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uniformly convergent on compact subsets. prove that $\lim_{k\rightarrow \infty}a_n(k) = a_n$ for each n. This is a homework problem of mine, and I think it's probably the most difficult problem I've faced so far in my time as a math student. I have a general strategy which I think might work but every time I try to write up the solution I end up teasing out more and more complexities which need to be dealt with, it's been a pretty frustrating process. My general strategy is this:

Proceed by Contradiction. Then there is some set of n's, possibly infinite, for which the sequence $a_n(k)$ doesn't converge to $a_n$. By the well-ordering principle of the integers this set has a least element, call it $n_0$. Restrict the domain of our functions to some compact ball centered at zero. We want to show that given arbitrary $\varepsilon>0$ and $K>0$, we can always find some $k>K$ such that $|\sum_{n=0}^{\infty}(a_n(k)-a_n)z^n|>\varepsilon$. In other words, we need to show that in fact the convergence isn't uniform on this compact ball.

To do this I show that the growth of the sequence of coeffecients $(a_n(k)-a_n)$ can be no faster than a certain exponential growth determined by the size of our compact ball, since the power series difference must converge (I think!). Then by choosing a point in our compact ball sufficiently close to zero I can make all those coeffecients greater than $n_0$ in our power series difference of negligible size compared to the $n_0$ coeffecient.

As for all those coeffecients less than $n_0$ there are a finite number forming a polynomial which becomes very close to $(a_0(k)-a_0)$ for points very close to zero. Our coeffecient $(a_{n_0}(k) - a_{n_0})$ stays propped up at a certain value of some minimum magnitude. But it is multiplied by $z^{n_0}$ for some very small value of z, thus it also gets very small for points near zero, I'm hoping that by not allowing K to grow as we bring our choice of z closer and closer to zero, that $(a_0(k)-a_0)$ will stay large enough so that it doesn't cancel out $(a_{n_0}(k) - a_{n_0})$, so if I can keep them different enough in magnitude and choose a point near enough to zero, I should be able to keep the whole thing's magnitude large enough to be bigger than epsilon.

Whew, that was a mouth full, I hope this question isn't too specific to be considered appropriate for this site. Does my strategy seem reasonable? Am I missing something and making things way too complicated? (very possible). Thanks.

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up vote 1 down vote accepted

Hint: the generalized Cauchy formula expresses the $a_n(k)$ in terms of contour integrals.

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Cauchy's Generalized Integral Formula? –  heat death Mar 19 '12 at 0:23
    
Yes, it's sometimes called that. The exact name will vary from text to text. –  Robert Israel Mar 19 '12 at 1:09
    
Gotta give mad late props on this one, very clever, completely simplified the problem. –  heat death Mar 21 '12 at 1:12
    
Cauchy's Theorem makes this SO MUCH easier than whatever you may try to do the same thing for real power series of a real variable... The next time someone asks you why to study complex numbers, think of this. –  GEdgar Jul 5 '12 at 17:49
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