Concavity of a function with a singular discontinuity in the 2nd derivative

Question

Suppose I have a function $f(x):[a,c]\rightarrow \mathbb{R}$ defined as $f(x)=\cases{f_1(x),\ a\leq x\leq b\\f_2(x),\ b<x\leq c}$ where $f_1(x),f_2(x)$ are two concave functions over the domains $[a,b],\ [b,c]$ respectively. Given that $f_1(b)=f_2(b)$ and that $f_1'(b)=f_2'(b)$ (where the derivatives at $b$ are naturally left for $f_1$ and right for $f_2$), how do I prove that $f(x)$ is concave? thanks a lot!

Answer 1

Here is a straightforward proof if $f_k$ are differentiable:

A differentiable $f$ is concave iff $f'$ is non increasing.

Since both $f_k'$ are non increasing, and $f'_1(b) = f'_2(b)$, we see that $f'$ is non increasing, hence concave.

This can be generalised in a number of ways, but the proofs a little messier.