Identity theorem
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In mathematics, especially in complex analysis, the identity theorem for holomorphic functions states: given a function f and g holomorphic on a connected open set D, if f = g on some neighborhood of z that is in D, then f = g on D. Informally, for this theorem we sometimes say holomorphic functions are hard while continuous functions are soft.
[edit] Proof
Any proof must use the connectedness. Obviously if D consisted of two disjoint open discs, the result could fail. Topologically, it is a matter of saying that f and g coincide on a set that is an open set and a closed set, since we are given that the set is not empty. Since f and g are both continuous functions, they coincide on a closed set.
Therefore the main issue is to show that they coincide on an open set. If f = g on a neighborhood of w, then, not just a point, we have:
- f(k)(w) = g(k)(w) for all
.
This is true on some open disc around w, also, since the Taylor series of a holomorphic function has non-zero radius of convergence. Since D is open there is such a disc contained in D. The union of all such open discs is an open set, and also the set of points in D where f coincides with g. QED.
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