Show that $\,f(x) = \sum_{n=0}^\infty a_nx^n$, for $x \in [0,1]$, is of bounded variation

Question

Let $\{a_n\} \subset \mathbb{R}$, be such that $\sum_{n=0}^\infty \lvert a_n\rvert < \infty$. Define $$f(x) = \sum_{n=0}^\infty a_nx^n \quad \text{for } x \in [0,1]$$ Prove that $f$ is of bounded variation.

Attempt:

Let $P = \{ x_0, \ldots , x_n \}$ be any partition of $[0,1]$. Then $$ \sum_{i=1}^n \left| \sum_{j=0}^\infty a_j x_i^j - \sum_{j=0}^\infty a_j x_{i-1}^j \right| = \sum_{i=1}^n \left| \sum_{j=0}^\infty a_j [ x_i^j - x_{i-1}^j ] \right|. $$ Since $\sum_{n=0}^\infty \lvert a_n\rvert < \infty$ and $x_i \in [0,1]$ $\forall i$, $$ \sum_{j=0}^\infty \left| a_j [ x_i^j - x_{i-1}^j ] \right| \leq \sum_{n=0}^\infty \lvert a_n\rvert < \infty. $$ Hence, $f(x)$ is of bounded variation since the elements of the sum are finite the index of the outer sum is also finite.

Is this a valid approach? What other ways are there to prove this?

Answer 1

Let $$ 0=x_0<\cdots<x_n=1, $$ then $$ \sum_{k=1}^n \lvert\, f(x_{k})-f(x_{k-1})\rvert=\sum_{k=1}^n\left|\sum_{j=0}^\infty a_jx_k^j-\sum_{j=0}^\infty a_jx_{k-1}^j\right| \le\sum_{k=1}^n \sum_{j=0}^\infty\left| a_j\right|(x_{k}^j-x_{k-1}^j) \\ = \sum_{j=0}^\infty \left| a_j\right|\sum_{k=1}^n(x_{k}^j-x_{k-1}^j)=\sum_{j=0}^\infty\lvert a_j\rvert<\infty. $$

Answer 2

Take $g(x) = \sum_{a_{k}>=0}a_{k}*x^{k}$ and $h(x) = \sum_{a_k < 0} |a_{k}|*x^{k}$ Then both g(x) and h(x) are increasing so they are of finite variance. Clearly f = g - h so it must be of finite variance as well (triangle inequality).