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(π*z*) 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(π*z*) 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*) ) / 2*i*

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.