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I guess this is probably asked before but I can not find it.

Let $X$ be a Banach space, and let $\ell x = 0$ for all $\ell \in X'$. Then $x = 0$

If all projections $\pi_\alpha x = 0$ and hence get "all coordinates" equal zero. But that probably not hold and I read you need Hahn-banach to prove it.

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up vote 2 down vote accepted

I think you have the right intuition, namely, if $x\neq 0$ you want to construct some functional $\ell\in X'$ such that $\ell x\neq 0$. The Hahn-Banach theorem makes this pretty easy to do.

Assume $x\in X$ is nonzero. Let $Y\subseteq X$ be the $1$-dimensional subspace spanned by $x$. Then define a linear map $L\colon Y\to \mathbb{R}$ by $L(\lambda x) = \lambda$. Here I'm assuming you're working over $\mathbb{R}$, but just replace $\mathbb{R}$ with $\mathbb{C}$ if you're working over $\mathbb{C}$. This $L$ is a continuous linear functional on $Y$, and $L(x) = 1$. The Hahn-Banach theorem tells us that $L$ extends to a continuous linear function $\ell\colon X\to \mathbb{R}$. This gives an $\ell\in X'$ such that $\ell(x) = 1$.

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This was what I was looking fore! great! – Johan Dec 26 '12 at 13:58

Hahn-Banach theorem says that if $X$ is any normed space then there exists $l\in X'$ such that $lx= \|x\|$ and $|lx|\leq \|x\|$ for all $x\in X$.[Rudin Corollary at page 59].

Now if $x\neq 0$, then we can not have $l(x)=0$ for all $l\in X'$.

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An interesting addendum. For general Banach space, the Hahn-Banach theorem, or some other consequene of the Axiom of Choice, is required to prove this. In plain ZF set theory it cannot be proved. A nice example:

$X = l^\infty/c_0$, a fairly concrete Banach space. Certainly $X$ has nonzero elements (for example, the equivalence class of $(1,1,1,\dots)$). But (in only ZF) one cannot show that $X$ has any nonzero linear functionals at all.

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Given $x\in X$ you know that there exist $f\in X'$ (use Hahn Banach Theorem) such that $$f(x)=\|x\|^2$$

Can you conclude?

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But is this linear? – Johan Dec 26 '12 at 13:55
This is not for all $x\in X$. This is for a fixed $x$ and $f$ is linear – Tomás Dec 26 '12 at 13:57

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