$\int_U |f(z)| |dz|^2$ Is this complex or topology?

Question

I am totally clueless. Can anyone help me ? What does that operator modulus and square means?

Answer 1

$\partial U$ is the boundary of $U$, and $d(z, \partial U)$ denotes the distance from $z$ to the boundary of $U$: $$ d(z, \partial U) = \inf \{ |z - w| : w \in \partial U \} \, . $$

If $0 \le \rho \le r < d(z, \partial U)$ then the disk $B(z, r)$ is contained in $U$, and an application of Cauchy's integral formula gives $$ |f(z)| \le \frac{1}{2 \pi} \int_{0}^{2 \pi} |f(z + \rho e^{it}| dt \,. $$ In other words: $|f|$ is subharmonic, so that its value at a point is $\le$ the average of the values in a circle around that point.

It follows (by integration with respect to $\rho \, d\rho$ over the interval $[0, r]$) that the value at a point is $\le$ the average of the values in a disk around that point, and that can be estimated above by the average value in $U$ (which is supposed to be finite):

$$ |f(z)| \le \frac{1}{\pi r^2} \int_{B_r(z)} |f(z)| |dz|^2 \le \frac{1}{\pi r^2} \int_{U} |f(z)| |dz|^2 \, . $$

(As Christian Blatter said, $|dz|^2$ is used here as a notation for the area element $dx \, dy$.)

Now take the limit for $r \to d(z, \partial U)$.