Let $q = q(z)\,dz^2$ be a quadratic differential on the unit disk $D(1)$, normalized so that $\int\_{D(1)}|q| = 1$. I have a fairly convoluted argument that for any smaller disk $D(r)$ with $r < 1$, there is an upper bound on the area:
$$
\int\_{D\_r} |q| \le r^2.
$$
(The optimal quadratic differential is just $dz^2$.) This result feels quite elementary, and there ought to be a better proof; does anyone have one?


(This fits into a larger context of more general surfaces and subsurfaces. I can prove similar statements there with non-explicit bounds, but this would be a nice warm-up example if there's a short proof.)



Bound areas of disks with respect to a quadratic differential
The family of averages of $|q|$ on $D(r)$ is indeed non-decreasing. To see this, let $A(r)$ be the average of $|q|$ on $D(r)$. Let $r<\rho$ Then $A(r)=v(r)$, where
$$v(z)=\frac{1}{\pi}\int\_{0}^{2\pi} \int\_{0}^{1}|q(zse^{it})|s \, ds \, dt.$$
Now, such a $v$ is subharmonic on $D(\rho)$ (see Ransford's book "Potential theory in the complex plane", theorem 2.4.8.). Clearly $v$ is also radial. By the maximum principle, we get
$$A(r)=v(r) \leq \sup\_{\partial D(\rho)} v = v(\rho)=A(\rho).$$


This also works if $|q|$ is replaced by any subharmonic function.


