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Let $\phi(z)=\frac{az+b}{cz+d}$ be a Möbius transformation, with $a,b,c,d\in \mathbb{R}$ and $ad-bc=1$ (so $\phi$ can be viewed as an element of $SL_2(\mathbb{R})$. Let $\mathbb{H}=\{z\in\mathbb{C}: \Im(z)>0\}$ be the upper half-plane, and fix a compact set $K\subset \mathbb{H}$. Suppose $z,\phi(z)\in K$. Can we provide a bound for $a,b,c,d$ depending only on the set $K$?

Since $K\subset\mathbb{H}$, in particular $K\cap \mathbb{R}=\emptyset$, so $K$ is at a positive distance from the real axis. Since it is also bounded, we deduce the existence of positive numbers $r$ and $M$ such that the inequalities

$$ r \le |w|\le M$$ $$r\le \Im (w)\le M$$

hold for every $w\in K$, and so they hold in particular for $z$ and $\phi(z)$. I've tried playing around with this but got nowhere. I also tried to use the fact that $$\Im \phi(z)=\frac{\Im z}{|cz+d|^2}$$ but I keep getting bounds along the lines of $(|cz|-|d|)^2\le \frac{M}{r}$ which I can't use to dominate $c$ or $d$. I've been able to solve the related problem $|\phi(i)-i|\le \varepsilon \implies |a|,|b|,|c|,|d|\le C_\varepsilon$, where the computations are much simpler, and tried using adequate Möbius maps to transfer the general case to this one, but couldn't get it to work.

EDIT: I was able to get

$$r\le |\Im \phi(z)|=\frac{|\Im z|}{|cz+d|^2}\le \frac{M}{c\Im z}\le \frac{M}{c^2r^2}$$

and so $c$ is bounded. But then

$$r\le |\Im \phi(z)|\le \frac{M}{|\Re (cz+d)|^2}=\frac{M}{|d+c\Re z|^2}$$

and thus $|d+c\Re z|\le \left(\frac{M}{r}\right)^{1/2}$, so $d$ must be bounded, since $c$ and $|z|$ are.

I still haven't been able to control $a$ and $b$. Getting lower bounds for $|c|$ and $|d|$ might be useful, since we know $ad-bc=1$, but I don't know if that is possible.

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  • $\begingroup$ In principle, we could provide bounds, yes. But these bounds (assuming you want the tightest ones possible) would be dependent on the shape of $K$. It's possible to construct arbitrarily nasty $K$ for which solving the bounds is increasingly difficult. So I'm unsure what exactly you're asking for here. $\endgroup$ – Myridium Jul 26 '17 at 21:10
  • $\begingroup$ @Myridium I'm sorry, but I don't think I understand what you're saying. I do not need the bounds to be at all sharp, the existence of the bound is enough. Regarding your shape comment, we could enclose any $K$ in a rectangle entirely contained within the half-plane, and use the bounds for the rectangles, so, as long as efficiency is irrelevant, it suffices to study rectangles. $\endgroup$ – Reveillark Jul 26 '17 at 21:17
  • $\begingroup$ "Suppose $z,g(z)\in K$." What does that mean? $\endgroup$ – zhw. Jul 28 '17 at 20:26
  • $\begingroup$ @zhw. Whoops, I meant $\phi(z)$, thanks! $\endgroup$ – Reveillark Jul 28 '17 at 20:31
  • $\begingroup$ So you just assume $z,\phi (z) \in K$ for one $z?$ $\endgroup$ – zhw. Jul 28 '17 at 20:35
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Let's first consider the case of $z=i \in K.$ Then

$$\tag 1 \phi(i) = \frac{ai+b}{ci+d} = \frac{ac+bd+i(ad-bc)}{c^2+d^2} = \frac{ac+bd+i}{c^2+d^2}.$$

So the imaginary part of $\phi(i)$ equals $1/(c^2+d^2).$ Because $K$ is compact, the imaginary parts of points in $K$ must be bounded above and below by positive constants (depending on $K$). This implies $c^2+d^2$ is also bounded above and below by positive constants depending on $K$.

On the other hand,

$$|\phi(i)| =\frac{|ai+b|}{|ci+d||}= \frac{ (a^2+b^2)^{1/2} }{ (c^2+d^2)^{1/2} }.$$

Since $|\phi(i)|$ is itself bounded above and below by positive constants (again depending on $K$), the same must be true of the numerator $(a^2+b^2)^{1/2},$ since we've already shown this for the denominator.

We're done in the case $z=i.$ But I think the general case is reducible to this case. After all, whatever $z,K$ are, we are just off by a translation and dilation, the former bounded and the latter bounded above and below by positive constants (again depending on $K$).

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