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eugene wrote:
It is going to be a silly question, but i'm a bit unsure:
If we know that the function f(z) is analytical in
D = {z: |z|<1} is it true that it is analytical in the closure of D^ = {z: |z|<=1} ?
No! Consider f : z |-> 1/(1-z). This is analytic on D, with power series f(z) =
1 + z + z^2 + z^3 + ...; but it has a pole at z = 1 and so can't possibly be
analytic on D^.
As far as i understand if f iz analytical at every point z from D, then it is
analytical in some neighborhood of z,
That's true.
and these neighborhoods cover the unit circle D^.
That isn't. They certainly cover D, the open unit disc you defined above; but
they're all subsets of D; none of them extend onto C = D^\D.
HTH,
Tom
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