The following is the mean value property of holomorphic function:
Let $f(z)$ be a holomorphic function on the disc $D(z_{0},R)$, then $$f(z_{0})=\dfrac{1}{2\pi}\int_{0}^{2\pi}f(z_{0}+re^{i\theta})d\theta,\ \ \text{for all}\ \ 0<r<R.$$
Basically it tells us the value of the holomorphic function at the center of the disc is the average mean of it on the boundary circle of any smaller disc.
I am wondering if we can extend this to a double integral, something like $$f(z_{0})=C\int_{D(z_{0}, R)}f(x+iy)dxdy=C\int_{0}^{2\pi}\int_{0}^{R}f(z_{0}+re^{i\theta})rdrd\theta,$$ for some constant $C$.
The reason I want something like this because I want to prove that if $f(z)$ is holomorphic function on $D(z_{0}, R)$, then $$|f(z_{0})|\leq \dfrac{C}{R}\Big(\int_{D(z_{0}, R)}|f(x+iy)|^{2}dxdy\Bigg)^{\frac{1}{2}}.$$ I have tried several method like using the power series expansion of $f\overline{f}$, which kinds of work, but somehow in the end you need to argue that you can interchange $\int_{0}^{R}$ with the infinite series, and I am not sure after so many manipulations if I can still argue the interchange by uniform and absolute convergence.
After several try, I thought perhaps working on series is not really a good approach, so I tried to use the Mean-Value property, but it only has a single integral.
If we can have a double integral like above, then I can directly replace $f:=f^{2}(z)$, and take square root.
Thank you!
Best Answer
Thanks to Conrad's inspiring suggestion (I was being really dumb), here is the solution.
I want to prove
As $g$ is holomorphic on $D(b,\epsilon)$, then so is $g^{2}(z)$. Hence, by the Mean-Value Property, we have $$g^{2}(b)=\dfrac{1}{2\pi}\int_{0}^{2\pi}g^{2}(b+re^{i\theta})d\theta.$$ Hence, $$\int_{D(b,\epsilon)}g^{2}(x+iy)dxdy=\int_{0}^{\epsilon}r\int_{0}^{2\pi}g^{2}(b+re^{i\theta})d\theta dr=\int_{0}^{\epsilon}r\cdot 2\pi g^{2}(b)dr=\pi\epsilon^{2}g^{2}(b).$$
Hence, $$g^{2}(b)=\dfrac{1}{\pi\epsilon^{2}}\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy$$ and thus $$|g(b)|=\dfrac{C}{\epsilon}\Bigg(\int_{D(b,\epsilon)}|g(x+iy)|^{2}dxdy\Bigg)^{\frac{1}{2}},$$ where $C=\frac{1}{\pi}$.
In particular, the Mean-value property can be extended to a double integral: $$\int_{D(z_{0},R)}f(x+iy)dxdy=\int_{0}^{R}r\int_{0}^{2\pi}f(z_{0}+re^{i\theta})d\theta dr=2\pi f(z_{0})\int_{0}^{R}rdr=\pi R^{2}f(z_{0}).$$