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.

Intersection of two disks


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.