Converge to Brownian Motion problem Consider the following sequence of SDEs:
$dX^n_t = \sin(nX^n_t)dt + dW_t, X^n_0 = 0\,\,\,$
Show that the solutions $X^n$ converge in finite dimensional distribution to Brownian Motion.
I have been working on this for a long time and can't get anywhere. Any help would be appreciated. Thank you.
 A: Surely not an exam proof, what is going to come.
We will use Theorem 5.4.1 from Ethier and Kurtz, Markov Processes, Wiley, 1986 (EK86); the important part is quoted below.
Let us first write down the generator $A_n$ of the process $X^{(n)}$
\begin{equation}
 A_n f(x) = \sin (nx) f'(x) + \frac{1}{2} f''(x) ,  x \in \mathbb{R}.
\end{equation}
The expected limit process, Brownian motion, has generator $A f(x) = \frac{1}{2}f''(x)$, $x \in \mathbb{R}$. Both definitions are understood for functions $f \in C_c^\infty(\mathbb{R})$, the compactly supported, infinitely often differentiable functions on $\mathbb{R}$.
Lemma 1: The sequence $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ is tight in the Skorohod space $D([0,\infty), \mathbb{R})$.
Lemma 2: For any $f \in C_c^\infty(\mathbb{R})$ we find a function $f_n$ such that
\begin{equation} \| f_n - f\|_\infty \to 0, \qquad \| A_n f_n - A f \|_\infty \to 0 \end{equation}
as $n \to \infty$.
Proof of your statement : Let $Y=(Y_t)_{t\geq 0}$ be a limit point of the sequence $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ which exists by Lemma 1. Then Lemma 2 (combined with Theorem 5.4.1 in EK86) tells us that $Y$ solves the martingale problem related to the generator $A$. That means that $Y$ is a Brownian motion. Since this holds for any limit point $Y$, we know that $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$ converges weakly in Skorohod space to Brownian motion. This implies converges in finite dimensional distributions by Theorem 3.7.5 in EK86.
Proof of Lemma 2:
Fix $f \in C_c^2(\mathbb{R})$ and let $a := \inf \text{supp } f$, $A:= \sup \text{supp } f$ such that $f(x) = 0$ for all $x \not \in [a,A]$.
\begin{equation}
 f_n (x) := f(x) - 2\int_a^x \exp (2n^{-1} \cos (ny) ) \left(-C + \int_a^y \exp(-2n^{-1}\cos(nz)) \sin(nz) f'(z) d z \right) .
\end{equation}
Here $C = C(n) = \int_a^A \exp(-2n^{-1}\cos(nz)) \sin(nz)f'(z) d z$.
It is easy to see that $A_n f_n(x) - Af(x) = 0$ for any $x \in \mathbb{R}$. So we are left with the verification of $\sup_{x} | f_n(x) -f(x)| \to 0$ as $n\to \infty$. Note that since $1-(2/n) \leq \exp(-n^{-1} \xi) \leq 1+(4/n)$ for any $\xi \in [-2,2]$, $n \geq 4$ we have
\begin{align}
 \int_a^y & \exp(-2n^{-1}\cos(nz)) 2\sin(nz) f'(z) d z \\
& = - \int_a^y \exp(-2n^{-1}\cos(nz)) f''(z) dz + f'(y) \exp(-2n^{-1}\cos(ny)) - f'(a) \exp(-2n^{-1}) \\
& = -\int_a^y f''(z) dz + f'(y) - f'(a) + \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\
& = \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right)
\end{align}
for any $y \in \mathbb{R}$. This allows to say that for $x \in [a,A]$:
\begin{align}
 f(x) - f_n(x) & = 2\int_a^x \exp(-n^{-1}\cos (ny)) dy \ \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\
 & \leq 4 |A-a| \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \to 0 
\end{align}
as $n \to \infty$. If we are not in the interval of support, i.e. $x \not \in [a,A]$, then we have for $x \geq A$:
\begin{align}
 f(x) -f_n(x) &= -f_n(x) = 2\int_a^x \exp (2n^{-1} \cos (ny) ) \left(-C + \int_a^y \exp(-2n^{-1}\cos(nz)) \sin(nz) f'(z) d z \right) \\
  & = |A-a| \cdot \mathcal{O}\left(n^{-1} \|f''\|_\infty + n^{-1} \|f'\|_\infty \right) \\
& \qquad + 2\int_A^x \exp (2n^{-1} \cos (ny) ) \left( -C + \int_a^A \cdots dz + \int_A^y \cdots dz \right) \, .
\end{align}
The first term vanishes as before uniformly in $x$, which leaves us with the terms of the last line. By definition of $C$ we are only left with the last term which is zero anyway, since we are not in the support of $f$ anymore. A similar, even easier reasoning applies for $x <a$. The previous lines include saying that $f_n$ is a bounded function.
Proof of Lemma 1:
Use Remark 5.4.2 in EK86 which requires standard conditions on $A$ which are fulfilled here and the compact containment condition for $\{X^{(n)}_t)_{t\geq 0} : n \in \mathbb{N}\}$. For the compact containment condition hold define two processes $Z^{i}$, $i = -1, 1$ solving $dZ_t^i = i\ dt + dB_t$. Then Theorem 5.2.18 in Karatzas and Shreve, Brownian Motion and Stochastic Calculus, 1991 allows to say we can construct probability spaces $(\Omega^n,\mathcal{A}^n,P^n)$ such that $Z_t^{-1} \leq X_t^n \leq Z_t^1$ for all $t \geq 0$ $P^n$ almost surely for all $n \in \mathbb{N}$.
Theorem 5.4.1 in EK86
Let $A \subset C_b(\mathbb{R}) \times C_b(\mathbb{R})$ and $A_n \subset \mathbb{B}_b(\mathbb{R}) \times \mathbb{B}_b(\mathbb{R})$, the bounded measurable functions, $n \in \mathbb{N}$. Suppose that for each $(f,g) \in A$ there is an $(f_n,g_n) \in A_n$ such that
\begin{equation}
 \| f_n -f \| \to 0, \quad \| g_n -g\| \to 0.
\end{equation}
If for each $n\in \mathbb{N}$, $X^{(n)}$ is a solution of the $A_n$ martingale problem with paths in the Skorohod space and $X_n \Rightarrow X$, then $X$ is a solution to the $A$-martingale problem.
Note: The authors of EK86 prefer to denote a generator $A: D(A) \to \mathbb{B}_b(E)$ by its graph $\{(f,Af): f \in D(A) \} \subset (\mathbb{B}_b(E),\mathbb{B}_b(E))$.
