$\newcommand\R{\mathbb R}$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: Implication (i)$\implies$(ii) is trivial.
By approximation, we may assume that the functions $f$ of interest are however smooth. So, in the case when $n_i=1$ for all $i$, implication (iii)$\implies$(i) follows by induction, using the mean value theorem. If $n_i>1$ for some or all $i$, implication (iii)$\implies$(i) follows then by a limit transition.
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, some nontrivial linear combination of the rows of $A$ is the zero row; that is, 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 $c\ge0$. $\quad\Box$.