# 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?

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I think the answer might have confused the definitions of closure and compactness – Omnomnomnom Dec 11 '13 at 21:18

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.

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Thanks. And why exactly we can find such sequence $x_n$? – Voldemort Dec 11 '13 at 21:53
Oh, I get it. It's because according to one of the definitions of unbounded linear operator, there exists $y_n$, with $\|y_n\|=1$ but $|f(y_n)|\to\infty$, now let $x_n=y_n/|f(y_n)|$ and we get what we want. – Voldemort Dec 11 '13 at 23:13

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.

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Why is $V/\ker f$ a finite dimensional space? – Voldemort Dec 11 '13 at 22:06
@Voldemort Because it's isomorphic to $\mathbb{R}$. – egreg Dec 11 '13 at 22:07
Btw, this answer was also given to the original question. – Carsten S Dec 12 '13 at 8:50
@egreg why is it isomorphic to $\mathbb R$ and not $\operatorname{Im}f$? – Arrow Jul 11 at 13:45
@Arrow The image of $f$ is $\mathbb{R}$, because $f$ is unbounded, so the image cannot be $\{0\}$. – egreg Jul 11 at 13:54

A correct proof is given by Beni Bogosel in his blog.

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