[proofplan] We first reduce to the case $M_0 = M_1 = 1$ by dividing by an exponential factor, then apply the maximum principle to a strip. Direct application of the maximum principle fails because $\mathcal{S}$ is unbounded; we surmount this by multiplying by a Gaussian-type damping factor $e^{\varepsilon(z^2 - 1)}$ that forces decay in $|y|$ while leaving the boundary bounds essentially unchanged. This converts the problem to one on a bounded rectangle, where the standard maximum principle applies. Letting $\varepsilon \to 0^+$ recovers the desired bound. [/proofplan] [step:Reduce to the normalised case $M_0 = M_1 = 1$ by dividing through an exponential factor] Define \begin{align*} G: \mathcal{S} &\to \mathbb{C} \\ z &\mapsto F(z) \cdot M_0^{z - 1} M_1^{-z}. \end{align*} Here $M_0^{z-1} := \exp((z-1) \log M_0)$ and $M_1^{-z} := \exp(-z \log M_1)$, with $\log$ the principal real logarithm; the case $M_0 = 0$ or $M_1 = 0$ forces $F \equiv 0$ on the corresponding boundary line, hence $F \equiv 0$ throughout $\mathcal{S}$ by the boundary uniqueness theorem for holomorphic functions on a strip (Schwarz reflection across the boundary line followed by the identity theorem), and the conclusion is immediate; assume henceforth $M_0, M_1 > 0$. The function $G$ is holomorphic on the interior $\{0 < \operatorname{Re}(z) < 1\}$ (product of $F$ with two nowhere-vanishing entire functions), continuous on $\mathcal{S}$, and bounded on $\mathcal{S}$ because \begin{align*} |M_0^{z - 1} M_1^{-z}| = M_0^{\operatorname{Re}(z) - 1} M_1^{-\operatorname{Re}(z)} \end{align*} is uniformly bounded for $\operatorname{Re}(z) \in [0, 1]$, and $F$ is bounded by hypothesis. On the boundary lines: \begin{align*} |G(iy)| &= |F(iy)| \cdot M_0^{-1} \cdot 1 \le M_0 \cdot M_0^{-1} = 1, \\ |G(1 + iy)| &= |F(1 + iy)| \cdot 1 \cdot M_1^{-1} \le M_1 \cdot M_1^{-1} = 1. \end{align*} It now suffices to prove $|G(z)| \le 1$ on $\mathcal{S}$, since the bound \begin{align*} |F(\theta + iy)| = |G(\theta + iy)| \cdot M_0^{1 - \theta} M_1^{\theta} \le M_0^{1 - \theta} M_1^{\theta} \end{align*} will follow. [/step] [step:Introduce a damping factor $H_\varepsilon(z) = G(z) e^{\varepsilon(z^2 - 1)}$ to control behaviour at $|y| \to \infty$] For $\varepsilon > 0$, define \begin{align*} H_\varepsilon: \mathcal{S} &\to \mathbb{C} \\ z &\mapsto G(z) \cdot e^{\varepsilon(z^2 - 1)}. \end{align*} The factor $e^{\varepsilon(z^2 - 1)}$ is entire, so $H_\varepsilon$ inherits holomorphy on the interior of $\mathcal{S}$ and continuity on $\mathcal{S}$ from $G$. Compute the modulus of the damping factor. For $z = x + iy$ with $0 \le x \le 1$: \begin{align*} \operatorname{Re}(z^2 - 1) = x^2 - y^2 - 1, \end{align*} so \begin{align*} |e^{\varepsilon(z^2 - 1)}| = e^{\varepsilon(x^2 - y^2 - 1)} \le e^{\varepsilon(1 - y^2 - 1)} = e^{-\varepsilon y^2}, \end{align*} where the inequality uses $x^2 \le 1$ for $x \in [0, 1]$. Two consequences: (i) On the boundary lines, $|H_\varepsilon(iy)| \le e^{-\varepsilon y^2} \cdot |G(iy)| \le e^{-\varepsilon y^2} \le 1$, and similarly $|H_\varepsilon(1 + iy)| \le 1$. (ii) Since $G$ is bounded on $\mathcal{S}$ — say $|G| \le K$ — we have $|H_\varepsilon(z)| \le K e^{-\varepsilon y^2}$ uniformly on $\mathcal{S}$, so $|H_\varepsilon(z)| \to 0$ as $|y| \to \infty$ uniformly in $x \in [0, 1]$. [/step] [step:Apply the maximum principle on a bounded rectangle to obtain $|H_\varepsilon(z)| \le 1$ on $\mathcal{S}$] Fix $z_0 = x_0 + iy_0$ in the interior of $\mathcal{S}$. We show $|H_\varepsilon(z_0)| \le 1$. By the bounded-modulus property (ii) of the previous step, $|H_\varepsilon(z)| \to 0$ as $|y| \to \infty$ uniformly in $x \in [0, 1]$. Choose $R > 0$ large enough that $R > |y_0|$ and \begin{align*} |H_\varepsilon(x + iy)| \le 1 \quad \text{for all } x \in [0, 1] \text{ and } |y| \ge R. \end{align*} Such an $R$ exists because $K e^{-\varepsilon R^2} \le 1$ for all $R \ge \sqrt{\log K / \varepsilon}$ (when $K > 1$), and for $K \le 1$ any $R > |y_0|$ works. Consider the closed rectangle \begin{align*} \mathcal{R} = \{x + iy : 0 \le x \le 1, \ -R \le y \le R\} \subset \mathcal{S}. \end{align*} Then $H_\varepsilon$ is continuous on $\mathcal{R}$ and holomorphic on the interior of $\mathcal{R}$ (which is contained in the interior of $\mathcal{S}$). By the maximum modulus principle for bounded domains, $|H_\varepsilon|$ attains its supremum on $\partial \mathcal{R}$. The boundary $\partial \mathcal{R}$ decomposes into four sides: - the left side $\{0 + iy : -R \le y \le R\}$, on which $|H_\varepsilon(iy)| \le 1$ by step 2(i); - the right side $\{1 + iy : -R \le y \le R\}$, on which $|H_\varepsilon(1 + iy)| \le 1$ by step 2(i); - the bottom side $\{x - iR : 0 \le x \le 1\}$, on which $|H_\varepsilon(x - iR)| \le 1$ by the choice of $R$; - the top side $\{x + iR : 0 \le x \le 1\}$, on which $|H_\varepsilon(x + iR)| \le 1$ by the choice of $R$. Therefore $|H_\varepsilon(z)| \le 1$ for all $z \in \mathcal{R}$. Since $z_0 \in \mathcal{R}$, $|H_\varepsilon(z_0)| \le 1$. Boundary points of $\mathcal{S}$ are handled by continuity: the bound $|H_\varepsilon| \le 1$ extends from the interior to the closure $\mathcal{S}$. [/step] [step:Send $\varepsilon \to 0^+$ to recover the bound on $G$] Fix $z = x + iy \in \mathcal{S}$. From the previous step: \begin{align*} |G(z)| \cdot |e^{\varepsilon(z^2 - 1)}| = |H_\varepsilon(z)| \le 1. \end{align*} Solving: \begin{align*} |G(z)| \le |e^{\varepsilon(z^2 - 1)}|^{-1} = e^{-\varepsilon(x^2 - y^2 - 1)} = e^{\varepsilon(y^2 + 1 - x^2)}. \end{align*} For each fixed $z = x + iy$, the right-hand side depends continuously on $\varepsilon$ and tends to $e^0 = 1$ as $\varepsilon \to 0^+$. Therefore \begin{align*} |G(z)| \le \lim_{\varepsilon \to 0^+} e^{\varepsilon(y^2 + 1 - x^2)} = 1. \end{align*} This holds for every $z \in \mathcal{S}$. [/step] [step:Translate back to $F$ to obtain the conclusion] Recall from step 1 that $G(z) = F(z) M_0^{z - 1} M_1^{-z}$, so \begin{align*} |F(z)| = |G(z)| \cdot |M_0^{z - 1} M_1^{-z}|^{-1} = |G(z)| \cdot M_0^{1 - \operatorname{Re}(z)} M_1^{\operatorname{Re}(z)}. \end{align*} Setting $z = \theta + iy$ with $\theta \in [0, 1]$ and using $|G(z)| \le 1$: \begin{align*} |F(\theta + iy)| \le M_0^{1 - \theta} M_1^\theta \quad \text{for all } y \in \mathbb{R}, \end{align*} which is the conclusion. [/step]