Carlson's theorem

Assume that satisfies the following three conditions. The first two conditions bound the growth of at infinity, whereas the third one states that vanishes on the non-negative integers. • is an entire function of exponential type, meaning that |f(z)| \leq C e^{\tau|z|}, \quad z \in \mathbb{C} for some real values , . • There exists such that |f(iy)| \leq C e^{c|y|}, \quad y \in \mathbb{R} • for every non-negative integer . Then is identically zero. ==Sharpness==

First condition The first condition may be relaxed: it is enough to assume that is analytic in , continuous in , and satisfies|f(z)| \leq C e^{\tau|z|}, \quad \operatorname{Re} z > 0for some real values , . Second condition To see that the second condition is sharp, consider the function . It vanishes on the integers; however, it grows exponentially on the imaginary axis with a growth rate of , and indeed it is not identically zero. Third condition A result due to relaxes the condition that vanish on the integers. Namely, Rubel showed that the conclusion of the theorem remains valid if vanishes on a subset of upper density 1, meaning that : \limsup_{n \to \infty} \frac{\left| A \cap \{0,1,\ldots,n-1\} \right|}{n} = 1. This condition is sharp, meaning that the theorem fails for sets of upper density smaller than 1. ==Applications==

Suppose is a function that possesses all finite forward differences \Delta^n f(0). Consider then the Newton seriesg(z) = \sum_{n=0}^\infty {z \choose n} \, \Delta^n f(0)where {z \choose n} is the binomial coefficient and \Delta^n f(0) is the -th forward difference. By construction, one then has that for all non-negative integers , so that the difference . This is one of the conditions of Carlson's theorem; if obeys the others, then is identically zero, and the finite differences for uniquely determine its Newton series. That is, if a Newton series for exists, and the difference satisfies the Carlson conditions, then is unique. ==See also==