Re: analytical function question
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Sun, 30 Apr 2006 18:49:13 +0100
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} ?
Well, it must at least be _defined_ 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, and these neighborhoods cover
the unit circle D^.
Where am i wrong?
The assertion "these neighborhoods cover the unit circle D^" is false.
For each _z_ in D, consider the open disk centered at _z_ with radius
equal to the distance r_z from _z_ to the circle { w | |w| = 1 }. Then
each disk D(z,r_z) is a neighborhood of _z_, but their union is just D
itself.
Best regards,
Jose Carlos Santos
.
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