Proving $\nu_{t/n} \to \delta_0$ weakly as $n \to \infty$ for convolution semigroup

Question

A convolution semigroup is a family of probability measures $(\nu_t)_{t \in I}$ on $\mathbb R^d$ with $I \subset [0,\infty)$ and $0 \in I$, for which $\nu_s * \nu_t = \nu_{s+t}$ for $s,t \in I$. I want to prove the following (from Achim Klenke's "Introduction to Probability Theory", Exercise. 14.4.2):

Let $\{\nu_t : t \geq 0\}$ be a convolution semigroup. Show that for $t \geq 0$, $\nu_{t/n}$ converges weakly to the Dirac measure $\delta_0$ as $n \to \infty$.

Idea 1: Write $\nu_t = \nu_{t/n}^{*n}$ so that $\nu_t((-\epsilon,\epsilon)^d) = \left(\nu_{t/n} * \cdots * \nu_{t/n}\right)((-\epsilon, \epsilon)^d)$ for every $\epsilon > 0$, and somehow use this to show $\lim_{n \to \infty} \nu_{t/n}(G) = 0$ for every open $G \subset \mathbb R^d$ not containing $1$, and $=1$ otherwise. But I’m having trouble doing that. I have an inductive formula that for $f : \mathbb R^d \to \mathbb R$ $\nu_{t/n}$-integrable, $$ \int f \, d\left(\nu_{t/n}^{*n}\right) = \int \cdots \int f(x_1 + \cdots + x_n) \,d\nu_{t/n}(x_n) \cdots d\nu_{t/n}(x_1) $$ but this doesn't seem to help.

Idea 2: Show that if $A \subset \mathbb R^d$ is measurable with $0 \not\in \overline A$, so that $\delta_0(\partial A) = \delta_0(A) = 0$, then $\lim_{n \to \infty} \nu_{t/n}(A) = 0$. For this, I thought I could show that if $\nu_t(A)$ increases as $t$ increases, then $\lim_{n \to \infty} \nu_{t/n}(A)$ exists since $\nu_{t/n}(A)$ decreases with $n$, and then I could try arriving at a contradiction if $\lim_{n \to \infty} \nu_{t/n}(A) > 0$. But the claim that $\nu_t(A)$ increases with $t$ is not true in general, e.g. if $\nu_t$ is a Gaussian normal distribution with mean $0$ and standard deviation $t$. Then for fixed $0 < a < b$ and values of $t$ sufficiently large, $\nu_t((a,b))$ decreases as $t$ increases.

Any thoughts on how this result can be proven?

Answer 1

Consider the characteristic function (i.e., Fourier transform)

$$ \hat{\nu}_t(\xi) = \int_{\mathbb{R}^d} e^{i \langle \xi, x \rangle} \, \nu_t(\mathrm{d}x). $$

Then for any $\xi \in \mathbb{R}^d$ and $s, t \geq 0$, we have $\hat{\nu}_{s+t}(\xi) = \hat{\nu}_{s}(\xi)\hat{\nu}_{t}(\xi)$.

Lemma. $\nu_t(\xi) \neq 0$ for all $\xi \in \mathbb{R}^d$.

Indeed, define the function $\psi_t$ by

$$ \psi_t(\xi) := \lim_{n\to\infty} |\nu_{t/n}(\xi)|^2 = \lim_{n\to\infty} |\nu_{t}(\xi)|^{2/n} = \mathbf{1}_{\{ \hat{\nu}_t(\xi) \neq 0\}}. $$

Noting that $|\nu_{t/n}(\cdot)|^2$ is also a c.f. and $\hat{\nu}_t$ is non-zero near the origin, we find that $\psi_t$ is continuous at $0$ and hence $\psi_t$ is a c.f. by Lévy's continuity theorem. Since any c.f. is uniformly continuous, we get $\psi_t \equiv 1$ and the conclusion follows. $\square$

Now we return to the original problem. The idea is to show that $\hat{\nu}_{t/n}$ is "the" $n$-th square root of $\hat{\nu}_t$ in some sense, hence converges to the constant function $1$. Then the desired conclusion will follow from Lévy's continuity theorem.

Fix $\xi \in \mathbb{R}^d$. Then the function $\mathbb{R} \ni \lambda \mapsto \hat{\nu}_t(\lambda\xi)/|\hat{\nu}_t(\lambda\xi)|$ is well-defined and continuous. This allows us to find a continuous function $\theta_t : \mathbb{R} \to \mathbb{R}$ such that

$$ \theta_t(0) = 0 \qquad\text{and}\qquad \frac{\hat{\nu}_t(\lambda\xi)}{|\hat{\nu}_t(\lambda\xi)|} = e^{i\theta_t(\lambda)}. $$

Then for each $n$,

$$ \omega_n(\xi) := \frac{\hat{\nu}_{t/n}(\lambda\xi)}{|\hat{\nu}_{t}(\lambda\xi)|^{1/n}e^{i\theta_t(\lambda)/n}} \quad\implies\quad \omega_n(\xi)^n = \frac{\hat{\nu}_{t/n}(\lambda\xi)^n}{\hat{\nu}_t(\lambda\xi)} = 1 $$

and the continuity together show that $\omega_n$ is constant with the value $\omega_n(0) = 1$. This shows that

$$ \hat{\nu}_{t/n}(\lambda\xi) = |\hat{\nu}_{t}(\lambda\xi)|^{1/n}e^{i\theta_t(\lambda)/n} \quad\text{for all}\quad \lambda\in\mathbb{R}. $$

From this, we get

$$ \lim_{n\to\infty} \hat{\nu}_{t/n}(\lambda\xi) = 1 $$

for any $\lambda\in\mathbb{R}$ and $\xi\in\mathbb{R}^d$, and therefore $\nu_{t/n} \to \delta_0$ as desired.

Answer 2

If you take the Fourier transform of $\nu_t$ you get that $\widehat{\nu_t}=(\widehat{\nu_{t/n}})^n$ and so, $|\widehat{\nu_t}|^{1/n}=|\widehat{\nu_{t/n}}|$. From this, and the continuity of $\nu_t$, you se that $\lim_n|\widehat{\nu}_{t/n}|$ converges to $1$ in a neighborhood of $0$.

Although $|\widehat{\nu}_t|$ is not necessarily a characteristic function, $|\widehat{\nu}_t|^2$ is, as $|\widehat{\nu}_t|^2$ is the Fourier transform of the measure $\nu_t*\tilde{\nu}_t$, where $\tilde{\nu}_t(A):=\nu_t(-A)$. Therefore, by the Lévy-Bochner continuity theorem, $\nu_{t/n}*(\widetilde{\nu_{t/n}})$ converges weakly to $\delta_0$. It follows from this that $\widehat{\nu_t}(s)\neq0$ for all $s\in\mathbb{R}^d$.

The following result from Complex variables will be very useful:

Theorem: If $\phi:\mathbb{R}^d\rightarrow\mathbb{C}$ is a continuous function with that $\phi(0)=1$ and $\phi(s)\neq0$ for all $s\in\mathbb{R}^d$, then there exists a unique function $f:\mathbb{R}^d\rightarrow\mathbb{C}$ such that $f(0)=0$ and $\phi(s)=\exp(f(s))$. Furthermore, for any $n\in\mathbb{N}$, there exists a unique function $g_n:\mathbb{R}^d\rightarrow\mathbb{C}$ such that $g_n(0)=1$ and $(g_n(s))^n=\phi(s)$ (in fact $g_n(s)=\exp\big(\frac1n f(s)\big)$).

The function $f$ is usually called the distinguished logarithm of $\phi$; $g_n$ is the distinguished $n$-th root of $\phi$. A proof of this can be found in Sato, K. Lévy Processes and Infinite Divisible Distributions, Cambridge University Press, pp. 33.

Fix $t_0>0$. Applying the theorem above to $\widehat{\nu_{t_0}}$ yields a continuous function $f_{t_0}:\mathbb{R}^d\rightarrow\mathbb{C}$, $f_{t_0}(0)=0$ such that $\widehat{\nu_{t_0}}(s)=\exp(f_{t_0}(s))$. Notice that for any $r\geq0$, $s\mapsto \exp(rf_{t_0}(s))$ is well defined continuous function.

For $r=\frac1n$, another application of the theorem above, along with the semigroup property of $\{\nu_t:t\geq0\}$ shows that for any $n\in\mathbb{N}$, $g_{t_0,n}(s)=\exp\big(\frac1n f_{t_0}(s)\big)$ is the characteristic function of $\nu_{t_0/n}$, that is, $$\widehat{\nu_{t_0/n}}(s)=\exp\big(\frac1n f_{t_0}(s)\big),\qquad s\in\mathbb{R}^d$$ Consequently $$\lim_{n\rightarrow\infty}\widehat{\nu_{t_0/n}}(s)=\lim_{n\rightarrow\infty}\exp\big(\frac1n f_{t_0}(s)\big)=1, \qquad s\in\mathbb{R}^d$$ which means that $\nu_{t_0/n}$ converges to $\delta_0$ in distribution.

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Observation: Although not needed for solving the OP, one can show that $s\mapsto\exp(rf_{t_0}(s))$ is also the characteristic function of a probability measure for any $r\geq0$.

Indeed, consider the case $t_0=1$ and denote $f:=f_1$. We have shown that $\widehat{\nu_{1/n}}(s)=\exp\big(\frac{1}{n} f(s)\big)$ and so, by the semigroup property, it follows that $\widehat{\nu_{m/n}}(s)=\exp\big(\frac{m}{n} f(s)\big)$ for all nonnegative rational number $\frac{m}{n}$. For any irrational $t>0$, choose a sequence of rational numbers $t_k\xrightarrow{k\rightarrow\infty}t$. Then, $$\widehat{\nu_{t_k}}(s)\xrightarrow{k\rightarrow\infty}\exp(t f(s)),\qquad s\in\mathbb{R}^d$$ and by Lévy-Bochner's continuity theorem, it follows that $\phi_t(s):=\exp(t f(s))$ is the characteristic function of some probability measure $\mu_t$.

Under additional continuity assumptions on the semigroup $\{\nu_t:t\geq0\}$, it may be proved that in fact $\nu_t=\mu_t$.