$\varphi(x)=0$ for every continuous linear form implies that $x=0$

Question

Let $(E,||~~||)$ a normed vector space and $x\in E$.

I want to show show that $$\varphi(x)=0~~\forall\varphi\in E'\Rightarrow~~x=0$$

Thank you in advance.

E is a vector space over $\mathbb{K}$, and E' is the space of continuous linear maps from E to $\mathbb{K}$, — Donnie Darko, Sep 28 '20 at 20:54
Should the proposition you want to prove be restated as follows?
$$\exists A \subseteq E : x \in A \wedge \phi(x) = 0 ~ \forall \phi \in E' \Rightarrow 0 \in A$$ — the_candyman, Sep 28 '20 at 21:06
@DonnieDarko For the future: usually you should write about your own attempts of solving the problem in your question. — supinf, Sep 28 '20 at 21:25
@supinf that what i do usualy, but this one it looks trivial but i cant prove it, — Donnie Darko, Sep 28 '20 at 21:33

score 1 · Accepted Answer · answered Sep 28 '20 at 21:16

1

This is usually done using the Hahn-Banach Theorem.

Some further hints: You start by defining a suitable linear functional on the linear hull that is spanned by $x$ (under the assumption that $x\neq 0$). Then you can extend this linear functional to a functional in $E'$.

answered Sep 28 '20 at 21:16

supinf

13,433

1

Yes sir .if we suppose that $x\neq 0$ then by a corollary of Hahn-Banach theorem there exists a continuous linear form $\phi$ such that $\phi(x)=||x||$ and $||\phi||=1$ hence the contradiction because $\phi(x)\neq 0$ – Donnie Darko Sep 28 '20 at 22:09

$\varphi(x)=0$ for every continuous linear form implies that $x=0$

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