Moment problem, ergodicity and spectral gap on the space of tempered distributions

Question

Let $\{ S_n \}_{n=0}^\infty$ be a collection of tempered distributions where $S_0:=1$ and $S_n$ is a tempered distribution on $\mathbb{R}^n$.

Just below formula [5] in p.122 of the Fröhlich paper, factorial growth bounds and Nelson-Symanzik positivity imply existence of a probability measure $d\mu$ on $\mathcal{S}'(\mathbb{R})$ having $\{S_n\}$ as moments: \begin{equation} S_n(\otimes_{i=1}^n f_i)=\int_{\mathcal{S}'} \prod_{i=1}^n \omega(f_i) d\mu(\omega) \end{equation}

A similar (or more general) result is presented in p.207-208 Theorem 3.7 of the Borchers-Yngvason paper.

Now in p.199 of the same paper, you can find the following sentence in the paragraph just above 2.1 Definition:

Moreover, a measure will not be uniquely determined by the functional it defines on $S(\mathscr{S})$.

Here, a functional on $S(\mathscr{S})$ corresponds to a collection of (Bosonic) Schwingers $\{S_n \}_{n=0}^\infty$ as mentioned above.

Therefore, my understanding is that uniqueness of the moment problem is generally NOT guaranteed.

In fact, 2.1 Definition in p.199 of the Borchers-Yngvason paper introduces a larger space $\mathscr{F}$ than $S(\mathscr{S})$. With a linear functional on $\mathscr{F}$, uniqueness of a measure is guaranteed as shown in p.200 Theorem 2.3 or p.201 Theorem 2.7.

In p.159 of Fröhlich paper, he somehow mentions both uniqueness and ergodicity, but I admit that I can't fully understand what he actually says there.

Meanwhile, I believe that relation between ergodicity and spectral gap is a well-understood topic, as presented in this MO post.

Therefore, I am led into these very rough and tentative questions:

Does ergodicity somehow ensures uniqueness of $\mu$ on $\mathcal{S}'$ as a solution of the moment problem for a give collection $\{ S_n \}_{n=0}^\infty$?

In this case, how does ergodicity relate to the notion of spectral gap?

Several ideas are mixed up here, and I admit that they are by no means organized...I deeply appreciate any insight.

Answer 1

Let $\mathscr{S}=\mathscr{S}(\mathbb{R}^d)$ be the space of real-valued Schwartz functions with its usual Frechet topology, and let $\mathscr{S}'=\mathscr{S}'(\mathbb{R}^d)$ be the space of real-valued temperate distributions equipped with the strong topology. For a Borel probability measure $\mu$ on $\mathscr{S}'$, we will say that it has finite moments of all orders iff for all $f\in\mathscr{S}$, and all $p\in[1,\infty)$, the map $\phi\mapsto \phi(f)$ is in $L^p(\mathscr{S}',\mu)$. By the multilinear Hölder inequality, this implies the map $\mathscr{S}'\rightarrow\mathbb{R}$, $$ \phi\longmapsto \phi(f_1)\cdots\phi(f_n) $$ is integrable with respect to $\mu$, for any $n$ and any choice of test functions $f_1,\ldots,f_n$ in $\mathscr{S}$. By a result of Fernique (based on the closed graph theorem and the continuity of the tensor product of distributions for the strong topology), the multilinear map $$ (f_1,\ldots,f_n)\longmapsto \int_{\mathscr{S}'}\phi(f_1)\cdots\phi(f_n)\ d\mu(\phi) $$ is automatically continuous. By the nuclear theorem, it is therefore given by $$ (f_1,\ldots,f_n)\longmapsto S_n(f_1\otimes\cdots\otimes f_n) $$ for a unique temperate distribution $S_n$ in $\mathscr{S}'(\mathbb{R}^{nd})$, which we will call the $n$-th moment of $\mu$. We will say that $\mu$ has the $n!$-growth property iff there exists a continuous seminorm $\|\cdot\|$ on $\mathscr{S}$, such that for all $n$, and all $f_1,\ldots,f_n$ in $\mathscr{S}$, $$ |S_n(f_1\otimes\cdots\otimes f_n)|\le n!\times \|f_1\|\cdots\|f_n\|\ . $$ An immediate consequence is that for any $n$ and test function $f$, $$ \int_{\mathscr{S}'} |\phi(f)|^n\ d\mu(\phi)\le 2^n n!\ \|f\|^n\ . $$ This follows from the Cauchy-Schwarz inequality, the growth condition and the elementary bound $$ (2n)!=\binom{2n}{n}\ n!^2\le 2^{2n}n!^2\ . $$

Theorem: Let $\mu$, $\nu$ be two such measures with the $n!$-growth property, and suppose they have the same moments $S_n$, then $\mu=\nu$.

Proof:

Fix a test function $f$, and consider the characteristic function $\Phi(z)=E(e^{izX})$ of the real-valued random variable $X=\phi(f)$ on $\Omega=\mathscr{S}'$ with the measure $\mu$. We will show it is analytic in the strip $|{\rm Im}\ z|<\frac{1}{2\|f\|}$. Note that for $z$ in this strip $e^{izX}$ is integrable since $$ E(|e^{izX}|)=E(e^{-({\rm Im}\ z)X})\le E(e^{|{\rm Im}\ z||X|}) $$ $$ =\sum_{n=0}^{\infty} \frac{|{\rm Im}\ z|^n}{n!}E(|X|^n) \le \sum_{n=0}^{\infty}\frac{|{\rm Im}\ z|^n}{n!} 2^n n! \|f\|^n<\infty\ . $$ Moreover, for $t$ real and $z$ complex such that $|z|<\frac{1}{2\|f\|}$, we have $$ \Phi(t+z)=E\left(e^{itX}\sum_{n=0}^{\infty}\frac{(iz)^n}{n!}X^n\right) $$ $$ =\sum_{n=0}^{\infty}\frac{(iz)^n}{n!}E(e^{itX}X^n) $$ and the exchange of series and integral is legitimate because $$ \sum_{n=0}^{\infty}\left|\frac{(iz)^n}{n!}\right|E(|e^{itX}X^n|) =\sum_{n=0}^{\infty}\frac{|z|^n}{n!}E(|X|^n) $$ $$ \le \sum_{n=0}^{\infty}\frac{|z|^n}{n!} 2^n n! \|f\|^n<\infty\ . $$ Moreover, at $t=0$ we have $$ \Phi(z)=\sum_{n=0}^{\infty} a_n z^n $$ with $$ a_n=\frac{i^n}{n!} S_n(f^{\otimes n})\ . $$ We then do the same for the random variable $Y=\phi(f)$ on $\Omega=\mathscr{S}'$ with the measure $\nu$.

We thus have two random variables whose characteristic functions are analytic on the intersection of the two strips, and have the same Taylor series at the origin. By the uniqueness of analytic continuation, they must coincide on this intersection and in particular on the entire real line, and thus at $z=1$. Hence, $$ \int_{\mathscr{S}'} e^{i\phi(f)}\ d\mu(\phi)=\int_{\mathscr{S}'} e^{i\phi(f)}\ d\nu(\phi)\ . $$ Since $f$ is arbitrary, the two measures have the same characteristic function, and by the Bochner-Minlos Theorem, we must have $\mu=\nu$.