Let X be a separable reflexive Banach space and f:X\to\mathbb{R} be a
Lipschitz function. Say that a point x in X is a *local supporting point
of* f if there exist x^\* in X^\* and an open neighborhood U of x
such that either x^\* (y-x)\leq f(y)-f(x) for all y in U or
x^\* (y-x)\geq f(y)-f(x) for all y in U.


Question: is true that the set of local supporting points of f is dense in X?


This question is obviously related to differentiability; it might be difficult. 


I would be very much interested to know whether it has been asked by others.



Local supporting points of Lipschitz functions
My guess is that you did not formulate question correctly --- in the present form the answer is NO.


One can take strictly saddle $f$ on $\mathbb R^2$, say $f(x,y)=\sqrt{1+x^2}-\sqrt{1+y^2}$.


