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Local chordal equicontinuity is equivalent to meromorphic normality on compact exhaustions

Statement

Assume the Axiom of Choice, and therefore in particular Countable Choice and Dependent Choice for the successive subsequence selections. Let $Ω$ be a plane domain and let $F$ be a family of meromorphic maps $Ω \to \hat{C}$ . Then $F is meromorphically normal ⟺ F ∣_{K} is equicontinuous for every compact K \subseteq Ω,$ where equicontinuity is taken with respect to the chordal metric on $\hat{C}$ . Equivalently, it is enough to check that equicontinuity on each compact stage of the canonical exhaustion.

Facts & Assumptions

Given: Choice, a plane domain $Ω$ , and a family $F$ of meromorphic sphere-valued maps.

[L1]

On a compact metric domain, compactness of the uniform closure is equivalent to equicontinuity and pointwise relative compactness (Ascoli–Arzelà in the uniform topology for nonempty compact metric domains).

[L2]

The canonical exhaustion $(K_{n})$ is compact, nested, and has interiors covering $Ω$ (Every plane domain has the canonical nested compact exhaustion by distance and radius cutoffs).

[L3]

In a metric space, compactness and sequential compactness are equivalent under Countable Choice and Dependent Choice (For a metric space, compact, countably compact, limit point compact, sequentially compact, and complete together with totally bounded are all equivalent, given countable choice and dependent choice).

[L4]

A chordally locally uniform meromorphic limit is meromorphic or identically $\infty$ (A chordally locally uniform meromorphic limit is meromorphic or identically infinity).

Proof

technique · direct

1.1

If the compact-set equicontinuity condition holds, then on each compact stage $K_{n}$ of [L2] the restricted family is equicontinuous. Pointwise relative compactness is automatic because the target sphere is compact, so [L1] makes the uniform closure on $K_{n}$ compact.

L1L2given

1.2

Using [L3], choose successively a subsequence converging uniformly on $K_{1}$ , then on $K_{2}$ , and so on, and take the diagonal subsequence. By [L2] that diagonal converges chordally locally uniformly on $Ω$ , and [L4] makes its limit meromorphic or identically $\infty$ . Thus compact-set chordal equicontinuity implies meromorphic normality.

L2L3L4givenchoose

1.3

Conversely, if $F$ is meromorphically normal, then every sequence of restrictions to a fixed $K_{n}$ has a uniformly convergent subsequence. Fact [L3] turns that sequential compactness into compactness of the restriction closure, and [L1] then gives equicontinuity on $K_{n}$ . Because every compact subset of $Ω$ lies in some stage of the exhaustion by [L2], the family is chordally equicontinuous on every compact subset.

L1L2L3given

2.1

The first two steps prove the forward implication and step 1.3 proves the reverse implication, so the two conditions are equivalent.

given∎

Depends on

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Sources

Matthias Weber, Complex Analysis, Ch. 5 §§5.1-5.2 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 2 §5.2 and Ch. 8 §3.2 (standard reference, not scraped)
Sheldon Axler, Paul Bourdon, and Wade Ramey, Harmonic Function Theory, Ch. 2 (standard reference, not scraped)