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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!

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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 $$ f(x) = |2 \sin (\frac x2 )\cos(\frac x2)| - x \le 2 |\sin (\frac x2 )| - x = 2 f(\frac x2) \, . $$

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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.

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  • $\begingroup$ Impressive counterexample! Thank you. Just one more question, what if the function is also assumed to be differentiable? Do you think that then the conditions are equivalent? $\endgroup$
    – Barreto
    Commented Aug 23, 2023 at 19:50
  • $\begingroup$ @MAB: I have added an example which is differentiable on $(0, \infty)$. $\endgroup$
    – Martin R
    Commented Aug 23, 2023 at 20:51
  • $\begingroup$ I made this followup question, this is what I wanted to really check: math.stackexchange.com/questions/4758737/… $\endgroup$
    – Barreto
    Commented Aug 25, 2023 at 23:08

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