Let $$f(x) = \frac {ax +b} {cx + d}$$ with $a,b,c,d \in \mathbb R, \neq 0$.
Suppose $f$ has fixed points $u,v, u \neq v$, and further suppose that $f(f(x)) = x$ wherever this is defined. Show that there exists a point $r = \frac {u +v } 2$ such that the domain and image of $f$ are both $\mathbb R – r$.
Source: Inspired by generalizing this problem. Note: Other solutions may exist. This question asks to verify or critique this solution below, which, after the lemma, uses geometrical symmetry.
Lemma: The function $f(x) = \frac {ax +b} {cx + d}, c \neq 0$ is either a constant or a rectangular hyperbola with asymptotes parallel to the $x$ and $y$ axes.
Proof: Write $$f(x) = S \cdot \frac 1 {x – V_a} + H_a$$ with $H_a = \frac a c, V_a = \frac {-d} c$ and $S$ as needed below and note $$\begin{align*}
f(x) &= \frac {ax + b} {cx + d} \\
&= \frac {a(x + \frac d c) – \frac {ad}{c} + b} {c(x + \frac d c)} \\
&= \frac a c + \frac {- \frac {ad} {c} + b}{c} \cdot \frac {1}{x + \frac {d} {c}}\\
&= S \cdot \frac 1 {x – V_a} + H_a.
\end{align*}$$
If $f$ is non-constant, then $S \neq 0$, giving a rectangular hyperbola with asymptotes $y = H_a$ and $x = V_a$.
Main Proof: Since $u \neq v$, then $f$ is non-constant. By the lemma, $f$ is a hyperbola with asymptotes parallel to the $x$ and $y$ axes. Such a curve is defined everywhere but its vertical asymptote, and has image everywhere but its horizontal asymptote. The hyperbola's center is the unique point on both of these asymptotes; thus it suffices to show that its center is $(\frac {u+v} 2, \frac {u+v} 2)$.
Since $f(f(x)) = x$, then the line $y = x$ is an axis of symmetry of $f$. A hyperbola has only two axes of symmetry: its major (aka transverse) axis and its minor (aka conjugate) axis. Since $f$ has fixed points but a hyperbola never intersects its minor axis, then the line $y = x$ must be the major axis of $f$.
A hyperbola intersects its major axis in exactly two points, which for $f$ must be $(u,u)$ and $(v,v)$. Its center is their midpoint, which is consequently $(\frac {u+v} 2, \frac {u+v} 2)$. This completes the proof.
Best Answer
I read your proof and it seems correct, i like that it is really geometric. The only thing to me is that you use facts about hyperbolas which, although they might be well known, do not seem trivial to check from the analytic definition that you give. To you in the first lemma of the proof, an hyperbola is just the image of a function with the structure $\frac{A}{x-B} + C$, but the properties you use later (existence of major-minor axes of symmetries, their properties, number of intersections) are not really clear from this definition (at least to me).
This might be purely subjective, but since the original statement is written in analytic terms (fixed points, image, domain, the property $f(f(x)) = x$...) it calls to me that an analytic solution might be more suitable if possible.
I provide here another, maybe less geometric solution, but to me more elementary if you just know basic calculus. I know this is out of the scope of your question, as you said, but maybe it will be interesting to another one.
Take as requested $a,b,c,d\in\mathbb{R}-\lbrace 0\rbrace$. We assume $f(x) = \frac{ax+b}{cx+d}$ has two distinct fixed points. Since $c,d\neq 0$, this function is well defined for every $x \neq -\frac{d}{c}$, hence its domain is $\mathbb R -\lbrace -\frac{d}{c}\rbrace$.
The condition $f(f(x)) = x$ implies that for every $x\neq \frac c d$: $$x = \frac{a\frac{ax+b}{cx+d} + b}{c\frac{ax+b}{cx+d}+d}$$ which implies that, for every $x \neq \frac{c}{d}$: $$(ca+cd)x^2 + (d^2 - a^2)x-(ab+bd) = 0$$ This can only happen if each coefficient in this polynomial is zero, in particular $c(a+d) = 0$, from where we get $a=-d$ since $c\neq 0$.
Now, the two possible fixed points are given by the formula $x = \frac{-dx+b}{cx+d}$, therefore both fixed points must satisfy $cx^2 +2dx -b = 0$ Hence:
$$x_{\pm} = \frac{-2d\pm\sqrt{4d^2 +4bc}}{2c}$$
Call $u = x_+, v = x_-$. Notice that we have $\frac{u+ v}{2} = -\frac{d}{c}$ as requested. It only rests to see that the image of $f$ is $\mathbb R - \lbrace -\frac d c \rbrace$. But setting $f(x) = y$ and isolating $x$ we can see that the inverse function of $f$ is: $$g(y) = \frac{-dy+b}{cy-a}$$ which is only not defined at $y=\frac a c = -\frac d c$.