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Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly

Statement

Let $Ω \subseteq C$ be open, let each $f_{n} : Ω \to C$ be holomorphic, and suppose $f_{n} \to f$ locally uniformly on $Ω$ in the sense of Locally uniform convergence on an open subset of the complex plane is compact convergence. Then $f$ is holomorphic and

$f_{n}^{(k)} ⟶ f^{(k)}$

locally uniformly on $Ω$ for every natural $k$ , with $f^{(0)} = f$ .

A locally uniform limit of holomorphic functions is holomorphic, and for every natural $k$ the $k$ th derivatives converge locally uniformly to the $k$ th derivative of the limit.

Facts & Assumptions

Given: An open set $Ω$ , holomorphic functions $f_{n} : Ω \to C$ , and locally uniform convergence $f_{n} \to f$ , equivalently uniform convergence on every compact subset by Locally uniform convergence on an open subset of the complex plane is compact convergence.

[L1]

A uniform limit of continuous complex-valued functions on a metric space is continuous (A uniform limit of continuous complex-valued functions is continuous).

[L2]

Uniform convergence of continuous integrands on a fixed rectifiable contour permits passage of the limit through the complex line integral (A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral).

[L3]

Every holomorphic function has zero integral around each contained filled triangle, including degenerate triangles (Goursat's triangle theorem: a holomorphic function integrates to zero around every triangle contained in its domain).

[L4]

A continuous function on an open subset of $C$ is holomorphic if and only if its integral around every contained filled triangle is zero (Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions).

[L5]

If $0 \leq r < R < S$ , $h$ is holomorphic on $D (a, S)$ , $M$ bounds $∣ h ∣$ on $∣ ζ - a ∣ = R$ , and $∣ z - a ∣ \leq r$ , then $∣ h^{(k)} (z) ∣ \leq k! R M / (R - r)^{k + 1}$ (Cauchy estimates on a smaller concentric disc).

[L6]

The boundary of a filled triangle is the union of its directed affine edge traces (Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter).

[L7]

Closed bounded subsets of Euclidean space, including $[0, 1]$ and closed complex discs, are compact (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[L8]

The continuous image of a compact metric space is compact (The image of a compact metric space under a continuous map is compact, and so is the image of any compact subset).

Proof

technique · direct

1.1givenL1L7

If $Ω$ is nonempty, fix $a \in Ω$ and choose $S > 0$ with $\overline{D} (a, S) \subseteq Ω$ ; [L7] makes this closed disc compact, the given convergence is uniform there, and [L1] makes $f$ continuous on it, so $f$ is continuous throughout $Ω$ .

1.2givenL2L3L6L7L8

Let $Δ \subseteq Ω$ be any filled triangle. By [L6], its boundary trace is a finite union of affine images of $[0, 1]$ , compact by [L7] and [L8]; convergence is therefore uniform on the trace, [L3] makes every $\int_{\partial Δ} f_{n}$ zero, and [L2] gives $\int_{\partial Δ} f = 0$ .

2.1step 1.1step 1.2L4

The continuity from step 1.1 and the vanishing triangle integrals from step 1.2 satisfy [L4], so $f$ is holomorphic throughout $Ω$ ; on the empty open set this conclusion is vacuous.

3.1step 2.1choose

Fix a natural derivative order $k$ and a point $a \in Ω$ , and choose radii $0 < r < R < S$ with $\overline{D} (a, S) \subseteq Ω$ ; by step 2.1 every difference $h_{n} : = f_{n} - f$ is holomorphic on $D (a, S)$ .

4.1step 3.1L5

Given $ε > 0$ , uniform convergence on the compact circle $∣ ζ - a ∣ = R$ gives $N$ such that $∣ h_{n} (ζ) ∣ < ε (R - r)^{k + 1} / (k! R)$ there for $n \geq N$ ; applying [L5] then gives $∣ f_{n}^{(k)} (z) - f^{(k)} (z) ∣ < ε$ for every $∣ z - a ∣ \leq r$ .

5.1step 4.1∎

Step 4.1 proves uniform convergence of the $k$ th derivatives on a neighbourhood of every point, hence local uniform convergence by the dictionary in the given data; when $k = 0$ it recovers the original convergence, and zero or eventually constant sequences require no exception.

Depends on

Used by

Dependency tree · two levels

58 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Lars Ahlfors, Complex Analysis, 3rd ed., Ch. 5 §1.1 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 2, Theorems 5.2 and 5.3 (standard reference, not scraped)
Matthias Weber, Complex Analysis, Theorem 2.4.4 (standard reference, not scraped)
Steven G. Krantz, A Guide to Complex Variables, §3.1.5 (standard reference, not scraped)