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A continuous function holomorphic off a single point is holomorphic
Statement
Let be open, let , and let be continuous on and holomorphic on . Then is holomorphic on , the point included.
Facts & Assumptions
Given: An open set , a point , and a function that is continuous on and holomorphic on .
If is open, , and is continuous and holomorphic on , then for every filled triangle of Filled complex triangles, their oriented three-edge boundary contours, diameter, and perimeter contained in ; the exceptional point may lie outside, inside, or on the boundary of (Goursat's triangle theorem remains valid for a continuous function holomorphic away from one point).
If is open and is continuous, then is holomorphic on if and only if whenever ; repeated or collinear vertices are permitted (Morera's theorem: vanishing triangle integrals characterize holomorphy among continuous functions).
Proof
The hypotheses of [L1] are exactly the given ones, so for every filled triangle , whether lies outside , inside it, or on its boundary.
The function is continuous on the open set and step 1.1 supplies the vanishing triangle integrals demanded by the right-hand side of [L2], so [L2] makes holomorphic on all of , including at .
Depends on
Used by
Dependency tree · two levels
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Sources
- L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 §3.1 (standard reference, not scraped)