Concave-Convex Decomposition of a continuous function

Question

In this paper http://www.stat.ucla.edu/~yuille/pubs/optimize_papers/cccp_nips01.pdf they have a theorem that says that a twice continuous differentiable function (an energy function, they say) with bounded Hessian can always be decomposed as a sum of a concave part and convex part. However, I am not able to see how one can decompose sin[x] in the interval $[0,4\pi]$. I have come up with a maybe incorrect solution $sin[x]=fcave+fvex$, with $fcave=1000\pi x^2$ and $fvex=sin[x]-1000\pi x^2$, with $fvex''=-sin[x]-2000\pi$ that is negative in all the reals. Is this correct, I am not convinced yet, can you tell me if sin[x] can be decomposed in this way indeed.

Answer 1

For sure: $$ \sin x = \sin^+(x)+\sin^-(x) = \max(0,\sin x)+\min(0,\sin x) \tag{1}$$ and now we just have to integrate twice the previous line to decompose $\sin x$ as a sum of a convex and a concave function.