I read that any continuous function can be represented as a sum of convex and concave function, meaning for all $f(x)$, $f(x) = g(x) + h(x)$ where $g$ is convex and $h$ is concave.
There could be infinitely many decompositions of that sort.
Anyone knows where I can see a proof or this, or knows of a proof of this?
Thanks.
Best Answer
You need more conditions than even absolute continuity. A characterisation of continuous, convex functions defined on open intervals is that they must be the indefinite integral of a monotonically non-decreasing function. (See here for example.) In particular, this means that continuous, convex functions are absolutely continuous.
So if you take any function that is continuous and not absolutely continuous (Cantor's stair case comes to mind), it cannot be decomposed as a sum of a convex and a concave function.
What's more important is that a convex function, by Alexandrov's theorem, must have a second derivative almost everywhere. Therefore your initial function must be even better than just absolutely continuous, it needs to also admit almost everywhere second derivatives.
And even assuming almost everywhere second derivatives is not enough, if you take the function $x \sin(1/x)$ which is analytic away from the origin, you cannot represent it as the difference of two convex functions.
Of course, there is still a gap between twice almost everywhere differentiable and David Speyer's everywhere twice-continuously differentiable condition. I am not entirely sure where the correct boundary lies, or if there is a correct boundary using just classical differentiability notions.