Is this monotonicity property equivalent to convexity?

Question

This is a follow-up of this question.

Let $\psi:[0,\infty) \to [0,\infty)$ be a strictly increasing $C^2$ (or $C^{\infty}$) function, satisfying $\psi(0)=0$.

Suppose that the function $f(r)=\psi'(r)+\frac{\psi(r)}{r}$ is non-increasing.

Must $\psi$ be concave?

The converse statement is true, i.e. $\psi$ concave implies $f$ non-increasing: Indeed, $$ f'(r)=\psi''(r)+\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r}), $$ $\psi'' \le 0$ by concavity, and since $\psi(r)=\int_0^r \psi'(t)dt \ge \int_0^r \psi'(r)dt=r\psi'(r)$, the term $\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r})$ is also non-positive.

Edit:

Here is a partial result-I can prove that $\psi''(0) \le 0$.

By our assumption $$ 0 \ge f'(r)=\psi''(r)+\frac{1}{r}(\psi'(r)-\frac{\psi(r)}{r}), $$ for every $r>0$. Using the mean value theorem (twice), we can rewrite this as $$ f'(r)=\psi''(r)+\psi''(s(r)) \le 0, \tag{1} $$ where $s(r)$ is some point in $(0,r)$.

In particular, taking the limit when $r \to 0$, we deduce that $\psi''(0) \le 0$.

Answer 1

$\psi$ can be non-concave. Here is a counterexample.

Let $$\psi(r) = \frac{r(r^2+2)}{r^2+1}.$$ Clearly, $\psi$ is smooth, and $\psi(0)=0$. Also, $\psi$ is strictly increasing since $$\psi'(r) = \frac{r^4+r^2+2}{(r^2+1)^2} > 0.$$

We have $$f(r) = \psi'(r) + \frac{\psi(r)}{r} = \frac{2(r^4+2r^2+2)}{(r^2+1)^2}$$ and $$f'(r) = - \frac{8r}{(r^2+1)^3}.$$ Thus, $f$ is non-increasing.

However $\psi$ is not concave since $$\psi''(r) = \frac{2r(r^2-3)}{(r^2+1)^3}.$$ For example, $\psi''(2) > 0$.

Answer 2

Note that $$\newcommand\d{\mathrm{d}} f(r) = \frac1r \, \frac{\d}{\d r} (r \, \psi(r)). $$ Hence, $$ g(r) := \frac{\d}{\d r} f(r) = \frac{\d}{\d r} \frac1r \, \frac{\d}{\d r} (r \, \psi(r))\le 0.$$ By solving backwards, we can compute $\psi$ from $g$. By evaluating $\psi''(r)$, we can see that we should choose $g$ such that $g(r_0)$ is near $0$ while $\int_0^{r_0} g(r) \, \d r < 0$. Starting with $g(x) = -\cos(x) -1 \le 0$ gives $$\psi(r) = -\frac{r^2}{3} - \frac{\sin(r)}{r} + \cos(r) + 5 \, r$$ which is not concave around $r = \pi$.

Finally, we have to modify this function, such that $\psi \ge 0$. Let $\hat r$ be the first root of $\psi''$ after $r = \pi$. Chop of $\psi$ at this point and extend the function after $\hat r$ in an affine way. This leads to a $C^2$ function with all the desired properties.

Answer 3

Partial answer .

Here we assume that $\lim_{r \to 0} \frac{\psi(r)}{r}=0$

Now assume that $\psi(r)$ is convex and $r\neq 0$.

So we have :

$$\psi(r)\geq \psi(0)+\psi'(0)r$$

Or :

$$\psi(r)\geq \psi'(0)r$$

Or :

$$\frac{\psi(r)}{r}\geq \psi'(0)$$

Or :

$$f(r)-\psi'(r)\geq \psi'(0)$$

But $f(r)$ is non increasing so $f(0)\geq f(r)$

So :

$$f(0)-\psi'(r)\geq \psi'(0)$$

But $\lim_{r \to 0} \frac{\psi(r)}{r}=0$ so $f(0)=\psi'(0)$

So :

$$-\psi'(r)\geq 0$$

Or :

$$\psi'(r)\leq 0$$

Wich is not the case since $\psi(r)$ is strictly increasing . So we deduce that $\psi(r)$ is not convex .