I am trying to get familiar with Rough Path Theory by reading 'Rough Path Theory and Stochastic Calculus' by Yuzuru Inahama. He argues in favour of Rough Paths as opposed to Ito integrals as the latter converge only in $L^2(\mu)$ sense. He states (statement 1):
"Each element of $L^2(\mu)$ is just an equivalence class with respect to μ and a single point set is of μ-zero set, the stochastic integral does not have an x-wise meaning."
Furthermore, he states (statement 2):
"Neither is it continuous in x. For example, let us consider Levy’s stochastic area for two-dimensional Brownian motion
$$ x = (x^1,x^2)\rightarrow \int_{0}^{1} (x_s^2 dx_s^1-x_s^1dx_s^2 ) $$
is not continuous".
My questions are: What is the simplest way of seeing statement 2? As far as statement 1 is concerned, is there a nice example, ideally with some relevance to other areas such as mathematical finance?
Many thanks
Some references for this same result (I couldn't find the article by Sugita referenced in the article), are in "A Course on Rough Paths" in Proposition 1.1 and Proposition 1.29 in Lyon's book "Differential Equations Driven by Rough Paths"
There exists no separable Banach space $B \subset C([0, 2\pi])$ with the following properties
1.Sample paths of Brownian motions lie in $B$ almost surely.
2.The map $(f, g) \mapsto\int_{0}^{\cdot}f(t)\dot{g}(t) dt$ defined on smooth functions extends to a continuous map from $B \times B$ into the space of continuous functions on $[0, 2\pi]$.
Proof
Fix iid standard Gaussians $(\xi_{n})_{n\in \mathbb{Z}}$ and orthonormal basis $e_{n}$ for the Cameron–Martin space of Brownian motion $W^{1,2}_{0}([0,2\pi])$
$$e_{0}(t)=\frac{t}{\sqrt{2\pi}},e_{n}(t)=\frac{\sin(kt)}{k\sqrt{pi}},e_{-n}(t)=\frac{\cos(kt)}{k\sqrt{pi}},n>0.$$ Then the processes $$X_{t}^{N}=e_{0}\xi_{0}+\sum_{n=1}^{N} e_{n}\xi_{n}+e_{-n}\xi_{-n}\text{ and } Y_{t}^{N}=e_{0}\xi_{0}+\sum_{n=1}^{N}e_{n}\xi_{-n}-e_{-n}\xi_{n},$$
are approximations to two Brownian motions $X_{t},Y_{t}$. On the other hand,
$$\int_{0}^{2\pi }X^{N}_{t}dY^{N}_{t}=\frac{\xi_{0}^{2}}{2}+\sum_{n=1}^{N}\frac{\xi_{n}^{2}+\xi_{-n}^{2}}{n}\to +\infty$$
and
$$\int_{0}^{2\pi }Y^{N}_{t}dX^{N}_{t}=\frac{\xi_{0}^{2}}{2}-\sum_{n=1}^{N}\frac{\xi_{n}^{2}+\xi_{-n}^{2}}{n}\to -\infty.$$
Example Here we do the two-dimensional counterexample in the OP. We follow On the Importance of the Levy Area for Studying the Limits of Functions of Converging Stochastic Processes. Application to Homogenization in pg.5. We consider $$X^{1,n}_{t}=B^{1}_{t}+\frac{\cos(n^{2}t)}{n},X^{2,n}_{t}=B^{2}_{t}+\frac{\sin(n^{2}t)}{n}.$$
This pair converges to the Brownian motion $(B^{1}_{t},B^{2}_{t})$. However, the Lévy area is
$$A(X^{1,n},X^{2,n})=\int X^{2,n}_{t} dX^{1,n}_{t}-\int X^{1,n}_{t} dX^{2,n}_{t}$$
$$=A(B^1,B^{2})+A(\frac{\sin(n^{2}t)}{n},\frac{\cos(n^{2}t)}{n})+\int B^{2}_{t}n\sin(n^{2}t)dt+\int \frac{\cos(n^{2}t)}{n} dB^{1}_{t}$$
$$-\int B^{1}_{t}(-n\cos(n^{2}t)dt)-\int \frac{\sin(n^{2}t)}{n} dB^{2}_{t}.$$
We have
$$A(\frac{\sin(n^{2}t)}{n},\frac{\cos(n^{2}t)}{n})=\pi t$$
and that two of the integrals diverge. So the Lévy area of the approximation doesn't converge to the limit Lévy area $A(B^1,B^{2})$, which is only defined in the $L^{2}$-sense.