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)?
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6$\begingroup$ Haven't had time to look too carefully into this question, but apparently it was on a general exam at Princeton: math.princeton.edu/generals/realan.txt $\endgroup$– VinceCommented Nov 5, 2010 at 18:27
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1$\begingroup$ You may be interested in ams.org/spmj/2004-15-03/S1061-0022-04-00817-9/… In general it would be more fruitful (I think) to search for Cantor function instead of Devil's Staircase. $\endgroup$– Willie WongCommented Nov 5, 2010 at 18:40
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6$\begingroup$ While the title is cute, "Fourier coefficients of the Cantor Function" would be more informative. $\endgroup$– j.c.Commented Nov 5, 2010 at 18:41
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1$\begingroup$ 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 : mathoverflow.net/questions/38751/… $\endgroup$– Pietro MajerCommented Nov 5, 2010 at 19:10
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2 Answers
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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.
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$\begingroup$ Now I'm surprised that I didn't notice that when I looked at that page a couple of hours ago. $\endgroup$ Commented Nov 5, 2010 at 22:23
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4$\begingroup$ 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. $\endgroup$ Commented Nov 5, 2010 at 23:00
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$\begingroup$ Thanks to everybody: both for the final answer and for the discussion. $\endgroup$ Commented Nov 6, 2010 at 18:18
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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.)