Re: dense rational functions with prescribed poles

On Wed, 29 Mar 2006 11:59:50 -0800, The World Wide Wade
<waderameyxiii@xxxxxxxxxxxxxxxxxxxx> wrote:

In article <e0ehm9$pgl$1@xxxxxxxxxxxxxxxxxxxxxx>,
israel@xxxxxxxxxxx (Robert Israel) wrote:

In article <25765172.1143654172917.JavaMail.jakarta@xxxxxxxxxxxxxxxxxxxxxx>,
eugene <jane1806@xxxxxxx> wrote:
On Wed, 29 Mar 2006 08:32:57 EST, eugene
<jane1806@xxxxxxx> wrote:

I have the following function: what condition should
me impose on the set S={\alpha_1,...,\alpha_n,...}
such that the set of rational functions with poles
from S to be dense in L^1 sense in L_1[-pi,pi].

Um, the polynomials are dense in L^1(-pi,pi).

I was actually interested in rational fucntions with poles from
\alpha_1,\alpha,...\alpha_n,... -polynomials are surely a particular
case of rational fucntions, but they don't have poles- i was interested
in non-trivial cases.
Thanks

Given f in L^1 and epsilon > 0, first approximate f with a polynomial
and then add delta/(x-alpha_1) where delta is very small (assuming
alpha_1 is not in the interval [-pi,pi]).

Nonconstant polynomials could be viewed as having poles at
infinity; conceivably he wants to disallow those unless infinity
is in S. If so, the condition would seem to be that S intersect
[-Pi,Pi] is empty. If infinity is in S, the polynomials are
dense. If not, choose any alpha in S and by Stone-Weierstrass,
polynomials in 1/(x-alpha) will be dense. That assumes alpha is
real. If alpha is complex, I think the conclusion is still true
by Runge's Theorem.

Seems right to me. (Or just look at a pole-pushing proof of
Runge's theorem: If alpha is not in the interval then the
closure of polynomials in 1/(a-alpha) is the same as the
closure of the polynomials.)


************************

David C. Ullrich
.

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