TITLE: Fourier transform of the Cantor function
QUESTION [9 upvotes]: 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).

REPLY [3 votes]: Let $\mu$ be the standard Cantor measure on the interval $(-1, 1)$. If we set $\mu(x)=\mu((-\infty, x))$, considering the self-similarity of $\mu$ on the first level, we easily obtain
$$
\mu(x)=\frac{1}{2}\Big(\mu(3x+2)+\mu(3x-2)\Big).
$$
Hence
$$(\mathcal F\mu)(3t)=\int \exp(3itx)\,d\mu(x)
=\frac{1}{2}\left(\int\exp(3itx)\,d\mu(3x+2)+\int\exp(3itx)\,d\mu(3x-2)\right) 
=\frac{1}{2}\left(\int\exp(it(y-2))\,d\mu(y)
+\int\exp(it(y+2))\,d\mu(y)\right) 
=\frac{1}{2}\Big(\exp(-2it)+\exp(2it)\Big)\times\int\exp(ity)\,d\mu(y) 
=\cos 2t\cdot(\mathcal F\mu)(t)$$
and
$$
(\mathcal F\mu)(t)
=\cos\frac{2t}{3}\times(\mathcal F\mu)\left(\frac{t}{3}\right)
=\cos\frac{2t}{3}\cdot\cos\frac{2t}{9}\times(\mathcal F\mu)\left(\frac{t}{9}\right)=\dots
=\prod_{n=1}^\infty \cos\frac{2t}{3^n},
$$
because the function $\mathcal F\mu$ is continuous at the origin and $(\mathcal F\mu)(0)=1$.