# Quadratic Variation of a continuous function

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.

• 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? – Graf Zahl Jun 1 '19 at 12:50
• 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. – Graf Zahl Jun 1 '19 at 13:24
• 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). – 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.