The following preliminary is pedantic but classic and the questions are "simples" for practitioners
Preliminaries and notations:
Let $T > 0$ a finite time and a probability space $(\Omega, \mathcal F, \mathbb F = (\mathcal F_t)_{t\in[0,T]},\mathbb P)$.
For $p\in[1,+\infty)$, let $\mathcal H^p(\mathbb P,\mathbb F)$ denotes the set of RCLL martingales M s.t : $$ \parallel M\parallel_{\mathcal H^p(\mathbb P,\mathbb F)}:=\left(\mathbb E(\sup_{s\in[0,T]}\mid M_s\mid^p)\right)^{1/p} < +\infty $$ For $p\in[1,+\infty)$, let $\mathcal L^p(M,\mathbb P,\mathbb F)$, where I assume $M$ is a local martingale, is the space of d-dimensional $\mathbb F$-predictable processes $v$ s.t.: $$ \parallel v\parallel_{\mathcal L^p(M,\mathbb P,\mathbb F)}:=\mathbb E_{\mathbb P}\left[\left(\int^T_0v_u^*d[M,M]_uv_u\right)^{p/2}\right] < +\infty $$
My questions :
- Let the mapping $v\rightarrow \int v^*dS$ from $(L^2(S,\mathbb P, \mathcal F),\parallel\cdot\parallel_{L^2(S,\mathbb P, \mathcal F)})$ to $(\mathcal H^2(\mathbb P, \mathcal F),\parallel\cdot\parallel_{\mathcal H^2(\mathbb P, \mathcal F)})$. Is it an isometry ?
- Then how to show that the following space : $$ I(L^2)=\left\{\int^T_0v^*_sdS_s: v\in L^2(S,\mathbb P, \mathcal F)\right\} $$ is closed in $L^2(\mathbb P)$?
My answers (probably wrong ?):
I have shown that $\parallel v\parallel^{1/2}_{L^2(S,\mathbb P, \mathcal F)}=\parallel\int^T_0 v^*dS\parallel_{\mathcal H^2(\mathbb P, \mathcal F)}$. I cannot kill the square root on $L^2$ side. So it is not an isometry. If someone could check if I am wrong or not ?
If we assume that my calculus (1) are good, to show that $I(L^2)$ is closed, $L^2(S,\mathbb P,\mathbb F)$ need to be closed ? but I miss on my topologic argument : Does $L^2(S,\mathbb P,\mathbb F)$ closed ? and why ?