Let $\{W_t\}_{t\in[0;T]}$ be a one-dimensional Brownian motion and let $\{\mathcal F_t\}_{t\in[0;T]}$ be the augmented filtration generated by this Brownian motion. Let $\{\sigma_t\}_{t\in[0;T]}$ be real-valued, progressively measurable and bounded. Is there a sequence of continuous and adapted processes $\{\tilde\sigma^n_t\}_{t\in[0;T]}$ such that $$ \lim_{n\to \infty} \mathbb E\Bigg( \int^T_0 \bigg( \int^t_0 \tilde\sigma^n_s \mathrm ds - \int^t_0 \sigma_s \mathrm dW_s \bigg)^2 \mathrm dt \Bigg) = 0? $$

## 1 Answer

Yes. Let $X_t := \int^t_0 \sigma_s \mathrm dW_s$. Due to Theorem V.6 from the book *Stochastic Integration and Differential Equations* (second edition) by P.E. Protter, there is a continuous and adapted process $\{\tilde X^n_t\}_{t\in[0;T]}$ such that
$$
\tilde X^n_t
=
\int^t_0 n \cdot \big( \tilde X^n_s - X_s \big) \mathrm ds.
$$
Hence we define $\tilde\sigma^n_s := n \cdot ( \tilde X^n_s - X_s )$, which is adapted and even continuous.

To prove the limit property, we first prove the following:

**Lemma 1.** For all $\beta,\delta \in (0;\infty)$, there exists $\nu \in (0;\infty)$ such that
$$
\mathbb P \bigg[ \sup_{0 \le s \le t \le \min(s+\nu,T)} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le \beta \bigg]
\ge 1 - \delta.
$$
**Proof:** We discretize the interval $[0;T]$ and consider events that its increments stay bounded. For all integer $0 \le k < N$ and all $\alpha \in (0;\infty)$, we define
$$
A^\alpha_{k,N}
:= \bigg\{ \max_{\frac{T}{N} k \le t \le \frac{T}{N} (k+1)} \bigg| \int_{\frac{T}{N} k}^t \sigma_u \mathrm dW_u \bigg| \ge \alpha \bigg\}.
$$
Due to the Burkholder–Davis–Gundy inequality (since $\sigma$ is bounded, say $\vert\sigma\vert \le \overline\sigma$),
$$
\mathbb E\bigg( \max_{T/N\cdot k \le t \le T/N\cdot (k+1)} \bigg| \int_{T/N\cdot k}^t \sigma_u \mathrm dW_u \bigg|^4 \bigg)
\le C_4 \mathbb E\bigg( \bigg\langle \int_{T/N\cdot k}^\cdot \sigma_u \mathrm dW_u \bigg\rangle_{T/N\cdot (k+1)}^2 \bigg)
\\=
C_4 \mathbb E\bigg( \bigg( \int_{T/N\cdot k}^{T/N\cdot (k+1)} \big(\sigma_u\big)^2 \mathrm du \bigg)^2 \bigg)
\le C_4 \bigg( \frac{T} N \cdot \overline\sigma^2 \bigg)^2
= N^{-2} C
$$
and due to the Markov inequality, we obtain
$$
\mathbb P \big( A^\alpha_{k,N} \big)
\le \alpha^{-4} \mathbb E\bigg( \max_{T/N\cdot k \le t \le T/N\cdot (k+1)} \bigg| \int_{T/N\cdot k}^t \sigma_u \mathrm dW_u \bigg|^4 \bigg)
\le \alpha^{-4} N^{-2} C
$$
Now we assume that $\omega \in \Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N}$ and assume $s, t \in [0;T]$ with $s \le t \le s + T/N$. Then we can find a $k \in \{0,\ldots,N-1\}$ such that $T/N\cdot k \le s \le T/N\cdot (k+1) \le t \le T/N\cdot (k+2)$ or $T/N\cdot k \le s \le t \le T/N\cdot (k+1)$. In the first case, we obtain
$$
\bigg| \bigg(\int^t_s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg|
\le \bigg| \bigg(\int_{T/N\cdot (k+1)}^t \sigma_u \mathrm dW_u\bigg)(\omega) \bigg|
+ \bigg| \bigg(\int_{T/N\cdot k}^{T/N\cdot (k+1)} \sigma_u \mathrm dW_u\bigg)(\omega) \bigg|
\\
\quad + \bigg| \bigg(\int_{T/N\cdot k}^s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg|
\le 3 \alpha.
$$
In the second case, we get the same result analogously.

Let $\omega \in \Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N}$ and $s, t \in [0;T]$ with $|s - t| \le \frac{T}{N}$. Then, $\bigg| \bigg(\int^t_s \sigma_u \mathrm dW_u\bigg)(\omega) \bigg| \leq 3 \alpha$ and so $$ \Omega \backslash \bigcup_{k=0}^{N-1} A^\alpha_{k,N} \subseteq \bigg\{ \max_{s,t\in [0;T], |s-t| \le \frac T N} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le 3 \alpha \bigg\}. $$ As a result, if $N$ is large enough, $$ \mathbb P \bigg[ \max_{s,t\in [0;T], |s-t| \le \frac T N} \bigg| \int^t_s \tilde\sigma_u \mathrm dW_u \bigg| \le 3 \alpha \bigg] \\\ge 1 - \sum_{k=0}^{N-1} \mathbb P \big( A^\alpha_{k,N} \big) \ge 1 - \frac{C}{\alpha^{4} N^{1}} \ge 1 - \delta, $$ which proves the statement.

Since $\tilde X^n$ always moves into the direction of $X$, we also have the following:

**Lemma 2.**
$$
\sup_{t \in [0;T]} \vert \tilde X^n_t \vert
\le
\sup_{t \in [0;T]} \vert X_t \vert
$$

Now since the increments of $X$ are bounded on an event of large probability due to Lemma 1, it is also straightforward to prove this:

**Lemma 3.** Let
$$
M^{\beta,\nu}:=\bigg\{\sup_{0 \le s \le t \le \min(s+\nu,T)} \bigg| \int^t_s \sigma_u \mathrm dW_u \bigg| \le \beta\bigg\}.
$$
Then for all $\omega\in M$, we have
$$
\sup_{t\in [0;T]} \big\vert \tilde X^{\beta/\nu}_t - X_t \big\vert
\le 3 \beta.
$$

Now we prove the main statement. Let $n:=\beta/\nu$. Due to the Minkovski inequality, $$ \sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) } \\\le \sqrt{ \mathbb E\bigg( \mathbb 1_{M^{\beta,\nu}} \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) } + \sqrt{ \mathbb E\bigg( \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \int^T_0 \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) } $$ The first summand can be bound directly by $3 \beta \sqrt T$ using Lemma 2. The second summand can be bound using Hölder inequality by $$ \mathbb E\bigg( \int^T_0 \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \big( \tilde X^n_s - X_s \big)^2 \mathrm dt \bigg) \\\le \sqrt{ \mathbb E\bigg( \int^T_0 \mathbb 1_{\Omega\backslash M^{\beta,\nu}} \mathrm dt \bigg) } \sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^4 \mathrm dt \bigg) } = \sqrt T \sqrt{1 - \mathbb P\big(M^{\beta,\nu}\big) } \sqrt{ \mathbb E\bigg( \int^T_0 \big( \tilde X^n_s - X_s \big)^4 \mathrm dt \bigg) } $$ The first factor can be made arbitrarily small if choosing $\nu$ small enough depending on $\beta$ due to Lemma 1, and the second factor is bounded due to the boundedness of $\sigma$ and Lemma 2.