0

Let $K$ be a real symmetric function about zero, Lipschitz continuous with compact support (say $supp K=[-1,1]$). Then $K$ is differentiable?

My idea is that it could be non-differentiable only on the boundaries of the support (i.e., on $\{-1,1\}$) and on the points on the non-differentiability of $K$. For short, it should be differentiable almost everywhere.

Celine Harumi
  • 2,821
  • 2
    What about $f(x) = |x|,$ or am I overlooking something? Also, regarding "on the points on the discontinuity of $K$", Lipschitz continuous functions have no points of discontinuity. – Dave L. Renfro Mar 19 '20 at 17:51
  • @DaveL.Renfro I think you are right. $K$ need not to be differentiable on its domain. And yes: "on the points o non-differentiability of K". I would say that these points were finite, but maybe Cantor set could break this argument. – Celine Harumi Mar 19 '20 at 17:53
  • 1
    The "symmetric about zero" condition is not significant, since one can simply reflect about the origin any appropriately pathological function defined on $[0,1]$ (with value $0$ at $x=0).$ As for "pathological", there exist Lipschitz functions defined on $[0,1]$ that additionally have a symmetric derivative at each point, but which fail to have a finite ordinary derivative on a set whose nonempty intersection with every open interval has Hausdorff dimension $1.$ See the 4th paragraph in this answer. – Dave L. Renfro Mar 19 '20 at 18:19
  • Thanks for the answer! This question leads me to another one (which is my real interest): https://math.stackexchange.com/questions/3587101/necessary-condition-verification-for-integrability-of-a-function-real-analysis – Celine Harumi Mar 19 '20 at 18:53

0 Answers0