Question concerning the mean-value property If $U$ is an open subset of $\mathbb C$, the mean square norm is defined as:
$$||f||_{L^2 (U)} = {\left(\int_U |f(z)|^2 dxdy\right)}^{1/2}$$
And the norm supremum is defined as:
$$||f||_{L^{\infty} (U)} = \sup_{z \in  U} |f(z)|$$
If $f$ is holomorphic in a neighborhood of $D_r(z_0)$, show that for any $0 < s < r,  \exists C > 0$, a constant, which depends on $s$ and $r$ such that:
$$||f||_{L^{\infty} (D_s(z_0))}\le C||f||_{L^{2} (D_r(z_0))}$$
And prove that if $\{f_n\}$ is a Cauchy sequence of holomorphic functions in the mean square norm, then the sequence $\{f_n\}$ converges uniformly on every compact subset of $U$ to a holomorphic function.
 A: Let $d > 0$. If $g$ is holomorphic on some neighbourhood of $\overline{D_d(z)}$, the mean-value theorem tells us that
$$g(z) = \frac{1}{2\pi} \int_0^{2\pi} g(z + \rho e^{i\varphi})\,d\varphi$$
for $0 \leqslant \rho \leqslant d$, hence
$$\lvert g(z)\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \lvert g(z + \rho e^{i\varphi})\rvert\,d\varphi. \tag{1}$$
Multiplying $(1)$ with $\rho$ and integrating from $0$ to $d$ we obtain
$$\lvert g(z)\rvert \frac{d^2}{2} \leqslant \frac{1}{2\pi} \int_0^d \int_0^{2\pi} \lvert g(z + \rho e^{i\varphi})\rvert\,\rho\,d\varphi\,d\rho = \frac{1}{2\pi} \iint_{\lvert w - z\rvert \leqslant d} \lvert g(w)\rvert\,dx\,dy.\tag{2}$$
Letting $g = f^2$ in $(2)$ and dividing by $d^2/2$, then  taking the square root, we obtain
$$\lvert f(z)\rvert \leqslant \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2}(D_d(z)). \tag{3}$$
Choose $d = r-s$ to see
$$\lvert f(z)\rvert \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_{r-s}(z))} \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}\tag{4}$$
for all $z \in D_s(z_0)$, whence
$$\lVert f\rVert_{L^{\infty}(D_s(z_0))} \leqslant \frac{1}{\sqrt{\pi}\,(r-s)} \,\lVert f\rVert_{L^2(D_r(z_0))}.$$
If $K \subset U$ is compact, choose a positive $d$ smaller than the distance from $K$ to $\mathbb{C}\setminus U$, so that $\overline{D_d(z)} \subset U$ for all $z\in K$. Estimating the right hand side of $(3)$ by $\frac{1}{\sqrt{\pi}\,d}\, \lVert f\rVert_{L^2(U)}$, we see
$$\lVert f\rVert_{L^{\infty}(K)} \leqslant \frac{1}{\sqrt{\pi}\,d} \,\lVert f\rVert_{L^2(U)}$$
for every square-integrable holomorphic $f \colon U \to \mathbb{C}$.
Thus if $(f_n)$ is an $L^2(U)$-Cauchy sequence of holomorphic functions, the restricted sequence $\bigl(f_n\rvert_K\bigr)$ is an $L^{\infty}(K)$-Cauchy sequence, hence $\bigl(f_n\rvert_K\bigr)$ converges uniformly. Since the compact subsets of $U$ cover $U$, it follows that $(f_n)$ converges pointwise on all of $U$, and by the above the convergence is uniform on all compact subsets of $U$. By Weierstraß' convergence theorem, the limit function is holomorphic on $U$.
