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, self-contained, complete (not technical or auxiliary like some other questions). $\endgroup$ 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 to--it'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. $\endgroup$ May 10 '13 at 5:39
  • $\begingroup$ I am sorry, I didn't see the deleted answers. $\endgroup$ May 10 '13 at 5:49

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.