# Prove for a linear functional with respect to weak topology $\forall_{x\in V}\;\;\; \langle f,x\rangle <\alpha$

Let $\mathbb E$ be a Banach space and $f:\mathbb E\to \mathbb{R}$ a linear functional. Show that $f$ is continuous with respect to $\mathbb E$ being equipped with the weak topology $\sigma(\mathbb E,\mathbb E^*)$ if and only if there exists weakly-open set $V\subset \mathbb E$ such that $0\in V$ and for some $\alpha\in \mathbb{R}$ $$\forall_{x\in V}\;\;\; \langle f,x\rangle <\alpha$$

I am thinking as follows: Since $V\in\mathbb{E}$ is open then $V^c$ is closed and consider a point $x\in V$. Since $\{x\}$ is compact and $V^c$ is open with respect to weak topology therefore open with respect to strong topology, then by Hanh Banach Geometric version 2 theorem there exists a closed hyperplane that strictly separates them. Therefore for $\alpha\in\mathbb{R}$ we get $$\forall_{x\in V}\;\;\; \langle f,x\rangle <\alpha$$

I am not sure whether I am right or wrong. Any help would be great. Thanks in advance.

$V$ is weakly open so there are $f_1,f_2,\cdots, f_n\in \mathbb E^*$ such that $\cap^{n}_{i=1}\left \{x: |f_i(x)|<r_i \right \}$ is open and contained in $V.$ Let $N=\cap^{n}_{i=1}$ker$f_i.$ Then, $N\subseteq V$ and if $x\in N$ then so is $tx$ for every $t\in \mathbb R^+.$ Therefore, $|f(tx)|<\alpha\Rightarrow |f(x)|<\alpha/t\Rightarrow f(x)=0,$ which implies now that $f=\sum_{i=1}^{n}c_if_i$ for some scalars $c_i$.
On the other hand, if $f$ is continuous then we may take $V=f^{-1}((-\infty,\alpha)).$