interchanging spatial integral and time integral in the Brownian context

Question

The problem is the following

My attempt is inspired in the following:

Consider $$F_n(x) = \int_{-\infty}^\infty h(a) u_n(x - a)\,da $$

By Itô's formula:

\begin{align} &F_n(W_t) = F_n(W_0) + \int_0^t F_n'(W_s)\, dW_s + \frac{1}{2}\int_0^t F''_n(W_s)\, ds \\ &\int_{-\infty}^\infty h(a) u_n(W_t - a)\,da = \int_{-\infty}^\infty h(a) u_n(W_0 - a)\,da + \int_0^t \int_{-\infty}^\infty h(a) u'_n(W_s- a)\,da\, dW_s \\ & \qquad \qquad \qquad\qquad \qquad+ \frac{1}{2} \int_0^t \int_{-\infty}^\infty h(a) u''_n(W_s - a)\,da\, ds\\ &\int_{-\infty}^\infty h(a) [u_n(W_t-a) - u_n(W_0-a) - \frac{1}{2} u''_n(W_s-a)] \,da = \int_0^t \int_{-\infty}^\infty h(a) u'_n(W_s-a)\,da\, dW_s \\ &\int_{-\infty}^\infty h(a) \bigg[\int_0^t u'_n(W_s - a)\, dW_s \bigg] \,da = \int_0^t \int_{-\infty}^\infty h(a) u'_n(W_s-a)\,da\, dW_s \\ \end{align} where $u_n(x) = \int_{-\infty}^x \int_{-\infty}^y \rho_n(z)\, dz \, dy$ Now since $u'_n (x-a)\to 1_{(0,\infty)}(x-a) = 1_{(a,\infty)}(x)$ we obtain that

$$\Bbb{E} \bigg[\bigg| \int_0^t \int_{-\infty}^\infty h(a) u'_n(W_s-a)\,da\, dW_s - \int_0^t\int_{-\infty}^\infty h(a) 1_{(a,\infty)}(W_s)\,da\, dW_s \bigg|^2 \bigg] = \int_0^t h(a)^2\Bbb{E}[( u'_n(W_s-a) -1_{(a,\infty)}(W_s))^2 ] \, ds \leq \int_0^t h(a)^2\Bbb{P}\bigg[|W_s-a| < \frac{2}{n} \bigg] \, ds \to 0 $$

So the right-hand side of 6.24 is done. How do we obtain the left hand side?

Answer 1

I think you already have it (I say this without knowing what $\rho_n$ is). You have shown that $$ \int_0^t \left( \int_{-\infty}^\infty u^\prime_n( W_s - a ) h(a) da \right)dW_s = \int_{-\infty}^\infty h(a) \left( \int_0^t u^\prime_n( W_s - a ) dW_s \right)da, $$ and that, as $n \to \infty$, the left hand side converges to $\int_0^t \left( \int_{-\infty}^\infty 1_{ ( a,\infty) }( W_s ) h(a) da \right)ds$. So if you show that the right hand side of the above display converges to the left hand side of the equation (6.24) on the scanned problem you will be done. This can be achieved by using the fact that $h$ is continuous with compact support.

Because $h$ is positive, we can apply Jensen's inequality to obtain that \begin{align*} &\left[ \int_{-\infty}^\infty h(a) \left( \int_0^t \left( u^\prime_n( W_s - a ) - 1_{(a,\infty)}(W_s)\right) dW_s \right)da \right]^2 \\ & \qquad \leq K(h) \int_{-\infty}^\infty h(a)\left( \int_0^t \left( u^\prime_n( W_s - a ) - 1_{(a,\infty)}(W_s)\right) dW_s \right)^2 da, \end{align*} where $K(h)=\int_{-\infty}^\infty h(a) da. $ If $C(h)$ is the support of $h$, and $h_\infty$ is an upper bound of $h$ (which exist because $C_h$ is compact and $h$ is continuous), taking expectation on both sides, using Fubini's theorem and Ito's isometry, it follows that \begin{align*} &\mathbf{E}\left[ \int_{-\infty}^\infty h(a) \left( \int_0^t \left( u^\prime_n( W_s - a ) - 1_{(a,\infty)}(W_s)\right) dW_s \right)da \right]^2 \\ & \qquad \leq |C(h)| h_\infty K(h) \sup_{ a \in C(h) }\int_0^t \mathbf{E}\left( u^\prime_n( W_s - a ) - 1_{(a,\infty)}(W_s)\right)^2 ds. \end{align*} The convergence to zero of the above sequence finishes the required proof. If $\rho_n$ is what i think it is, this convergence should follow by standard arguments.