Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17
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The filled difference quotient is continuous at its exceptional point and holomorphic away from it

Statement

Let U⊆C be open, let f:U→C be holomorphic, and fix z∈U. Define

g(ζ)={f(ζ)−f(z)ζ−z,ζ≠z,f′(z),ζ=z.

Then g is continuous on U and holomorphic on U∖{z}. No holomorphy at the filled point z is asserted.

Facts & Assumptions

Given: An open set U, a holomorphic f:U→C, and a fixed point z∈U.

[L1]

The derivative f′(z) is the limit of (f(ζ)−f(z))/(ζ−z) as ζ→z through ζ≠z (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions).

[L2]

Sums, differences, and quotients with nonzero denominator of holomorphic functions are holomorphic (Linearity, product, reciprocal, and quotient rules for complex derivatives).

[L3]

Proof

technique · direct
1.1L1

By [L1], the off-point quotient tends to f′(z)=g(z) as ζ→z, which is exactly continuity of g at z; this also covers constant f.

1.2givenL2

On U∖{z}, the numerator and denominator are holomorphic and the denominator is nonzero, so [L2] makes g holomorphic there.

2.1step 1.1step 1.2L3∎

By [L3], step 1.2 also makes g continuous away from z; together with step 1.1 this proves continuity on all of U, without claiming differentiability at the filled point.

Depends on

Used by

Dependency tree · two levels

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Sources