As we all know that

If $\{\varphi_n\}$ is a sequence of characteristic functions of probability measures $\{ \mu_n \}$ on $\mathbb{R}$. And $\lim\varphi_n(t)$ exists for each $t\in \mathbb{R}$. Set $\varphi(t)=\lim\varphi_n(t)$, then if $\varphi$ is continuous at $t=0$, $\varphi$ is a characteristic function for some probability measure $\mu$, and $ \mu_n $ converge weakly to $\mu$.

I want to know whether the conclusion is valid for infinite Hilbert space case, e.g.

Assuming $\{ \mu_n \}$ is a sequence of probability measures in a separable Hilbert space $H$. $\{\varphi_n\}$ are the corresponding characteristic functions defined by $$ \varphi(\lambda) = \int_{H} e^{i\langle \lambda,x \rangle} \mu(\mathrm{d}x) ,\ \ \lambda\in H $$

$\varphi_n$ pointwisely converges to a continuous function $\varphi$.

Is $\varphi$ a characteristic function for some probability measure? If not, is there any counterexample?

What is the critical difference between finite and infinite dimension cases?

Are there any references?

  • 1
    $\begingroup$ $\mu_n=\delta_{e_n}$: Then $\varphi_n\to\varphi\equiv 1$ pointwise, but $\mu_n$ doesn't approach $\delta_0$ in any reasonable sense. $\endgroup$ – Christian Remling Apr 29 '14 at 19:12
  • $\begingroup$ It is not really about a critical difference between finite and infinite dimension but rather about spaces which are nuclear versus not. On a space of Schwartz distributions like $\mathcal{S}'$ or $\mathcal{D}'$ the statement of the Levy Continuity Theorem is exactly like in finite dimensions. $\endgroup$ – Abdelmalek Abdesselam Nov 26 '16 at 15:42

As Christian Remgling's example $\mu_n:=\delta_{e_n}$ shows, the convergence of the characteristic function of $\mu_n$ to some characteristic function does not even guarantee tightness.

It's worth pointing out that a sequence of characteristic functions can converge pointwise to a continuous positive definite function which is not the characteristic function of a random variable. Indeed, consider the map $x\mapsto \exp\left(-\lVert x\rVert^2/2\right)$. Let $(\eta_j)_{j\geqslant 1}$ be a sequence of i.i.d. standard Gaussian random variables and $X:=\sum_{j\geqslant 1}\eta_je_j$, where $(e_j)_{j\geqslant 1}$ is an orthonormal basis of $H$. By uniqueness theorem in separable Hilbert spaces, $\phi$ would be the characteristic function of $X$. But the sequence $\left(\sum_{j=1}^n\eta_je_j\right)_{n\geqslant 1}$ is not tight.

However, there exists a characterization of tightness of a family of measures on a Hilbert space which involve the characteristic functional and Hilbert Schmidt operators. In Araujo and Giné's book The central limit theorem in Banach spaces, we encounter the following result:

Theorem 1.4.17. Let $H$ be a separable Hilbert space, and $\Gamma$ a set of probability measures on the Borel $\sigma$-algebra of $H$. The set $\Gamma$ has a compact closure for the weak-$^*$ topology if and only if for all $\varepsilon>0$, we can find a family of $\{A_{\mu}^\varepsilon\}_{\mu\in\Gamma}$ Hilbert-Schmidt operators on $H$ such that for a Hilbert basis $\{e_j\}$, the following properties hold:

  1. $\displaystyle\sup_{\mu\in\Gamma}\sum_{j=1}^{+\infty}\lVert A_\mu^\varepsilon(e_j)\rVert^2<\infty$;
  2. $\displaystyle\lim_{N\to +\infty}\sup_{\mu\in\Gamma}\sum_{j=N}^{+\infty}\lVert A_\mu^\varepsilon(e_j)\rVert^2=0$;
  3. for all $v\in H$, $\mu\in\Gamma$, $$\left|1-\int_He^{i\langle v,x\rangle}d\mu(x)\right|\leqslant \lVert A\mu^{\varepsilon}(v)\rVert+\varepsilon.$$
  • $\begingroup$ Above Remling and Giraudo give the $\mu_n=\delta_{e_n}$ example. But we notice that $\varphi=1$ is also a characteristic function for some probability, here is $\delta_0$. My new question is whether the characteristic function $\varphi_n$ converges to a continuous function that is not a characteristic function for any measure. $\endgroup$ – Wieshawn Apr 30 '14 at 1:45
  • $\begingroup$ @wxlu It's the role of the function $x\mapsto \exp\left(-\lVert x\rVert^2\right)$. $\endgroup$ – Davide Giraudo Apr 30 '14 at 8:32

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