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I have to prove the following statement:

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of real numbers, such that $\exp(itX_n)$ converges for every $t\in\mathbb{R}$. Show that the sequence $(X_n)$ converges.

The problem I see is, that the complex logarithm is not continous. Hence, I have to work around, but I dont know how. I hope someone can help.

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Hint: if $\exp(itX_n)$ converges to $z$, then $z\ne 0$ and $\exp(itX_n)/z$ converges to $1$. Now you can take logarithms (at least for $n$ sufficiently large) because the complex logarithm is continuous in a neighborhood of $1$. –  whuber Oct 28 '12 at 14:59
    
why should $z\neq0$ hold? that is not obvious for me... –  johanMozart Oct 28 '12 at 15:10
    
I am thinking about a prove along following lines: (i) show that the limit of $\exp(itX_n)$ is of the form $\exp(itX_\infty)$ (ii) show that $\exp(itX_n)$ is continous for all t. then it should be possible to conclude that $(X_n)$ converges too. But (i) seems to be difficult to show (at least for me). –  johanMozart Oct 28 '12 at 15:15
    
for me, as a non-mathematician, this seems to be standard. can't noone help? :) –  johanMozart Oct 28 '12 at 16:33
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Observe that if convergence of $\exp(i t X_n)$ is only postulated for $t\in\mathbb Q$, the sequence $(X_n)$ may indeed diverge. –  Hagen von Eitzen Oct 28 '12 at 21:26
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3 Answers

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Since I've been chewing on this for a while now, allow me to first construct a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent for all $t\in\mathbb Q$, before I shall solve the original problem below.

Counterexample for $\mathbb Q$ (incomplete space)

Let $(q_n)_{n\in\mathbb N}$ be an enumeration of $\mathbb Q\setminus 0$. Observe that $\bigcap_{k=1}^n q_k\mathbb Z=r_n\mathbb Z$ for some $r_n\in \mathbb Q_{>0}$. By letting $y_1=0$, and selecting $y_n\in r_n\mathbb Q$ with $y_n>y_{n-1}+1$ we obtain a diverging sequence $(y_n)_{n\in\mathbb N}$. If $t\in\mathbb Q$, then $ty_n\in\mathbb Z$ for almost all $n$: If $t=0$, this is trivial; if $t\ne 0$, we find $N$ with $\frac1t=q_N$ and have $y_n\in q_N\mathbb Z$ for all $n\ge N$.

Therefore, by letting $X_n=2\pi y_n$ we find a divergent sequence $(X_n)_{n\in\mathbb N}$ such that $(e^{itX_n})_{n\in\mathbb N}$ is convergent (in fact, is eventually constant $=1$) for all $t\in \mathbb Q$.

Proof for the case $\mathbb R$ (complete space)

What's different with $\mathbb R$ instead of $\mathbb Q$? First let's see that $(X_n)_{n\in\mathbb N}$ is bounded: For $c>0$ and $N\in\mathbb N$ consider the set $$A_{c,N}=\{t\in\mathbb R\mid\forall n,m\ge N\colon t(X_n-X_m)\in[-c,c]+2\pi\mathbb Z\}.$$ If $t\notin A_{c,N}$, then there are $n,m\ge N$, $k\in\mathbb Z$ with $2k\pi+c<t(X_n-X_m)<2(k+1)\pi-c$, which also holds for $t'$ if
$$|t'-t|<\frac{\min\{t(X_n-X_m)-2k\pi-c,2(k+1)\pi-c-t(X_n-X_m)\}}{|X_n-X_m|}.$$ Therefore the complements of the $A_{c,N}$ are open and the $A_{c,N}$ themselves are closed. The convergence of $(e^{itX_n})_{n\in\mathbb N}$ for all $t\in\mathbb R$ implies that for any $c>0$ we have $$\mathbb R=\bigcup_{N\in\mathbb N}A_{c,N}$$ Specifically, we can consider $c=\frac{2\pi}5$. By the Baire category theorem, there exists an $N$ such that $A_{c,N}$ contains an open interval $(t_0-\epsilon,t_0+\epsilon)$ with $\epsilon>0$. Without loss of generality, $t_0\ne0$ and $\epsilon< |t_0|$. Let $$M=\max\left\{|X_1|,\ldots,|X_N|\right\}+\frac\pi{\epsilon}.$$ Then $(X_n)_{n\in\mathbb N}$ is bounded by $M$. To prove this, assume that $|X_n|>M$ for some $n$. Then clearly $n>N$ and $|X_n-X_N|>\frac\pi{\epsilon}$. Because $n>N$, there exists $k\in\mathbb Z$ such that $|t_0(X_n-X_N)-2k\pi|\le c$. Let $t_1=t_0+\frac\pi{(X_n-X_N)}$ so that $|t_1-t_0|<\epsilon$ and hence $t_1\in A_{c,N}$. Then there is $k'\in\mathbb Z$ with $|t_1(X_n-X_N)-2k'\pi|\le c$. But $|t_1(X_n-X_N)-t_0(X_n-X_N)|=\pi$ makes this impossible because $2c<\pi<2\pi-2c$: the left inequality rules out $k=k'$, the other rules out $|k-k'|\ge 1$. We conclude that $(X_n)_{n\in\mathbb N}$ is bounded.

Now the final step is easy: If $|X_n|\le M$ for all $n$, then consider the case $0<t<\frac\pi{2M}$. For such $t$, all $tX_n$ are in $(-\frac\pi2,\frac\pi2)$. Since the map $(-\frac\pi2,\frac\pi2)\to \mathbb C$, $x\mapsto e^{ix}$ is an embedding, convergence of $(e^{itX_n})_{n\in\mathbb N}$ implies convergence of $(tX_n)_{n\in\mathbb N}$ and finally convergence of $(X_n)_{n\in\mathbb N}$.$_\blacksquare$

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Hi Hagen, yes this is what I had in my mind, but which I was not able to perform correctly!!! Thank you! –  johanMozart Oct 30 '12 at 19:27
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For the counterexample, $X_n=2\pi (n!)$ without enumeration of $\mathbb Q$. –  Did Nov 8 '12 at 11:28
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Here's a solution that takes advantage of a little Fourier analysis.

Set $y_n = \int_{-\infty}^\infty e^{-t^2/2} e^{i t x_n}\,dt$. Since $|e^{i t x_n}| = 1$ and $e^{-t^2/2}$ is integrable, the dominated convergence theorem implies that $y_n$ converges to some finite value $y$. On the other hand, computing the integral above shows that $y_n = \frac{1}{\sqrt{2 \pi}} e^{-x_n^2/2}$. (This is the fact that a Gaussian is its own Fourier transform.) If we define $f : [0,\infty) \to \mathbb{R}$ by $f(x) = \frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, we have $y_n = f(|x_n|)$. Now $f$ is continuous and strictly decreasing, so it has a continuous inverse. Hence $|x_n| = f^{-1}(y_n) \to f^{-1}(y)$. Call this limit $x$.

If $x=0$ then we have $x_n \to 0$. Otherwise, let $s_n$ be the sign of $x_n$ and consider $z_n := \sin(\frac{\pi}{2} \frac{x_n}{x})$. On the one hand $z_n$ is the imaginary part of $e^{i \pi x_n / 2x}$ and hence $z_n$ converges to some $z$. On the other hand, since $\sin$ is an odd function we have $$s_n = \frac{z_n}{\sin(\frac{\pi}{2} \frac{|x_n|}{x})}.$$ Since $|x_n| \to x$, the denominator converges to 1, and so $s_n$ converges to some $s = \pm 1$. Thus $x_n = s_n |x_n| \to s x$.

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Hi Nate... I am not sure if I understand correctly your approach. However, I am missing something in your solution: You nowhere used that $\exp(itx_n)$ converges. We have to show that $x_n$ converges, if the sequence of the complex exponentials converges. The crucial point here is the existence of a limit of the form $\lim_{n\rightarrow\infty}\exp(itx_n)=\exp(itx_\infty)$, where the $x_\infty$ does not depend on $t$. However, I have a sketch of the prove in my mind, which I described above... –  johanMozart Oct 30 '12 at 19:22
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I used the fact that $\exp(itx_n)$ converges when I invoked the dominated convergence theorem; its hypothesis is that the integrands converge pointwise. I didn't assume anything about what the limit function looks like. –  Nate Eldredge Oct 31 '12 at 0:23
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Actually, we don't need to require the converge of $\{e^{itx_n}\}$ for each $t$ in the real line, but only for $t\in A$, where $A$ is of positive Lebesgue measure.Let $l(t):=\lim_{n \to +\infty}e^{itx_n}$ for $t\in A$.

  • We show that $\{x_n\}$ is bounded. Suppose not. By the dominated convergence theorem, for all $A'\subset A$, $\int_{A'}e^{itx_n}dt\to \int_{A'}l(t)dt$. But it also converges to $0$ by Riemann-Lebesgue lemma. We deduce that $l=0$ on $A'$, a contradiction.

  • Assume that for a subsequences of $\{x_n\}$, say $\{x_{n'}\}$ and $\{x_{n''}\}$ converge respectively to $x'$ and $x''$. We have to show that $x'=x''$. We have by dominated converge theorem that $\int_{A'}e^{itx'}dt=\int_{A'}e^{itx''}dt$ for all $A'\subset A$ measurable. This gives that $e^{itx'}=e^{itx''}$ for all $t\in A$. If $x'\neq x''$, then defining $S:=\frac 1{x'-x''}A$, we would have $e^{is}=1$ for $s$ is a set of positive Lebesgue measure, a contradiction, as $S$ should contain points which are not of the form $2k\pi$, $k\in\Bbb Z$.

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