# Strict convexity and uniqueness of functionals

Is it true that if $x$ is a norm-one vector in a strictly convex Banach space then there exists a unique bounded linear functional $f$ on that space such that $f(x)=1=\|f\|$?

It seems unlikely to me but I've seen it mentioned without a proof.

It's false. For a strictly convex norm, given $f$, $x$ is unique (if it exists); the dual notion, where, given $x$, $f$ is unique, is called "smoothness" in convex geometry.
For a concrete example, take $\mathbb{R}^2$ with the norm whose unit ball is the intersection of two disks of radius $2$, centred at $(0,\pm 1)$. The boundary of the unit ball contains no line segments, so it's strictly convex, but it has pointy bits, where there are multiple supporting functionals.