エクセンダールSDEもんだい3.14 - やがてめぐになる

A solution of Exercise 3.14 in the Øksendal SDE book is given in this article. It seems this exercise to be complex and require careful observation.

Let $(\mathcal{F}_t)$ be the filtration generated by the Brownian motion $(B_t)$ .

We see obviously that $h$ is $\mathcal{F}_t$ -measurable if it is a random variable as in the assertion. Then we show the converse.

Suppose $h$ is $\mathcal{F}_t$ -measurable. We now follow the hint: a)-c).

a) Define $h_n=h\boldsymbol{1}_{\{|h|\leq n\}}$ for $n=1,2,\dots$ . Then we see that $\lim_{n\to\infty}h_n=h$ a.s. If, for every $n$ , we have a sequence $(g^n_k)_{k=1}^\infty$ of bounded random variables which converges to $h_n$ in $L^1(P)$ , taking an appropriate $m_n$ yields that $E[ |h_n-g^n_{m_n}| ]\leq1/n$ .
Then let $g_n:=g^n_{m_n}$ . We obtain
\begin{align*}
P(|h-g_n|>\varepsilon)
&\leq P(|h-h_n|>\varepsilon/2) + \frac{2E[|h_n-g_n|]}{\varepsilon} \\
&\leq P(|h-h_n|>\varepsilon/2) + \frac{2}{n\varepsilon}
\end{align*}
for $\varepsilon>0$ . Since $h_n$ tends to $h$ as $n\to\infty$ in probability, so does $g_n$ in probability. Thus $g_n$ must have an a.s. convergent subsequence with limit $h$ .
Then it suffices to construct a sequence of as in the assertion which converges to any bounded $h$ in $L^1(P)$ .

b) We now show that
\begin{align*}
\mathcal{F}_t = \bigvee_{n=1}^\infty \mathcal{H}_n =: \mathcal{H} .
\end{align*}
For $s\leq t$ , take a sequence $(t_j)_{j\geq 1}$ from $\{t^{(n)}_j\leq t;n,j=0,1,2,\dots\}$ converging to $s$ . Since $B_{t_j}\to B_s$ and every $B_{t_j}$ is $\mathcal{H}$ -measurable, $B_s$ is also $\mathcal{H}$ -measurable. Thus, we see that $\mathcal{F}_t \subset \mathcal{H}$ . And then we get the inverse inclusion by definitions of $\mathcal{(H_n)}$ and $\mathcal{H}$ .
By Corollary C.9, it follows that $h=E[h|\mathcal{F}_t] = \lim_{n\to\infty}E[h|\mathcal{H}_n]$ a.s. and in $L^1(P)$ .

c) By the Doob-Dynkin lemma (Lemma 2.1.2), we have
\begin{align*}
h_n = G_n(B_{t_1},\dots,B_{t_k}),
\end{align*}
where $h_n=E[h|\mathcal{H}_n]$ for $n=1,2,\dots$ . Since $h_n$ are bounded, we see that $G_n\colon \mathbb{R}^k\to\mathbb{R}$ are also bounded.

Let $A$ be the set of all continuous functions $g\colon \mathbb{R}^k \to \mathbb{R}$ , which are univariate functions and whose support are compact when regarding as ones $\mathbb{R}\to\mathbb{R}$ .
We now consider the following:
Claim Any bounded function $G\colon \mathbb{R}^k\to\mathbb{R}$ admits a sequence $(F_j)_{j\geq 1}$ in the subabgebra generated by $A$ such that
\begin{align*}
& \sup_{j\geq 1,~ x\in\mathbb{R}^k}|F_j(x)|<\infty, \quad \lim_{j\to\infty} F_j \to G, \quad \text{a.e.}
\end{align*}

We prepare Lemma 1 and 2 before showing Claim.
Lemma 1 There exists a sequence $(F_j)_{j=1,2,\dots}$ in $C_0(\mathbb{R}^k)$ *1 such that $\sup_{x\in\mathbb{R}^k}|F_j(x)|\leq K$ for some $K>0$ independent of $j$ , and $\lim_{j\to\infty}F_j=G$ a.e.

Proof. Let $G_N:=G\boldsymbol{1}_{\{|x|\leq N\}}$ for $N=1,2,\dots$ . Then every $G_N$ is in $L^p(\mathbb{R^k})$ for $p\geq 1$ under the Lebesgue measure and tends to $G$ a.e. as $N\to\infty$ .
For $N=1,2,\dots$ , by density of $C_0(\mathbb{R}^k)$ in $L^1(\mathbb{R}^k)$ , one can take an $L^1$ -convergent sequence $(F^{(N)}_j)_{j\geq 1}$ of $C_0(\mathbb{R}^k)$ with limit $G_N$ . Moreover, we may assume that
\begin{align*}
\sup_{x\in\mathbb{R}^k}|F^{(N)}_j(x)|\leq K:=\sup_{x\in\mathbb{R}^k}|G(x)|+1 \quad \text{for $j,N\in\mathbb{N}$}.
\end{align*}
Indeed, defining $\tilde{F}^{(N)}_j\in C_0(\mathbb{R^k})$ as
\begin{align*}
\tilde{F}^{(N)}_j(x) &=
\begin{cases}
K & ,\text{if $F^{(N)}_j(x)>K$}, \\
-K & ,\text{if $F^{(N)}_j(x)<-K$}, \\
F^{(N)}_j(x) & ,\text{otherwise},
\end{cases}
\end{align*}
leads that $\sup_{x\in\mathbb{R}^k}|\tilde{F}^{(N)}_j(x)|\leq K$ and
\begin{align*}
\int_{\mathbb{R}^k} |\tilde{F}^{(N)}_j(x)-G_N(x)|dx
&\leq \int_{\mathbb{R}^k} |F^{(N)}_j(x)-G_N(x)|dx,
\end{align*}
and then it follows that $\lim_{N\to\infty}\tilde{F}^{(N)}_j=G_N$ in $L^1(\mathbb{R}^k)$ . Then the same argument as in a) yields the assertion. □

Lemma 2 For $j\geq 1$ , let $R_j=\prod_{l=1}^k [a^{(j)}_l,b^{(j)}_l]$ be a $k$ dimensional compact rectangle containing the support of $F_j$ and let $A_j$ be the set of all continuous functions $g\colon R_j \ni x\mapsto g(x_l)\in\mathbb{R}$ , $l=1,2,\dots,k$ . Then each $F_j$ admits a sequence $(\tilde{F}^{(j)}_m)_{m\geq 1}$ of the subalgebra generated by $A_j$ such that $\lim_{m\to\infty}\sup_{x\in R_j}|\tilde{F}^{(j)}_m(x)-F_j(x)|=0$ . Furthermore, we can extend the domain of each $\tilde{F}^{(j)}_m$ to whole $\mathbb{R}^k$ satisfying $\lim_{m\to\infty}\sup_{x\in \mathbb{R}^k}|\tilde{F}^{(j)}_m(x)-F_j(x)|=0$ .

Proof. We refer to Miyajima [1], Theorem 7.65, for the Stone-Wierstrass theorem:

Theorem Let $X$ be a compact Hausdorff space and $A$ be a subset of $C(X)$ . Then, the subabgebra generated by $A$ is dense in $C(X)$ if and only if the following (1) and (2) hold:
(1) For any $x\in X$ , there exists an $f\in A$ such that $f(x)\neq 0$ .
(2) For any different $x,y\in X$ , there exists an $f\in A$ such that $f(x)\neq f(y)$ .

The theorem is stated in other books or media as well, e.g. [2-4].

It is clear that each $A_j$ satisfies (1) and (2) in the theorem. Thus, we see that the former holds.
Next, for the latter. We extend each of the domain of $\tilde{F}^{(j)}_m$ by putting the same values on $R^m_j\setminus R_j$ , where $R^m_j:=\prod_{l=1}^k [-1/m + a^{(j)}_l,b^{(j)}_l + 1/m ]$ , as ones on the boundary of $R_j$ along each axis (Fig. 1).

f:id:megu0210:20190310183942j:plain:w500

Fig. 1 Extention of the domain of $\tilde{F}^{(j)}_m$ ( $k=2$ )

f:id:megu0210:20190310191157j:plain:w500

Fig. 2 Shape of $f^{(j,m)}_l(x_l)$

And then, rewrite $\tilde{F}^{(j)}_m$ as the product of the extended $\tilde{F}^{(j)}_m$ above and $f^{(j,m)}_1f^{(j,m)}_2\cdots f^{(j,m)}_k$ , where each $f^{(j,m)}_l$ is defined as in Fig. 2. Then we see that each of the extended $\tilde{F}^{(j)}_m$ is in the subalgebra generated by $A$ , which is defined before Claim, and $\lim_{m\to\infty}\sup_{x\in\mathbb{R}^k}|\tilde{F}^{(j)}_m(x)-F_j(x)|=0$ . □

Proof of Claim. By Lemma 1 and 2, we may assume that $\sup_{j,m\geq 1,~x\in\mathbb{R}^k}|\tilde{F}^{(j)}_m(x)|\leq K+1$ and that $\sup_{x\in\mathbb{R}^k}|\tilde{F}^{(j)}_j(x)-F_j(x)|\leq 1/j$ , by renumbering if necessary. Then we have
\begin{align*}
|\tilde{F}^{(j)}_j(x)-G(x)|
&\leq |F_j(x)-G(x)| + \frac{1}{j} \\
& \to 0, \quad \text{as $j\to\infty$}
\end{align*}
almost everywhere. □

We are now back to showing the exercise. By Claim, there exists, for $n=1,2,\dots$ , a sequence $(F^{(n)}_j)_{j\geq 1}$ in the subabgebra generated by $A$ converging to $G_n$ a.e. By the dominated convergence theorem, we see that $E[|G_n(B_{t_1},\dots,B_{t_k}) - F^{(n)}_j(B_{t_1},\dots,B_{t_k})|]\to 0$ as $j\to\infty$ . Then as repeated above, we assume that
\begin{align*}
E[|G_n(B_{t_1},\dots,B_{t_k}) - F^{(n)}_n(B_{t_1},\dots,B_{t_k})|]\leq \frac{1}{n}.
\end{align*}
Then, letting $n\to\infty$ yields
\begin{align*}
E[|h-F^{(n)}_n(B_{t_1},\dots,B_{t_k})|]
&\leq E[|h-h_n|] + \frac{1}{n} \\
&\quad \to 0.
\end{align*}
As mentioned in a), we constructed a desired convergent sequence. We complete the proof.

References

[1] Miyajima, S., Functional Analysis, Yokohama Publishers, 2007. In Japanese.
[2] Rudin, W., Functional Analysis, McGraw-Hill, 1991.
[3] Yosida, K., Functional Analysis, Springer, 1980.
[4] Wikipedia, Stone–Weierstrass theorem

Hmm, too heavy exercise...x(

*1:It denotes the space of all continuous functions with compact support.