Give an example of a bounded harmonic function \(u\) on the unit disk whose harmonic conjugate \(v\) is unbounded.
Solution
Let \[ \Phi(z)=\frac{1+z}{1-z}. \] Since \(\Phi\) maps \(\Delta\) into the right half-plane, choose there the holomorphic logarithm with argument in \((-\pi/2,\pi/2)\) and write \[ F(z)=\Log\Phi(z)=A(z)+iB(z). \] Set \[ \boxed{u(z)=B(z),\qquad v(z)=-A(z).} \]
1. The function \(u\) is harmonic and bounded on \(\Delta\).
For \(|z|<1\), \[ \operatorname{Re}\Phi(z) =\frac{1-|z|^2}{|1-z|^2}>0. \] Hence \(F=\Log\Phi\) is holomorphic on \(\Delta\). Its imaginary part \(B=\operatorname{Im}F\) is therefore harmonic. Because the chosen argument of every point in the right half-plane lies in \((-\pi/2,\pi/2)\), \[ |u(z)|=|B(z)|<\frac{\pi}{2}. \] Thus \(u\) is bounded.
2. The function \(v=-A\) is a harmonic conjugate of \(u\).
Since \[ u+iv =B-iA =-i(A+iB) =-iF, \] and \(F\) is holomorphic, the function \(u+iv\) is holomorphic. Hence \(v\) is a harmonic conjugate of \(u\).
3. The harmonic conjugate \(v\) is unbounded.
Along the real radius \(0<r<1\), \[ \Phi(r)=\frac{1+r}{1-r}>0, \] so \[ v(r) =-\operatorname{Re}\Log\Phi(r) =-\log\frac{1+r}{1-r}. \] As \(r\to1^-\), \[ \frac{1+r}{1-r}\longrightarrow\infty, \] and therefore \[ v(r)\longrightarrow-\infty. \] Thus \(v\) is unbounded on \(\Delta\).
4. Every harmonic conjugate of \(u\) is unbounded.
On the connected disk, any two harmonic conjugates of the same harmonic function differ by a real constant. Adding a constant cannot make the unbounded function \(v\) bounded.