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!



kth finite difference always positive when kth derivative is?
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 $k$.


