Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $q^{ij}$, $i,j\in\mathbb{N}$, be predictable real-valued stochastic processes. Let $(e^i)$, $i\in\mathbb{N}$ be an ONB of a separable Hilbert space $H$. Assume that $Q=\sum_{i,j=1}^\infty q^{ij}e_i\otimes e_j$ is a predictable, positive definite, self-adjoint, nuclear operator. If you do not 'like' predictability, forget about it, and replace the concept with jointly measurable. How would I go about proving that there exists a choice of both eigenvalues and eigenvectors which are predictable? A more general result is welcomed. Thank you in advance.

share|improve this question

2 Answers 2

Let's consider, say, the whole set of pairs (eigenvalue, projection on the eigenspace) as a point process in a Polish space (with respect to strong topology on projections), prove its measurability, then extract its points one-by-one in a measurable way (which can be done directly for any locally finite point process in a Polish space), obtain bases for the eigenspaces by simple orthogonalization, and convert them into a single sequence.

Perhaps the most difficult part here is to prove that the point process is measurable. To achieve this I would use a well-known theorem of Lusin and Souslin (see Kehris "Classical descriptive set theory" (15.1) and (15.2)):

An injective Borel mapping of standard Borel spaces has Borel image and is an isomorphism onto its image.

Note that we have a double characterization of nuclear operators - as elements of the Polish space $S^1(H)$ of nuclear operators, and as suitable integer-valued measures on $\mathbb{R} \times P(H)$, where $P(H)$ is the Polish space of projections on $H$. That the conditions on the point processes define a Borel subset can be verified directly (the measurable structure on the space of measures is generated by evaluations, and it is standard Borel). Another easy thing to verify directly is that the map "point process $\mapsto$ operator" is Borel (here it is convenient to generate the measurable structure on $S^1$ by evaluations $A \mapsto \left\langle Ax,y\right\rangle,x,y\in H$). Hence by the Lusin-Souslin theorem, the inverse mapping "operator $\mapsto$ point process" is Borel.

share|improve this answer

I stumbled upon a proof in a book by Zabczyk and Peszat, entitled "Stochastic Partial Differential Equations driven by Levy Noise." I will provide the proof for the convenience of others.

This can be proved using the Kuratwoski-Ryll-Nardzesk section theorem:

Assume that $K$ is a compact metric space. Let $(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{P}})$ be a probability space, and let $\psi: K\times \tilde{\Omega}\rightarrow \mathbb{R}$ be a function such that, for every $x\in K$, $\psi(x,\cdot)$ is measurable and, for every $\tilde{\omega}\in \tilde{\Omega}$, $\psi(\cdot,\tilde{\omega})$ is continuous. Then there exists a $K$-valued random variable $\xi: \tilde{\Omega}\rightarrow K$ such that $$ \psi(\xi(\tilde{\omega}),\tilde{\omega})=\sup_{k\in K} \psi(x,\tilde{\omega}), $$ for all $\tilde{\omega}\in \tilde{\Omega}$.

Let $Q_t(\omega)$ be a predictable process taking values in the space of positive nuclear operators. Also, let $Q$ be self-adjoint. Being nuclear, the random operator is also compact. Using the above theorem, given $T<\infty$, we can apply the above theorem to $\tilde{\Omega}=\Omega\times[0,T],$ $\tilde{\mathcal{F}}=\mathcal{P}_{[0,T]}$(the predictable sigma-algebra), and $\tilde{P}=T^{-1}\mathbb{P}\otimes dt$, to the set $K:=\{x\in U: |x|_H\le 1\}$ endowed with the weak topology (so compact by Alagou) and to the function $$ \psi(x,\omega,t)=\psi(x,\tilde{\omega})=(Q_t(\omega)x,x)_H.$$ Then there exists a $H$-valued process $g^1_t(\omega)$, $t\in [0,T]$, $\omega\in \Omega$ such that $$(Q_t(\omega)g^1_t(\omega),g^1_t(\omega))_H=\sup (Q_t(\omega)y,y)=\gamma^1_t(\omega), $$ where the supremum is taken over all $y\in H$ with $|y|_H\le 1$. Using the random variables $\gamma^1, \gamma^2,\ldots, \gamma^n$ and $g^1,g^2,\ldots,g^n$, we can repeat this procedure for the operator-valued process $$ Q_t(\omega)-\sum_{j=1}^n \gamma^j(\omega,t)g^j(\omega,t)\otimes g^j(\omega,t). $$ Note that the first eigenvalue is either $||Q_t(\omega)||_{op}$ or $-||Q_t(\omega)||_{op}$ and that the process ends because $\lim{n\rightarrow\infty} \lambda_n(t,\omega)=0$, where, for each $(t,\omega)$, $\lambda_n(t,\omega)$ are the eigenvalues of $Q_t(\omega)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.