# A path with zero increments and positive area

I am studying rough paths from the 2007 St Flour lecture notes and I came across the example at the end of chapter one of the sequence of paths $$X(n):[0,2\pi]\to \mathbb R^2$$ given by $$X_t(n) = \frac{1}{n} (\cos(n^2t),\sin(n^2t))$$, which for each $$n$$ enclose an area of $$\pi$$ but converge to the zero path $$\mathbf{0}$$ in the uniform norm $$\|\cdot\|_{\infty,[0,2\pi]}$$. To compute the area in the sense I'm referring to, the area of a loop ($$\frac{\pi}{n^2}$$) must be multiplied by the winding number (number of loops, which is $$n^2$$).

It can also be shown that the $$2$$-variation $$\|\cdot\|_{2,[0,2\pi]}$$ of $$X(n)$$ is bounded. By using Lemma 1.13 and the fact that $$X(n)\to \mathbf{0}$$ in the uniform norm, it can be shown that $$X(n)$$ in fact converges to $$\mathbf{0}$$ in the $$p$$-variation norm $$\|\cdot\|_{\mathcal{V}^p([0,2\pi])}:= \|\cdot\|_{p,[0,2\pi]} + \|\cdot\|_{\infty,[0,2\pi]}$$ for any $$p>2$$. Hence, the book concludes that the map sending $$X\in \mathcal{V}^p([0,2\pi])$$ (where $$\mathcal{V}^p([0,2\pi])$$ is the space of paths with finite $$p$$-variation) to $$A^X_{2\pi}:=\frac{1}{2}[\int_{0 is not continuous in the $$p$$-variation norm for any $$p>2$$. The object $$A^X_t$$ gives the area enclosed by the curve described by $$X$$ until time $$t$$, if we complete the curve to a close curve by a straight line and provided the winding number is taken into account if $$X$$ loops into itself.

I have heard from an expert that "the limit of the paths in the $$2$$-variation topology is a path with zero increments and positive area". This sounds confusing. I think this is what he meant:

The $$2$$-variation topology is a topology on $$2$$-rough paths, and is introduced in Chapter 3. A $$2$$-rough path $$\mathbb X_{s,t}:=(1, X^1_{s,t}, X^2_{s,t})$$ is a mapping from the set $$\triangle_T :=\{(s,t)| s,t \in[0,T] \:\text{ and } s\leq t\}$$ into $$V^{\otimes 0}\oplus V^{\otimes 1}\oplus V^{\otimes 2}$$, where $$V$$ is a vector space, whose entries satisfy certain condition. Take a smooth and finite one-variation path $$X:[0,T]\to V$$, then $$X_{s,t} :=(1, X_t-X_s, \int_{s turns out to be a $$2$$-rough path (the entries in fact correspond to the iterated integrals (i.e. $$X^k_{s,t} := \int^t_{s}X^{k-1}_{s,u}\otimes \dot{X}_u du$$). The second entry $$X^2_{s,t}:=\int_{s can be decomposed as $$X^2_{s,t}:=S^X_{s,t}+A^X_{s,t}$$, where $$A^X_{s,t}$$ has the same interpretation as the "enclosed area" as before (now starting at $$s$$, not at $$0$$). It can be shown that $$S^X_{s,t}\to 0$$ in the uniform norm in the above example, so that $$X^2_{s,t}$$ tends to $$\pi\:(\:= A^{X(n)}_{s,t},\: \forall n)$$.

I think that what the expert meant is that, in the $$2$$-variation topology, the $$2$$-rough paths associated with each $$X_n$$ converge to $$(1,0,\pi)$$, whereas the $$2$$-rough path associated with the zero path is $$(1,0,0)$$.

Can someone shed light on the matter? If they don't understand rough paths, maybe they can try to answer the question independently of rough path theory considerations. Can a path have zero increments but enclose a positive area? Can the limit of the sequence of paths I have described in a relevant topology be considered an example of such a pathological path? Or am I right and what Lyons meant is that we can construct a ($$2$$-) rough path with zero increments and positive area, i.e. $$(1,0,\pi)$$?

• I think such a path does indeed not exist in the classical sense and what is meant is that the sequence converges in the sense of rough paths. Rhw idea behind rough paths is that you don't only keep track of the path itself but also its iterated integrals. This additional information allows the existence of such non-trivial, pathological objects. Jun 22 at 6:59