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 $(0,+\infty)$?

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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. $f(x) = (x^2+10x)/(x+1)$.

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$f(x) = x + \sqrt{x}$ works, too. –  t.b. May 31 '11 at 19:36
Thanks a lot. I can image the shape of the class of functions for counterexamples now. –  Weimin Miao May 31 '11 at 19:49

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.

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How can this proof be extended to the non-differentiable case? It's full of derivatives! –  TonyK May 31 '11 at 18:48
(Also, I think you mean the derivative is non-positive.) –  TonyK May 31 '11 at 18:51
Am I missing something, or should there be a factor of two in front of the first term of the numerator of the derivative? –  J. J. May 31 '11 at 18:58
I think there is a mistake in your proof. The derivative of the fraction should be $$\frac{2f'(2x)f(x)-f'(x)f(2x)}{f^2(x)}.$$. So there is a gap in your proof. –  Weimin Miao May 31 '11 at 19:00
Yes, this proof is not correct. The derivative of $f(2x)$ is $2f'(2x)$. –  Jim Belk May 31 '11 at 19:25