Weak convergence of stochastic integrals

Question

Suppose that all random variables here live on a common probability space which supports a Brownian motion. Given a sequence of continuous processes $(H^n_s)_{s \in [0,T]}$ which have a unifom bound in $L^2:$ $$ \sup_{n} \sup_{0\le s\le T} \mathbb{E}[|H^n_s|^2] < \infty $$ and which converge weakly, i.e. there is a continuous process $(H_s)_{s \in [0,T]}$ such that: $$ H^n \Rightarrow H $$ as random variables on $C([0,T]; \mathbb{R})$, I would like to prove that: $$ \int_0^{\cdot} H^n_sdW_s \Rightarrow \int_0^{\cdot} H_sdW_s $$ My problem is that I fall in the trap of using the $L^2$ theory. So for example if I could bound $$ \sup_{n} \mathbb{E}[\sup_{0\le s\le T} |H^n_s|^2] < \infty $$ I think I would be able to prove this by using some far too complicated results, such as Skorokhod's representation theorem. But I feel like there should be a simple straight foreward argument, which maybe doesn't need any $L^2$ boundedness at all.

Answer 1

It isn't true and the question isn't even well posed. The weak limit of the $H^n$ is only unique up to equality in law, but the law of the stochastic integral doesn't only depend on the law of the integrand. If you just take $H^n_t = W_t$ for all $n$, you have $H^n \to W$ weakly as well as $H^n \to -W$. But $\int_0^\cdot W_s\,dW_s \ne \int_0^\cdot -W_s \,dW_s$ in law. (We have $\int_0^t W_s\,dW_s = \frac{1}{2} (W_t - t)$ which isn't symmetric in law; for instance, when $t=1$, it has positive probability to be greater than 1 but zero probability to be less than $-1$.)

Alternatively, take $H^n_s = (-1)^n W_s$. Then every $H^n$ is a Brownian motion, so they are all equal in law, hence converge weakly (to a Brownian motion). By Ito's formula we have $X^n_t := \int_0^t H^n_s\,dW_s = \frac{(-1)^n}{2}(W_t^2 - t)$. These oscillate between two different laws so they do not converge weakly.

If you drop continuity of $H^n$ and let $T=\infty$ you can get a more vivid example of this phenomenon: let $H^n_t = 1$ until $W_t$ hits $(-1)^n$, and $0$ thereafter. Then all $H^n$ are equal in law (by symmetry of Brownian motion) but $X^n_\infty = (-1)^n$ almost surely.

If you have weak convergence of the joint distributions $(H^n, W)$ then I think it will probably be true.