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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?

Yiorgos S. Smyrlis
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mathjacks
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  • Your expression $\sum_{i=1}^n \left| a_j [ x_i^j - x_{i-1}^j ] \right|$ doesn't make any sense. What does $j$ mean when it isn't the index of a summation? – Erick Wong Feb 14 '15 at 18:07
  • Sorry, fixed it. – mathjacks Feb 14 '15 at 18:12
  • Thanks. It's not enough to just bound each $i$ separately. This gives an upper bound of $n \sum_{j=0}^\infty |a_j|$ for the variation, and that's unbounded (since $n$ can be arbitrarily large). – Erick Wong Feb 14 '15 at 19:40
  • @ErickWong Thanks, that makes sense. Although, how else would you tackle this? This is the only definition of bounded variation that I'm aware of. – mathjacks Feb 14 '15 at 22:24

2 Answers2

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

Yiorgos S. Smyrlis
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  • Thanks for this. Is this essentially the same as my approach? I see in the comments to my question that it apparently is not enough to bound each $i$ separately because $i$ can be infinitely large. – mathjacks Feb 14 '15 at 22:23
  • @mathjacks This solution uses the stronger fact that $\sum_{k=1}^n (x^j_k - x^j_{k-1}) = 1$. You only used $|x^j_k - x^j_{k-1}| \le 1$. Yiorgos's bound is for the total variation independent of $n$. – Erick Wong Feb 15 '15 at 07:42
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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).

Salcio
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