If $f$ is convex and decays exponentially, then $f'$ decays exponentially Does the following statement about exponential convergence of real functions hold true and does anybody have a proof or counterexample? I would believe that an elementary answer should be possible but high level reasoning is fine as well.  
Let $f :[0, \infty ) \rightarrow \mathbb{R} $  be a smooth function satisfying that there exist positive constants $\alpha , \beta > 0$  such that for all $x \in [0,\infty )$: 


*

*$0 \leq f(x) \leq \alpha e^{ - \beta x} $

*$f^{\prime} (x) < 0 $ 

*$f^{\prime \prime } (x) > 0 $


Then there exist constants $ \gamma , \delta > 0 $ such that for all $x \in [0,\infty)$ we have  $0 \leq - f^{\prime} (x)  \leq \gamma e^{- \delta x} $ . In other words for an exponentially bounded monotoneously decreasing convex function the absolute value of its derivative is already exponentially bounded. 

Suppose $0\le f(x)\le e^{-x}.$ Let $x\ge 1.$ Look at the chord connecting $ (x-1,f(x-1))$ to $ (x,f(x)).$ The slope of this chord equals
$$\frac{f(x)-f(x-1)}{1} \ge \frac{0-f(x-1)}{1}\ge -e^{-(x-1)}.$$
But by the MVT, this slope equals $f'(c)$ for some $c\in (x-1,x).$ Because $f'$ is increasing, we have
$$f'(x) \ge f'(c) \ge -e^{-(x-1)} = - e\cdot e^{-x}.$$
Thus $0 > f'(x) \ge - e\cdot e^{-x}$ for all $x\ge 1.$ 
So we have the desired result in this case, and I expect the general result to follow in a like manner.
