Suppose \(f\) is holomorphic on the open unit disc \(D(0,1)\) and continuous on \(\overline{D(0,1)}\). Assume \(\abs{f(\xi)} < 1\) for \(\xi \in \partial D(0,1)\). Show that there exists a unique point \(a \in D(0,1)\) such that \(f(a) = a\).
Solution
1. There is \(q<1\) such that \(\abs{f(z)}\le q\) for every \(z\in D(0,1)\).
Since the boundary circle is compact and \(f\) is continuous there, \[ q\coloneqq\max_{\abs{\xi}=1}\abs{f(\xi)}<1. \] The maximum modulus principle gives \(\abs{f(z)}\le q\) throughout the disk.
2. The function \(g(z)=f(z)-z\) has exactly one zero in the unit disk, counted with multiplicity.
Choose \(r\) with \(q<r<1\). On \(\abs{z}=r\), step 1 gives \[ \abs{f(z)}\le q<r=\abs{-z}. \] The functions \(f\) and \(-z\) are holomorphic on an open neighborhood of the closed disk \(\abs{z}\le r\). Therefore the Rouché theorem, applied to \[ g(z)=(-z)+f(z), \] shows that \(g\) and \(-z\) have the same number of zeros in \(\abs{z}<r\), counted with multiplicity. Thus \(g\) has exactly one zero there.
If \(g(a)=0\) anywhere in the unit disk, then \(a=f(a)\), so step 1 gives \[ \abs{a}=\abs{f(a)}\le q<r. \] Hence every zero of \(g\) in the unit disk already lies in \(\abs{z}<r\), and there is exactly one such zero in the whole unit disk.
3. The unique zero of \(g\) is the unique fixed point of \(f\) in the unit disk.
For \(a\in D(0,1)\), \[ g(a)=0 \quad\Longleftrightarrow\quad f(a)=a. \] By step 2 there is exactly one such zero, so there is exactly one fixed point.
4. Q.E.D.
Step 3 proves the required existence and uniqueness.