Lemma 28.2.5 (Hadamard’s Three Lines Lemma).label Let $S = \bracs{z \in \complex| \text{Re}(z) \in [0, 1]}$, $E$ be a Banach space over $\complex$, and $f \in H(S; E) \cap BC(\ol{S}; E)$. For each $s \in [0, 1]$, let

\[M(\theta) = \sup_{t \in \real}\norm{f(s + it)}_{E}\]

then for each $s \in [0, 1]$, $M(s) \le M(0)^{s} M(1)^{1-s}$.

Proof, [Theorem VI.3.7, Con78]. Assume without loss of generality that $M(0), M(1) > 0$. Let

\[g: \complex \to \complex \quad z \mapsto M(0)^{z} M(1)^{1 - z}\]

then $g$ is a non-vanishing entire function, and for each $z \in \complex$,

\[|g(z)| = M(0)^{\text{Re}(z)}M(1)^{\text{Re}(1 - z)}\]

so $|g|^{-1}$ is bounded on $\ol S$ by Proposition 5.16.2. Let

\[h: \ol S \to E \quad z \mapsto \frac{f(z)}{g(z)}\]

then $h \in H(S; E) \cap BC(\ol S; E)$ with $\norm{h(z)}_{E} \le 1$ for all $z \in \partial S$. By the Maximum Modulus Theorem, $\norm{h(z)}_{E} \le 1$ for all $z \in S$. Thus for every $z \in S$,

\[f(z) \le M(0)^{\text{Re}(z)}M(1)^{1-\text{Re}(z)}\]

Therefore $M(s) \le M(0)^{s} M(1)^{1-s}$ for every $s \in [0, 1]$.$\square$

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