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I'm trying to prove below result.

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 $E$.

The space $E$ is assumed to be locally convex. However, another statement "$\ker f$ is closed iff $f$ is continuous" holds without such assumption. I would like to ask if

  • Above theorem holds without local convexity.

  • My proof contains some mistakes.

Thank you so much!

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 $H :=\ker f$ is not dense in $E$. Then there exist $a \in E$ and a convex open neighborhood $U$ of $a$ such that $U \subseteq H^c$. WLOG, we assume $f(a)>0$. By Hahn-Banach theorem, there exist $r \in \mathbb R$ and a continuous linear functional $g$ such that $$ g(x) < r \le g(y) \quad \forall (x,y) \in H \times U. $$

Because $H$ is a non-trivial vector space, $g(x) = 0$ for all $x\in H$. This implies $r>0$ and thus $g(a) > 0$. Let $(x_d)$ be a net in $H$ such that $x_d \to b \in E$.

Assume the contrary that $b \notin H$. WLOG, we assume $f(b) < 0$. Then there is $\lambda \in (0, 1)$ such that $c := \lambda a+(1-\lambda)b \in H$. Clearly, $x'_d := \lambda a+(1-\lambda)x_d \to c$. Then $$ 0 <\lambda g(a) =g(x'_d) \to g(c) =0, $$ which is a contradiction. Hence $b \in H$ and thus $H$ is closed. This completes the proof.

Analyst
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