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Not too sure where to start. Seems like it should be an easy adaptation of a theorem, but have't been able to find it in my notes. Thank you!

Kristaps
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1 Answers1

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Consider the subspace $Z=\operatorname{span}(x-y)$. Now let $f\in Z'$ be defined by $f(\alpha(x-y))=\alpha\|x-y\|$. Note that the norm $\|\cdot\|$ is a sublinear functional on $X$, and on $Z$ we have $\|\alpha(x-y)\|=|\alpha|\|x-y\|\geq f(\alpha(x-y)).$ Thus, on $Z$, $f$ is dominated by the norm, so by Hahn-Banach we can extend $f$ to some $f'\in X'$ such that $f|_Z=f'|_Z$. In particular $f(x-y)=f'(x-y)$, but by assumption $f'(x-y)=0$. Thus $\|x-y\|=0$, so $x=y$.

K.Power
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