How can I push the derivative through the following sum?

Question

Let $f$ be a probability density and $F$ the corresponding distribution function. In other words $f$ is non-negative, integrates to 1, and $F' = f$. I'm interested in the following: $$ \frac{d}{dx}\sum_{i\in\mathbb{Z}}F(i+x)-F(i)\stackrel{?}{=}\sum_{i\in\mathbb{Z}}f(i+x) $$ I'm doubtful that this result holds in general. But what if I add assumptions such as that $f$ be Lipschitz-continuous?

Answer 1

It suffices to show that $\sum_{i\in\mathbb{Z}}F(i+x)-F(i)$ is a primitive function of $\sum_{i\in\mathbb{Z}}f(i+x)$.

\begin{align} & \int_0^x \sum_{i\in\mathbb{Z}}f(i+t) dt \\ =& \sum_{i\in\mathbb{Z}} \int_0^x f(i+t) dt \tag{Fubini-Tonelli Theorem} \\ =& \sum_{i\in\mathbb{Z}} (F(x+i)-F(i)) \end{align}

The result follows from the Fundamental Theorem of Calculus.