Wielandt's theorem

Remarks

The gamma function satisfies both conditions: the functional equation, and $\abs{\Gamma(z)} \le \Gamma(\operatorname{Re} z) \le 1$ on the strip, by Euler's integral. For the proof, put $v = F - F(1)\Gamma$; it satisfies condition 1 and $v(1) = 0$. The equation $v(z) = v(z + 1)/z$ extends $v$ to a meromorphic function on $\C$, and the possible poles at $0, -1, -2, \ldots$ are removable because $v(1) = 0$, so $v$ is entire and bounded on $0 \le \operatorname{Re} z \le 1$. Then $u(z) = v(z) v(1 - z)$ is entire, satisfies $u(z + 1) = -u(z)$, and is bounded on that strip, so it is bounded on $\C$ and constant by Liouville's theorem; since $u(1) = 0$, $u = 0$ and hence $v = 0$. Condition 2 cannot be dropped: $F(z) = \Gamma(z)\big(1 + \sin(2\pi z)\big)$ satisfies condition 1 but is not a multiple of $\Gamma$, and it is unbounded on the strip.

Going deeper

For a student

The theorem is a quick tool for proving identities: to show that a function $F$ equals $\Gamma$, check the functional equation, boundedness on the strip $1 \le \operatorname{Re} z \le 2$, and $F(1) = 1$. For example, the Euler–Gauss limit formula and the Legendre duplication formula can be proved this way. Compare with the Bohr–Mollerup theorem, which works on the real line.