Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|cite|improve this question
Haven't had time to look too carefully into this question, but apparently it was on a general exam at Princeton: –  Vince Nov 5 '10 at 18:27
You may be interested in… In general it would be more fruitful (I think) to search for Cantor function instead of Devil's Staircase. –  Willie Wong Nov 5 '10 at 18:40
While the title is cute, "Fourier coefficients of the Cantor Function" would be more informative. –  j.c. Nov 5 '10 at 18:41
btw, if you just need the Fourier coefficients of a continuous function with suitable bad features, maybe the Weierstrass function could do. Here there was a related question :… –  Pietro Majer Nov 5 '10 at 19:10

2 Answers 2

up vote 18 down vote accepted

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.

share|cite|improve this answer
Now I'm surprised that I didn't notice that when I looked at that page a couple of hours ago. –  Michael Hardy Nov 5 '10 at 22:23
The calculation of characteristic function simply follows from the observation that for Cantor distributed $X$, its ternary expansion is given by $X = \sum_{k\geq1} X_k 3^{-k}$, where $X_k$ are iid and takes values 0 or 2 with equal probability. –  mr.gondolier Nov 5 '10 at 23:00
Thanks to everybody: both for the final answer and for the discussion. –  ccarminat Nov 6 '10 at 18:18

I might start by thinking about the Riemann--Stieltjes 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 well-understood way on the first $n$ cumulants. Then I'd try to use self-similarity 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 odd-order cumulants are zero because of symmetry.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.