Fourier transform of the Cantor function

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.

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 $(\mathcal F\mu)(0)=1$.