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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.

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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 $f(x,y)=\sqrt{1+x^2}-\sqrt{1+y^2}$.

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  • $\begingroup$ It works even on the entire plane. $\endgroup$
    – 002
    Commented Jan 8, 2010 at 2:00
  • $\begingroup$ @Leonid, sure :) $\endgroup$
    – Anton Petrunin
    Commented Jan 8, 2010 at 2:06
  • $\begingroup$ Um, it's not Lipschitz then. Maybe the stated question can be salvaged by allowing an error term such as O(|y-x|^2)... Under given conditions f is known to be Frechet differentiable on a dense set (Preiss), but I don't know if one can make the error of linear approximation anything less than o(|y-x|). $\endgroup$
    – 002
    Commented Jan 8, 2010 at 3:54
  • $\begingroup$ @Leonid, no, you can not (by the same example). $\endgroup$
    – Anton Petrunin
    Commented Jan 8, 2010 at 5:53
  • $\begingroup$ Anton, as it was stated the question your example is correct. The question was motivated by the concept of $\varepsilon$-Frechet differentiability (and the error of linear approximation); as you pointed out it has to be reformulated. $\endgroup$
    – Anonymous
    Commented Jan 8, 2010 at 13:03

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