I am reading Folland's Real Analysis book and there is a problem which says a linear functional $f$ is bounded if and only if the kernel of $f$ is closed. One way is trivial since the inverse image of a closed set is closed since $f$ is bounded and hence continuous. For the other direction I was able to explicitly construct a sequence in the the kernel that would converge outside the kernel if $f$ was not bounded. Is it possible to make a more general statement regarding this? Any advice or comment is very much appreciated. Sorry if it is a stupid question.
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Yep. Choose $v \in V$ such that $f(v) = \lambda$. Then
$$ f^{-1}(\{ \lambda \}) = v + \ker f. $$
If this set is closed then $\ker f$ is also closed because $\ker f$ is the image of $f^{-1}(\{ \lambda \})$ under the translation map $w \mapsto w - v$ which is a homeomorphism of $V$ and so $f$ is continuous.
