Let \(G \subset \mathbb{C}\) be a bounded region.
Show that if \(f \in H(G)\) and \(B = B(a,r) \subset G\), then \[ |f(a)| \leq \frac{1}{|B|} \int_B |f(z)|\,dx\,dy, \] where \(|B|\) denotes the area of \(B\).
Show that the space \[ A^1(G) = \{f \in H(G) : \|f\|_1 := \int_G |f(z)|\,dx\,dy < \infty\} \] endowed with the metric \(d(f,g) = \|f - g\|_1\) is complete.
Note: You may use the result in part (i) even if you did not prove this.
Solution
For (i), the subharmonic mean-value inequality for \(|f|\) gives \[ |f(a)|\le \frac1{2\pi}\int_0^{2\pi}|f(a+\rho e^{it})|\,dt \] for every \(0<\rho<r\). Multiply by \(2\rho\) and integrate from \(0\) to \(r\). Polar coordinates give \[ r^2|f(a)|\le \frac1\pi\int_{B(a,r)}|f(z)|\,dA(z), \] which is the stated inequality because \(|B|=\pi r^2\).
For (ii), let \((f_n)\) be Cauchy in \(L^1(G)\). Then it converges in \(L^1(G)\) to some \(f\in L^1(G)\). Fix a compact \(K\Subset G\) and choose \(r>0\) such that \(B(z,r)\subset G\) for every \(z\in K\). Applying (i) to \(f_n-f_m\) gives \[ \sup_K|f_n-f_m|\le \frac1{\pi r^2}\|f_n-f_m\|_1. \] Thus \((f_n)\) is uniformly Cauchy on every compact subset of \(G\), so it converges locally uniformly to a holomorphic function \(g\). A subsequence of the \(L^1\)-convergent sequence converges to \(f\) almost everywhere, while the same subsequence converges pointwise to \(g\); hence \(f=g\) almost everywhere. Therefore \(g\in A^1(G)\) and \(\|f_n-g\|_1\to0\). So \(A^1(G)\) is complete.