Is there an interesting or useful necessary and sufficient condition for a rational function $f \in \mathbb{C}(X)$ to be expressible as the difference $g(X+1)-g(X)$ of another rational function $g \in \mathbb{C}(X)$?
As an example, every polynomial $f$ is expressible in such a way since we have the Faulhaber's formula.
Alternately, we can observe that $(X+1)^n - X^n = \sum\limits_{k=0}^{n-1}{n\choose k}X^k$ and so given a polynomial $f(X) = a_nX^n + \ldots + a_1X + a_0$ we may solve the system of equations obtained by equating the $k$-th terms of $f$ with those of $\sum\limits_{k=0}^{n}\left(\sum\limits_{i=k+1}^{n+1}b_i{i \choose k}\right)X^k$. The associated matrix consists of the upper triangular matrix with entries $A_{ij} = {j \choose i-1}, 1 \leq i \leq j \leq n+1$, and its determinant is clearly nonzero since it is the product of the diagonal entries ${i \choose i-1} = i$, so the determinant is $(n+1)!$.
The solution for the $b_i$, $1 \leq i \leq n+1$ will give a polynomial $g(X) = b_{n+1}X^{n+1} + \ldots b_1X + b_0$ such that $f(X) = g(X+1)-g(X)$ ($b_0$ is arbitrary).
Another example is the rational function $f(X) = \frac{1}{X(X+1)}$ where $g(X) = -\frac{1}{X}$.
As a non-example, consider the function $f(X) = \frac{1}{X}$. Suppose there exist polynomials $p(X), q(X) \in \mathbb{C}[X]$ such that $\frac{p(X+1)}{q(X+1)}-\frac{p(X)}{q(X)} = \frac{1}{X}$. Assume moreover that $p$ and $q$ are coprime after dividing by any common factors.
Then we have $Xp(X+1)q(X)-Xp(X)q(X+1)=q(X+1)q(X)$ so that $q(X)$ divides $Xq(X+1)$ and $q(X+1)$ divides $Xq(X)$. This means we have polynomials $\alpha(X), \beta(X)$ such that $$Xq(X+1) = \alpha(X)q(X)$$ $$Xq(X) = \beta(X)q(X+1)$$
By degree considerations, we have $\deg \alpha = \deg \beta = 1$.
Then $X^2q(X) = \alpha(X)\beta(X)q(X)$ which means $X^2 = \alpha(X)\beta(X)$ and finally this forces $\alpha(X) = cX$ and $\beta(X) = \frac{1}{c}X$ for some $c \neq 0$. Substituting in the equations above we obtain $q(X+1) = cq(X)$ which is only possible if $q$ is constant because any zero would lead to infinitely many zeros.
But this is impossible because $\frac{p(X)}{q(X)}$ would then be a polynomial, and its forward difference cannot equal $\frac{1}{X}$.
I believe this proof can generalize to other cases where $f(X) = \frac{1}{g(X)}$ where $g$ is irreducible (that would just be the linear functions since $\mathbb{C}$ is algebraically closed).
But I have not found a "nice" condition on $f$ which could be necessary and sufficient. The continuous analogue of the problem is easy, we know that $f$ has a rational antiderivative if and only if its partial fraction decomposition does not contain denominators of degree $1$.
Every rational fraction $f\in \mathbb{C}(X)$ can be written uniquely as a sum $$f(X) = P(X) + R(X)$$ where $P$ is a polynomial and $R$ is the polar part. Now the polar part is uniquely expressed as a finite sum $$R(X)= \sum_{k\ge 1, \alpha \in \mathbb{C}}\frac{c_{k,\alpha}}{(X-\alpha)^k}$$
The map $g(X) \mapsto g(X+1) -g(X)$ takes polynomial (polar) part to polynomial (polar) part. So $f$ is in the image of this map if and if only if its polar part is. It is easy to see that the necessary and sufficient condition is $$\sum_{n\in \mathbb{Z}} c_{k,\alpha +n}=0$$ for all $k\ge 1$ and $\alpha \in \mathbb{C}$. (finitely many linear conditions).
ADDED: Consider now the case $f\in \mathbb{Q}(X)$. The decomposition in partial fraction of $f$ happens in a finite (Galois) extension $K$ of $\mathbb{Q}$. Assume that there exists $g\in \mathbb{C}(X)$ so that $f(X) = g(X+1) - g(X)$. Since the linear conditions above are satisfied then we can find $g\in K(X)$ so that $f(X) = g(X+1)-g(X)$. Now average all the equalities $f(X) =g^{\sigma}(X+1)-g^{\sigma}(X)$ for all $\sigma \in Gal(K/\mathbb{Q})$. We get $\bar g\in \mathbb{Q}(X)$ so that $f(X) = \bar g(X+1) - \bar g(X)$.
ADDED:
The linear necessary and sufficient conditions follow from the following combinatorial lemma: for a sequence $(a_n)_{n\in \mathbb{Z}}$ of finite support there exists a sequence of finite support $b= (b_n)_{n\in \mathbb{Z}}$ so that $a= \Delta b$ if and only if $\sum a_n =0$. Moreover, in that case $b$ is uniquely defined. This is the discrete analogue of the result: a function with compact support on $\mathbb{R}$ has an antiderivative with compact support if and only if its total integral is $0$. In our case we want $b_{n+1}-b_n =a_n$ for all $n$. The unique solution with finite support is given by $b_n=\sum_{k<n}a_k$ for all $n$.
ADDED: As @Steven Stadnicki: noticed, if a solution $g$ exists, it is unique up to a constant. Moreover, if $f$ has coefficients in a subfield $K$, so can $g$ be taken with coefficients in $K(X)$.
We give now some results in particular cases:
Assume that $f(X) = \frac{P(X)}{Q(X)}$ where $Q$ has simple integral roots and $\deg Q \ge \deg P +2$. Then $f$ is a difference. Indeed, it is enough to check that the sum of residues of $f$ is $0$, and this follows from Cauchy's formula.
Assume that $\deg f =-1$. Then $f$ is not a difference. This can be seen in two ways: one, since the sum of residues is not $0$; another way is to notice that $\deg f = \deg g -1$, but $g$ is at most of degree $-1$ ( basically we look at the rate of decrease at $\infty$.
ADDED:
If the equation $f(X)= h(X+2)-h(X)$ has a solution $h$, then the equation $f(X)=g(X+1)-g(X)$ has a solution $g(X)= h(X+1)+h(X)$.