1
$\begingroup$

Let $f: I\to \mathbb{R}$ be a convex function.

Why does the second derivative of $f$ exist almost everywhere?

By searching I knew that is Alexandrov theorem, but I didn't find the proof...

My try

For $x_1<x<x_2$ where function is defined , we have $$\frac{f(x)-f(x_1)}{x-x_1}\le\frac{f(x_2)-f(x_1)}{x_2-x_1}\le\frac{f(x_2)-f(x)}{x_2-x}$$ Considering $x_1\to x^-$ and $x_2\to x^+$, we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)\le f_+'(x)$. (in this process we only need $\frac{f(x)-f(x_1)}{x-x_1}\le\frac{f(x_2)-f(x)}{x_2-x}$)

Considering $x\to x_1$ and $x\to x_2$, we can conclude that $$f_+'(x_1)\le \frac{f(x_2)-f(x_1)}{x_2-x_1} \le f_-'(x_2)$$

So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)\le f_+'(x)$.

Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.

How to move on? Any hints? Thank you in advance!

$\endgroup$
  • $\begingroup$ I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere. $\endgroup$ – SmileyCraft Dec 24 '18 at 15:29
  • $\begingroup$ @SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that... $\endgroup$ – Zero Dec 24 '18 at 15:50
  • 2
    $\begingroup$ Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable. $\endgroup$ – SmileyCraft Dec 24 '18 at 15:55
  • $\begingroup$ @SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion $\endgroup$ – Zero Dec 24 '18 at 16:04
  • $\begingroup$ This is not correct , convex functions can be nondifferentiable on a dense subset of $R$. $\endgroup$ – Red shoes Dec 25 '18 at 2:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.