ChapterZero

Uniform convergence of power series

Filed under: Mathematics — Alex @ 11:06 am 6/22/2005

Proposition 25
Suppose $a_0,a_1,\ldots$ is a bounded sequence. We’ve seen that if is a point, |z|<1 , then the sequence $s_0(z), s_1(z), \ldots$ converges, where

$\displaystyle s_0(z) = a_0, s_1(z) = a_0 + a_1z, s_2(z) = a_0 + a_1 z + a_2 z^2, \ldots.$

Suppose now that is a function with initial set that contains all points , |z|<1 , and for each , |z|<1 , $s_0(z), s_1(z), \ldots$ converges to f(z) . Show that if 0<r<1 and c>0 , there is a positive integer such that if n>N and $|z|\leq r$ ,

$\displaystyle |f(z) - s_n(z)| < c$ .

Soln
For each , |z|<r let N_z be the least integer satisfying $n > N \Rightarrow |s_n(z) - f(z)| < c$ . Let be the collection of all N_z .

Claim: is finite. Assume is not finite, then for every $n \in \N$ , there is a non-empty set $V_n = \{z \,:\, |z|\leq r \text{ and } N_z > n \}$ . Notice $V_{n+1} \subseteq V_n$ , and each V_n contains infinitely many elements. From here we can use the same ‘boxing in’ process that we used to show power series converge: namely, since V_n is infinite, we can find a rectangle of definite size (which we get to choose) which contains infinitely many of V_n , and then a rectangle of even smaller size which contains infinitely many of $V_n \cap V_{n+1}$ . Since we get to choose the size of the rectangles, we can make them converge to a point as $n \rightarrow \infty$ . This point is in $\big\cap_\N V_n$ , and therefore has the property that $\forall n \in N \,:\, N_z > n$ , which is not possible. Therefore is finite.

Since is finite, it contains its maximum which satisfies: if n>M and $|z|\leq r$ ,

$\displaystyle |f(z) - s_n(z)| < c$ .

The fact that $|z| \leq r <1$ is used in making the rectangles to ensure that the converged to satisfies |z| < 1 ; otherwise you couldn't claim a contradiction.

Uniform convergence of power series

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