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I am studing some old exams and came across this one that has me stumped. Suppose $f:[0,1] \rightarrow \mathbb{R}$ is absolutely continuous. Show

$$\lim_{n \rightarrow \infty} \sum_{k=0}^{2^{n}-1} \left| f( k/2^{n}+1/2^{n} )-f( k/2^{n} )\right|^{2} =0 $$

I can prove this if the derivative of $f$ is bounded but I cannot solve the general case.

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Boy,am I rusty in serious analysis-I couldn't remember how to even begin this problem.I better brush up.......... –  Mathemagician1234 Mar 22 '12 at 5:53
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up vote 4 down vote accepted

This result is proved using some extra elementary facts in Theorem 1.10, Page 8, of "Introduction to Stochastic Calculus with Applications" by Fima Klebaner.

Every absolutely continuous function is continuous. This one happens to be continuous on a compact set, so also uniformly continuous (although, I think that might be directly implied by the absolute continuity. I can't recall offhand).

Anyway, it's a standard result that any absolutely continuous function $f$ has finite variation, $V_{f} = M < \infty$.

Then the quadratic variation can be written like this: $$ [f](t) = \lim_{\delta_{n}\to{0}} \sum_{i=0}^{n-1}\quad \biggl( f(t_{i+1}^{n}) - f(t_{i}^{n})\biggr)^{2} $$ $$ \leq \lim_{\delta_{n}\to{0}} \max_{i}|f(t_{i+1}^{n}) - f(t_{i}^{n})|\cdot{}\sum_{i=0}^{n-1}\quad |f(t_{i+1}^{n}) - f(t_{i}^{n})|$$ $$\leq \lim_{\delta_{n}\to{0}} \max_{i}|f(t_{i+1}^{n}) - f(t_{i}^{n})|\cdot{} M$$

The last simplification is because the variation is the supremum taken over all partitions of the quantity $\sum_{i=0}^{n-1}\quad |f(t_{i+1}^{n}) - f(t_{i}^{n})|$.

Now, since $f$ is uniformly continuous on $[0,1]$ and thus on $[0,t]$ for any $t\in [0,1]$, the last limit goes to zero: $$ \lim_{\delta_{n}\to{0}} \max_{i}|f(t_{i+1}^{n}) - f(t_{i}^{n})| = 0. $$

The particular choice of partition in your case involves $k/2$ and $1/2$ raised to the power $n$, but it doesn't matter. The $t_{i}^{n}$ partition points, and the notation $\delta_{n}$ for a $\delta_{n}$-fine partition proves it more generally.

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So we really only needed that f is BV and continuous on a compact set –  Digital Gal Mar 22 '12 at 14:26
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