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Let $f$ be continuous and $\alpha$ of bounded variation (rectifiable path) in the interval $[a,b]$. Define $F:[a,b]\to\mathbb{R}$ by doing $F(x) = \int_a^x f d\alpha$. Prove that $F$ is of bounded variation and that its total variation won't exceed $M\cdot V$, where $M = \sup\{|f(x)|; x\in [a,b]\}$ and $V$ is the total variation of $\alpha$ in $[a,b]$

If we recall to the definition of a Stieltjes integral:

$$F(x) = \int_a^x f d\alpha = \lim_{|P|\to 0} \sum_{i}^k f(\zeta_i)[\alpha(t_i)-\alpha(t_{i-1})]$$

I don't know what 'total variation' means. Is it just the $\alpha(b)-\alpha(a)$? I think that since $f\le \sup{|f|}$ I may have something. I also thought about doing a telescope sum on $\alpha(t_i)-\alpha(t_{i-1})$ but it didn't work. I know I must arrive at something that is less than $\sup\{f\}\cdot V$, being $V$ whatever it means by 'total variation'.

kimchi lover
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Guerlando OCs
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  • Using your notation, the total variation of $\alpha$ is the number $\displaystyle\sup_P\sum_i^k|\alpha(t_i)-\alpha(t_{i-1})|$. – sranthrop Aug 23 '17 at 02:22

1 Answers1

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Let $P = \{t_0 , t_1, \ldots t_n\}$ be any partition of $[a,b]$.

Then

$$var(P,F) = \sum_{i=1}^n |F(t_i) - F(t_{i-1})|= \sum_{i=1}^n\left|\int_{t_{i-1}}^{t_i}fd\alpha\right|$$ $$ \le \sum_{i=1}^n \int_{t_{i-1}}^{t_i}|f|d\alpha \le M\sum_{i=1}^n \int_{t_{i-1}}^{t_i}d\alpha = M \sum_{i=1}^{n} \alpha(t) - \alpha(t_{i-1})$$

$$\le MV$$ by definition, since

$$V = \sup \sum_{i=1}^n \alpha(t_i) - \alpha(t_{i-1})$$

where the sup is taken over all partitions of $[a,b]$.

Note that we use that the weighting function $\alpha$ is monotonically increasing here, which is usually stipulated. We also don't need that $f$ is continuous here, but just integrable and bounded.

Since $P$ was an arbitary we may conclude that $F$ is $BV$ and its total variation won't exceed $MV$.

David Bowman
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  • The question does not stipulate that $\alpha$ is increasing, and this theorem is true for any $\alpha$ of bounded variation. – RRL Aug 23 '17 at 03:03
  • Here it is proved that $\left|\int_{t_{j-1}}^{t_j} f , d\alpha \right| < \sup_{t \in [t_{j-1},t_j]} f(t) ,V_{t_{j-1}}^{t_j}(\alpha)$. Using that result you can generalize this. – RRL Aug 23 '17 at 03:12
  • Fair enough. If $\alpha$ were montonically increasing then its variation would be indeed trivial. My reasoning for that comment was simply that I'd only ever seen the Stieltjes integral defined in the context of monotonically increasing $\alpha$. Otherwise we wouldn't even have a triangle inequality (which is where I suppose my proof breaks down). – David Bowman Aug 23 '17 at 03:14
  • If the integrator is assumed to be monotonic then many things are easy, but the RS integral extends to a much broader class not restricted only to functions of bounded variation. – RRL Aug 23 '17 at 03:23