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?
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$\begingroup$ 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). $\endgroup$– Włodzimierz HolsztyńskiCommented May 10, 2013 at 22:08
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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).
answered May 10, 2013 at 5:08
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$\begingroup$ The OP was basically told this already in a deleted answer. It is far past the time that this thread should have been closed. $\endgroup$ Commented May 10, 2013 at 5:39
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$\begingroup$ I am sorry, I didn't see the deleted answers. $\endgroup$ Commented May 10, 2013 at 5:49