# Bounds on the second derivative of a convex function.

Edit: This question was answered in https://mathoverflow.net/questions/265256/how-bad-can-the-second-derivative-of-a-convex-function-be

Let $$a and suppose that $$f:(a,b)\to \mathbb{R}$$ is a increasing convex function. From the Alexandrov’s theorem, $$f$$ is twice differentiable almost everywhere. My question is the following

Does $$f''\in L_{loc}^1(a,b)$$?

If the answer to the previous question is negative, consider the set

$$A=\{x\in (a,b):\ \mbox{there is no}\ \delta>0,\ \mbox{such that}\ f\in L^1(x-\delta,x+\delta)\}.$$

What is the measure of $$A$$?

No ideas up to now. Any reference is appreciated.

For an accessible reference, take a look here.

• I know that this is an old question, but I would like to drop a comment. I guess that we have something like $f'(d) - f'(c) \ge \int_c^d f''(x) \, \mathrm{d}x \ge 0$ for $[c,d] \subset (a,b)$. This should give your assertion. – gerw Nov 22 '18 at 12:36
• Or speaking differently: The second derivative of a convex function can be identified with a positive distribution, i.e., a measure. Then, your $f''$ should be the absolutely continuous part of this measure. Again, it belongs to $L^1_{\text{loc}}(a,b)$. – gerw Nov 22 '18 at 12:36
• @gerw take a look in mathoverflow.net/questions/265256/… – Tomás Dec 3 '18 at 20:01
• I didn't know that this question was crossposted. You should link both questions. – gerw Dec 3 '18 at 20:21