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

The radial harmonic function with value zero at $|z|=r$ and one at $|z|=1$ isIndeed $\log |z|$ is harmonic away from zero, and the displayed expression has the required boundary values. Uniqueness of the Dirichlet problem identifies it as the harmonic measure of an annulus.

The concentric normalization of an eccentric annulus here usesThis Möbius transformation preserves the unit circle and sends the inner circle's real endpoints $0,4/5$ to $-1/2,1/2$. Its image circle is symmetric about the real axis and hence is $|w|=1/2$. Equivalently, expanding $|T(z){|}^2=1/4$ gives $|z{|}^2=(4/5)\mathrm{Re}z$, precisely the inner-circle equation. Thus $T$ sends the domain to $1/2<|w|<1$. Conformal invariance of harmonic measure and i giveThe zero of $T$ lies in the removed inner disk, so this logarithm is harmonic throughout $\Omega$.

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 $u_n$.