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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!

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  • $\begingroup$ For function $f(x)=10 (1 - \exp(-1/(x + 0.1)^2)) - 9) + \exp(x)$ first 13 derivatives at zero are positive, while the first difference is negative. See my answer. $\endgroup$
    – Anixx
    Commented Oct 22, 2016 at 18:32
  • $\begingroup$ Strictly positive for all x is the first example in my answer. $\endgroup$
    – Anixx
    Commented Oct 22, 2016 at 18:43
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    $\begingroup$ So, there are incompatible answers based on different interpretations of the question. I read it as asking that the function be $k$ times differentiable on all of $\mathbb{R}$, and the $k$th derivative strictly positive everywhere. Zak, can you edit the question to clarify what you want? Also, the expression $\Delta(g,h)$ defines a function of $x$, if I read it correctly. Are you asking whether $\Delta(f, h_1, \dots, h_k)(x) > 0$ for all $x$, some $x$, etc? And for every $h_1, \dots, h_k > 0$? $\endgroup$
    – Nate Eldredge
    Commented Oct 22, 2016 at 19:59
  • $\begingroup$ @NateEldredge Yes of course, I tried to be more clear. I've never seen $f > 0$ mean there exists an x such that $f(x) >0$, so that interpretation wasn't on my radar. $\endgroup$
    – George Shakan
    Commented Oct 23, 2016 at 2:09

2 Answers 2

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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 $k$.

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    $\begingroup$ @Zak this is simply wrong answer, consider for instance, function $f(x)=\exp(x)$ if $x<1/2$ and $f(x)=-exp(x)$ if $x>1/2$. It has all derivatives at zero positive, yet its all differences are negative. Or a function from my answer. $\endgroup$
    – Anixx
    Commented Oct 22, 2016 at 18:25
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    $\begingroup$ @Anixx: Well, in general, when somebody says a function is positive, they mean positive everywhere. If they meant positive at a specific point, they'd have said so. $\endgroup$
    – Nate Eldredge
    Commented Oct 22, 2016 at 18:57
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    $\begingroup$ @Nate Eldredge when somebody says a derivative is positive, it rarely means the derivative is positive everywhere. $\endgroup$
    – Anixx
    Commented Oct 22, 2016 at 18:58
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    $\begingroup$ @Anixx: I guess our experience differs. Anyway, OP's comment above clarifies that "positive everywhere" was intended. $\endgroup$
    – Nate Eldredge
    Commented Oct 22, 2016 at 18:59
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    $\begingroup$ @Anixx $f$ is defined on $\mathbb{R}$ $\endgroup$
    – Fedor Petrov
    Commented Oct 22, 2016 at 19:17
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No, this is not true. For instance, for function $(10(1-e^{-1/x^2})-9)+\exp(x)$ and $f(0)=2$ all the consecutive derivatives are positive at $x=0$, while the first difference is not.

For function

$f(x)=10 (1 - \exp(-1/(x + 0.1)^2)) - 9) + \exp(x)$

first 13 derivatives at zero are positive, while the first difference is negative.

Function $f(x)=e^{\tan (2x)}$ has all derivatives positive everywhere, where defined, yet its first difference at $0$ is negative.

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  • $\begingroup$ @Nate Eldredge I fixed the function, this is the graphic: wolframalpha.com/input/… $\endgroup$
    – Anixx
    Commented Oct 22, 2016 at 18:54

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