2
$\begingroup$

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 $X_2$

  • 0
    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.2012-06-16
  • 0
    @ 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.2012-06-16
  • 0
    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.2012-06-16

1 Answers 1