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 $(0,+\infty)$?
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 $$ f(x) \;=\; \begin{cases}2x & \text{if }0\leq x \leq 1 \\ x+1 & \text{if }1 \leq x.\end{cases} $$ Then $f$ satisfies the hypotheses you have given, but $$ \frac{f(2)}{f(1)} = \frac{3}{2} \qquad\text{and}\qquad \frac{f(4)}{f(2)} = \frac{5}{3} > \frac{3}{2}\text{,} $$ 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 $f' \geq 0$ is non-increasing, and your ratio is $$ \frac{\int_0^{2x} f'(y) dy}{\int_0^x f'(y) dy}. $$ The derivative of the fraction is $$\frac{f'(2x) f(x) - f'(x) f(2x)}{f(x)^2}. $$ Now $0 \leq f(x) \leq f(2x)$ and $0 \leq f'(2x) \leq f'(x)$ 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.