Apparent contradiction in convexity of compositions of functions

Question

I offer a proposition with both a proof and a counterexample. Thus, either the proof is incorrect, or the counterexample is not actually a counterexample, or both. Which is it?

Proposition. Given a function $h(x)$ which is twice differentiable, strictly convex, and strictly decreasing, there does not exist a strictly increasing, twice differentiable function $g(y)$ such that $f(x) \equiv (g \circ h)(x)$ is concave.

Proof. Suppose $g(y)$ exists. By the properties of concave functions and the chain rule,

$$0 \geq f''(x) = (g \circ h)''(x) = [g'(h(x)) h'(x)]' = g'(h(x)) \underbrace{h''(x)}_{\gt 0} + {g''(h(x))} \underbrace{[h'(x)]^2}_{\gt 0}$$

For the statement to hold, we need $g'(h(x)) \leq 0 $ and $g''(h(x)) \leq 0$. Thus $g'$ must be weakly decreasing (and concave), a contradiction.

Counterexample. Consider $h(x) = \exp (-x)$ and $g(y) = \log y$. $h$ is twice differentiable, convex, and strictly decreasing. $g$ is strictly increasing and twice differentiable. Finally, the function $f = (g \circ h)(x) = - x$ is linear, and therefore concave.

What's going on?

Answer 1

In your proof, you write

For the statement to hold, we need $g'(h(x)) \leq 0 $ and $g''(h(x)) \leq 0$. Thus $g'$ must be weakly decreasing (and concave), a contradiction.

But, it really should read

For the statement to hold, we need $g'(h(x)) \leq 0 $ or $g''(h(x)) \leq 0$. Thus $g'$ must either be weakly decreasing (a contradiction) or $g$ is concave. Oh well, I guess $g$ is concave, as per my counterexample.

Answer 2

For the second derivative of $f\left(x\right)$, $g^\prime \left(h\left(x\right)\right) \leq 0 $ and $g^\prime \left(h\left(x\right)\right) \leq 0$ is sufficient condition of $f\left(x\right)^{\prime \prime} \leq 0$ but not necessary.