Note: This is a strengthening of the following result, motivated by the need for strong convergence in applications.
Let $W$ be a one dimensional standard Brownian motion, and let $X$ be the solution to the SDE
$$dX_t = \sigma(X_t) \, dW_t \, , \, X_0 = 0$$
with $\sigma: \mathbb R \to \mathbb R$ Lipschitz continuous.
For each $c > 0$, define the process $Y^c$ on $[0, 1]$ by
$$X^c_t := c^{-1/2} X_{ct}.$$
Similarly define
$$W^c_t := c^{-1/2} \, W_{ct}.$$
Question: Is it true that as $c \to 0^+$, we have
$$\mathbb E[\sup_{0 \leq t \leq 1} |X^c_t – \sigma(0) W^c_t|] \to 0?$$
Best Answer
Yes, this is true. Since $(X_{t}^{c})_{t\geq{0}}$ and $(W_{t}^{c})_{t\geq{0}}$ are both martingales for fixed $c$, applying Cauchy- Schwarz and then Doob's inequality tells us that: $$ \mathbb{E}(\sup_{0\leq{t}\leq{1}}|X_{t}^{c}-\sigma(0)W_{t}^{c}|)\leq{\big(\mathbb{E}(\sup_{0\leq{t}\leq{1}}|X_{t}^{c}-\sigma(0)W_{t}^{c}|^{2})\big)^{1/2}}\leq{\sqrt{2}\big(\mathbb{E}|X_{1}^{c}-\sigma(0)W_{1}^{c}|^{2}\big)^{1/2}} $$ By our SDE for $(X_{t})_{t\geq{0}}$ we have that: $$ X_{1}^{c}=c^{1/2}X_{c}=c^{1/2}\int_{0}^{c}\sigma(X_{u})dW_{u}=c^{1/2}\int_{0}^{1}\sigma(X_{cr})dW_{cr}=\int_{0}^{1}\sigma(X_{cr})dW^{c}_{r} $$ Since $(W^{c}_{t})_{t\geq{0}}$ is a standard Brownian motion for any $c>0$, by the Ito isometry: $$ \mathbb{E}(X_{1}^{c}-\sigma(0)W_{1}^{c})^{2}=\mathbb{E}\Big(\int_{0}^{1}(\sigma(X_{cr})-\sigma(0))dW^{c}_{r}\Big)^{2}=\int_{0}^{1}\mathbb{E}(\sigma(X_{cr})-\sigma(0))^{2}dr $$ If $L>0$ is the Lipschitz constant of $\sigma$, we have that: $$ \int_{0}^{1}\mathbb{E}(\sigma(X_{cr})-\sigma(0))^{2}dr\leq{L^{2}\int_{0}^{1}\mathbb{E}X_{cr}^{2}dr}\leq{L^{2}\sup_{0\leq{r}\leq{1}}\mathbb{E}X_{cr}^{2}}\leq{2L^{2}\mathbb{E}X_{c}^{2}} $$ The last inequality follows by Doob's inequality, since $(X_{t})_{t\geq{0}}$ is a martingale. Finally, recall that by the Gronwall- type argument in 1 we have that: $$ \mathbb{E}X_{c}^{2}\leq{\frac{\sigma^{2}(0)}{L^{2}}\big(e^{2L^{2}c}-1\big)} $$ Putting all this together, we have that: $$ \mathbb{E}(\sup_{0\leq{t}\leq{1}}|X_{t}^{c}-\sigma(0)W_{t}^{c}|)\leq{2|\sigma(0)|\sqrt{e^{2L^{2}c}-1}}\rightarrow{0} \hspace{10pt}\text{as}\hspace{10pt}c\rightarrow{0} $$