Hadamard three-lines theorem

Remarks

Proof idea: assume $M(0), M(1) > 0$ and put $g(z) = f(z) M(0)^{z - 1} M(1)^{-z}$, so that $\abs{g} \le 1$ on both boundary lines. For $\varepsilon > 0$ the function $g(z) e^{\varepsilon z^2}$ tends to $0$ as $\abs{\operatorname{Im} z} \to \infty$ in $S$, since $g$ is bounded. The maximum modulus principle on long rectangles gives $\abs{g(z)} e^{\varepsilon (x^2 - y^2)} \le e^{\varepsilon}$, and $\varepsilon \to 0$ gives $\abs{g} \le 1$. The boundedness hypothesis is needed. The function $f(z) = \exp(i e^{-i\pi z})$ satisfies $\abs{f(x + iy)} = \exp(\sin(\pi x) e^{\pi y})$. It has modulus $1$ on both lines but is unbounded on the line $\operatorname{Re} z = \frac{1}{2}$. This theorem is the key step in the Riesz–Thorin interpolation theorem.

Going deeper

For a student

The theorem is the strip form of the Hadamard three-circle theorem and a special case of the Phragmén–Lindelöf principle. Its main use is in interpolation: the Riesz–Thorin theorem bounds an operator on $L^{p}$ for intermediate $p$ by applying it to an analytic family of functions. The same idea appears in complex interpolation of Banach spaces and in convexity bounds for the Riemann zeta function in the critical strip.