# Concavity of decreasing, continuous function

Suppose I have some continuous, decreasing and real-valued function $$f$$ who I would like to verify whether it is concave on the region $$\mathbb{R}_{\geq 0}$$. Since it is continuous, I only must check the concavity condition in the "midpoints", i.e. that:

$$f\left(\frac{\theta_1 + \theta_2}{2}\right) \geq \frac{f(\theta_1)+f(\theta_2)}{2} \quad \forall \theta_1,\theta_2 \in \mathbb{R}_{\geq 0}$$

Now, suppose for concreteness that $$f(\theta_1)=0$$ and that I manage to prove that the above inequality holds at least if we fix $$\theta_1=0$$, or that

$$f\left(\frac{\theta}{2}\right) \geq \frac{f(\theta)}{2} \quad \forall \theta \in \mathbb{R}_{\geq 0}$$

Is this sufficient to conclude concavity in the required region? Can we use the decreasing property to prove it? Sketching some functions suggests this is true, although I am still not able to verify it formally. Thanks in advance!

A counterexample is $$f(x) = |\sin(x)| - x$$. $$f$$ is continuous and decreasing on $$\Bbb R$$ but not concave. It satisfies $$f(0) = 0$$ and $$f(x) = |2 \sin (\frac x2 )\cos(\frac x2)| - x \le 2 |\sin (\frac x2 )| - x = 2 f(\frac x2) \, .$$
A differentiable counterexample can be constructed as follows: First define $$g: \Bbb R \to \Bbb R$$ as $$g(u) = \frac{\sin(2 \pi u)}{2 \pi} + u \, .$$ $$g$$ is strictly increasing with $$g(u+1) =g(u) + 1$$ and $$\lim_{u \to -\infty} g(u) = - \infty$$. Then define $$f: [0, \infty) \to \Bbb R$$ as $$f(x) =-2^{g(\log_2 x)}$$ for $$x > 0$$, and $$f(0) = 0$$. $$f$$ is strictly decreasing with $$f(2x) = 2f(x)$$, but $$f$$ is not concave.
• @MAB: I have added an example which is differentiable on $(0, \infty)$. Commented Aug 23, 2023 at 20:51