$\ell x = 0$ for all $\ell \in X'$ for banach spaces 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.
 A: 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 $\ell(x) = 1$.
A: 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 $l\in X'$. 
A: 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.
A: 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?
