Good day, I studyng about analytic functions and I see this link
How to show set of all bounded, analytic function forms a Banach space?.
My question is: Is it valid if the functions are continuous in the adherence of the region?
the banach space of analytic functions
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0Yes, in the same way that $C([0,1])$ is a closed subspace of $L^\infty([0,1])$, the space of analytic functions continuous on the boundary is a closed subspace of the space of analytic functions bounded on the boundary (all with the $\sup$ norm) – 2017-03-01
1 Answers
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Yes, for any open set $\Omega \subset \mathbb C$ the space of bounded continuous functions on the closure of $\Omega$ that are analytic in $\Omega$ form a Banach space (with the supremum norm). To prove it you note that if a sequence of such functions converges uniformly, the limit is also such a function.