Given two real analytic functions, $g(x)$ and $f(x)$, on an open interval $I\subset \mathbb{R}$, it is obvious that $g(x) \leq f(x)$ does *not* imply $g\_n \leq f\_n$ (here $g\_n = [x^n] g(x)$ denotes the $n$-th Taylor coefficient of $g$). However, does the inequality in the coefficients hold if we restrict to the class of univariate [holonomic](https://en.wikipedia.org/wiki/Holonomic_function) functions? Let me illustrate with a simple example: 


Suppose we are given a linear recurrence inequality, 
$$
f\_{n+1} \leq \alpha f\_n + \beta f\_{n-1},\qquad n\geq 1,
$$ 
and real $\alpha, \beta$, and $f\_0, f\_1$ are given. We approach this with the standard generating function $F(x) = \sum\_{x=0}^\infty f\_n x^n$, then (maybe assuming $x\in (0,\mu)\in \mathbb{R}^+$) we obtain 
$$
F(x) \leq \dfrac{f\_0 + xf\_1 - \alpha x f\_0}{1-\alpha x - \beta x^2}.
$$
The question is if we are allowed to compare the Taylor coefficients on the left and on the right hand sides of this inequality,
$$
f\_n = [x^n] F(x) \overset{?}{\leq} [x^n] \dfrac{f\_0 + xf\_1 - \alpha x f\_0}{1-\alpha x - \beta x^2}.
$$



linear recurrence inequality
The answer is no. E.g., let 
$\alpha = -1,\beta = 0,f\_0= 0,f\_1= 0,f\_2= -2,f\_3= 1$. Then $f\_{n+1} \leq \alpha f\_n + \beta f\_{n-1}$ for $n=1,2$, whereas 
$$
f\_3 = 1 \not\leq 0= \left(\alpha ^2+\beta \right)f\_1 +\alpha \beta f\_0=[x^3] \dfrac{f\_0 + xf\_1 - \alpha x f\_0}{1-\alpha x - \beta x^2}.
$$


