Past exam of the complex analysis: conformal maps and geometry course of the University of Oxford / 2023 / 3 / b

For $0<|z|<R$, set $w=R^2/\bar{z}$, which lies outside the closed disk. On $|\zeta |=R$, direct algebra gives the Poisson kernel identityUsing $d\theta =d\zeta /(i\zeta )$, the proposed Poisson formula for a holomorphic function becomesThe first term is $f(z)$ by the Cauchy integral formula; the second vanishes by the Cauchy integral theorem, since its pole lies outside the disk. At $z=0$ the kernel is one, and the same identity is the mean value property for holomorphic functions. Therefore

For $|z|<r$, the Schwarz integral on the unit disk, rescaled to radius $r$, isOn each smaller closed disk $|z|\le s<r$, its kernel and all its derivatives in $z$ are uniformly bounded. Since the boundary values of the real harmonic function $u$ are continuous on this circle, differentiation under the integral sign proves that $g$ is holomorphic. Taking the real part gives the Poisson integral of those boundary values. It agrees with $u$ by uniqueness of the Dirichlet problem on the radius-$r$ disk. Hence $g$ is holomorphic and $\mathrm{Re}g=u$ on $|z|<r$. Its imaginary part is the harmonic conjugate normalized by $\mathrm{Im}g(0)=0$.

Fix $r<1$ and construct $g_n$ from the boundary values of $u_n$ on the circle $|z|=r$ as in ii. By the assumed locally uniform convergence, those boundary values converge uniformly to $u$. For each $s<r$, the kernel is bounded on $|z|\le s$, so the same integral shows that $g_n$ converges uniformly there toThe Weierstrass convergence theorem makes $g$ holomorphic. Since $\mathrm{Re}{g}_n=u_n$, passing to the limit gives $u=\mathrm{Re}g$ on $|z|<r$. The real part of a holomorphic function is harmonic. As $r<1$ was arbitrary, $u$ is harmonic throughout the unit disc. This proves the locally uniform limit of harmonic functions theorem without presuming convergence of the derivatives of $u_n$.