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For positive integer $k$, let $\mu_k=\dfrac{1}{2}\left(\delta(x)+\delta\left(x-\dfrac{2}{3^k}\right)\right)$. Let $dC_k=\mu_1\ast\cdots\ast\mu_k$. We have that $dC_k$ converges weakly to $\mu_C$, where $C$ is the Cantor function.

Now we want to show that $\hat{\mu}_C(y)=e^{ay}\prod_{k=1}^\infty\cos(y/3^k)$ for some $a\in\mathbb{C}$. For this, we also need to show that the infinite product converges.

I want to use the weak convergence. Is it true that $\hat{dC_k}(y)$ converges to $\hat{\mu}_C(y)$?

Kunal
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    By dominated convergence (the integrand is bounded by 1 in modulus) $E[e^{i\sum_{j=1}^k\frac{X_j}{3^j}}] \to E[e^{i\sum_{j=1}^\infty\frac{X_j}{3^j}}]$, where the $X_i$ sequence is from my answer to your previous question. – Chris Janjigian Nov 29 '13 at 03:36
  • @ChrisJanjigian I really don't see how that expectation $E$ is related to what I'm asking. Could you explain it please? Thanks. – Kunal Nov 29 '13 at 04:36
  • Remember that $\mu_k$ is the law of $\sum_{j=1}^k \frac{X_j}{3^j}$. That means that $E\left[e^{i t\sum_{j=1}^k \frac{X_j}{3^j}} \right] = \int e^{itx}\mu_k(dx)$, which is the definition of the Fourier transform of the measure $\mu_k$. – Chris Janjigian Nov 29 '13 at 14:40
  • @ChrisJanjigian I get it that $|e^{itx}|=1$, but don't we also have to bound the $\mu_k$ term? Also, the integral is over $\mathbb{R}$, so the constant function $1$ is not integrable? – Kunal Nov 29 '13 at 21:13
  • The whole point of the probabilistic approach to this problem is that you can realize all of these integrals with respect to a single measure (which is why I wrote it with expectations). $1$ is integrable with respect to that measure. What's really happening is that that expectation is with respect to the infinite product measure you get from Kolmogorov's extension theorem applied to the $X_j$ sequence. That measure is a probability measure, so $1$ is integrable. – Chris Janjigian Nov 30 '13 at 03:25
  • @ChrisJanjigian Thanks Chris. Sorry to bother you again, but how can I turn $E\left[e^{it\sum_{j=1}^k\frac{X_j}{3^j}}\right]$ into the cosine formula? – Kunal Nov 30 '13 at 03:46
  • Using the fact that $X_j$ is $0$ with prob. $1/2$ and $2/3^j$ with prob. $1/2$, I get that the expectation for each term $X_j/3^j$ is $\frac12\left(1+e^\frac{2yi}{3^n}\right)$. The whole expectation is the product of each one. – Kunal Nov 30 '13 at 03:52
  • Factor out $e^{\frac{it}{3^j}}$ to get $\frac{1}{2}\left(1 + e^{i\frac{2t}{3^j}}\right) = e^{\frac{it}{3^j}}\cos\left(\frac{t}{3^j}\right)$. The limit is $e^{\frac{it}{2}}\prod_{k=1}^\infty \cos\left(\frac{t}{3^j}\right)$ – Chris Janjigian Nov 30 '13 at 19:01
  • @ChrisJanjigian, you've solved the problem so far. May be you can organize it in the answer – Norbert Dec 01 '13 at 07:17

1 Answers1

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In my answer to your previous question, I showed that $dC_k \rightharpoonup dC$, where $dC$ is the Cantor measure. One way to proceed from here to the convergence of the Fourier transforms of these measures is to try to realize the entire $dC_k$ sequence and $dC$ on the same measure space so that we can apply dominated convergence directly.

If you look at the argument I gave in the previous thread, we implicitly did this. If you take as your measure space the natural one for the coin flip sequence $X_j$, then the law of $\sum_{j=1}^k \frac{X_j}{3^j}$ is $dC_k$ and the law of $\sum_{j=1}^\infty \frac{X_j}{3^j}$ is $dC$. This means that if we call $P$ the law of the coin flip sequence, then \begin{align*} E\left[e^{it \sum_{j=0}^k \frac{X_j}{3^j}} \right] &= \int e^{it \sum_{j=0}^k \frac{X_j(\omega)}{3^j}} dP(\omega) \\ &= \int e^{itx} dC_k(x) \\ &= \widehat{dC}_k(t), \end{align*}

where the first equality is by definition and the second equality comes from the argument I gave in my previous answer. Similarly, \begin{align*} E\left[e^{it \sum_{j=0}^\infty \frac{X_j}{3^j}} \right] &= \int e^{it \sum_{j=0}^\infty \frac{X_j(\omega)}{3^j}} dP(\omega) \\ &= \int e^{itx} dC(x) \\ &= \widehat{dC}(t). \end{align*}

$dP$ is a probability measure (remember, it is the law of a sequence of $\{0,2\}$ valued fair coin flips), so $1$ is integrable. I showed in my last answer that $\sum_{j=0}^k \frac{X_j(\omega)}{3^j} \to \sum_{j=0}^\infty \frac{X_j(\omega)}{3^j}$. We can therefore apply dominated convergence to the functions $e^{it \sum_{j=0}^k \frac{X_j(\omega)}{3^j}}$ to get the desired limit.

If you are not comfortable with that approach, here is another option. $e^{itx}$ is a bounded continuous function and by the Portmanteau theorem, a sequence of probability measures converges weakly if and only if for all bounded continuous functions the expectations converge. Again, we already showed weak convergence in the previous question. It does matter here that our measures are all probability measures, by the way. That result fails for general measures and general weak convergence.

All that is left is to compute the characteristic functions. Notice that

\begin{align*} \widehat{\mu}_j(t) &= \frac{1}{2}\left(1 + e^{i\frac{2t}{3^j}}\right) \\ &= e^{i\frac{t}{3^j}}\underbrace{\frac{1}{2}\left(e^{-i\frac{t}{3^j}} + e^{i\frac{t}{3^j}}\right)}_\textrm{$\cos \left( \frac{t}{3^j}\right)$} \end{align*}

so that \begin{align*} \widehat{dC}_k (t) = e^{i\sum_{j=1}^k \frac{t}{3^j}} \prod_{j=1}^k \cos\left(\frac{t}{3^j} \right). \end{align*}

Taking limits (and the limits exist by what we did previously), we find that \begin{align*} \widehat{dC}(t) &= e^{i\sum_{j=1}^\infty \frac{t}{3^j}} \prod_{j=1}^\infty \cos\left(\frac{t}{3^j} \right) \\ &= e^{i\frac{t}{2}} \prod_{j=1}^\infty \cos\left(\frac{t}{3^j} \right). \end{align*}

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