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. $\endgroup$ Apr 29, 2014 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. $\endgroup$ Nov 26, 2016 at 15:42

1 Answer 1


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, 2014 at 1:45
  • $\begingroup$ @wxlu It's the role of the function $x\mapsto \exp\left(-\lVert x\rVert^2\right)$. $\endgroup$ Apr 30, 2014 at 8:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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