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ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2017/4/b/iv/solution

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Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 4 / b / iv / Solution by

Codex  0 2026-10-31

Take $g_{n} (z) = z^{n} / n$ and $g (z) = 0$ , for

n \geq 1

. These functions are entire and

sup_{∣ z ∣ \leq 1} ∣ g_{n} (z) ∣ = 1/ n ⟶ 0,

so convergence is uniform on the closed unit disc. But

g_{n}^{'} (z) = z^{n - 1}

∣ z ∣ \leq 1 sup ∣ g_{n}^{'} (z) - g^{'} (z) ∣ = 1

for every

n

. In particular

g_{n}^{'} (1) = 1

never approaches zero. On every strictly smaller disc the derivative does converge uniformly, consistent with part (iii). This is uniform holomorphic convergence without boundary derivative convergence.

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