Bounded analytic function in $D$ invariant under a Mobius transformation

Question

Suppose that $\varphi(z)=\frac{az+b}{cz+d}$ is a Mobius transformation. What is the form of $\varphi$ such that one can have a bounded analytic function $f$ in the unit disc with \begin{equation*} f \circ \varphi=f? \end{equation*} I guess the answer is $ad-bc=1$, but I don't know how to start the proof. Can anyone have any suggestions?

Answer 1

For the problem to be non-trivial we assume that we require a nonconstant $f$ as otherwise any Mobius automorphism of the disc $\varphi(z)=e^{i\theta}\frac{z-a}{1-\bar a z}, |a|<1$ works.

Then we claim that $\varphi$ must have finite order so $\varphi^{(n)}(z)=z$ for some $n \ge 1$

(for example, the order two automorphisms are $-\frac{z-a}{1-\bar a z}$, while $\alpha z$ is of order $n$ for any $\alpha^n=1$ primitive root of order $n$ of unity)

One way is easy since if we pick $\psi$ an arbitrary disc automorphism and $\varphi$ has order $n \ge 2$ (for the identity the problem is trivial of course), then $f(z)=\psi(z) \cdot\psi\circ\varphi(z)\cdot...\cdot\psi\circ\varphi^{(n-1)}(z)$ is a Blaschke product of order $n$ hence bounded by $1$ and $f \circ \varphi=f$ by construction. More generally we can pick any bounded $g$ on the unit disc and repeat the construction above to get an $f \circ \varphi=f$, but the Blaschke products of order $n$ as above are "minimal".

Conversely since $f \circ \varphi=f$ implies $f \circ \varphi^{(n)}=f$ for any integral $n$ and $f$ nonconstant, it follows that if $\varphi$ has a fixed point the orbit of any $z$ is finite since it lies at a fixed distance in the hyperbolic metric from the fixed point, so it doesn't go to the boundary so there is $n(z) \ge 1, \varphi^{(n(z))}(z)=z$. But now a cardinality argument shows that there is an $n \ge 1$ for which the relation above holds for uncountably many $z$, so by the identity theorem $\varphi^{(n)}(z)=z$ so $\varphi$ has finite order.

The only case when there is no fixed point is when $\varphi(z)=\frac{z-a}{1-\bar a z}$ and $a \ne 0$. Then $\varphi^{(n)}(z)$ accumulate to the boundary and $f \circ \varphi=f$ with nonconstant $f$ immediately leads to a contradiction as the image of $f$ is open (take a normally convergent subsequence of $\varphi^{(n)}(z)$, it converges to some constant $\alpha$ of modulus $1$ etc)

(incidentally, disc automorphisms which are not the identity cannot have more than one fixed point by easy geometric reasonings using the invariance of hyperbolic metric distance under them so two fixed points would mean the full geodesic joining them being fixed points etc- or algebraically as Mobius transforms can have at most two fixed points and the disc automorphisms special form shows that if there are two fixed points their product has absolute value $1$, so at most one can be inside the unit disc)