$f(z)=\sum_{k=0}^\infty a_kz^{k}$ with $\sum |a_k|<\infty$ is of bounded variation on circle $|z|=1$

Question

Let $f(z)=\sum_{k=0}^\infty a_kz^{k}$ be a power series. Show that if $\sum |a_k|<\infty$, 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 $f(x)=\sum a_k^+x^k-\sum a_k^-x^k$). Problem copied from Wheeden & Zygmund Measure and Integral.

For example working in general with $a_k\in \mathbb{C}$, for all $k$, I was able to prove that $$S_\Gamma=\sum_{j=1}^m\left|f(z_j)-f(z_{j-1})\right|<M$$ for any partition $\Gamma=\{z_0,\cdots,z_m\}$ of the following radius: line joining $0$ and $1$, line joining $-1$ and $0$, line joining $0$ and $i$, line joining $-i$ and $0$. The interesting part, occurs when the radius is a line joining $0$ and other point in the circle $|z|=1$, let $z_m=x_m+iy_m$ be such point. Consider a partition $\Gamma=\{z_0,\cdots,z_m\}$ of this radius, where $z_j=x_j+iy_j$ and $z_0=0+i0$. If we can make $S_\Gamma<\infty$, we are done

$$S_\Gamma=\sum_{j=1}^m|f(z_j)-f(z_{j-1})| = \sum_{j=1}^m\left|\sum_{k=0}^\infty a_k\left((x_j+iy_j)^k-(x_{j-1}+iy_{j-1})^k\right)\right| = \sum_{j=1}^m\left|\sum_{k=0}^\infty a_k\sum_{l=0}^k \binom{k}{l}\left( (x_j)^{k-l}(iy_j)^k-(x_{j-1})^{k-l}(iy_{j-1})^k \right)\right|\leq $$ $$ \leq \sum_{j=1}^m \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right|$$ since $|i^k|=1$.

Interchanging the order of summation we have $$S_\Gamma \leq \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}\sum_{j=1}^m\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right| = \sum_{k=0}^\infty |a_k|\sum_{l=0}^k \binom{k}{l}x_m^{k-l}y_m^k =$$ $$= \sum_{k=0}^\infty |a_k|(x_m+y_m)^k $$ in order to get convergence is enough that $|x_m+y_m|^k\leq 1$, since the initial condition is $|z_m|=1$ we have $x_m^2+y_m^2=1$. Replacing we get $|x_m+y_m|=|x_m+\sqrt{1-x_m^2}|$, but the function $|x+\sqrt{1-x^2}|>1$ on $0<x<1$. For this reason I cannot verify the convergence of the serie, so I can't conclude if it's of bounded variation.

Is my bounding approach too excessive? so that loses its property of convergence? I think the problem could be the last two equalities, since I'm assuming $\left| x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k \right| = x_j^{k-l}y_j^k-x_{j-1}^{k-l}y_{j-1}^k $ which could be something that is not true all the time. Any comment will be appreciated.

Answer 1

Your approach is too complicated. It suffices to consider the case of a ray joining $z=0$ and $z=1$:

Let $\Gamma=\{z_0,\cdots,z_m\}$ be a partition of the ray joining $z=0$ and $z=e^{i \alpha}$ on the unit circle. Then $z_j = x_j e^{i \alpha}$ with real numbers $$ 0 = x_0 < x_2 < \cdots < x_m = 1 \, . $$ Also $$ \sum_{j=1}^m\left|f(z_j)-f(z_{j-1})\right| = \sum_{j=1}^m\left|F(x_j)-F(x_{j-1})\right| $$ where $F$ is defined as $F(z) = \sum_{n=0}^\infty a_n e^{i n\alpha} z^n$.

The coefficients of $F$ satisfy the same condition: $ \sum_{n=0}^\infty |a_n e^{i n\alpha}| = \sum_{n=0}^\infty |a_n| < \infty$. This shows that it suffices to show that $f$ is of bounded variation on the interval $[0, 1]$.

For a proof of that case, see for example Show that $\,f(x) = \sum_{n=0}^\infty a_nx^n$, for $x \in [0,1]$, is of bounded variation.