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Let $\{\mu_n:n\in\mathbb N\}$ and $\mu$ be distributions on $\mathbb R$, and let $\{\phi_n:n\in\mathbb N\}$ and $\phi$ be their respective characteristic functions.

We can easily show using a direct application of the so called continuous mapping theorem that if $\mu_n\to\mu$ weakly, then $\phi_n\to\phi$ pointwise. I also know that if $\mu_n\to\mu$ weakly, then for any bounded set $B\subset\mathbb R$, we have that $\phi_n\to\phi$ uniformly. However, my attempts to understand why this is true have so far been in vain.

I've tried to find upper bounds for \begin{align*} \sup_{t\in[a,b]}\left|\int_{-\infty}^{\infty}e^{itx}~\mu_n(dx)-\int_{-\infty}^{\infty}e^{itx}~\mu(dx)\right|, \end{align*} but it always ends up being too small or too big. Any hint would be greatly appreciated.

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Hints

  1. Show that the sequence $(\mu_n)_{n \in \mathbb{N}}$ is tight, i.e. for any $\varepsilon>0$ there exists a compact set $K \subseteq \mathbb{R}$ such that $$\mu_n(K^c) \leq \varepsilon$$ for all $n \in \mathbb{N}$.
  2. The tightness entails that $(\phi_n)_{n \in \mathbb{N}}$ is uniformly equicontinuous. To prove this, note that $$|\phi_n(\xi)-\phi_n(\eta)| \leq \int_K |1-e^{\imath \, (\eta-\xi) \cdot x}| \, \mu_n(dx) + \int_{K^c} |1-e^{\imath \, (\eta-\xi) \cdot x}| \, \mu_n(dx).$$ Now use that the first term is small for $|\eta-\xi|<\delta$ and the second one is bounded by $2 \mu_n(K^c)$.
  3. It follows from the uniformly equicontinuity and the pointwise convergence that $\phi_n \to \phi$ locally uniformly; hence, uniformly on compact sets.
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  • $\begingroup$ In 2. the set $K$ depends of $n$, so how are they uniformly equicontinuous? $\endgroup$ – Yiorgos S. Smyrlis Feb 27 '14 at 7:54
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    $\begingroup$ @YiorgosS.Smyrlis No, the set $K$ does not depend on $n$ since $(\mu_n)_n$ is tight. $\endgroup$ – saz Feb 27 '14 at 7:57

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