As we all know on real line $\mathbb{R}$, the following is valid

A $\mathbb{C}$-valued function $\varphi$ is a characteristic function of a probability measure on $\mathbb{R}$ if and only if $\varphi$ is a continuous, positive definite function such that $\varphi(0)=1$.

But on separable Hilbert space, Theorem 2.13 in Stochastic Equations in infinite Dimensions (by Da Prato) says that

A $\mathbb{C}$-valued function on $H$ $\varphi$ is a characteristic function of a probability measure on $H$ if and only if

(i) $\varphi$ is a continuous, positive definite function such that $\varphi(0)=1$.

(ii) $\forall \varepsilon >0, $ $\exists$ a nonnegative nuclear operator $S$ such that $$ 1-\mathrm{Re}\ \varphi(\lambda) \leq \varepsilon, \mathrm{for\ all}\ \lambda \text{ satisfying }\langle S\lambda,\lambda \rangle \leq1 $$

What confused me is that why (ii) condition is added in the infinite case? Is there any intuitive or intelligible explanation for it?

  • $\begingroup$ alternatively if you want the theorems like Bochner and Levy continuity to read exactly like in finitely-many dimensions, ie., without extra hypotheses like (ii), then you need to abandon Hilbert and Banach spaces and work with duals of nuclear spaces, like $S'(\mathbb{R}^d)$ or $\mathbb{R}^{\mathbb{N}}$. $\endgroup$ – Abdelmalek Abdesselam Dec 8 '14 at 20:26

A good way to see why (ii) is needed in the infinite dimensional case is to find an example where (i) holds but not (ii). (in the finite dimensional case, (ii) is a consequence of (i)). Also, it could be good to check (ii) in the case of a Gaussian random variable.

Define $\phi(x):=\exp(-\lVert x\rVert^2_H)$: it satisfies (i). Now we shall see that (ii) does not hold with $\varepsilon:=1/2$. Suppose that there exists a non-negative nuclear operator $S$ such that if $\langle Sx,x\rangle\leqslant 1$, then $\phi(x)\geqslant 1/2$. Notice that $S$ has the form $$S(x)=\sum_{j=1}^{+\infty}\lambda_j\langle x,a_j\rangle b_j$$ where $\lambda_j\geqslant 0$, $\sum_{j=1}^{+\infty}\lambda_j<+\infty$, $a_j,b_j\in H$ and $M:=\sup_j\lVert a_j\rVert+\lVert b_j\rVert$ is finite. Fix an integer $N\geqslant 1$. If $x$ belongs to the orthogonal of the vector space generated by $a_j,j\leqslant N$, then the condition $\langle Sx,x\rangle$ will be satisfied once $M^2\lVert x\rVert^2\sum_{j\geqslant N}\lambda_j\leqslant 1$. If we pick $x$ such that the equality holds, then we should have $$\exp\left(-\frac 1{M^2\sum_{j\geqslant N}\lambda_j}\right)\geqslant \frac 12.$$ If we choose $N$ large enough we get a contradiction.

The problem is that for infinite dimensional spaces $\langle Sx,x\rangle$ may be smaller than $1$ for $x$ having a large norm.

  • 1
    $\begingroup$ To slightly expand on Davide's answer: you can embed $H$ into a larger Hilbert space $H'$ such that, after making the natural identifications, $\phi$ is the characteristic function of a probability measure on $H'$. The corresponding random variable is of the form $\sum_n e_n \xi_n$ with $\xi_n$ an iid $N(0,1)$ sequence and $e_n$ an o.n.b. of $H$. This of course doesn't converge in $H$ (all coefficients are of about the same size). $\endgroup$ – Martin Hairer May 1 '14 at 15:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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