A linear form $f$ is continuous respect to weak$^{\star}$ topology if and only if there exists $x \in E$ such that $f(x')=x'(x)$ for all $x'\in E'$.

Question

Let $E$ normed vector space. Show that a linear form $f:(E',\sigma(E',E))\rightarrow \mathbb{K}$ is continuous if and only if there exists $x \in E$ such that $f(x')=x'(x)$ for all $x'\in E'$.

Remark: Here $E'$ is the dual space of $E$, and $\sigma(E',E)$ is the weak$^{\star}$ topology.

The problem: The direction $\Leftarrow)$ is immediate of the definition of $\sigma(E',E)$, my problem is the direction $\Rightarrow)$, note that this dirrection can be reduced to show that if $f$ is continuous with respect to the topology $\sigma(E',E)$, then $f\in J(E)$ where $J$ is the canonical application $$\begin{array}{rclrcl} J:E &\longrightarrow & E'' \\ x &\longmapsto& J(x):&E' &\rightarrow & \mathbb{K}\\ &&&x' &\mapsto & x'(x). \end{array}$$

Note that if $E$ is reflexive then the result is trivial, but $E$ is any normed vector space.

Answer 1

Suppose $f:(E',\sigma(E',E))\to \mathbb{K}$ is continuous, then there exists some $x_1,\cdots, x_m\in E$ such that $\lvert f(x')\rvert \leq \sum_{i=1}^m \lvert x_i(x')\rvert\forall x'\in X'$. Hence $\ker f\subset \cap_{i=1}^m\ker x_i$, thus there exists some scalars $\alpha_1,\cdots, \alpha_m$ such that $ f=\sum_{i=1}^m \alpha_i x_i\in X$.