When zeros at natural numbers implies zero everywhere

Suppose a function f(z) equals 0 at z = 0, 1, 2, 3, …. Under what circumstances might you be able to conclude that f is zero everywhere?

Clearly you need some hypothesis on f. For example, the function sin(πz) is zero at every integer but certainly not constantly zero.

Carlson’s theorem says that if f is analytic and bounded for z with non-negative real part, and equals zero at non-negative integers, then f is constantly zero.

Carlson’s theorem doesn’t apply to sin(πz) because this function is not bounded in the complex plane. It is bounded on the real axis, but that’s not enough. The identity

sin(z) = ( exp(iz) – exp(-iz) ) / 2i

shows that the sine function grows exponentially in the vertical direction.

Liouville’s theorem says that if a function is analytic and bounded everywhere then it must be constant. Carleson’s theorem does not require that the function f be bounded everywhere but in the right half-plane.

In fact, the boundedness requirement can be weakened to requiring f(z) be O( exp(k|z|) ) for some k < π. This, in combination with having zeros at 0, 1, 2, 3, …. is enough to conclude that f is zero.

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