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Let $(E;\Vert \cdot \Vert )$ be a normed vector space. Show that given $x, y \in E$, $x \neq y$, there exists $f \in E'$ ($E'$ is the bounded dual space of $E$) such that $f(x) \neq f(y)$.

I started trying to show that there always is an injective map on the dual, which would imply this result, but it happens to be case that some bounded dual spaces do not have any injective maps.

Any help would be greatly appreciated. Thanks

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    $\begingroup$ Hahn–Banach Theorem $\endgroup$ – Yiorgos S. Smyrlis Oct 29 '16 at 12:39
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Let $Z$ be the subspace $\{\lambda(x-y)\in E:\lambda\in \mathbb{R}\}$ of $E$, generated by the vector $x-y\neq 0_E$ and consider the functional $f:Z\rightarrow \mathbb{R}:~f(\lambda(x-y))=\lambda\|x-y\|$. Then $f$ is clearly linear and bounded ($Z$ is finite dimensional), so there is a linear bounded extension $\tilde{f}: E\rightarrow \mathbb{R}$ of $f$, according to the Hahn - Banach theorem. Therefore $\tilde{f}\in E'$ and $$\tilde{f}(x)-\tilde{f}(y)=\tilde{f}(x-y)=f(x-y)=\|x-y\|\neq 0.$$

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