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$V$ is a locally convex space. I can't manage to prove that a subspace $M$ is necessarily dense if $\forall f\in V^*(f(M)=\lbrace{0\rbrace} \implies f(V)=\lbrace{0\rbrace})$.

All I have is $\exists x_0 \exists I \subset \mathbb{N} \text{ finite} \quad \exists \varepsilon > 0 \quad | (\cap_I B_i(x_0, \varepsilon))\cap M = \emptyset$ where $B_i$ stands for a "semi"-ball for seminorm $i$

The context is the analytic Hahn-Banach theorem, but I don't really see how that helps. I also have proved that for any $x\neq0$ there exists $g\in V^*$ such that $g(x)=1$, which I thought may be useful but not quite getting there.

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I’ll assume that $V$ is over a field $\Bbb K$, where $\Bbb K=\Bbb R$ or $\Bbb C$.

All I have is $\exists x_0 \exists I \subset \mathbb{N} \text{ finite} \quad \exists \varepsilon > 0 \quad | (\cap_I B_i(x_0, \varepsilon))\cap M = \emptyset$ where $B_i$ stands for a "semi"-ball for seminorm $i$

We can finish your proof as follows. Put $A=\cap_I B_i(x_0, \varepsilon)$. Since the set $A$ is non-empty, open and convex, the set $M$ is non-empty and convex, and the sets $A$ and $B$ are disjoint, by Hahn-Banach separation Theorem there exist a continuous linear map $f : V \to\Bbb K$ and a number $t\in\Bbb R$ such that $\operatorname{Re}f(a) < t\le \operatorname{Re} f(b)$ for all $a\in A$ and $b\in M$. Since $M$ is a subspace, $f(M)=\Bbb K$ or $f(M)=\{0\}$. The first assumption contradicts to the inequality $t\le \operatorname{Re} f(b)$ for all $b\in M$. Thus $f(M)=\{0\}$, whereas $f(a)\ne 0$ for each point $a\in A$.

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