Let $f:[0,1]\to[0,1]$ be the classical devil's staircase. Has anybody ever computed (or studied) the fourier coefficient of $f(x)$?
Related question: is the fourier series of $f(x)x$ normally convergent (with respect to uniform norm)?
Let $f:[0,1]\to[0,1]$ be the classical devil's staircase. Has anybody ever computed (or studied) the fourier coefficient of $f(x)$? Related question: is the fourier series of $f(x)x$ normally convergent (with respect to uniform norm)? 


The Fourier transform of the derivative $\mu$ of the Devil staircase is explicitely stated on the wikipedia page of the Cantor distribution, in the table at the right, under the heading "cf" (characteristic function). Its value is $$ \int_0^1 e^{itx} d\mu(x) = e^{it/2}\ \ \prod_{k=1}^\infty \cos(t/3^k)$$ Just multiply by $1/it$, add $1/it$, and you get the Fourier transform of the Devil staircase. A word on the proof. The Cantor distribution is the weak limit of the functions obtained by summing the indicator functions of the 2^n intervals generating the Cantor set at the nth step (after renormalization). The Fourier transform of these sums can be computed explicitely. Then let n goes to infinity. 


I might start by thinking about the RiemannStieltjes integral $\varphi(t) = \int_0^1 e^{itx} \; df(x)$. Since $f$ is cumulative probability distribution, the $n$th moment of that distribution would be $E(X^n) = \varphi^{(n)}(t)$ where $X$ is a random variable so distributed. The $n$th moment depends in a wellunderstood way on the first $n$ cumulants. Then I'd try to use selfsimilarity together with the law of total cumulance to figure out what the cumulants are. Having written that, I see at this article that I knew the cumulants several years ago; I think I added them to that Wikipedia article. (The oddorder cumulants are zero because of symmetry.) 

