$f$ convex strictly decreasing function , is $f'(x+\delta)-f'(x)$ convex

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).

• 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.
– gerw
Feb 3, 2016 at 7:31
• Thanks @gerw what you mentioned is correct, I was deceived by some examples in mind.
– MrX
Feb 4, 2016 at 6:45

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 $\delta=0,1+0,2\,k$, $k=0,1,2,3,4$.

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

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

• Thanks for the nice counter-examples.
– MrX
Feb 4, 2016 at 6:45