Question

Prove or disprove that uniform limit of holomorphic functions is holomorphic.

Answer 1

Answer:

Let U be an open subset of C.

Let {fn}n∈N be a sequence of analytic functions fn:U→C.

Let {fn} converge locally uniformly to f on U.

Then f is analytic.

Proof

By Equivalence of Local Uniform Convergence and Compact Convergence, fn converges to f locally uniformly on U.

Then for any z∈U, there is an ϵ>0 so that:

Bϵ(z)⊂U

and fn converges uniformly on Bϵ(z).

Let γ be any simple closed curve in Bϵ(z).

Since fn→f uniformly on γ (because γ⊂Bϵ(z)), we have:

limn→∞∫γfn(z)dz=∫γf(z)dz

Since each fn is analytic, we have that:

∀n∈N:∫γfn(z)dz=0

So we conclude also that

∫γf(z)dz=0

Since γ was arbitrary, we have by Morera's Theorem that f is analytic in Bϵ(z).

Since z was arbitrary, f is analytic on all of U.

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Step-by-step explanation:

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