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!



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

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.

  • $\begingroup$ 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

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