Let \(\mathcal U\) be the set of positive harmonic functions \(u\) on the unit disk \(\Delta\), normalized by \(u(0)=1\). Show that \(\mathcal U\) is compact: every sequence in \(\mathcal U\) has a subsequence converging uniformly on compact subsets of \(\Delta\).
Solution
Let \((u_n)\) be a sequence in \(\mathcal U\).
1. For each \(n\), choose a harmonic conjugate \(v_n\) so that\[ F_n=u_n+iv_n \]is holomorphic on \(\Delta\) and \(v_n(0)=0\).
The disk is simply connected, so every harmonic function on it has a harmonic conjugate. Adding a real constant to the conjugate does not change holomorphicity, so normalize by \(v_n(0)=0\). Since \(u_n(0)=1\), \[ F_n(0)=1. \] Also \(\Re F_n=u_n>0\).
2. The family \((F_n)\) is uniformly bounded on every compact subset of\(\Delta\).
By the estimate of E-HM213A17-59, applied to \(F_n\), \[ |F_n(z)| \le |F_n(0)|\frac{1+|z|}{1-|z|} =\frac{1+|z|}{1-|z|}. \] Thus for every \(0<r<1\), \[ \sup_n\sup_{|z|\le r}|F_n(z)| \le\frac{1+r}{1-r}<\infty. \]
3. A subsequence \((F_{n_k})\) converges locally uniformly on \(\Delta\) toa holomorphic function \(F\).
By Montel’s theorem, the local bounds from 2 make \((F_n)\) a normal family. Hence a subsequence converges uniformly on every compact subset to a holomorphic limit \(F\).
4. The corresponding subsequence \((u_{n_k})\) converges locally uniformlyto \(u=\Re F\).
Taking real parts preserves uniform convergence, so \[ u_{n_k}=\Re F_{n_k}\longrightarrow\Re F=u \] uniformly on each compact subset. The real part of a holomorphic function is harmonic, hence \(u\) is harmonic.
5. The limit \(u\) belongs to \(\mathcal U\).
Since every \(u_{n_k}>0\), local uniform convergence gives \(u\ge0\). Also \[ u(0)=\lim_{k\to\infty}u_{n_k}(0)=1. \] Thus \(u\) is not identically zero. By the minimum principle for harmonic functions, a nonnegative harmonic function that vanishes at an interior point would be identically zero; therefore \(u>0\) throughout \(\Delta\). Hence \(u\in\mathcal U\).
6. Therefore \(\mathcal U\) is compact for the topology of uniformconvergence on compact subsets.
Starting from an arbitrary sequence in \(\mathcal U\), <1>3–<1>5 produced a subsequence converging locally uniformly to another member of \(\mathcal U\). This is the asserted compactness.