If a linear functional is not bounded, then it has a non-closed kernel. In this answer, the OP gave a proof to the proposition that if a linear functional $f$ is not bounded, then it has a non-closed kernel. However, he only showed that if $f$ is not bounded, then a unbounded sequence exists in the kernel, which doesn't necessarily prove that the kernel is non-closed, since closed sets obviously can have unbounded sequences. Am I wrong? If that proof was wrong, then what's the correct proof of this proposition?
 A: Assuming we are working with normed spaces here, pick a fixed vector $v$ with $f(v)=1$. If $f$ is not bounded, there is a sequence $(x_n)$ with $x_n\to0$ and $f(x_n)=1$. Thus $v-x_n\in\operatorname{ker} f$. But $v-x_n\to v\notin\operatorname{ker}f$, so $\operatorname{ker}f$ is not closed.
Addendum: The answer is already accepted, but for future reference, here is how to generalize the above to a topological vector space. It requires a little lemma, which is not hard to prove (I leave the proof to the reader): If $f$ is an unbounded linear functional, then every neighbourhood of the origin contains some $x$ with $f(x)=1$. Assuming this, as in the above case start with a fixed vector $v$ with $f(v)=1$. If $U$ is a neighbourhood of $v$, then $v-U$ is a neighbourhood of $0$, so it contains some $x$ with $f(x)=1$. Thus $v-x\in U\cap\operatorname{ker}f$, and we're done.
A: For a linear map between normed spaces, boundedness is equivalent to continuity at $0$, which is equivalent to continuity everywhere.
If $f\colon V\to\mathbb{R}$ (or $\mathbb{C}$) is a linear map and $f$ is continuous (that is, bounded), then $\ker f$ is the inverse image of $\{0\}$, so it's closed.
If $\ker f$ is closed, then $f$ induces a map from $V/\ker f$ (which is a normed space) to $\mathbb{R}$. If $f$ is nonconstant (otherwise the result is trivial), then the induced map is an isomorphism and it is continuous because such is any linear map between finite dimensional normed spaces. Therefore $f$ is the composition of a continuous linear map with the projection $V\to V/\ker f$, which is continuous by definition.
A: A correct proof is given by Beni Bogosel in his blog.
