We can simplify the answer of @zeb.

$C(n,f)=\forall h>0, \Delta^n_h(f) \geq 0$

a) Let $h>0$, $\Delta_h f (x)=f(x+h)-f(x)$.

By induction, for $n=1$ $f$ is increasing so $\Delta_h f\geq 0$

Suppose $\forall g \in C^{\infty}(\mathbb R)$, if $g^{(n)}\geq 0$ then $\Delta_h^n g \geq 0$ : (*)

Let $n \in\mathbb N^*$, $f \in C^{\infty}(\mathbb R),f^{(n+1)}\geq 0$ so $f^{(n)}$ increasing and $\Delta_h f^{(n)}= (\Delta_hf)^{(n)} \geq 0$

by using (*) so $0\leq\Delta^n_h(\Delta_h f)=\Delta^{n+1}_hf$

b) Suppose $\forall h>0, \Delta_h^n f \geq 0$

We have $\lim \limits_{e\rightarrow 0} \dfrac{\Delta_e ^{(n)}f(x)}{h}=f^{(n)}(x)$

So $f^{(n)} \geq 0$

$C(n,f)=\forall h>0, \Delta^n_h(f) \geq 0$

a) Let $h>0$, $\Delta_h f (x)=f(x+h)-f(x)$.

By induction, for $n=1$ $f$ is increasing so $\Delta_h f\geq 0$

Suppose $\forall g \in C^{\infty}(\mathbb R)$, if $g^{(n)}\geq 0$ then $\Delta_h^n g \geq 0$ : (*)

Let $n \in\mathbb N^*$, $f \in C^{\infty}(\mathbb R),f^{(n+1)}\geq 0$ so $f^{(n)}$ increasing and $\Delta_h f^{(n)}= (\Delta_hf)^{(n)} \geq 0$

by using (*) so $0\leq\Delta^n_h(\Delta_h f)=\Delta^{n+1}_hf$

b) Suppose $\forall h>0, \Delta_h^n f \geq 0$

We have $\lim \limits_{e\rightarrow 0} \dfrac{\Delta_e ^{(n)}f(x)}{e^n}=f^{(n)}(x)$

So $f^{(n)} \geq 0$