TITLE: kth finite difference always positive when kth derivative is?
QUESTION [6 upvotes]: Let $f : \mathbb{R} \to \mathbb{R}$ such that the $k^{\rm th}$ derivative of $f$ is strictly positive for every $x \in \mathbb{R}$. Define the forward difference operator to be:
$$\Delta(g,h) = g(x+h) - g(x),$$
and for $h_1, \ldots , h_k > 0$, 
$$\Delta(g, h_1, \ldots , h_k) = \Delta( \Delta(g , h_1 , \ldots , h_{k-1}), h_k).$$
Is it true that for all $x \in \mathbb{R}$ $$\Delta(f , h_1 , \ldots , h_k)(x) > 0?$$
For $k = 2$ this is true by convexity. I would like to have a reliable reference for this if it is true. If it is not true, is it true when $h_1 = \ldots = h_k$? 
Any relevant information would be much appreciated, even if you feel it is only indirectly related, please leave a comment!

REPLY [6 votes]: Yes, this is true. For the function $\Delta(f,h)$ its $(k-1)$-st derivative is strictly positive, since by Lagrange theorem it equals $$f^{(k-1)}(x+h)-f^{(k-1)}(x)=hf^{(k)}(x+\theta h)>0,$$ for some $\theta\in (0,1)$. Then induct on $k$.