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?
Best Answer
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 $\hat{\nu}_{s+t}(\xi) = \hat{\nu}_{s}(\xi)\hat{\nu}_{t}(\xi)$.
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.