Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

share|improve this answer
    
It works even on the entire plane. –  002 Jan 8 '10 at 2:00
    
@Leonid, sure :) –  Anton Petrunin Jan 8 '10 at 2:06
    
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|). –  002 Jan 8 '10 at 3:54
    
@Leonid, no, you can not (by the same example). –  Anton Petrunin Jan 8 '10 at 5:53
    
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. –  Anonymous Jan 8 '10 at 13:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.