Let \(f\) be analytic in a region \(U\), and suppose \[ \int_U |f(z)|\,|dz|^2<\infty. \] Prove that there is a constant \(M\) such that for every \(z\in U\), \[ |f(z)|\le \frac{M}{d(z,\partial U)^2}. \]
Solution
Set \[ L=\int_U |f(w)|\,|dw|^2<\infty. \] Then one may take \[ \boxed{M=\frac{L}{\pi}}. \]
1. Fix \(z\in U\) and let\[ 0<r<d(z,\partial U). \]Then the disk \(D(z,r)\) is contained in \(U\).
By the definition of distance to the boundary, no point at distance less than \(d(z,\partial U)\) from \(z\) can lie outside the component of \(U\) containing \(z\). Hence every disk of radius strictly smaller than that distance lies in \(U\).
2. The analytic mean-value formula gives\[ f(z)=\frac1{\pi r^2}\int_{D(z,r)}f(w)\,|dw|^2. \]
1. Combining thetwo real identities yields the displayed complex-valued formula.
3. Consequently\[ |f(z)| \le \frac{L}{\pi r^2}. \]
Taking absolute values in 2 and using the triangle inequality gives \[ |f(z)| \le \frac1{\pi r^2} \int_{D(z,r)}|f(w)|\,|dw|^2. \] Since \(D(z,r)\subset U\), the last integral is at most \(L\).
4. Letting \(r\uparrow d(z,\partial U)\) gives\[ |f(z)| \le \frac{L}{\pi\,d(z,\partial U)^2}. \]
The bound in 3 holds for every radius \(r<d(z,\partial U)\). Taking the limit from below yields the stated estimate. Thus the required constant is \(M=L/\pi\), independent of \(z\).