• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a complete filtration of $\mathcal A$
  • $H$ be a separable $\mathbb R$-Hilbert space
  • $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H$
  • $X$ be an $L^2$-bounded almost surely continuous $H$-valued $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$ with $X_0=0$ almost surely and $$X^n:=\langle X,e_n\rangle_H\;\;\;\text{for }n\in\mathbb N$$

We can show that $([X^m,X^n]_te_m\otimes e_n)_{(m,\:n)\in\mathbb N^2}\subseteq\text{HS}(H)$ is summable for all $t\ge0$ almost surely and $$A_t:=\sum_{(m,\:n)\in\mathbb N^2}[X^m,X^n]_te_m\otimes e_n\;\;\;\text{for }t\ge0$$ is self-adjoint for all $t\ge0$ almost surely.

How can we show that $A_t$ is a nonnegative operator, i.e. $$\langle A_tx,x\rangle_H\ge0\;\;\;\text{for all }x\in H\;,\tag1$$ for all $t\ge0$ almost surely?

Let $x\in H$. Basic results about summability and the quadratic covariation yield $$\langle A_tx,x\rangle_H=\sum_{m\in\mathbb N}\sum_{n\in\mathbb N}\left[\langle X,\langle x,e_m\rangle_He_m\rangle_H,\underbrace{\langle X,\langle x,e_n\rangle_He_n\rangle_H}_{=:\:Y^n}\right]_t\tag2$$ for all $t\ge0$ almost surely. Now, $$\sum_{n\in\mathbb N}Y^n=\langle X,x\rangle_H\tag3\;,$$ but I wasn't able to conclude something useful from $(3)$ for $(2)$.

  • $\begingroup$ Will a mimic of Bochner's theorem work? Have you try? $\endgroup$ – Henry.L Jul 24 '17 at 12:29

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.