5
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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(\Omega\times[0,T],\mathfrak{F_t},\mathbb{P}\times m) \rightarrow & L^2(\Omega,\mathfrak{F},\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 norms are defined as $$ \|\xi_t\|^2_2 \triangleq \mathbb{E}[\int_0^T\xi_t^2d\mu(t)]\, and \, \|X\|^2_2\triangleq \mathbb{E}[X^2], $$ where $\xi_t$ is in the domain of $F_T$ and $X$ in it's codomain.

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.

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  • $\begingroup$ Would you mind explicitly defining the two $L^2$ norms you have in mind here? $\endgroup$ – Cameron Williams Oct 17 '16 at 15:15
  • $\begingroup$ Hi Cameron, I have put in those details, thanks. $\endgroup$ – AIM_BLB Oct 17 '16 at 15:19
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    $\begingroup$ Unless $\mu$ (or $m,$ you should choose one) is $d<X>_t$ I don't see why the integral should be defined. But if this is the case, then the operator is bounded because you have an isometry (it is actually the Ito isometry). $\endgroup$ – Kore-N Oct 21 '16 at 6:00
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    $\begingroup$ mathoverflow.net/questions/252359/… $\endgroup$ – Jonas Dahlbæk May 25 '17 at 11:32

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