[proofplan] We introduce an auxiliary barrier function using an exponent $\gamma$ with $\beta < \gamma < \alpha$. The key idea is that $e^{-\varepsilon z^\gamma}$ decays uniformly on the entire closed sector (including the boundary rays), unlike $e^{-\varepsilon z^\alpha}$ which is neutral on the rays. On truncated sectors $\Omega_R$, the [Maximum Modulus Principle](/theorems/491) bounds $|f(z) e^{-\varepsilon z^\gamma}| \le M$, since the barrier suppresses the growth $Ce^{R^\beta}$ on the circular arc (because $\gamma > \beta$) while preserving the bound $M$ on the rays. Sending $R \to \infty$ and then $\varepsilon \to 0$ yields $|f(z)| \le M$. [/proofplan] [step:Analyse the decay of $e^{-\varepsilon z^\gamma}$ in the sector for $\beta < \gamma < \alpha$] Choose $\gamma$ with $\beta < \gamma < \alpha$. For $z \in \overline{\Omega}$ with $z = re^{i\theta}$ where $r > 0$ and $|\theta| \le \pi/(2\alpha)$, define the principal branch \begin{align*} z^\gamma &= r^\gamma e^{i\gamma\theta}. \end{align*} This is well-defined and holomorphic on $\Omega$ since $|\theta| < \pi/(2\alpha) < \pi$ (using $\alpha > 1/2$), so the sector lies in the slit plane $\mathbb{C} \setminus (-\infty, 0]$. The real part of $z^\gamma$ is \begin{align*} \operatorname{Re}(z^\gamma) &= r^\gamma \cos(\gamma\theta). \end{align*} Since $\gamma < \alpha$ and $|\theta| \le \pi/(2\alpha)$, we have $|\gamma\theta| \le \gamma\pi/(2\alpha) < \alpha\pi/(2\alpha) = \pi/2$. This strict inequality gives $\cos(\gamma\theta) > 0$ on the entire closed sector, including the boundary rays. Define the uniform lower bound \begin{align*} \delta_0 &:= \cos\!\bigl(\gamma\pi/(2\alpha)\bigr) > 0. \end{align*} Then for all $z \in \overline{\Omega}$: \begin{align*} |e^{-\varepsilon z^\gamma}| &= e^{-\varepsilon r^\gamma \cos(\gamma\theta)} \le e^{-\varepsilon r^\gamma \delta_0} \le 1. \end{align*} [guided] A natural first attempt would be to use the barrier $e^{-\varepsilon z^\alpha}$, matching the exponent to the sector opening. However, on the boundary rays $\theta = \pm\pi/(2\alpha)$, one has $\cos(\alpha\theta) = \cos(\pi/2) = 0$, so $|e^{-\varepsilon z^\alpha}| = 1$ — the barrier is neutral there and provides no decay. This makes it impossible to control $|f \cdot e^{-\varepsilon z^\alpha}|$ on the circular arc near the boundary rays. The fix is to use an exponent $\gamma$ strictly between $\beta$ and $\alpha$. Write $z = re^{i\theta}$ in polar form. For $z$ in the closed sector, $|\theta| \le \pi/(2\alpha)$, so \begin{align*} z^\gamma = r^\gamma e^{i\gamma\theta}, \qquad \operatorname{Re}(z^\gamma) = r^\gamma \cos(\gamma\theta). \end{align*} Since $\gamma < \alpha$, the angle $\gamma\theta$ satisfies $|\gamma\theta| \le \gamma\pi/(2\alpha) < \alpha\pi/(2\alpha) = \pi/2$. This strict inequality is the crucial point: it means $\cos(\gamma\theta) \ge \cos(\gamma\pi/(2\alpha)) =: \delta_0 > 0$ uniformly on the entire closed sector, including the boundary rays. Therefore $|e^{-\varepsilon z^\gamma}| = e^{-\varepsilon r^\gamma \cos(\gamma\theta)} \le e^{-\varepsilon \delta_0 r^\gamma}$. This decays like $e^{-c r^\gamma}$ as $r \to \infty$ everywhere in $\overline{\Omega}$. Since the growth assumption gives $|f(z)| \le Ce^{|z|^\beta}$ with $\beta < \gamma$, the product $|f(z)| \cdot |e^{-\varepsilon z^\gamma}| \le Ce^{r^\beta - \varepsilon \delta_0 r^\gamma} \to 0$ as $r \to \infty$, because the $r^\gamma$ term dominates $r^\beta$. [/guided] [/step] [step:Define the auxiliary function and bound it on the boundary of the truncated sector $\Omega_R$] For $\varepsilon > 0$ and $R > 0$, define the truncated sector \begin{align*} \Omega_R &= \{z \in \Omega : |z| < R\} \end{align*} and the auxiliary function \begin{align*} \varphi_\varepsilon: \overline{\Omega} &\to \mathbb{C} \\ z &\mapsto f(z) \cdot e^{-\varepsilon z^\gamma}. \end{align*} The function $\varphi_\varepsilon$ is holomorphic on $\Omega$ (as a product of holomorphic functions) and continuous on $\overline{\Omega}$. The boundary $\partial\Omega_R$ consists of three parts: **Boundary rays** ($\theta = \pm\pi/(2\alpha)$, $0 \le r \le R$): Since $|e^{-\varepsilon z^\gamma}| \le 1$ on the closed sector and $|f(z)| \le M$ by hypothesis (1): \begin{align*} |\varphi_\varepsilon(z)| &= |f(z)| \cdot |e^{-\varepsilon z^\gamma}| \le M \cdot 1 = M. \end{align*} **Circular arc** ($|z| = R$, $|\theta| \le \pi/(2\alpha)$): Using hypothesis (2) and the uniform decay bound: \begin{align*} |\varphi_\varepsilon(Re^{i\theta})| &\le Ce^{R^\beta} \cdot e^{-\varepsilon R^\gamma \delta_0} = Ce^{R^\beta - \varepsilon \delta_0 R^\gamma}. \end{align*} Since $\gamma > \beta$, the exponent $R^\beta - \varepsilon \delta_0 R^\gamma \to -\infty$ as $R \to \infty$. In particular, for $R$ large enough (depending on $\varepsilon, C, \beta, \gamma, \delta_0$), we have $|\varphi_\varepsilon| \le M$ on the entire arc. [guided] We work on the bounded domain $\Omega_R$ where the [Maximum Modulus Principle](/theorems/491) applies. The boundary $\partial\Omega_R$ has three pieces: 1. The two straight segments along the rays $\operatorname{arg}(z) = \pm\pi/(2\alpha)$ from $0$ to $R$. 2. The circular arc $\{|z| = R, |\operatorname{arg}(z)| \le \pi/(2\alpha)\}$. On the straight segments, the barrier satisfies $|e^{-\varepsilon z^\gamma}| = e^{-\varepsilon r^\gamma \delta_0} \le 1$, and the boundary hypothesis gives $|f(z)| \le M$, so $|\varphi_\varepsilon(z)| \le M$. On the circular arc, we combine the growth bound with the uniform decay: \begin{align*} |\varphi_\varepsilon(Re^{i\theta})| &\le Ce^{R^\beta} \cdot e^{-\varepsilon \delta_0 R^\gamma} = Ce^{R^\beta - \varepsilon \delta_0 R^\gamma}. \end{align*} Since $\gamma > \beta$, the term $\varepsilon \delta_0 R^\gamma$ dominates $R^\beta$ for large $R$. Concretely, $R^\beta - \varepsilon \delta_0 R^\gamma = R^\beta(1 - \varepsilon \delta_0 R^{\gamma - \beta}) \to -\infty$ as $R \to \infty$. So for $R$ sufficiently large, $Ce^{R^\beta - \varepsilon \delta_0 R^\gamma} < 1 \le M$, and the bound $|\varphi_\varepsilon| \le M$ holds on the entire circular arc. This is where the choice $\gamma > \beta$ is consumed: if we had used $\gamma = \beta$ (or $\gamma < \beta$), the exponential decay would not overpower the growth, and the argument would fail. [/guided] [/step] [step:Apply the Maximum Modulus Principle on $\Omega_R$ to establish $|\varphi_\varepsilon(z)| \le M$ on $\overline{\Omega}$] The domain $\Omega_R$ is bounded, and $\varphi_\varepsilon \in \mathcal{O}(\Omega_R) \cap C(\overline{\Omega_R})$. For $R$ large enough, we have shown $|\varphi_\varepsilon(z)| \le M$ on all of $\partial\Omega_R$. By the [Maximum Modulus Principle](/theorems/491), the maximum of $|\varphi_\varepsilon|$ on $\overline{\Omega_R}$ is attained on $\partial\Omega_R$, so \begin{align*} |\varphi_\varepsilon(z)| &\le M \quad \text{for all } z \in \overline{\Omega_R}. \end{align*} Since $R$ can be taken arbitrarily large, this gives $|\varphi_\varepsilon(z)| \le M$ for all $z \in \overline{\Omega}$. [guided] The [Maximum Modulus Principle](/theorems/491) states that if $g$ is holomorphic on a bounded open set $U$ and continuous on $\overline{U}$, then $\max_{\overline{U}} |g| = \max_{\partial U} |g|$. We apply this to $g = \varphi_\varepsilon$ and $U = \Omega_R$. We verify the hypotheses: $\Omega_R$ is bounded (contained in the disk $|z| < R$), $\varphi_\varepsilon$ is holomorphic on $\Omega_R$ (product of holomorphic functions $f$ and $e^{-\varepsilon z^\gamma}$), and continuous on $\overline{\Omega_R}$ (since both factors extend continuously to the closure). The hypotheses are satisfied. From the previous step, $|\varphi_\varepsilon| \le M$ on $\partial\Omega_R$ for $R$ large enough. The Maximum Modulus Principle then gives $|\varphi_\varepsilon(z)| \le M$ for all $z \in \overline{\Omega_R}$. Now fix any $z_0 \in \overline{\Omega}$. For any $R > |z_0|$ that is large enough for the boundary estimate to hold, we have $z_0 \in \overline{\Omega_R}$, hence $|\varphi_\varepsilon(z_0)| \le M$. Since $z_0$ was arbitrary, $|\varphi_\varepsilon(z)| \le M$ on all of $\overline{\Omega}$. [/guided] [/step] [step:Send $\varepsilon \to 0$ to conclude $|f(z)| \le M$ on $\overline{\Omega}$] We have established $|\varphi_\varepsilon(z)| = |f(z)| \cdot e^{-\varepsilon r^\gamma \cos(\gamma\theta)} \le M$ for all $z \in \overline{\Omega}$ and every $\varepsilon > 0$. Fix any $z_0 = r_0 e^{i\theta_0} \in \overline{\Omega}$. Then \begin{align*} |f(z_0)| &\le M \cdot e^{\varepsilon r_0^\gamma \cos(\gamma\theta_0)} \end{align*} for every $\varepsilon > 0$. As $\varepsilon \to 0^+$, the right-hand side converges to $M \cdot e^0 = M$. Therefore \begin{align*} |f(z_0)| &\le M \end{align*} for all $z_0 \in \overline{\Omega}$, which is the desired conclusion. [guided] The final step is standard in Phragmen-Lindelof arguments. We have $|f(z)| \le M e^{\varepsilon r^\gamma \cos(\gamma\theta)}$ for every $\varepsilon > 0$. The factor $e^{\varepsilon r^\gamma \cos(\gamma\theta)}$ depends on $\varepsilon$ and on $z$, but for any fixed $z$, it converges to $1$ as $\varepsilon \to 0^+$. Since the inequality holds for all $\varepsilon > 0$, taking the limit gives $|f(z)| \le M$. Note that we cannot take $\varepsilon \to 0$ before fixing $z$ — the convergence is pointwise, not uniform. But pointwise convergence is all we need, since we want the bound $|f(z)| \le M$ for each fixed $z$. [/guided] [/step]