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The filled difference quotient is holomorphic in each variable separately

Statement

Let $Ω \subseteq C$ be open, let $f : Ω \to C$ be holomorphic and let $g$ be its filled difference quotient (The filled difference quotient of a holomorphic function is jointly continuous). Then for each fixed $z \in Ω$ the map $ζ \mapsto g (ζ, z)$ is holomorphic on the whole of $Ω$ , the point $ζ = z$ included; and by symmetry, for each fixed $ζ \in Ω$ the map $z \mapsto g (ζ, z)$ is holomorphic on $Ω$ .

Facts & Assumptions

Given: An open $Ω \subseteq C$ , a holomorphic $f : Ω \to C$ and its filled difference quotient $g$ .

[L1]

With $U$ open, $f$ holomorphic on $U$ and $z \in U$ fixed, the function equal to $(f (ζ) - f (z)) / (ζ - z)$ for $ζ \neq z$ and to $f^{'} (z)$ at $ζ = z$ is continuous on $U$ and holomorphic on $U ∖ {z}$ ; no holomorphy at the filled point is asserted (The filled difference quotient is continuous at its exceptional point and holomorphic away from it).

[L2]

If $U \subseteq C$ is open, $p \in U$ and $h : U \to C$ is continuous on $U$ and holomorphic on $U ∖ {p}$ , then $h$ is holomorphic on $U$ (A continuous function holomorphic off a single point is holomorphic).

[L3]

The filled difference quotient $g$ of a holomorphic $f$ on $Ω$ is $(f (ζ) - f (z)) / (ζ - z)$ off the diagonal and $f^{'} (z)$ on it, and it satisfies $g (ζ, z) = g (z, ζ)$ (The filled difference quotient of a holomorphic function is jointly continuous).

Proof

technique · direct

1.1

Fix $z \in Ω$ . By [L3] the map $ζ \mapsto g (ζ, z)$ is exactly the function of [L1] for that $z$ , so it is continuous on $Ω$ and holomorphic on $Ω ∖ {z}$ .

givenL1L3

2.1

Applying [L2] with $U = Ω$ , $p = z$ and $h = g (\cdot, z)$ , step 1.1 upgrades that function to a holomorphic function on all of $Ω$ .

step 1.1L2

3.1

By the symmetry $g (ζ, z) = g (z, ζ)$ of [L3], the map $z \mapsto g (ζ, z)$ for fixed $ζ$ is the map of step 2.1 with the roles of the two arguments exchanged, hence holomorphic on $Ω$ as well.

step 2.1L3∎

Depends on

Used by

Dixon's glued function is entire and vanishes at infinity Lemma

Dependency tree · two levels

23 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

J. Lebl, Complex Analysis, Ch. 4 §4.2 (standard reference, not scraped)