)The weak topology $\sigma(E,E^*)$ on $E$ is defined as the weakest topology on $E$ which makes continuous the map $f:E\to\mathbb{R},$ for any $f\in E^*.$

2)The weak-* topology $\sigma(E^{**},E^*)$ on $E^{**}$ is by definition the weakest topology on $E^{**}$ which make continuous the map $J^*(f):\xi\in E^{**}\to\langle\xi,f\rangle\in\mathbb{R}$ for any $f\in E^*.$
Here I am considering the canonical map $J^*:E^*\to E^{***}$, starting on $E^*.$

3)A characteristic property of weak-* topology is that, for any topological space $(X,\tau),$ we have:
$F:(X,\tau)\to(E^{**},\sigma(E^{**},E^*))$ is continuous iff $J^*(f)\circ F:(X,\tau)\to\mathbb{R}$ is continuous, $\forall f\in E^*.$
Here and below I consider $\mathbb{R}$ with its usual topology.

4)By point 3) the canonical map $J:(E,\sigma(E,E^*))\to(E^{**},\sigma(E^{**},E^*))$ is continuous iff $J^*(f)\circ J:(E,\sigma(E,E^*))\to\mathbb{R}$ is continuous, $\forall f\in E^*.$

5)As you have noted, for any $f\in E^*,$ the map $J^*(f)\circ J:x\in E\to(J^*(f)\circ J)(x)\in\mathbb{R}$ coincides with $f:E\to\mathbb{R}$.

Finally by points 1), 4) and 5) you get the continuity of $J:(E,\sigma(E,E^*))\to(E^{**},\sigma(E^{**},E^*)).$

P.S. Certainly I could be more succint but I would highlight the single point or the argument. I hope it helps.