Hadamard three lines lemma

\mathbf{Problem:} Let stripe S=\{z\in \mathbb{C}|0<\Re(z)<1\}, f is analytic in S and continuous to the boundary of S. Suppose |f(z)|\leq e^{C|z|^{2-\delta}} for some C>0 and \delta>0. Define \displaystyle M(x)=\sup\limits_{y}|f(x+iy)|. Then M(t)\leq M(0)^{1-t}M(1)^{t} for 0\leq t\leq 1, which means logM(t) is a convex function.

\mathbf{Proof:} First let us prove if M(0)\leq 1 and M(1)\leq 1, then |f|\leq 1.

Fix \epsilon>0, define F(z)=e^{\epsilon z^2}f(z) on S, then |F(z)|\to 0 as z\to \infty. By the maximum principle, \sup_S|F(z)| must occur on the edge of S.

|F(z)|\leq e^{\epsilon z^2} for z\in S

Let \epsilon \to 0, we get |f(z)|\leq 1 for z\in S.

For the general case, define F(z)=f(z)M(0)^{z-1}M(1)^{-z}, then |F(z)|\leq 1 on \partial S. And F has growth rate no more than some e^{C|z|^{2-\delta}}, so apply the above result, we know that \sup_S |F(z)|\leq 1, which means \sup_S |f(z)|\leq M(0)^{1-z}M(1)^{z}.

Let z=t\in [0,1], we get the conclusion.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} f can not grow too fast, otherwise this theorem fails.

Consider \displaystyle f(z)=e^{e^{\pi(z-\frac 12)i}} on S. Easy to verify that |f(z)|\leq 1 on \partial S. But on \Re(z)=\frac 12, f grows extremely fast.

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