Complex Power Series Boundary Property

skeboy · #1 $Old$ July 2nd, 2009, 05:42 PM

In Shilov's Elementary Real and Complex Analysis, he writes on page 216:

'As we know, a (complex) power series a0 + a1 (z - z0) + a2 (z - z0)^2 + ... with radius of convergence p may or may not converge at points on the boundary of its region of convergence, i.e., at points of the circle |z - z0| = p. However, if the series converges at a boundary point z1, then it converges uniformly on the whole segment going from the center of the circle z0 to the boundary point z1. To see this, we need only consider the case z0 = 0, z1 = t1 > 0 (here z1 = t1 is real, as opposed to the general case where it is complex). Why?"

He then goes on to prove the special case with z0 = 0 and z1 = t1 > 0. Why does having proved this special case imply the general case for any point on the boundary of a region of convergence centered at any complex point? I think it may have something to do with shifting the power series and dividing/multiplying, but I'm not sure quite how to make it work.

halbard · #2 $Old$ July 5th, 2009, 05:48 AM

Suppose you have a complex power series $\sum_{n=0}^\infty b_n(z-z_0)^n$ with radius of convergence $R>0$ which converges at some point $z_1$ on the boundary of the disk $D=D(z_0,R)$ . By shifting and rotating the axes, you obtain another power series $\sum_{n=0}^\infty a_nz^n$ with radius of convergence $R$ which converges at the point $z=R$ . (If $z_1=z_0+R\mathrm e^{\mathrm i\theta}$ then $a_n=b_n\mathrm e^{\mathrm in\theta}$ .)

Now you need Abel's test: Let $f_n$ be a sequence of complex functions on a set $A$ and let $g_n$ be a decreasing sequence of non-negative functions on $A$ . If the series $\sum_{n=0}^\infty f_n$ converges uniformly on $A$ and if there is a constant $M$ such that $|g_n(x)|\leq M$ for every $x\in A$ and for all non-negative integers $n$ , then $\sum_{n=0}^\infty f_ng_n$ converges uniformly on $A$ .

In this case, let $A=[0,R]$ , and for $x\in A$ let $f_n(x)=a_nR^n$ and $g_n(x)=(x/R)^n$ .

By hypothesis, the series $\sum_{n=0}^\infty a_nR^n$ is convergent, so $\sum_{n=0}^\infty f_n(x)$ converges uniformly on $A$ . Also $|g_n(x)|\leq 1$ for all $n$ and for all $x\in A$ .

By Abel's test, the series $\sum_{n=0}^\infty f_n(x)g_n(x)=\sum_{n=0}^\infty a_nR^n(x/R)^n=\sum_{n=0}^\infty a_nx^n$ converges uniformly on $[0,R]$ .

Thus the original series is uniformly convergent on the radius of $D$ from $z_0$ to $z_1$ .

Will this do?