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Suppose $f:\mathbb{R_+} \to \mathbb{R}$ is a continuous and strictly increasing function. Define $g(x) = \frac{f(x+1)-f(1)}{f(x)-f(0)}$. For which $f$ the function $g$ satisfies $g(x) \geq g(1)$ for all $x \geq 0$?

Comment: this is part of a larger project where the existence of a solution boils down to the condition $g(x) \geq g(1)$ above. I am looking for necessary and sufficient conditions on $f$ which guarantee this.

Examples:

  1. $f(x)=\frac{1}{a} \left[1-e^{-a x} \right]$ implies that $g(x) = e^{-a}$ independently of $x$ and therefore the condition is satisfied.
  2. $f(x)=x^a$ for $a > 0$ implies that $g(x) = \frac{(x+1)^a-1}{x^a}$ so that $g(x) \geq 1$ if and only if $a \geq 1$.
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$$g(x) = \frac{f(x+1)-f(1)}{f(x)-f(0)}$$

has a minimum for $x=1$. Differentiating gives

$$g'(x) = \frac{f(x)f'(x+1)-f(0)f'(x+1)-f'(x)f(x+1)+f(1)f'(x)}{(f(x)-f(0))^2}$$

This must be 0 if $x=1$. This gives:

$$f(1)f'(2)-f(0)f'(2)-f'(1)f(2)+f(1)f'(1)=0$$

Rearranging gives:

$$\frac{f'(2)}{f'(1)}=\frac{f(2)-f(1)}{f(1)-f(0)}$$

$$\frac{f'(2)}{f'(1)}=g(1)$$

This is a necessary condition if $f$ is differentiable.

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  • 2
    $\begingroup$ This is nice but it assumes that $f$ is differentiable. $\endgroup$ – Clive Newstead Jul 16 '15 at 15:05

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