I want to show the following: Let $X$ be a topological vector space (and Hausdorff), $f:X \rightarrow \mathbb{K}$ a linear functional. Then the following are equivalent:
1.$f$ is continuous
2.$f$ is continuous at $0$
3.$ker(f)$ is closed
From 1 $\Rightarrow$ 2 is obviuos. So let's start with 2 $\Rightarrow$ 3. Let $f$ be continuous at $0$. We have to show that $ker(f)$ is closed. $ker(f)=\{x \in X: f(x)=0\}=f^{-1}(\{0\})$. Since $\{0\}$ is closed in $\mathbb{K}$ and $f$ is continuous in $0$, $f^{-1}(\{0\})=ker(f)$ is closed.
I do know the following: Let $X_1,X_2$ be topological spaces, $g:X_1 \rightarrow X_2$. $g$ is continuous iff for every closed subset $S \subseteq X_2$, the set $g^{-1}(S)$ is closed.
Question 1: Since the lin. functional $f$ is assumed to be only continuous at $0$, can I still use the above theorem to conclude that $ker(f)$ is closed?
Question 2: I do not really know how to conclude 3 $\Rightarrow$ 1. The only thing that I can think of are the characterizations of continuous maps between topological spaces. I.e. preimage of open sets are open and preimage of closed sets are closed. I assume that I need to combine one of them with the fact that $f$ is a functional, i.e. maps to $\mathbb{K}$. But I can't figure out how to do this part.
Edit: $2 \Rightarrow 1$: Since in topological spaces a map $f: X \rightarrow Y$ is continuous in $x$ if: for every neighborhood $V$ of $f(x)$ there exists a neighborhood $U$ of $x$ such that $f(U) \subseteq V$.
Let $x \in X$ be arbitrary. To verify that $f$ is continuous at $x$, we need to show that there exists a neighborhood $U$ of $x$ s.t. for a neighborhood $V$ of $f(x)$: $U$ $\subseteq$ $f(V)$.
So let $V$ be a neighborhood of $f(x)$, then by translation invariance $V-f(x)$ is a neighborhood of $0_{\mathbb{K}}$. Since $f$ is linear it maps $0_X$ to $0_{\mathbb{K}}$, now by the definition of continuity (in $0_X$) we can find a neighborhood $U$ of $0$ s.t. $f(U) \subseteq V-f(x)$. Now $U+x$ is a neighborhood of $x$, and using the linearity again $f(x+U)=f(x)+f(U) \subseteq V $, since $f(U)\subseteq V-f(x)$.
So the only thing remaining would be to either show that $3 \Rightarrow 1$ or $3 \Rightarrow 2$.
