Proof of : $\cap_{f\in E'}Ker(f)= \left\{0 \right\}$

Question

Let $(E,N)$ be a normed vector space. The dimension of $E$ could be infinite.

Let $E'= \left\{f:E \rightarrow \mathbb{K} ,\quad f \quad linear \quad and \quad continuous \right\}$.

do we have $\cap_{f\in E'}Ker(f)= \left\{0 \right\}$ ?

Answer 1

Yes. That's a consequence of the Hahn-Banach theorem. Take $v\in E\setminus\{0\}$. Consider the linear form $\alpha\colon\mathbb{R}v\longrightarrow\mathbb R$ defined by $\alpha(\lambda v)=\lambda$. Then $\alpha$ is continuous and therefore you can extend it to an element $f\in E'$. But then $f(v)=\alpha(v)=1\neq0$. So, $v\notin\ker f$.

Answer 2

If $\mathbb K = \mathbb R$ or $\mathbb C$, then this is True by Hahn-Banach Theorem. I am not certain but you have more general version of Hahn-Banach for other fields such as p-adic fields.