Is $f(2x)/f(x)$ nonincreasing for concave functions with $f(0)=0$? I have a question about concave functions.

Let $f:R_+\rightarrow R_+$ be any nonidentically zero, nondecreasing, continuous, concave function with $f(0)=0$. Do we have that the ratio function $f(2x)/f(x)$ is nonincreasing on $(0,+\infty)$?
No. Let $f$ be the function
$$
f(x) \;=\; \begin{cases}2x &amp; \text{if }0\leq x \leq 1 \\ x+1 &amp; \text{if }1 \leq x.\end{cases}
$$
Then $f$ satisfies the hypotheses you have given, but
$$
\frac{f(2)}{f(1)} = \frac{3}{2} \qquad\text{and}\qquad \frac{f(4)}{f(2)} = \frac{5}{3} &gt; \frac{3}{2}\text{,}
$$
so $f(2x)/f(x)$ increases from $x=1$ to $x=2$.

Edit: There are also differentiable counterexamples, e.g. $f(x) = (x^2+10x)/(x+1)$.
Edit: This proof has an error. See if you can spot it without looking at the comments!

Suppose $f$ is differentiable. Then $f&#39; \geq 0$ is non-increasing, and your ratio is
$$ \frac{\int_0^{2x} f&#39;(y) dy}{\int_0^x f&#39;(y) dy}. $$
The derivative of the fraction is
$$\frac{f&#39;(2x) f(x) - f&#39;(x) f(2x)}{f(x)^2}. $$
Now $0 \leq f(x) \leq f(2x)$ and $0 \leq f&#39;(2x) \leq f&#39;(x)$ so the derivative is non-negative. Hence in this case, the ratio function is non-increasing.

This proof can probably be extended to the non-differentiable case.