Throughout this post, let $(a,b,c,d)$ refer to the entries of some unspecified element: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{GL}_2(\Bbb Z)$$By a “linear fractional equivalence” (LFE) I mean a map: $$F(x)=\frac{ax+b}{cx+d}$$Over all real $x$ where it can be defined. For any real $x$ and positive real $\gamma$, let: $$Q_{x,\gamma}=\left\{(p,q)\in\Bbb Z^2:\left|x-\frac{p}{q}\right|<\frac{1}{q^\gamma},q>0,\mathrm{gcd}(p,q)=1\right\}$$And define the irrationality measure: $$\begin{align}\mu:\Bbb R&\to[1,\infty]\\x&\mapsto\sup\{\gamma>0:|Q_{x,\gamma}|=\aleph_0\}\end{align}$$
Clearly, any LFE $F$ will map rationals to rationals. A comment left on a MO post suggested that $\mu(F(x))=\mu(x)$ for all $x$ where the LFE $F$ is defined: I found this to be an interesting claim, and with some help from others I think I’ve found a proof. I’d really appreciate feedback on the correctness of said proof!
Proof of LFE invariance:
Take any LFE $F(x)=\frac{ax+b}{cx+d}$ and note that, if $x=p/q$ as a coprime ratio, then $F(x)=\frac{ap+bq}{cp+dq}$ is also a coprime ratio since any divisor of both the numerator and denominator must divide $d(ap+bq)-b(cp+dq)=\pm p$ and $c(ap+bq)-a(cp+dq)=\mp q$, so any divisor must be $\pm1$ since $p,q$ are coprime. Since every rational $x$ has $\mu(x)=1$ and every $F(x)$ will also be rational, it suffices to prove the claim for irrational $x$, which satisfy $\mu(x)\ge2$.
Fix any irrational $\xi$ and $0<\varepsilon<1$. Put $\mu:=\mu(\xi)\ge2$, and let $p,q$ refer to elements $(p,q)\in Q_{\xi,\mu-\varepsilon}$, which is an infinite set containing pairs with arbitrarily large $q$. Indeed, consider $(p,q)$ with $q$ large enough such that $F$ is well-defined on the interval $\left[\frac{p-1}{q},\frac{p+1}{q}\right]$, which contains $\xi$ as $\mu-\varepsilon>1$. Let $D>0$ be a (finite) upper bound for $|F’|$ on this interval. By the mean value theorem: $$|F(\xi)-F(p/q)|\lt\frac{D}{q^{\mu-\varepsilon}}$$We have $F(p/q)=\frac{ap+bq}{cp+dq}=\frac{r}{s}$ in lowest terms, up to a change of sign in $\pm(cp+dq)$ taking $s$ positive. Note that $|p|\lt\max\{|q\xi-1|,|1+q\xi|\}\le Mq$ where $M:=1+|\xi|$(again using $\mu-\varepsilon>1$). Then we can bound $s\lt|c|Mq+|d|q=Aq$ for a constant $A=|d|+|c|M$.
If I want $s^b<q^a$ for any positive constants $a,b,k$, I can settle for $s^b<A^bq^b<kq^a$ precisely when $\frac{A^b}{k}<q^{a-b}$. No matter what choice of $b\in(0,a)$ I make, I can always pick infinitely many $q$ large enough, from $Q_{\xi,\mu-\varepsilon}$, that will satisfy the above inequality. Then for any $0<\varepsilon’$ I can have $\frac{D}{q^{\mu-\varepsilon}}<\frac{1}{s^{\mu-\varepsilon-\varepsilon’}}$, hence: $$\left|F(\xi)-\frac{r}{s}\right|<\frac{1}{s^{\mu-\varepsilon-\varepsilon’}}$$
That proves that $Q_{F(\xi),\mu-\varepsilon-\varepsilon’}$ always contains infinitely many such $(r,s)$: by the arbitrary nature of $\varepsilon,\varepsilon’$, we conclude that $\mu(F(\xi))\ge\mu$. However, $F^{-1}$ is defined at $F(\xi)$ and is itself an LFE, so by this symmetry we have $\mu\ge\mu(F(\xi))$.
That concludes $\mu(F(\xi))=\mu(\xi)$ for all irrational $\xi$ as desired.
Is that right? Feel free to comment more elegant solutions, or reference other classes of function $T$ such that $\mu(T(x))=\mu(x)$. Many thanks!