Carlson's theorem

Sharpness

First condition

The first condition may be relaxed: it is enough to assume that f is analytic in Re z > 0, continuous in Re z ≥ 0, and satisfies$${\displaystyle |f(z)|\le C{e}^{\tau |z|},\mathrm{Re}z>0}$$for some real values C, τ.

Second condition

To see that the second condition is sharp, consider the function f(z) = sin(πz). It vanishes on the integers; however, it grows exponentially on the imaginary axis with a growth rate of c = π, and indeed it is not identically zero.

Third condition

A result due to Rubel (1956) relaxes the condition that f vanish on the integers. Namely, Rubel showed that the conclusion of the theorem remains valid if f vanishes on a subset A ⊂ {0, 1, 2, ...} of upper density 1, meaning that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}}$ ${\displaystyle \frac{\left|A\cap \{0,1,\dots ,n-1\}\right|}{n}=1.}$

This condition is sharp, meaning that the theorem fails for sets A of upper density smaller than 1.

Applications

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