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In a previous post, I proved that

Let $E$ be a locally convex t.v.s. and $f \not \equiv 0$ a linear functional on $E$. Then $\ker f$ is either closed or dense in $E$.

Then I have found a proof without using Hahn-Banach theorem. However, this proof is for n.v.s. Below, I try to adopt that proof to a general t.v.s.

Theorem: Let $E$ be a t.v.s. and $f \not \equiv 0$ a linear functional on $E$. Then $\ker f$ is either closed or dense in $E$.

Could you please have a check on my attempt?

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Analyst
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1 Answers1

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Assume that $\ker f$ is not closed. Then there is a net $(x_d)_{d \in D}$ in $\ker f$ and $a \notin \ker f$ such that $x_d \to a$. Let $x'_d := x_d-a$. Then $x'_d \to 0$ and $f(x'_d) = f(x_d)-f(a)=f(a) \neq 0$ for all $d\in D$. Let $y_d : = x'_d / f(a)$. Then $y_d \to 0$ and $f(y_d) = 1$ for all $d \in D$.

Given $y \in E$, let $y'_d := y- y_d f(y)$. By linearity of $f$, we have $f(y'_d) = 0$ and thus $y'_d \in \ker f$ for all $d\in D$. Moreover, $y'_d \to x$. This completes the proof.

Analyst
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