1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Being continuous is not enough. You need to look at the number of points where f is not differentiable (for exmple f(x) = abs(x) is not differentiable at 0.) if the function is cotinuous and almost everywhere differentiable (i.e. is not differentiable at at most countably many points) then q.v. is 0. The q.v. can only be non-0 if the function is not differentiable at uncountably many points. $\endgroup$ – Dimitri Vulis Apr 26 at 14:22
3
$\begingroup$

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.

| improve this answer | |
$\endgroup$
0
$\begingroup$

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.)

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.