Give an example of an unbounded analytic function on the unit disk with the property that \[ \lim_{z\to w}|f(z)|=1 \] for every \(w\in S^1\) except \(w=1\).
Solution
Take \[ \boxed{ f(z)=\exp\!\left(\frac{1+z}{1-z}\right).} \]
1. The function \(f\) is analytic on \(\Delta\).
The denominator \(1-z\) does not vanish for \(|z|<1\), so \[ \Phi(z)=\frac{1+z}{1-z} \] is holomorphic on \(\Delta\). Composing with the entire exponential function shows that \(f=e^\Phi\) is holomorphic on \(\Delta\).
2. The function \(f\) is unbounded on \(\Delta\).
Along the real radius \(0<r<1\), \[ \Phi(r)=\frac{1+r}{1-r}\longrightarrow+\infty \qquad(r\to1^-). \] Therefore \[ |f(r)| =\exp\!\left(\frac{1+r}{1-r}\right) \longrightarrow\infty. \] Hence \(f\) is unbounded.
3. If \(w\in S^1\) and \(w\neq1\), then\[ \operatorname{Re}\frac{1+w}{1-w}=0. \]
For \(|w|=1\), \[ \overline w=\frac1w. \] Thus \[ \overline{\frac{1+w}{1-w}} =\frac{1+\overline w}{1-\overline w} =\frac{w+1}{w-1} =-\frac{1+w}{1-w}. \] So \((1+w)/(1-w)\) is purely imaginary.
4. For every \(w\in S^1\setminus\{1\}\),\[ \lim_{z\to w}|f(z)|=1. \]
Because \(1-w\neq0\), the rational function \(\Phi\) extends continuously to a neighborhood of \(w\). Hence \[ \Phi(z)\longrightarrow\Phi(w) \qquad(z\to w,\ z\in\Delta). \] Using \(|e^\zeta|=e^{\operatorname{Re}\zeta}\) and 3, \[ |f(z)| =e^{\operatorname{Re}\Phi(z)} \longrightarrow e^{\operatorname{Re}\Phi(w)} =1. \] This gives the required boundary behavior at every point except \(1\).