Integration by parts formula for Wiener integral

Question

Hi I need an help understanding "integration by parts" in Wiener integral. I've defined this integral as in the following: let $T=[0,t]\subset \mathbb R$ we want to define $\int_T f(s) dB_s$ where $(B_s)_s$ is a Brownin motion on $(\Omega, \mathcal F, \mathbb P)$.

1. Let $g: T\to \mathbb R$ be a step function, so of the form $g=\sum_{i=1}^n c_i \chi_{[t_{i-1},t_{i}]}$. Then we define $W(g):=\int_T g(s) dB_s:=\sum_{i=1}^n c_i ({B_{t_{i}}-B_{t_{i-1}}})$. It is easy to see that $\mathbb E(W(g))=0$ and that $W(g)\in L^2(\Omega).$
2. Let $f: T\to \mathbb R$ be in $L^2(T)$. Since step functions are dense in $L^2(T)$, we define $W(f)$ by density. That means that we take $f_n\to f\, \text{in } L^2(T)$ with $(f_n)_n$ step functions, consider $(W(f_n))_n\subset L^2(\Omega)$ and define $\int_T f(s) dB_s=W(f):=\lim_{n\to \infty}W(f_n)\, \text{in } L^2(\Omega)$.

Now I have this exercise: Let $f\in C^1(T).$ Prove that it holds a sort of integration by parts formula $$\int_T f(s) dB_s=f(t)B(t)-\int_T f'(s) B_s ds,$$ almost surely in $\Omega.$

I think that the idea is to find a suitable approximation for $f$ with step functions in order to have that equality. So I can divide the interval $T$ with a partition $\{t_0=0<t_1<\dots <t_n=T\}$ and take as step function $f^{(n)}:=\sum_{i=1}^n f(t_{i-1}) \chi_{[t_{i-1},t_{i}]}$. As $n$ becomes larger, also the partition of the interval $T$ becomes more dense. Now I can write

$W(f^{(n)})= \sum_{i=1}^n f(t_{i-1}) ({B_{t_{i}}-B_{t_{i-1}}})=\sum_{i=1}^n (f(t_{i}) B_{t_{i}} - f(t_{i-1}) B_{t_{i-1}})-\sum_{i=1}^n (f(t_{i})-f(t_{i-1})) B_{t_{i}}.$

But this is equal to $f(t)B_t-\sum_{i=1}^n f'(\alpha_i)(t_{i}-t_{i-1}) B_{t_{i}},$ for some numbers $\alpha_i\in [t_{i-1},t_i]$.

So we have $$W(f^{(n)})=f(t)B_t-\sum_{i=1}^n f'(\alpha_i)(t_{i}-t_{i-1}) B_{t_{i}}.$$

Now, my question is: is it correct to say that, for almost every $\omega\in \Omega$, we have $W(f^{(n)})(\omega)\rightarrow f(t)B_t(\omega)- \Big(\int_T f'(s)B_s ds \Big)(\omega)$ as $n\to \infty$ ?

If this were true we would have finished. But I have another doubt: I haven't proved that $f^{(n)}\to f$ in $L^2(T)$, but I don't know how to show it exactly, but it seems true to me.

Can someone help me? Any suggestion would be really appreciated.

Answer 1

First of all: Your proof is correct.

Concerning your question how to prove $f^{(n)} \to f$ in $L^2(T)$: We have

$$\begin{align*} \|f^{(n)}-f\|_{L^2(T)}^2 &= \sum_{i=1}^n \int_{t_{i-1}^n}^{t_i^n} (f(t_{i-1}^n)-f(t))^2 \, dt \\ &\leq \sup_{s,t \leq T, |s-t| \leq |\Pi_n|} |f(s)-f(t)|^2 \int_0^T \, dt \tag{1} \end{align*}$$

where

$$|\Pi_n| := \max_{i} |t_{i+1}^n-t_i^n|$$

denotes the mesh size of the partition $\Pi_n=\{0=t_0^n< \ldots < t_n^n = T\}$. Since $f$ is uniformly continuous on $[0,T]$, it follows from $(1)$ that $\|f^{(n)}-f\|_{L^2(T)}$ converges to $0$ as $n \to \infty$.