I have got the following problem, but no idea where to start, so maybe one of you can help me out. We have got a continuous function $f:[0,1]\to \mathbb{R}$ with $f(0)\neq f(1)$. The task is to construct a sequence of partitions $\Delta_n :=\lbrace 0=t_0,t_1,\dots, t_{k_n}=1 \rbrace $ such

a) the corresponding sequence of mesh sizes tends to zero,

b) the quadratic variation of $f$ along $\Delta_n$ is zero, this means $\lim_{n\to \infty} \sum_{i=1}^{k_n}(f(t_i)-f(t_{i-1}))^2=0$.

I understood, that for $f$ Lipschitz this holds for any partition sufficing a), but since this condition is not given I do not have really an idea how to find the partition.

I am grateful for any hint (or counterexample, if the statement is just wrong and a prerequisite was forgotten in the exercise).


If $f$ is continuous and of bounded variation, then any sequence of partitions $\Delta_n$ with $\|\Delta_n\| \to 0$ as $n \to \infty$ will suffice.

We have

$$\sum_{j=1}^{k_n}|f(t_j) - f(t_{j-1})|^2 \leqslant \sup_{1 \leqslant j \leqslant k_n}|f(t_j)-f(t_{j-1})|\sum_{j=1}^{k_n}|f(t_j) - f(t_{j-1})| \leqslant \sup_{1 \leqslant j \leqslant k_n}|f(t_j)-f(t_{j-1})|V_0^1(f)$$

Since $f$ is uniformly continuous on the compact interval $[0,1]$, for any $\epsilon >0$ there exists $\delta > 0$ such that for all $x,y \in [0,1]$ with $|x-y| < \delta$ we have $|f(x) - f(y)| < \epsilon/V_0^1(f)$.

If $\|\Delta_n\| \to 0$ then there exists $N$ such that $\|\Delta_n\| < \delta$ for all $n \geqslant N$.

Whence, for all $n \geqslant N$, we have $\sup_{1 \leqslant j \leqslant k_n}|f(t_j)-f(t_{j-1})| < \epsilon/V_0^1(f),$ and

$$\sum_{j=1}^{k_n}|f(t_j) - f(t_{j-1})|^2 \leqslant \epsilon$$

For something more exotic -- where quadratic variation over partitions converges to $0$ only for some but not all partition sequences (if such a thing is possible in general) -- you would have to consider continuous functions of unbounded variation.

  • $\begingroup$ If I understood you correctly, you are assuming that $f$ has bounded variation and is uniformly integrable. Can this be deduced or are those additional assumptions? $\endgroup$ – Graf Zahl Jun 1 '19 at 12:50
  • $\begingroup$ I don't know why I cannot edit my further comment, so I will add here that I did not mean "uniformly integrable" but "uniformly continuous". My fault. $\endgroup$ – Graf Zahl Jun 1 '19 at 13:24
  • $\begingroup$ Comments get locked and cannot be edited after a few minutes. I assume bounded variation and continuous (as with your hypothesis). If $f$ is continuous it is also uniformly continuous since $[0,1]$ is compact (closed and bounded). $\endgroup$ – RRL Jun 1 '19 at 17:34

So, here is a way that should work.

Let $\varepsilon>0$ and let $\Delta=\{0=t_0,t_1,\dots,t_k=1\}$ be a partition. (We later use the notation $S(\Delta)=\sum_{i=1}^{k}(f(t_i)-f(t_{i-1}))^2$ and $|\Delta|:=\sup_{i=1,\dots,k}{|t_i-t_{i-1}}|$). We can find a partition $\tilde{\Delta}\supset \Delta$ such that $|\tilde{\Delta}|\leq \varepsilon$. For this partition, we want to construct a further partition $\Delta_{\varepsilon}$ such that $\Delta_{\varepsilon}\supset \tilde{\Delta}$ and $S(\Delta_{\varepsilon})<\varepsilon$. By the continuity of $f$ and the intermediate value theorem, we can find $s_i\in(t_{i-1},t_i)$ such that $f(s_i)=\frac{1}{2}(f(t_{i-1})+f(t_i))$. Hence we obtain a new partition $\bar{\Delta}$ with $|\bar{\Delta}|<\varepsilon$ and $S(\bar{\Delta})=\frac{1}{2}S(\tilde{\Delta})$. Now we repeat the last step until we obtain the desired result.

Note: I never used $f(0)\neq f(1)$ although it was given in the exercise.


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