$\newcommand\R{\mathbb R}$The purpose of this answer is to provide a short, simple, and self-contained proof of the theorem in zeb's answer. 

For (say) a nonempty open interval $I\subseteq\R$ and sufficiently smooth functions $f\colon I\to\R$, let 
\begin{equation*}
	Lf:=\sum_{i\in[m]}\sum_{j\in[n_i-1]_0}c_{i,j}f^{(j)}(a_i),
\end{equation*}
where $m$ and the $n_i$'s are natural numbers, $[m]:=\{1,\dots,m\}$, $[m]_0:=\{0,\dots,m\}$, 
the $c_{i,j}$'s are real numbers such that $c_{i,n_i-1}\ne0$, $f^{(j)}$ is the $j$th derivative of $f$, and the $a_i$'s are distinct points in $I$. 
Let also 
\begin{equation*}
	k:=-1+\sum_{i\in[m]}n_i, \tag{0}
\end{equation*}
and consider the divided difference 
\begin{equation*}
	L_0f:=[a_1^{,n_1},\dots,a_m^{,n_m};f],
\end{equation*}
where $a^{,n}:=\underbrace{a,\dots,a}_{n}$. 
Finally, let $C_{k-1}^+$ denote the set of all $(k-1)$th-order convex functions and let $p_r(x):=x^r$. 

Then the mentioned theorem can be restated as follows: 

> **Theorem 1:** The following three statements are equivalent to one another: 
> 
> (i) $Lf\ge0$ for all $f\in C_{k-1}^+$.
> 
> (ii) $Lp_r=0$ for all $r\in[k-1]_0$ and $Lp_k\ge0$.
>
> (iii) $L=cL_0$ for some real $c\ge0$. 

*Proof:* Implications (iii)$\implies$(i) and (i)$\implies$(ii) are trivial. 

It remains to check implication (ii)$\implies$(iii). The part ($Lp_r=0$ for all $r\in[k-1]_0$) of statement (ii) can be rewritten as the system of equations 
\begin{equation*}
	\sum_{(i,j)\in K}a_{r,(i,j)}c_{i,j}=-c_{m,n_m-1}p_r^{(n_m-1)}(a_m)\text{ for }r\in[k-1]_0 \tag{1}
\end{equation*}
with the "unknowns" $c_{i,j}$ for $(i,j)\in K$, where 
\begin{equation*}
	K:=\{(i,j)\colon i\in[m]\ \&\ j\in[n_i-1]_0\}\setminus\{(m,n_m-1)\}
\end{equation*}
and $a_{r,(i,j)}:=p_r^{(j)}(a_i)$. The matrix $A:=(a_{r,(i,j)}\colon r\in[k-1]_0,(i,j)\in K)$ of system (1) is square (by (0)) and nonsingular -- otherwise, for some real $b_r$'s not all of which are $0$, the polynomial $P:=\sum_{r\in[k-1]_0}b_rp_r$ of degree $\le k-1$ would have $k$ roots (namely, $a_1^{,n_1},\dots,a_m^{,n_m-1}$, counting the multiplicities). 

So, for each given value of $c_{m,n_m-1}$, system (1) has a unique solution $(c_{i,j}\colon(i,j)\in K$. On the other hand, $L_0p_r=0$ for all $r\in[k-1]_0$. So, $L=cL_0$ for some real $c$. Moreover, $L_0p_k>0$ and hence the condition $Lp_k\ge0$ in (ii) yields $c\ge0$. $\quad\Box$.