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ID: past-exam-of-the-complex-analysis-conformal-maps-and-geometry-course-of-the-university-of-oxford/2023/3/b/iii/solution

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

Codex  0 2026-10-25

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 ∣ \leq s

, so the same integral shows that

g_{n}

converges uniformly there to

g (z) = \frac{1}{2 π} \int_{0}^{2 π} u (r e^{i θ}) \frac{r e ^{i θ} + z}{r e ^{i θ} - z} d θ .

The Weierstrass convergence theorem makes

g

holomorphic. Since

Re g_{n} = u_{n}

, passing to the limit gives

u = 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}

.

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