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Let $f:[0,1] \to [0,1]$ be the Cantor function.
Extend $f$ to all of $\mathbb R$ by setting $f(x)=0$ on $\mathbb R \setminus [0,1]$. Calculate the Fourier transform of $f$ $$ \hat f(x)= \int f(t) e^{-ixt} dt $$ where $dt$ is the Lebesgue measure on $\mathbb R$ divided by $2\pi$, and the integral is over $\mathbb R$.

I think this MO post says the result is $$ \hat f (x)= \frac{1}{ix}-\frac{1}{ix}e^{ix/2}\prod_{k=1}^{\infty} \cos(x/3^k). \tag{1} $$

To get this, I approximate $f$ by simple function $$ f_n(x)= \sum_{i=1}^n \sum_{j=1}^{2^{i-1}} \frac{2j-1}{2^i}\chi_{E_{n,k}} $$ where $E_{n,k}$ is the $k$th set removed during the $n$th stage of the Cantor process. Then $$ \hat f_n(x) = \sum_{i=1}^n \sum_{j=1}^{2^{i-1}} \frac{2j-1}{2^i}\int_{E_{n,k}} e^{-ixt} dt \tag{2} $$

But I don't see how, in the limit, (2) simplifies to (1).

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1 Answer 1

up vote 2 down vote accepted

Since the excised intervals vary in their length while the remaining intervals are not, it seems easier to focus on the remaining intervals.

Let $I(n,k)$ for $n \geq 1$ and $0 \leq k \leq 2^n - 1$ be the remaining $2^n$ intervals after the $n$-th stage of the construction of the Cantor set. Then $|I(n, k)| = 3^{-n}$, and we can approximate $f$ by

$$f_n(x) = \int_{0}^{x} \left( (3/2)^n \sum_{k=0}^{2^n - 1} \chi_{I(n,k)}(t) \right) \, dt $$

To see this really approximates $f$, observe that $f_n$ increases only on $C(n) = \bigcup_{k=0}^{2^n-1}I(n,k)$ and on each subinterval $I(n,k)$, $f_n$ increases by exactly $2^{-n}$, as we can check:

$$ \int_{I(n,k)} (3/2)^n \chi_{I(n,k)}(t) \, dt = \frac{1}{2^n}.$$

Thus $f_n$ coincides exactly with the $n$-th intermediate function appearing in the construction of the Cantor-Lebesgue function $f$. Then $f_n \to f$ uniformly, and we have

$$ \begin{align*} \int f(t) \, e^{-ixt} \, dt &= \lim_{n\to\infty} \int f_n(t) \, e^{-ixt} \, dt \\ &= \lim_{n\to\infty} \left( \left[ -\frac{1}{ix} f_n(t) e^{-ixt} \right]_{0}^{1} + \frac{1}{ix} \int f_n'(t) \, e^{-ixt} \, dt \right) \\ &= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \lim_{n\to\infty} \int f_n'(t) \, e^{-ixt} \, dt \\ &= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \left(\frac{3}{2}\right)^n \lim_{n\to\infty} \sum_{k=0}^{2^n - 1} \int_{I(n,k)} e^{-ixt} \, dt \end{align*}$$

Now, direct calculation shows that

$$\int_{a}^{a+\beta h} e^{-ixt} \, dt + \int_{a+(1-\beta)h}^{a+ h} e^{-ixt} \, dt = 2 \cos\left(\frac{1-\beta}{2} hx\right)\frac{\sin(\frac{\beta}{2}hx)}{\sin(\frac{1}{2}hx)} \int_{a}^{a+h} e^{-ikt} \, dt. $$

Thus plugging $h = 3^{-n}$ and $\beta = \frac{1}{3}$, we have

$$\int_{I(n+1,2k)} e^{-ixt} \, dt + \int_{I(n+1,2k+1)} e^{-ixt} \, dt = 2 \cos\left(\frac{x}{3^{n+1}}\right)\frac{\sin\left(\frac{x}{2\cdot 3^{n+1}}\right)}{\sin\left(\frac{x}{2\cdot 3^{n}}\right)} \int_{I(n,k)} e^{-ikt} \, dt. $$

Inductively applying this relation allows us to calculate the limit above, which I leave because I have to go out.

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Your answer is quite good. Studying it I was able to make links between the simple function approach and the probabilistic approach. Thank you. –  Nicolas Essis-Breton Oct 26 '12 at 15:58

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