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I want to show, that the following spaces are Banach spaces: $X_1:=\{M=(M_t)_{0\le t \le T} ;\mbox{ M is an adapted RCLL process }\}$ with the norm $\|M\|_{X_1}:=\|\sup_{0\le t\le T}|M_t|\|_{L^2(P)}$ where $P$ is a probability measure on a probability space. And $ X_2:=\{M=(M_t)_{0\le t \le T}; \mbox{M is a optional process}\}$ with the norm $\|M\|_{X_2} :=(E(\int_0^T|M_s|^2ds))^{\frac{1}{2}}$

For $X_1$ any Cauchy sequence would be a Cauchy sequence uniformly in $t$ in $L^2$. Hence there is a limit. But how do I show that this limit is again in $X_1$? Also I have no idea how to prove this for $X_2$

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To show that the limit $M$ of the Cauchy sequence $M^{(n)}$ is indeed in $X$, you can extract an almost everywhere converging subsequence $M^{(n_k)}$ (i.e. such that $M_t^{(n_k)}\to M_t$ almost everywhere for all $t\in \Bbb Q\cap [0,T]$ (use diagonal extraction). This will help you to show that $M$ is right continuous and admits a left limit at each point. I guess adaptedness follows from here. –  Davide Giraudo Jun 16 '12 at 14:33
    
@ Davide: I would appreciate if you could turn your comment into an answer. (Especially about the diagonalization procedure. About the RHS of the definition of the norm of space $X_2$, please note the expectation. I do not think that it is a r.v. –  user20869 Jun 16 '12 at 15:49
    
Sorry, I didn't notice it (so it's also a random variable, a constant one, but of course it's well-defined). I will remove this useless comment. –  Davide Giraudo Jun 16 '12 at 16:09
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1 Answer 1

Let $\{M^{(n)}\}$ a Cauchy sequence in $X_1$. Then $\sup_{0\leq t\leq T}|M_t^{(n)}-M_t^{(m)}|\to 0$ when $m,n\to +\infty$ in $L^2(P)$. In particular, for each $t$ fixed in $[0,T]$, the sequence $\{M_t^{(n)}\}$ is Cauchy in $L^2(P)$. Since this space is complete, we can define $M_t$ as the limit in $L^2(P)$ of this sequence. Now, have have several things to check:

  • We have that for all $\varepsilon >0$, we can find $n_0$ such that if $m,n\geq n_0$ and $t\in [0,T]$ then $\lVert M_t^{(n)}-M_t^{(m)}\rVert_{L^2(P)}\leq \varepsilon$ (since $|M_t^{(n)}-M_t^{(m)}|\leq \sup_{0\leq s\leq T}|M_s^{(n)}-M_s^{(m)}|$). Taking the limit $m\to +\infty$, we get that $\lVert M-M^{(n)}\rVert_{X_1}\to 0$. Indeed, you can check, using the fact that converge in $L^2$ implies convergence of a subsequence almost everywhere, that for each subsequence it's true for a further subsequence.
  • We check that $M\in X_1$. We look at right continuity. We can find an increasing sequence $\{n_k\}$ of integers such that $\sup_{0\leq t\leq T}|M^{(n_k)}_t-M_t|\to 0$ almost everywhere (say in $\Omega'$. Fix $t_0\in [0,T]$. For $s\geq t_0$ and $\omega\in\Omega'$, we have \begin{align} |M_s(\omega)-M_{t_0}(\omega)|&\leq |M_s(\omega)-M_s^{(n_k)}(\omega)|\\ &+|M_s^{(n_k)}(\omega)-M_{t_0}^{(n_k)}(\omega)|+|M_{t_0}^{(n_k)}(\omega)-M_{t_0}(\omega)|\\ &\leq 2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|+|M_s^{(n_k)}(\omega)-M_{t_0}^{(n_k)}(\omega)|. \end{align} Now, we fix $\varepsilon>0$, and pick $k$ such that $2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|\leq \varepsilon$. Then we conclude using the right continuity of $M^{(n_k)}$. We have now to see left limit. It follows a similar argument, plus the Cauchy criterion: fix $t_0\in [0,T]$. For $s_1,s_2\leq t$ we have $$|M_{s_1}(\omega)-M_{s_2}(\omega)|\leq 2\sup_{0\leq t\leq T}|M_t(\omega)-M_t^{(n_k)}(\omega)|+|M_{s_1}^{(n_k)}(\omega)-M_{s_2}^{(n_k)}(\omega)|.$$
  • We have to check that $\{M_t\}$ is adapted. To see that, note that the set $\{\omega,\{M_t^{(n_k)}(\omega)\}\mbox{ doesn't converge}\}$ can we written as a countable intersection of countable union of elements of $\mathcal F_t$.
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@ davide: Thanks for your comment. I have some question: First bullet: if $\| M_t^{n}-M_t^{m}\|_{L^2}\le \epsilon$ why is it true that $\| M_t^{n}-M_t\|_{X_1}\to 0$ in the $X_1$ norm? Second bullet: Here we use that $L^p$ convergence implies convergence a.e. of a subsequence (if finite measure) to obtain such a subsequence, right? Third bullet: Sorry, here I do not understand at all what you mean. I upvote your answer, however, to get the bounty it would be nice, if you could add the completeness about the second space. Again thanks. –  user20869 Jun 18 '12 at 7:11
    
I've added detailed, and can add more I you want/need. In the third bullet, I show that $M_t$ is $\mathbb F_t$ measurable. It's almost everywhere limit of such functions, and we have to show that the set on which we don't have convergence is $\mathbb F_t$ measurable for all $t$. –  Davide Giraudo Jun 18 '12 at 14:56
    
Thank you for your patience. But still, I have some troubles to understand why in the first bullet the convergence is valid in the $X_1$ norm. Is my conclusion in the comment above about the second bullet correct? Could you please write down in the third bullet the decomposition into a countable intersection of countable unions of elements of $\mathcal{F}_t$? Thank you. Do you have an idea how to show the completeness for $X_2$? –  user20869 Jun 19 '12 at 7:10
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