Let us consider Takagi-function defined by

$T(x) \colon\!= \sum_{n=0}^{\infty}s(2^nx)/2^n$,

where $s(x) \colon\!\!= \underset{n \in {\Bbb Z}}{\mathrm{min}} \,|x-n|$.

$T(x)$ has its period $1$, so I am trying to develop its Fourier expansion

$S(T)(x) \colon\!=\sum_{n= -\infty}^{\infty}\left(a_n\,sin(2 \pi n x) + b_n\,cos(2 \pi n x)\right)$.

Q: Where do we have $T(x) = S(T)(x)$?

For the case of Weierstrass function $W(x) \colon\!= \sum_{n=0}^{\infty}a^n cos(b^n \pi x)$ with $ab > 1 + 3\pi/2$, $ 0 < a < 1$ and $b$ an odd integer, it is already expanded in Fourier series with period $2$ by shape.

By Carlson, we know $T(x) = S(T)(x)$ except for sets of Lebesgue measure $0$. I would like to know whether $T(x) = S(T)(x)$ holds everywhere for Takagi-function as well.


1 Answer 1


Yes. The triangle wave $s(x):=\min_{k\in\mathbb{Z}}\big|x-k\big|$ has an absolutely convergent Fourier series

$$s(x)=\frac{1}{4}-\frac{2}{\pi^2}\sum_{k=0}^\infty\frac{1}{(2k+1)^2}\cos\big(2\pi (2k+1)x\big)\, ,\qquad x\in\mathbb{R}.$$

Therefore $$T(x):=\sum_{n=0}^\infty2^{-n}s(2^nx)= \frac{1}{4}\sum_{n=0}^\infty2^{-n}-\frac{2}{\pi^2}\sum_{n=0}^\infty\sum_{k=0}^\infty \frac{1}{2^n(2k+1)^2}\cos\big(2\pi 2^n(2k+1)x\big)\, .$$ By absolute convergence we can reorder the double sum. Since every positive integer $m\ge1$ writes uniquely as $m=2^n(2k+1)$ for nonnegative integers $n$ and $k$, we then obtain an absolutely convergent Fourier series for $T(x)$:

$$T(x)=\sum_{m=0}^\infty a_m\cos(2\pi m x)$$ with $a_0=1/2$ and for $m\ge 1$ $$a_m:=-\frac{2^{\nu(m)+1}}{\pi^2m^2}\, , $$ where $2^{\nu(m)}$ is the maximum power of $2$ that divides $m$.

  • $\begingroup$ (I added it to the wiki article on $T(x)$ ) $\endgroup$ Mar 11, 2014 at 17:28

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