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I know that the second derivative of a convex function exists almost everywhere (Alexandrov's theorem).

And I know that the first derivative exists everywhere except countably many points (This question).

So, does the second derivative exist everywhere except countably many points? I'd appreciate a source I can cite, or a counterexample.

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If $g(t)$ is a nondecreasing real-valued function, then $f(x)=\int_a^x g(t)\,dt$ is convex. If $g$ is also continuous, then $f'(x)=g(x)$ everywhere. If $g(t)$ is the Cantor function, then $g$ is nondifferentiable at uncountably many points.

Gap: I do not have a reference or proof handy for showing that $g$ is nondifferentiable at uncountably many points in the Cantor set.

For a reference on the nondifferentiability set of the Cantor function, see Darst's "The Hausdorff Dimension of the Nondifferentiability Set of the Cantor Function is $[\ln(2)/\ln(3)]^2$" (JSTOR link), which I found in the references on the Wikipedia page linked above.

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  • $\begingroup$ I see "integrable" is redundant there. Here is a reference: math.stackexchange.com/questions/1318407/… In the question it is assumed that the integrand is differentiable, but in the accepted answer this is not used. Continuity of the integrand is not even used. $\endgroup$ – Jonas Meyer Mar 20 '17 at 12:55
  • $\begingroup$ I think continuity of the integrand is used: "$F(x)$ is continuous due to the continuity of $f(x)$." $\endgroup$ – David K Mar 20 '17 at 13:00
  • $\begingroup$ @DavidK: It is cited unnecessarily, as I have commented there now. The indefinite integral of a locally Lebesgue integrable function is absolutely continuous. The indefinite integral of a locally bounded integrable function is locally Lipschitz. Every nondecreasing function is continuous a.e.. And $g$ in this case is continuous anyway. Note that at that answer they do not indicate why continuity of $f$ implies continuity of $F$; if one were to work it out I believe local boundedness of $f$ would be a more natural property to use, and that follows from monotonicity. $\endgroup$ – Jonas Meyer Mar 20 '17 at 13:03

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