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?

$\begingroup$ This is a nice question, even if not too difficult: it's pure, simple, meaningful, selfcontained, complete (not technical or auxiliary like some other questions). $\endgroup$ – 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}$$
(I am willing to provide the details if asked toit's a simple matter).

$\begingroup$ The OP was basically told this already in a deleted answer. It is far past the time that this thread should have been closed. $\endgroup$ – Bill Johnson May 10 '13 at 5:39

$\begingroup$ I am sorry, I didn't see the deleted answers. $\endgroup$ – Włodzimierz Holsztyński May 10 '13 at 5:49