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Let $f: X\to \mathbb R$ be an unbounded linear functional, where $X$ is a Banach space. Then $\ker f$ is dense in $X$.

I tried to prove contrapositive of the statement.

Suppose that $\ker f$ is not dense in $X$. Then there exists a $p\in X$ not in closure of $\ker f, f(p)$ is non zero. It follows that range $f= \mathbb R.p$.

But I got stuck while trying to prove that $f$ should be bounded:

$|f(x)|=\|r_x.p\|$ for some $r\in \mathbb R$ for all $x: \|x\|=1$. From here, $|f(x)|\le |r_x|\|p\|$. Not sure, how to bound $r_x$ from here.

Koro
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    does this help: https://math.stackexchange.com/questions/123282/linear-functional-on-a-banach-space-is-discontinuous-then-its-nullspace-is-dense – daw Feb 26 '23 at 20:06
  • @daw: yes, it does. Thank you so much :-). – Koro Feb 26 '23 at 20:08
  • @Koro A linear functional is bounded iff kernel is closed iff kernel is not dense. – SoG Feb 27 '23 at 18:46

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