I prove the following result:
Let a Fréchet space $X$ be with an increasing set of seminorms $p_1,p_2,\ldots$ and $E\subseteq X$ be a subspace. $E$ satisfies the condition that $\ker f=X$ whenever $f\in X^*$, $E\subseteq \ker f$. Here $\ker$ denotes the null space of a linear operator and $X^*$ is the continuous dual of $X$. The topology on $X$ is generated by open balls of all radius in all seminorms $p_m$. Then, $E$ is dense in $X$.
Proof. Suppose not. Then there exists $x_0\in X, m\in \mathbb Z, \epsilon>0$ such that the ball $B(x_0,m;\epsilon)=\{\xi\in X:p_m(x-\xi)<\epsilon\}$ is disjoint from $E$. Consequently, $B(x_0,n;\epsilon)\cap E=\emptyset$ for all $n\geq m$.
Define $\tilde p_n(x)=\sup_{y\in E} p_n(x-y)$. It's easy to verify that $\tilde p_n$ is a seminorm. Define in the one-dimensional subspace spanned by $x_0$ the linear functional $f_n$ by $f_n(kx_0)=k\tilde p_n(x_0)$. By Hahn-Banach theorem, there exists an extension to $X^*$, such that $|f_n(x)|\leq \tilde p_n(x)$ for all $x\in X$. In particular, $\tilde p_m(y)=0$ for all $y\in E$, so $f(y)=0$ for all $y\in E$. Contradiction. $\square$
Note that the space $X$ doesn't even have to be Hausdorff. Is my proof correct?
Now, I am thinking about the following things. Let $X$ be a general topological vector space, and let $E$ be a subspace such that whenever $f\in X^*$ and $E\subseteq \ker f$, $\ker f=X$.
- If $X$ is Hausdorff, is it always the case that $E$ is dense in $X$?
- It seems that it is not even necessary for $X$ to be Hausdorff. Actually, even if the topology is trivial, we may conclude $E$ is dense. If we have the discrete topology on $X$, the every linear functional is continuous, so we may also conclude that $E$ is dense.
In the end, under what circumstances is the condition "whenever $f\in X^*$ and $E\subseteq \ker f$, $\ker f=X$" sufficient for $E$ to be dense?