# Duality map in strictly convex Banach spaces

Say $(B,\|\cdot\|)$ is a finite dimensional, strictly convex Banach space. Is it true that the map $\phi:B^*\rightarrow B$ which takes a linear functional $f$ with $\|f\|=1$ into the unique unit norm vector $u$ such that $f(u)=1$ is Lipschitz?

• This is a nice question, even if not too difficult: it's pure, simple, meaningful, self-contained, complete (not technical or auxiliary like some other questions). – Włodzimierz Holsztyński May 10 '13 at 22:08

There are $n$-dim Banach spaces for which $\phi$ is not Lipschitz in every dimension $n\ge 2$.
It is enough to provide an example in dimension $2$. The higher dimensional examples are obtained by rotating the lower dimensional examples.
In dimension $2$ the required norm, in $\mathbb R^2$, can be given by:
$$\|(x\ y)\| := (x^4+y^4)^{\frac 14}$$