Let $E$ be a Banach space and let $F(y)$ denote the duality map: $$F(y)=\{y^*\in E^*/ \langle y^*,y\rangle =\|y\|^2=\|y^*\|^2\}$$ where $E^*$ is the dual space of $E$.

Are there any sufficient condition for $F$ to be locally Lipschitz apart the case where $E$ is a Hilbert space?


Global Lipschitz continuity

$F$ is globally Lipschitz if and only if the norm of $E$ is uniformly smooth with quadratic modulus of smoothness, meaning that there exists a constant $k>0$ such that the inequality $$ \|x+y\|+\|x-y\|\le 2+k\|y\|^2 $$ holds for all unit vectors $x$ and all $y\in E$. This is an old result which one can usually find in books dealing with moduli of smoothness and convexity. For a proof, see the paper by M. Zemek, Strong monotonicity and Lipschitz-continuity of the duality mapping, Acta Universitatis Carolinae. Vol. 32 (1991), No. 2, 61-64.

For example, the spaces $L^p$ and $\ell^p$ have quadratic modulus of smoothness for $2\le p<\infty$.

Local vs global Lipschitz continuity

If the duality map $F:E\to E^*$ is locally Lipschitz on $E$, then it is globally Lipschitz on $E$. Indeed, being locally Lipschitz implies being Lipschitz in some neighborhood of $0$; in particular, on some sphere $S_r = \{x: \|x\| = 1\}$. By construction, $F$ is degree $1$ homogeneous: $F(tx)=tF(x)$ for $t>0$. It is an elementary exercise to prove that a degree $1$ homogeneous map is globally Lipschitz if and only if it is Lipschitz on some sphere $S_r$.

Local vs global Lipschitz continuity, on a sphere

A more interesting question is to ask when $F$ is locally Lipschitz on the unit sphere $S_1$. Now the preceding argument with a neighborhood of zero doesn't apply, and it a duality map can be locally Lipschitz on $S_1$ without being globally Lipschitz. I suggest looking up the literature on locally uniformly smooth Banach spaces: it should be true that $F$ is locally Lipschitz on $S_1$ iff the norm has a quadratic modulus of smoothness in some neighborhood of every point of $S_1$, with multiplicative constant $k$ depending in the point.


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