TITLE: If a linear functional is not bounded, then it has a non-closed kernel.
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: 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.