# Using normal families to bound a complex integral

I am trying to prove that

$$\int_{\partial T(Q)} |F'(z)| \,ds(z) \lesssim \int\int_{T(Q)} |F'(z)| |\varphi'(z)|^2 \log \frac{1}{|z|} \,dx\, dy$$

This is an estimate on page $6$ of this paper by Peter Jones, and the author states that this follows "by normal families and" and knowing that $\varphi'$ has Bloch norm at most $6$. The symbol $\lesssim$ means $\le$ with constants that should not depend on $F$, $\varphi$ or $Q$.

For notation, $F$ is a Riemann map (i.e. conformal equivalence) between the disk and some other simply connected region whose boundary has finite length (so that $F'$ is in the Hardy class $H^1$). Then $\varphi$ is a logarithm of $F'$, sot hat $e^{\varphi} = F'$; earlier in the paper, it's noted that $\varphi$ has Bloch norm at most $6$, i.e. $$|\varphi'(z)| \le 6 (1 - |z|^2)^{-1}$$

We can decompose the circle $T$ into dyadic arcs $I$ of length $\pi 2^{-n}$, and define "squares"

$$Q = Q_I = \left\{z : \frac{z}{|z|} \in I, 1 - \pi 2^{-n} \le |z| \le 1\right\}$$ (so that $Q$ is part of a sector of the circle, and its "side-lengths" are all comparable). Let $z_{Q}$ be the center of this region, and $T_Q$ it upper half - i.e. $\{z \in Q : 1 - \pi 2^{-n} \le |z| \le 1 - \pi 2^{-n - 1}\}$.

Finally, assume that there exists a $z \in T_Q$ with $|\varphi(z) - \varphi(z_Q)| \ge \epsilon$. I believe this is necessary - in the extreme case, when $\varphi$ is constant, the right-hand integral is $0$ while the left is not.

My question: I don't see what is meant by "by normal families" here, or what normal family is being considered here. If anyone could clarify this statement, or point out some general theorem this is referencing (or refer me to a similar, but wordier proof), I would greatly appreciate it.

Remark This is somewhat similar to an inequality proved earlier in the paper, namely

$$\int\int_{\mathbb D} |F'(z)| |\varphi'(z)|^2 \log \frac{1}{|z|} dx dy \le 4 \int_{T} |F'(e^{i\theta})| \,d\theta$$

which follows from Green's formula via

$$\int_T |G(e^{i\theta}) - G(0)|^2 d\theta = \int\int_{\mathbb{D}} |G'(z)|^2 \log \frac{1}{|z|} \,dx\, dy$$ where $G$ is a square root of $F'$. However, the inequality here goes in the other direction.

## 1 Answer

Let's begin by simplifying things a bit. Anything pinched between two comparable positive constants can be treated as a constant. For example, $\log \frac{1}{|z|} \approx 2^{-n}$ on $T(Q)$.

Also, since $|\varphi'|\lesssim 2^n$, the oscillation of $\operatorname{Re}\varphi$ on $T(Q)$ is bounded by a constant. Hence, $|e^\varphi| =e^{\operatorname{Re}\varphi}$ is pinched between two comparable constants. Since it appears on both sides, cancel it out. It remains to show that $$\int_{\partial T(Q)} 1\, ds(z) \lesssim 2^{-n} \iint_{T(Q)} |\varphi'(z)|^2 \, dx\, dy$$ which further simplifies to $$1 \lesssim \iint_{T(Q)} |\varphi'(z)|^2 \, dx\, dy \tag2$$ To show (2), we must relate the integral on the right to the oscillation of $\varphi$ (which is at least $\epsilon$). We need some facts:

1. a ball in the Bloch space is a normal family (the Bloch seminorm controls continuity on compact subsets)
2. the Bloch seminorm is invariant under Möbius transformations of the unit disk
3. The right hand side of (2) is invariant under conformal changes of the variable $x+iy$.

Move the center of $T(Q)$ to the origin by a Möbius transformation. The image of $T(Q)$ contains a fixed domain $\Omega \ni 0$, independent of $Q$. It is also contained in a fixed domain $\Omega'$, compactly contained in the unit disk. If (2) fails, there is a sequence $\varphi_k$ which converges locally uniformly in the unit disk, so that the integrals $\iint_{\Omega} |\varphi_k'(z)|^2 \, dx\, dy$ tend to zero. The limit of such a sequence must be a constant function. But then the oscillation of $\varphi_k$ on $\Omega'$ tends to zero, which contradicts the lower oscillation bound $\epsilon$.