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


$$Y^c\_t := c^{-1/2} X\_{ct}$$


**Question:** Is it true that as $c \to 0$, $Y^c$ converges in law to $\sigma(0) B\_t$ for a Brownian motion $B$?



Blow up limits for SDE
The answer to your question is yes, at least for the one- dimensional marginals. That said, I'm confident that the convergence you want actually holds on the level of stochastic processes (i.e. $c^{-1/2}X\_{ct}\xrightarrow{(d)}B\_t$ as functions on $C[0,T]$ for any $T>0$), but this requires some additional work. That said, here's the argument that proves convergence in distribution of the one dimensional marginals.


Suppose $(X\_{t})\_{t\geq{0}}$ solves the stochastic differential equation:
$$
dX\_{t}=\sigma(X\_{t})dB\_{t}, \hspace{5pt} X\_{0}=0
$$
where $\sigma$ is Lipschitz with Lipschitz constant $K>0$. That is $|\sigma(x)-\sigma(y)|\leq{K|x-y|}$ for all $x,y\in{\mathbb{R}}$. First, we will show that $\sup\_{t\in{[0,T]}}\mathbb{E}X\_{t}^{2}<\infty$ for all $T>0$.


To this effect, consider the stopping time $T\_{n}:=\inf\{t\geq{0}:|X\_{t}|=n\}$. For each $n\in{\mathbb{N}}$ we define the function $f\_{n}:[0,\infty)\rightarrow{\mathbb{R}}$ as follows:
$$
f\_{n}(t)=\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}
$$
Since $|X\_{t\wedge{T\_{n}}}|\leq{n}$ by the definition of $T\_{n}$, $|f\_{n}(t)|\leq{n^{2}}$ for all $t>0$. By Ito's formula applied to $X\_{t}$,
$$
X\_{t\wedge{T\_{n}}}^{2}=\int\_{0}^{t\wedge{T\_{n}}}\sigma(X\_{s})X\_{s}dB\_{s}+\int\_{0}^{t\wedge{T\_{n}}}\sigma^{2}(X\_{s})ds
$$  

Taking expectations we have that:
$$
\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}=\mathbb{E}\int\_{0}^{t\wedge{T\_{n}}}\sigma^{2}(X\_{s})ds\leq{\int\_{0}^{t}\mathbb{E}\sigma^{2}(X\_{s\wedge{T\_{n}}})ds}
$$  

Using the fact that $\sigma$ is Lipschitz we have that for any $x\in{\mathbb{R}}$, $|\sigma(x)|\leq{|\sigma(0)|+ K|x|}$.This in turn implies that $\sigma^{2}(x)\leq{2\sigma^{2}(0)+2K^{2}|x|^{2}}$. Applying this to the inequality above gives us:
$$
\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}\leq{\int\_{0}^{t}\big(2\sigma^{2}(0)+2K\mathbb{E}X\_{s\wedge{T\_{n}}}^{2}\big)ds}=2\sigma^{2}(0)t+2K^{2}\int\_{0}^{t}\mathbb{E}X\_{s\wedge{T\_{n}}}^{2}ds
$$
In other words, for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$, $f\_{n}$ satisfies the following inequality:
$$
f\_{n}(t)\leq{2\sigma^{2}(0)t+2K^{2}\int\_{0}^{t}f\_{n}(s)ds}
$$
By a Gronwall- type argument, it follows that for all $n\in{\mathbb{N}}$ and $t\in{[0,\infty)}$ we have that:
$$
f\_{n}(t)\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
By Fatou's lemma,
$$
\mathbb{E}X\_{t}^{2}\leq{\liminf\_{n\rightarrow{\infty}}\mathbb{E}X\_{t\wedge{T\_{n}}}^{2}}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}
$$
Hence, we have that $\mathbb{E}X\_{t}^{2}\leq{\frac{\sigma^{2}(0)}{K^{2}}\left(e^{2K^{2}t}-1\right)}<\infty$ for all $t>0$. From here, the proof is routine. We're interested in the process $Y^{c}\_{t}=c^{-1/2}X\_{ct}$ where $c>0$ is small.
$$
X\_{ct}=\int\_{0}^{ct}\sigma(X\_{s})dB\_{s}
$$
By Brownian scaling we have that $(c^{-1/2}B\_{ct})\_{t\geq{0}}\overset{d}{=}(\widetilde{B}\_{t})\_{t\geq{0}}$ where $(\widetilde{B}\_{t})\_{t\geq{0}}$ is a Brownian motion. Hence, "changing variables" in the stochastic integral above, we have that:
$$
X\_{ct}=c^{1/2}\int\_{0}^{t}\sigma(X\_{cu})d\widetilde{B}\_{u}
$$
Fix $\varepsilon>0$. Then:
\begin{align\*}
\mathbb{E}(c^{-1/2}X\_{ct}-\sigma(0)\widetilde{B}\_{t})^{2}&=\mathbb{E}\left(\int\_{0}^{t}\left(\sigma(X\_{cu})-\sigma(0)\right)d\widetilde{B}\_{u}\right)^{2}=\int\_{0}^{t}\mathbb{E}(\sigma(X\_{cu})-\sigma(0))^{2}du \\
&\leq{K^{2}\int\_{0}^{t}\mathbb{E}X\_{cu}^{2}du} \leq{K^{2}\varepsilon^{2}t+K^{2}\int\_{0}^{t}\mathbb{E}X\_{cu}^{2}1\_{(|X\_{s}|>\varepsilon \hspace{2pt} \text{for some $s\in{[0,ct]}$})}du}
\end{align\*}
Since the process $(X\_{t})\_{t\geq{0}}$ is continuous and square integrable, the second term on the last line goes to $0$ as $c$ tends to $0$. Since $\varepsilon>0$ is arbitrary, we conclude that for any $t>0$,
$$
c^{-1/2}X\_{ct}\xrightarrow{(d)}\sigma(0)B\_{t} \hspace{5pt} \text{as $c\rightarrow{0}$}
$$
**Bonus:** It turns out "upgrading" our argument to prove that:
$$
c^{-1/2}X\_{ct}\xrightarrow{(d)}\sigma(0)B\_t \hspace{5pt} \text{as functions on $C[0,T]$ for all $T>0$}
$$  

is not that hard. Observe that it suffices to prove two things:


1. Convergence of the finite- dimensional marginals of the processes $(c^{-1/2}X\_{ct})\_{t\geq{0}}$ to those of $(\sigma(0)B\_{t})\_{t\geq{0}}$.
2. Precompactness of the measures on $C[0,T]$ induced by the processes $(c^{-1/2}X\_{ct})\_{t\geq{0}}$.


The argument for (1) is analogous to the one- dimensional case that I've written out above, so I won't bother writing it out in detail. For precompactness, we make use of a neat theorem from Billingsley's "Convergence of Probability Measures." Namely, theorem 8.3 of this book gives us a condition for a collection of measures $(\mathbb{P}\_{n})\_{n\geq{1}}$ on $C[0,T]$ to be precompact:


**Theorem:** A sequence of measures $(\mathbb{P}\_{n})\_{n\geq{1}}$ on $C[0,T]$ is tight if:


1. For any $\varepsilon>0$ there exist $M>0$ and $n\_{0}\in{\mathbb{N}}$ such that for all $n\geq{n\_{0}}$,
$$
\mathbb{P}\_{n}(\{x:|x(0)|\geq{M}\})\leq{\varepsilon}
$$
2. For all $\varepsilon>0$ there exists $\delta>0$ and $n\_{0}\in{\mathbb{N}}$ such that for all $n\geq{n\_{0}}$,
$$
\delta^{-1}\mathbb{P}\_{n}(\{x:\max\limits\_{t\leq{s}\leq{t+\delta}}|x(s)-x(t)|\geq{\varepsilon}\})\leq{\varepsilon}
$$
for every $t\in{[0,T]}$.


Observe that the processes $Y^{c}\_{t}=c^{-1/2}X\_{ct}$ are all $0$ at $0$, so the first criteria is satisfied. For the second criteria, repeating our Gronwall- type argument for the 4th moment in place of the 2nd moment tells us that:
$$
C(T):=\max\_{t\in{[0,T]}}\mathbb{E}X\_{t}^{4}<\infty
$$
for any $T>0$. Using a Tchebyshev- type argument we have:
\begin{align\*}
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})&=\mathbb{P}\left(\max\limits\_{t\leq{s}\leq{t+\delta}}|\int\_{t}^{t+s}\sigma(X\_{cr})d\widetilde{B}\_{r}|\geq{\varepsilon}\right) \\ 
&\leq{\frac{\mathbb{E}\Big(\max\limits\_{t\leq{s}\leq{t+\delta}}|\int\_{t}^{t+s}\sigma(X\_{cr})d\widetilde{B}\_{r}|^{4}\Big)}{\varepsilon^{4}}} \\ 
&\leq{C\hspace{2pt}\mathbb{E}\Big(\int\_{t}^{t+\delta}\sigma(X\_{cr})d\widetilde{B}\_{r}\Big)^{4}\varepsilon^{-4}} \\
&\leq{C'\hspace{2pt}\mathbb{E}\Big(\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\Big)^{2}\varepsilon^{-4}}
\end{align\*}
where $C,C'>0$ are absolute constants. The inequality on the third line follows by Doob's inequality, using the fact that the process $\Big(\int\_{t}^{t+s}\sigma(X\_{cr})\widetilde{B}\_{r}\Big)\_{s\geq{0}}$ is a martingale. The inequality on the fourth line follows by applying the Burkholder- Davis- Gundy inequality (<https://almostsuremath.com/2010/04/06/the-burkholder-davis-gundy-inequality/>). Using the fact that $\sigma$ is Lipschitz,
$$
\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\leq{\int\_{t}^{t+\delta}\big(2\sigma^{2}(0)+2K^{2}X\_{cr}^{2}\big)dr}\leq{2\delta\sigma^{2}(0)+2\delta{K^{2}}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{2}dr}
$$
$$
\Big(\int\_{t}^{t+\delta}\sigma^{2}(X\_{cr})dr\Big)^{2}\leq{8\delta^{2}\sigma^{4}(0)+8\delta^{2}{K^{4}}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{4}dr}
$$
Thus:
\begin{align\*}
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})\leq{8C'\varepsilon^{-4}\delta^{2}\Big(\sigma^{4}(0)+K^{4}\mathbb{E}\max\_{t\leq{r}\leq{t+\delta}}X\_{cr}^{4}\Big)}
\end{align\*}
Finally, since $(X\_{t})\_{t\geq{0}}$ is a martingale, by Doob's inequality,
$$
\mathbb{E}\big(\max\_{r\in{[t,t+\delta]}}X^{4}\_{cr}\big)\leq{\mathbb{E}\big(\max\_{r\in{[0,T]}}X^{4}\_{r}\big)}\leq{C\hspace{2pt}\mathbb{E}X\_{T}^{4}}\leq{C\cdot{C(T)}}
$$
Thus, we see that taking $\delta<C''\varepsilon^{4}$ for some constant $C''>0$ we have that:
$$
\mathbb{P}(\max\limits\_{t\leq{s}\leq{t+\delta}}|c^{-1/2}X\_{cs}-c^{-1/2}X\_{ct}|\geq{\varepsilon})\leq{\delta\cdot{\varepsilon}}
$$
for any $t\in{[0,T]}$. This completes our proof.


