Higher-order convexity

Question

Let $f \in C^\infty(\mathbb R)$.

$f^{(1)}=f'\geq 0$ iff $\forall (a,b) \in\mathbb R^2,a\leq b$ then $f(a) \leq f(b)$
$f^{(2)}=f''\geq 0$ iff $\forall (a,b)\in\mathbb R^2,\forall t\in [0,1], f(ta+(1-t)b)\leq tf(a)+(1-t)f(b)$

Let $n \in\mathbb N,n>2$, $f^{(n)} \geq 0$ iff $C(n,f)$

Is there examples of condition on $f$ : $C(n,f)$, which is expressed even if $f$ is any real function, as for the examples $n=1$ and $n=2$ ?

Given distinct real numbers $x_0,\dots,x_n$, you should want to find constants $a_0,\dots,a_n$ where $a_0f(x_0)+\dots+a_nf(x_n)=0$ whenever $f$ is a polynomial of degree less than $n$ and where $a_0x_0^n+\dots+a_nx_n^n>0$. Since the Vandermonde matrix is invertible, the constants $a_0,\dots,a_n$ are unique up to a positive constant factor. In this case, we should have $f^{(n)}\geq 0$ precisely when $a_0f(x_0)+\dots+a_nf(x_n)\geq 0$ whenever $x_0,\dots,x_n$ are distinct and $a_0,\dots,a_n$ are constructed according to the above conditions. I do not have a proof that this works though. — Joseph Van Name, Commented Mar 25, 2023 at 15:34
The theory of divided differences gives a complete answer to this question. See my previous answer to one of your questions for details: mathoverflow.net/a/407420/2363 — zeb, Commented Mar 25, 2023 at 17:34
I specified my question, there is surely not only one possible — Dattier, Commented Mar 26, 2023 at 5:33

2 revs · Accepted Answer · 2023-03-27 08:18:08Z

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

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

1 Answer 1