Ex2-18

==== Problem 2.18 ====
Let f(z)=\sum_{k=0}^\infty a_kz^{k} be a power series. Show that if \sum_{k=0}^\infty |a_k|<\infty, then f(z) is of bounded variation on every radius of the circle |z|=1. (If, e.g., the radius is 0\leq x\leq 1 and the a_k are real, then f(x)=\sum_{k=0}^\infty a_k^+x^k-\sum_{k=0}^\infty a_k^-x^k).  
==== Solution. ====
Let \Gamma=\{z_0,\cdots,z_m\} be a partition of the ray joining z=0 and z=e^{i \theta} on the unit circle for some \theta\in[0,2\pi). Define z_k = x_k e^{i \theta} with real numbers 0=x_0<x_1<\cdots<x_n=1. It follows that\begin{align}
S_\Gamma&=\sum^n_{j=1}\biggr |f(z_j)-f(z_{j-1}) \biggr| \\
&=\sum^n_{j=1}\sum^\infty_{k=0}\biggr | a_kz_j^k-a_{k}z_{j-1}^k \biggr|\\
&\leq \sum^\infty_{k=0} |a_k| \sum^n_{j=1} \biggr | z_j^k-z_{j-1}^k \biggr|\\
&\leq \sum^\infty_{k=0} |a_k| \sum^n_{j=1} \biggr | \left(x_j^k-x_{j-1}^k\right)e^{i\theta} \biggr|\\
&= \sum^\infty_{k=0} |a_k|\\
&<\infty
\end{align}since \sum^n_{j=1}\left(x_j^k-x_{j-1}^k\right)=x^k_{n}-x^k_0=1 . As S_\Gamma<\infty, this implies V<\infty and hence f(z) is of bounded variation on every radius of the circle |z|=1.\blacksquare