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Assume you have a strictly decreasing convex differentiable function $f(x)$, $x \in \Bbb R^+$, I am wondering if the increment of the first derivative is also convex; i.e., $$g(x) = f'(x+\delta) - f'(x)$$ where $\delta$ is any positive number.

What I concluded :

I can say that $f'(x)$ is a strictly increasing function, also since $f(x)$ is strictly decreasing, $f'(x)$ is always negative, meaning that it increases and approaches zero as $x \to \infty $, now I can visualize $f'$ as concave and the difference : $f'(x+\delta)-f'(x)$ to be decreasing but not sure how to show its convexity (if it is).

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    $\begingroup$ The convexity of $g$ would amount to $g''(x) = f'''(x+\delta) - f'''(x) \ge 0$. I do not see any reason why this should hold true. $\endgroup$ – gerw Feb 3 '16 at 7:31
  • $\begingroup$ Thanks @gerw what you mentioned is correct, I was deceived by some examples in mind. $\endgroup$ – MrX Feb 4 '16 at 6:45
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To complete gerw's comment and to avoid leaving the question unanswered, let me give an explicit counterexample.

The question (letting $g=f'$) is equivalent to the following one: If $g\colon(0,\infty)\to\mathbb{R}$ is continuous, strictly increasing and bounded above, is $g(x+\delta)-g(x)$ convex for $\delta>0$? Here is a counterexample: $$ g(x)=\frac{x-\sin x}{1+x-\sin x}. $$ We have $$ g'(x)=\frac{1-\cos x}{(1+x-\sin x)^2}\ge0,\quad 0\le g(x)\le1. $$ $g$ is increasing, bounded above and $\lim_{x\to\infty}g(x)=1$, but it is not concave. Here is the graph of $g(x+\delta)-g(x)$ for $\delta=0,1+0,2\,k$, $k=0,1,2,3,4$. enter image description here

Clearly $g(x+\delta)-g(\delta)$ is not convex.

If you want a counterexample in terms of $f$ let $$ f(x)=\int_0^x(g(t)-1)\,dt. $$

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  • $\begingroup$ Thanks for the nice counter-examples. $\endgroup$ – MrX Feb 4 '16 at 6:45

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