1
$\begingroup$

Let $X_n$ be a sequence of Ito diffusions $$dX_n(t)=b_n(t) \, dt+\sigma_n(t) \, dW(t), \qquad 0\leq t\leq T$$ with $b_n$ uniformly bounded and $\sigma_n$ uniformly elliptic. Then Krylov's estimation gives the following: for any bounded measurable function $f$ with compact support $\mathcal K$, $$E\int_0^T|f(t,X_n(t)|dt\leq \left(\int_0^T\int_\mathcal K|f(t,x|^2 \, dt \, dx \right)^\frac{1}{2}.$$

Problem: If there exists $X$ such that $$\lim_{n\rightarrow\infty}E \left[\sup_{0\leq t\leq T}|X_n(t)-X(t)|^2 \right]=0,$$ then $$E\int_0^T|f(t,X(t)|dt\leq \left(\int_0^T\int_\mathcal K|f(t,x|^2 \, dt \, dx \right)^\frac{1}{2}.$$

The above is what I read in a book. I tried to prove it with Lusin Theorem. For any $\varepsilon >0$, it follows from Lusin Theorem that there exists a closed set $A$ with $Leb(A)<\varepsilon$ and $f$ is continuous on $A$. \begin{align*} &E\int_0^T|f(t,X(t)| \, dt \\&=E\int_0^T|f(t,X(t)| I\{X(t)\in A\} \, dt+E\int_0^T|f(t,X(t)|I\{X(t)\in A^c\} \, dt. \end{align*} It is easy to get the upper bound for the first part. But I don't think the second part is negligible.

Another method may be to get the representation of $X$ and then apply the Krylov estimation. However, I am not sure how to get the form of $X$.

Any thoughts on how to prove this? Thanks!

$\endgroup$
1
$\begingroup$

Since

$$\lim_{n \to \infty} \mathbb{E} \left( \sup_{0 \leq t \leq T} |X_n(t)-X(t)|^2 \right) = 0$$

there exists a subsequence $(X_{n(k)})_{k \geq 1}$ such that

$$\sup_{0 \leq t \leq T} |X_{n(k)}(t)-X(t)| \xrightarrow[]{k \to \infty} 0 \tag{1}$$

almost surely. To prove the inequality $$\mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) \leq \left( \int_0^T \int |f(t,x)|^2 \, dx \, dt \right)^{1/2} \tag{2}$$ we first prove this inequality for a nice class of functions and then use a density argument:

If $f: [0,T] \times \mathbb{R}^d \to \mathbb{R}$ is a function which is continuous with respect to the 2nd variable and which satisfies $f(t,x)=0$ for all $t \in [0,T]$, $|x| \geq R$ for some $R>0$, then it follows from the dominated convergence theorem and the Krylov estimate for $X_n$ that

\begin{align*} \mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) &= \lim_{k \to \infty} \mathbb{E} \left( \int_0^T |f(t,X_{n(k)}(t)| \, dt \right) \\ &\leq \left( \int_0^T \int |f(t,x)|^2 \, dx \, dt \right)^{1/2}, \end{align*}

i.e. $(2)$ holds for any such $f$. Now if $f$ is a function of the form $$f(t,x) = 1_A(t) 1_C(x) \tag{3} $$ for some Borel set $A$ and a closed set $C$ with $C \subseteq B(0,R)$ for some $R>0$, then we can find a sequence $(g_n)_{n \in \mathbb{N}}$ of continuous functions supported in $B(0,R+1)$ such that $$1_C(x) = \inf_{n \in \mathbb{N}} g_n(x), \qquad x \in \mathbb{R}^d, \tag{4}$$ (see Urysohn's lemma for details). If we define $$f_n(t,x) := 1_A(t) g_n(x)$$ then it follows from the first part of this proof and $(4)$ that

\begin{align*} \mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) &\stackrel{(4)}{\leq} \mathbb{E} \left( \int_0^T |f_n(t,X(t))| \, dt \right) \\ &\leq \sqrt{\int_0^T \int |f_n(t,x)|^2 \, dx \, dt} \end{align*}

for all $n \in \mathbb{N}$. Taking the infimum over all $n \geq 1$ yields, by (4) and the monotone convergence theorem, that

$$ \mathbb{E} \left( \int_0^T |f(t,X(t))| \, dt \right) \leq \sqrt{\int_0^T \int |f(t,x)|^2 \, dx \, dt},$$

i.e. $(2)$ holds for functions of the form $(3)$. Now an application of the (functional) monotone class theorem (see e.g. Theorem 1 here) gives that $(2)$ holds for any bounded measurable function $f: [0,T] \times B(0,R) \to \mathbb{R}$. As $R>0$ is arbitrary, this proves the assertion.

$\endgroup$
  • $\begingroup$ In order to use Theorem 1, we still need to verify that if $f$ is bounded and there is a sequence of nonnegative functions $f_n$ increasing pointwise to $f$ with $f_n$ satisfying (2), then $f$ satisfies (2). To prove this, note that $f_n$ increasing pointwise to $f$, then $f_n(t,X)$ converges to $f(t,X)$ almost surely. Therefore, $$\mathbb E\left(\int_0^T|f(t,X(t))|dt\right)\leq \mathbb E\left(\int_0^T|f(t,X(t))-f_n(t,X_n(t))|dt\right)+\sqrt{\int_0^T\int|f_n(t,x)|^2dtdx}.$$ Letting $n\rightarrow\infty$, we get that $f$ satisfies (2). Is my computation right? $\endgroup$ – SHAN Apr 7 at 12:50
  • $\begingroup$ @SHAN Well, yes; alternatively you can just apply the monotone convergence theorem, i.e. use $$\mathbb{E} \int_0^T |f(t,X(t))| \, dt = \sup_n \mathbb{E} \int_0^t |f_n(t,X(t))| \, dt$$ $\endgroup$ – saz Apr 7 at 13:13
  • $\begingroup$ Yes. Thank you for your help. $\endgroup$ – SHAN Apr 8 at 2:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.