Re: Continuity of complex function question,
- From: José Carlos Santos <jcsantos@xxxxxxxx>
- Date: Sun, 30 Apr 2006 21:44:04 +0100
James wrote:
I am trying to show that
H(z) = int_[0,1] h(t)/(t-z) dt
(where h(t) is a continuous function on [0,1])
is continuous.
It is even analytic.
I am stuck at one little place :
First of all, h continuous on [0,1] means |h(t)| <= M for some M.
No it doesn't mean that. It just implies it.
Let |z-z'| < D.
So, | H(z) - H(z') | <= M int_[0,1] (z-z')/[(t-z)(t-z')] <= D*M int_[0,1] 1/[(t-z)(t-z')]
since 1/(t-z) - 1/(t-z') = (z-z')/[(t-z)(t-z')].
But what do I do with int_[0,1] 1/[(t-z)(t-z')]?
Well, you failed to mention what is the domain of H, but I'll assume
that it is C\[0,1]. Let _d_ be the distance from _z_ to [0,1]. Suppose
now that D was chosen such that the distance from each element of the
open disk centered at _z_ with radius D to [0,1] is greater than d/2.
Then, in particular, the distance from z' to [0,1] is greater than d/2.
So, for each _t_ in [0,1], |(t - z)(t - z'| >= d^2/2 and therefore
int_[0,1] |1/(t - z)(t - z') dt <= 2/d^2.
Best regards,
Jose Carlos Santos
in particular,
.
- References:
- Continuity of complex function question,
- From: James
- Continuity of complex function question,
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