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Let $\psi:[0,\infty) \to [0,\infty)$ be a strictly increasing $C^1$ function, satisfying $\psi(0)=0$.

Suppose that for every $r>0$, $$\psi'(r)+\frac{\psi(r)}{r} \le 2\psi'(0). \tag{1}$$

Is it true that $\psi$ is concave?

The converse statement is true:

$\psi(r)=\int_0^r \psi'(t)dt \le \int_0^r \psi'(0)dt=r\psi'(0)$, where $\psi'(t) \le \psi'(0)$ by concavity.

Edit: (A conceptual explanation of Stefano's answer:)

Inequality $(1)$ is "linear" in $\psi$, and of we replace it by equality, then the only solution (defined at $r=0$) is $\psi(r)=cr$. Now, the idea is to find another function which satisfies the inequality, and "add" it to the linear solution.

In the end, we did not really need full concavity. We can have "convex regions" alongside "concave regions" as long as the convex ones do not dominate the concave too much.

Asaf Shachar
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1 Answers1

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The statement seems in general false to me. Consider as counterexample the function $$ \psi(r) \dot = \sin r + r. $$

Obviously one has $\psi \colon [0,\infty) \to [0, \infty)$, $\psi \in C^1$ and $\psi(0) = 0$. Moreover $\psi'(r) = \cos r +1$, from which one can easily deduce that $\psi$ is strictly increasing.

A direct computation gives $$ \psi'(r) + \frac{\psi(r)}{r} = \cos r + 1 + \frac{\sin r}{r} + 1 \leq 4 = 2 \psi (0) \qquad \forall r \geq 0. $$

However, $\psi''(r) = -\sin r$ which has no definite sign, hence $\psi$ is not concave.

Stefano
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