The Setup

Let $\xi_t$ be a process adapted to the filtration $\mathfrak{F_t}$ of the semi-martinagale $X_t$, such that both are square integrable. Then is the map \begin{align} F_T: L^2(\mathfrak{F_t},\mathbb{P}\times m) \rightarrow & L^2(\Omega,\mathbb{P}),\\ \xi_t \mapsto & \int_0^T \xi_tdX_t \end{align} where $m$ is the Lebesgue measure and $(\Omega,\mathfrak{F},\mathfrak{F}_t,\mathbb{P})$ is a stochastic base for the process $X_t$.

The Question

For any given $T>0$ is the map $F_T$ a continuous operator, as it is clearly $\mathbb{R}$-linear? If not is it at least closed or bounded?

Note: Ultimately, my goal is to obtain some sort of boundedness result.

  • $\begingroup$ If I unterstand your notations correctly, if $X_t = B_t$, then the map is even an isometry, namely the Ito Isometry. If $X_t$ is just a semi-martingale, this will maybe not be an isometry, but it should still be bounded. $\endgroup$ – Matthias Ludewig Oct 17 '16 at 15:37
  • $\begingroup$ Yes, I notices this also but how to formally prove this? $\endgroup$ – AIM_BLB Oct 17 '16 at 15:38

You want an inequality like $E(\int \xi_t dX_i)^2 < cE\int \xi^2_t dt$ (this is part question as I am not sure what the norm on the rhs is ), however, if $X_t = \int \sigma(t) dW_t$ where $\sigma $ is deterministic and W Brownian motion you get $E(\int \xi_t dX_i)^2 = E\int \xi^2_t \sigma^2(t) dt$ . As $X$ is fine as long as $\int \sigma^2(t) dt < \infty$ but $\sigma $ does not have to be bounded the map will not be bounded.


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