Quadratic Variation, Continuity, and Bounded Variation
Summary
The discussion corrects the claim that continuity is equivalent to zero quadratic variation. A continuous path need not have zero quadratic variation: Brownian motion is continuous while its quadratic variation over time is nonzero. The answer also gives a continuous oscillatory function as a counterexample to the claim that every continuous function has zero quadratic variation.
A sufficient condition for zero quadratic variation is that a function be both continuous and of bounded variation on a compact interval. The argument bounds the sum of squared increments by the largest increment multiplied by total variation. Uniform continuity makes the largest increment shrink as partitions become finer, while bounded variation keeps the other factor finite. This result has a specific assumption: continuity alone does not establish the conclusion, and the counterexamples show why path regularity and variation must be distinguished. The material is a mathematical clarification relevant to stochastic calculus, rather than a trading strategy or empirical analysis.
Key ideas
- Continuity alone does not imply zero quadratic variation.
- Brownian paths are continuous and have nonzero quadratic variation.
- A continuous function of bounded variation has zero quadratic variation under increasingly fine partitions.
- The bounded-variation argument controls squared increments using the largest increment and total variation.
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# Proof that $f$ is continuous if and only if it has 0 quadratic variation?
# Proof that $f$ is continuous if and only if it has 0 quadratic variation?
I understand that $f$ continuous $\Rightarrow Q(f) = 0$ where this is defined over a bounded interval [0,T] as then we may use uniform continuity and the mean value theorem. But I am not sure how the converse implication can be shown?
## Answer by Kevin (score 4)
https://quant.stackexchange.com/a/53626
The statement is not quite correct. Brownian motion has continuous sample paths but not zero quadratic variation. In fact, $[B_t]=t$, which is finite yet unbounded. The sample paths of Brownian motion have infinite variation which is why we need the Itô integral in the first place and can't simply use the Riemann-Stieltjes integral.
You need $f$ to be of bounded variation, e.g. if $f$ is smooth, that would imply bounded variation. In this case, continuity and bounded variation imply zero quadratic variation.
The key step is given here and more details are here.
Basically, you need a partition $(t_j)_{j=1,...,n}$ of $[0,T]$ and observe that \begin{align*} \sum_{j=1}^n |f(t_j)-f(t_{j-1})|^2 &= \sum_{j=1}^n |f(t_j)-f(t_{j-1})||f(t_j)-f(t_{j-1})|\\ &\leq \sup_{j=1,...,n}|f(t_j)-f(t_{j-1})|\cdot\sum_{j=1}^n |f(t_j)-f(t_{j-1})| \\ &= \sup_{j=1,...,n}|f(t_j)-f(t_{j-1})|\cdot V_T(f), \end{align*} where $V_T(f)$ is the variation of $f$ over $[0,T]$. The continuity of $f$ implies that $f$ is uniformly continuous over the compact set $[0,T$] and the supremum converges to zero for finer partitions. If $V_T(f)$ is bounded, then the quadratic variation of $f$ tends to zero, as well.
## Answer by ir7 (score 1)
https://quant.stackexchange.com/a/53630
Direct implication is not true. Function
$$f(x) = \left\{\matrix{x^2\sin\left(\frac{1}{x^4}\right) & x \not= 0\\0 & x = 0}\right.$$
is continuous, but its quadratic variation over interval $[0,1]$ is non-zero. You can find a proof here.
(Bounded variation and differentiability of functions of type $y=x^a \sin (1/x^b)$ can be found here and here.)Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.