Let $f : \mathbb{R} \to \mathbb{R}$ such that the $k^{\rm th}$ derivative of $f$ is strictly positive. 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 $$\Delta(f , h_1 , \ldots , h_k) > 0?$$

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!

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!