- Prove that if a sequence of harmonic functions on the open disk converges uniformly on compact subset of the disk, then the limit is harmonic.
Solution
Let \((u_n)\) converge uniformly on compact subsets of the open disk \(D\) to \(u\). First take the harmonic functions to be real-valued.
1. On each smaller disk, the limit is the Poisson integral of its boundary values.
Fix \(a\in D\) and \(r>0\) with \(\overline{D(a,r)}\subset D\). Uniform convergence on this closed disk makes \(u\) continuous there. For \(|z-a|<r\), the Poisson formula for \(u_n\) gives \[ u_n(z)=\frac1{2\pi}\int_0^{2\pi} \frac{r^2-|z-a|^2}{|re^{it}-(z-a)|^2} u_n(a+re^{it})\,dt \] [@SS03]. The kernel is nonnegative and has integral \(2\pi\). Therefore replacing the boundary values of \(u_n\) by those of \(u\) changes this integral by at most \(\sup_{|w-a|=r}|u_n(w)-u(w)|\), which tends to zero. Taking limits gives the same formula for \(u\).
2. This Poisson integral is harmonic in the smaller disk.
Define \[ H(z)=\frac1{2\pi}\int_0^{2\pi} \frac{re^{it}+(z-a)}{re^{it}-(z-a)} u(a+re^{it})\,dt, \qquad |z-a|<r. \] The integrand is holomorphic in \(z\). On any compact subdisk \(|z-a|\leq s<r\), the denominator has modulus at least \(r-s\), and both the integrand and its \(z\) derivative are uniformly bounded in \(t\). The continuous boundary values of \(u\) are bounded. Differentiation under the integral is consequently justified, so \(H\) is holomorphic.
For \(\xi=re^{it}\) and \(w=z-a\), multiplication by \(\overline\xi-\overline w\) shows that \[ \operatorname{Re}\frac{\xi+w}{\xi-w} =\frac{r^2-|w|^2}{|\xi-w|^2}. \] Step 1 therefore says \(u=\operatorname{Re}H\) on \(D(a,r)\). The real part of a holomorphic function is harmonic [@SS03]. Since \(a\) was arbitrary, \(u\) is harmonic throughout \(D\).
For complex-valued harmonic functions, apply the same argument separately to the real and imaginary parts; both inherit compact-uniform convergence.