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I am working on part (iii) of exercise 3.3.17. in Durrett's Probability book.

See this question: Relationship between the weak law of large numbers and characteristic functions

Basically, my issue is this. $\varphi$ is the ch.f. of a random variable. We know that $n(\varphi(t/n)-1)\to iat$ for all $t$ as $n\to\infty$ through the integers. I am trying to show that $$ \frac{\varphi(h)-1}{h}\to ia $$ as $h\downarrow 0$ in an arbitrary way, thereby proving that $\varphi'(0)$ exists. I know that $\varphi$ is uniformly continuous. Following the solution suggested in the above link, I considered the following approach.

Let $h_{n}$ be an arbitrary sequence of real numbers descending to $0$. For an arbitrary $\delta >0$, we may write $h_{n}=t_{n}/m_{n}$ for sufficiently large $n$, where $m_{n}\in\mathbb{N}$ and $|t_{n}-1|<\delta$. Hence, for sufficiently large $n$,

$$ \begin{aligned} \left|\frac{\varphi\left(\frac{t_{n}}{m_{n}}\right)-1}{\frac{t_{n}}{m_{n}}}-ia\right|&=\frac{1}{t_{n}}\left|m_{n}\left(\varphi\left(\frac{t_{n}}{m_{n}}\right)-1\right)-iat_{n}\right|\\ &\leq\frac{1}{1-\delta}\left|m_{n}\left(\varphi\left(\frac{t_{n}}{m_{n}}\right)-\varphi\left(\frac{1}{m_{n}}\right)\right)\right|\\ &\qquad+\frac{1}{1-\delta}\left|m_{n}\left(\varphi\left(\frac{1}{m_{n}}\right)-1\right)-ia\right|\\ &\qquad+\frac{1}{1-\delta}\left|ia(1-t_{n})\right| \end{aligned} $$

Question: The last two terms I can control, my question is can I use the uniform continuity of $\varphi$ to prove that $$ \left|m_{n}\left(\varphi\left(\frac{t_{n}}{m_{n}}\right)-\varphi\left(\frac{1}{m_{n}}\right)\right)\right|\to 0\qquad\text{ as }n\to\infty? $$

ervx
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  • NateEldredge suggested to prove locally uniform convergence (in $t$) of $n^{-1} (\varphi(t/n)-1)$, right? Using this (stronger) type of convergence, it is not difficult to see that the expression in your question converges to $0$ as $n \to \infty$. – saz Nov 27 '16 at 07:44
  • I'm having trouble understanding this point since both $t_{n}$ and $m_{n}$ depend on $n$. So, I guess I'm unsure how to take $t_{n}\to 1$ without also having $m_{n}\to \infty$. – ervx Nov 27 '16 at 16:28

1 Answers1

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As @NateEldredge pointed out in his answer to the question you mentioned, you have to show that the convergence

$$\lim_{n \to \infty} n (\varphi(t/n)-1) = iat$$

holds locally uniformly (in $t$). Using the locally uniform convergence, it's not difficult to answer your question. Note that by the triangle inequality

$$\begin{align*} \left| m_n \left[ \varphi \left( \frac{t_n}{m_n} \right)- \varphi \left( \frac{1}{m_n} \right) \right] \right| &\leq 2 \sup_{t \in [1-\delta,1+\delta]} \left| m_n \left[ \varphi \left( \frac{t}{m_n} \right)-1 \right] -iat \right|+|ia(t_n-1)| \\ &\leq 2 \sup_{t \in [1-\delta,1+\delta]} \left| m_n \left[ \varphi \left( \frac{t}{m_n} \right)-1 \right] -iat \right| + |a| \, \delta . \end{align*}$$

Becuase of the locally uniform convergence, we know that the first term on the right-hand side converges to $0$ as $n \to \infty$ (for fixed $\delta>0$). Hence,

$$\limsup_{n \to \infty} \left| m_n \left[ \varphi \left( \frac{t_n}{m_n} \right)- \varphi \left( \frac{1}{m_n} \right) \right] \right| \leq |a| \, \delta.$$

Letting $\delta \to 0$ finishes the proof.

Community
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saz
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  • Thanks so much for your help. I was able to prove that $(\varphi(t/n))^{n}\to e^{iat}$ uniformly in $t$ on compact subsets. Is there a way to use this to prove that $n(\varphi(t/n)-1)=iat$ converges locally uniformly in $t$? – ervx Nov 28 '16 at 03:52