*Note: This is a strengthening of the following [result](https://mathoverflow.net/questions/438492/blow-up-limits-for-sde), 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


$$\mathbb E[\sup\_{0 \leq t \leq 1} |X^c\_t - \sigma(0) W^c\_t|] \to 0?$$



Strong blow up limits for SDE
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](https://mathoverflow.net/a/438499/80052) 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:
$$
\mathbb{E}(\sup\_{0\leq{t}\leq{1}}|X\_{t}^{c}-\sigma(0)W\_{t}^{c}|)\leq{2|\sigma(0)|\sqrt{e^{2L^{2}c}-1}}\rightarrow{0} \hspace{10pt}\text{as}\hspace{10pt}c\rightarrow{0}
$$


