\(\displaystyle \|f\|_{(\infty, s)} = \text{sup}_{z \in D_s(z_0)}|f(z)|\) and \(\displaystyle \|f\|_{(1, r)} = \int_{D_r(z_0)} |f(z)|dx dy\).
Solution
1. Mean Value Property for holomorphic functions:
1.1. Let \(w \in D_s(z_0)\) and set \(\rho = r - s > 0\).For any \(\zeta \in D_\rho(w)\), the triangle inequality gives:\[ |\zeta - z_0| \le |\zeta - w| + |w - z_0| < \rho + s = (r - s) + s = r. \]Thus the disc \(D_\rho(w)\) is entirely contained in \(D_r(z_0)\).
1.2. By Cauchy’s Integral Formula, for any radius \(t \in (0, \rho)\):\[ f(w) = \frac{1}{2\pi} \int_0^{2\pi} f(w + t e^{i\theta}) \, d\theta. \]
2. Area mean value formula:
2.1. Multiplying both sides of the circular mean value identity by \(t\) and integrating from \(t = 0\) to \(t = \rho\):\[ \int_0^\rho t f(w) \, dt = \frac{1}{2\pi} \int_0^\rho \int_0^{2\pi} f(w + t e^{i\theta}) t \, d\theta \, dt. \]
2.2. Evaluating the left-hand side gives \(\frac{\rho^2}{2} f(w)\), and changing to Cartesian coordinates on the right-hand side gives:\[ \frac{\rho^2}{2} f(w) = \frac{1}{2\pi} \iint_{D_\rho(w)} f(z) \, dx \, dy \implies f(w) = \frac{1}{\pi \rho^2} \iint_{D_\rho(w)} f(z) \, dx \, dy. \]
3. Norm estimate:
3.1. Taking absolute values and using the sub-mean value property:\[ |f(w)| \le \frac{1}{\pi \rho^2} \iint_{D_\rho(w)} |f(z)| \, dx \, dy. \]
3.2. Since \(D_\rho(w) \subseteq D_r(z_0)\) and the integrand is non-negative:\[ \iint_{D_\rho(w)} |f(z)| \, dx \, dy \le \iint_{D_r(z_0)} |f(z)| \, dx \, dy = \|f\|_{(1, r)}. \]
3.3. Substituting \(\rho = r - s\):\[ |f(w)| \le \frac{1}{\pi (r - s)^2} \|f\|_{(1, r)}. \]Taking the supremum over all \(w \in D_s(z_0)\):\[ \|f\|_{(\infty, s)} \le \frac{1}{\pi (r - s)^2} \|f\|_{(1, r)}. \]
4. Conclusion:\(\|f\|_{(\infty, s)} \le \frac{1}{\pi (r - s)^2} \|f\|_{(1, r)}\) for all \(0 < s < r\). Q.E.D.