Bound areas of disks with respect to a quadratic differential

Question

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.)

Answer 1

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.

Answer 2

Am I right that in fact, your question can be re-stated in the following way: if $q(z)$ is a holomorphic function in $D(1)$ such that the average of $|q|$ over this disc is equal to 1, then the average of $|q|$ over $D(r)$ is at most 1? If yes, the first idea that comes to my mind would be to use that $|q|$ is subharmonic. Then you can proceed in two ways:

1) For a subharmonic function, if you average it w.r.t. a measure and then diffuse this measure, the average changes non-decreasingly. So -- you can launch a Brownian motion with the initial condition chosen w.r.t. the uniform measure on $D(r)$, and stop it in a (Markovian) stopping time chosen in such a way that the new distribution is uniform on $D(1)$ (a Skorokhod-like problem). There is a way of doing so, but it could be overcomplicated to write down.

2) The previous part brings into attention that the family of averages on $D(r)$ should be non-decreasing in $r$. And to show this, it suffices to show that an average on the radius $r$ circle is non-decreasing in $r$. And to compare these two averages, the previous arguments work quite well: you launch a BM from any point of an inner circle, and let it stop once it intersects the outer one. The rotational symmetry implies that the exit points are also distributed uniformly; and whatever you gain during this random walk (that is, the integral of $\Delta |q|\ge 0$ w.r.t. the expectation of the occupation measure) is the difference between the two.