1
$\begingroup$

I'm trying to solve Problem 2.4.12 on page 64 of Karatzas-Shreve's book "Brownian motion and stochastic calculus": enter image description here

My attempt is to use triangle inequality (denoting $\Omega=C[0,\infty)$) $$|\int_\Omega f_ndP_n-\int_\Omega fdP|\leq |\int_\Omega (f_n-f)dP_n|+|\int_\Omega fdP_n-\int_\Omega fdP|\quad\star,$$ and estimate the first term with dominated convergence, the second using weak convergence of the measures.

For $\epsilon>0$, since $(P_n)_n$ is tight (by Prohorov thm) I can choose a compact $K\subset \Omega$ such that $P_n(K)\geq 1-\epsilon$. Fix $m\geq1$, then by dominated convergence for $n$ large enough I have $$ |\int_\Omega (f_n-f)dP_m|=|\int_K (f_n-f)dP_m|+|\int_{K^c} (f_n-f)dP_m| \leq \epsilon+\epsilon\sup_{n\geq1,\omega\in K^c}|f_n(\omega)-f(\omega)|.$$

  1. To conclude I need $f$ to be bounded. Does this follow from the fact that $f$ is the compact-limit of uniformly bounded functions? (Note that I need boundedness also to use weak convergence on the second term of $\star$)

  2. Is it correct that if $\forall m\geq1$ $|\int_\Omega (f_n-f)dP_m|\to_{n\to\infty}0$, then $ |\int_\Omega (f_n-f)dP_n|\to_{n\to\infty}0?$ To be fair I think it's true only if the first convergence is uniform in $m$.

Thanks in advance :)

$\endgroup$
  • $\begingroup$ You know that the sequence is uniformly bounded, so the same bound holds also for the pointwise limit $\endgroup$ – Bob Jul 28 '18 at 21:23
  • $\begingroup$ Re your 2nd question: What's wrong about doing the estimate for $m=n$? $\endgroup$ – saz Jul 29 '18 at 5:39
  • $\begingroup$ @saz: by fixing $m\geq1,\epsilon>0$ I found $N(\epsilon,m)\geq1$ such that $\forall n\geq N(\epsilon,m),$ $\int_\Omega(f_n-f)dP_m<\epsilon$. My problem is that $N(\epsilon,m)$ depends on $m$. How do I know that $\sup_m N(\epsilon,m)$ is finite? $\endgroup$ – Demetrio Masciurett Jul 29 '18 at 8:58
  • $\begingroup$ @Bob: thanks, since I have point-wise convergence and boundedness I even don't need to separate the integral in $K$ and $K^c$: I can apply dom convergence directly on $\Omega$, right? $\endgroup$ – Demetrio Masciurett Jul 29 '18 at 9:45
1
$\begingroup$

Finally got it:

$|\int_\Omega (f_n-f)dP_n|\leq \int_{K_\epsilon} |f_n-f|dP_n +\int_{K^c_\epsilon} |f_n-f|dP_n \leq \sup_{\omega\in K_\epsilon}|f_n(\omega)-f(\omega)|+\epsilon||f_n-f||_\infty.$

Since $\epsilon$ is arbitrary, letting $n\to \infty$ the last term goes to zero by uniform compact convergence of $(f_n)_{n\geq 1}$ to $f$.

$\endgroup$

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.