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In complex analysis, a branch of mathematics, the Hadamard three-lines theorem is a result about the behaviour of holomorphic functions defined in regions bounded by parallel lines in the complex plane. The theorem is named after the French mathematician Jacques Hadamard.

Statement

Let f(z) be a bounded function of z = x + iy defined on the strip

$$\{x+iy:a\leq x\leq b\},$$

holomorphic in the interior of the strip and continuous on the whole strip. If

$$M(x)=\sup _{y}|f(x+iy)|,\,$$

then log M(x) is a convex function on [a, b].

In other words, if x=ta+(1-t)b with $$0\leq t\leq 1$$, then

$$M(x)\leq M(a)^{t}M(b)^{{1-t}}.\,$$

Proof

Define F(z) by

$$F(z)=f(z)M(a)^{{{z-b \over b-a}}}M(b)^{{{z-a \over a-b}}}.$$

Thus |F(z)| ≤ 1 on the edges of the strip. The result follows once it is shown that the inequality also holds in the interior of the strip.

After an affine transformation in the coordinate z, it can be assumed that a = 0 and b = 1. The function

$$F_{n}(z)=F(z)e^{{z^{2}/n}}e^{{-1/n}}$$

tends to 0 as |z| tends to infinity and satisfies |Fn| ≤ 1 on the boundary of the strip. The maximum modulus principle can therefore be applied to Fn in the strip. So |Fn(z)| ≤ 1. Since Fn(z) tends to F(z) as n tends to infinity. it follows that |F(z)| ≤ 1.
Applications

The three-line theorem can be used to prove the Hadamard three-circle theorem for a bounded continuous function g ( z ) {\displaystyle g(z)} g(z) on an annulus { z : r ≤ | z | ≤ R } {\displaystyle \{z:r\leq |z|\leq R\}} \{z:r\leq |z|\leq R\}, holomorphic in the interior. Indeed applying the theorem to

$$f(z)=g(e^{{z}}),\,$$

shows that, if

$$m(s)=\sup _{{|z|=e^{s}}}|g(z)|,\,$$

then$$\log \,m(s)$$ is a convex function of s.

The three-line theorem also holds for functions with values in a Banach space and plays an important role in complex interpolation theory. It can be used to prove Hölder's inequality for measurable functions

$$\int |gh|\leq \left(\int |g|^{p}\right)^{{1 \over p}}\cdot \left(\int |h|^{q}\right)^{{1 \over q}},$$

where $${1 \over p}+{1 \over q}=1$$, by considering the function

$$f(z)=\int |g|^{{pz}}|h|^{{q(1-z)}}.$$