MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here is a question which (up to some translation) I have been asked by an electrical engineer. Let $f:\mathbb{R}\to[0,1]$ be a smooth function with $f(x+1)=f(x)$. I would like to approximate $f$ in some sense by a characteristic function $\chi_A$, where $A\subset\mathbb{R}$ is also $\mathbb{Z}$-periodic. The idea is that $A$ should be a disjoint union of very short intervals, such that for any moderately short interval $J$ centred at $x$, the measure of $A\cap J$ should be close to $f(x)$ times the length of $J$. One possibility is to fix a large integer $N$ and put $$ A = \bigcup_{k\in\mathbb{Z}} [k/N,(k+f(k/N))/N], $$ but more complicated schemes could also be used. The quality of approximation should probably be measured by a norm of the form $$ \|g\| = \left(\sum_n a_n|c_n(g)|^2 \right)^{1/2}, $$ where $c_n(g)$ is the $n$'th Fourier coefficient, and the $a_n$ are positive constants depending on the application.

Is there any literature on this kind of question?

share|cite|improve this question
I don't see how your norm captures the described approximation behavior. Something like $1/t sup_{x} |\int_{x}^{x+t} f(x) - \chi_A(x) dx|$ for some not too small $t > 0$, seems like the more appropriate choice. (One could metrize this topology, since it corresponds to some weak topology). – Helge Sep 15 '11 at 16:15
@Helge: the stated norm is what we really want to control. The preceding paragraph describes how the engineers propose to attempt to make the norm small, but there may be better ways. – Neil Strickland Sep 15 '11 at 16:22
I am doubtful approximation in that norm is possible, if the a_n are not decaying very quickly to $0$. Consider $f \equiv 1/2$. Then $\hat{f}(0) = 1/2$ and $\hat{f}(k) = 0$ for all other $k$. It is easy to see that $\|\hat{\chi}_{A}\|_2 = |A|^{1/2}$ and that $\hat{\chi}_{A} (0) = |A|$. Hence, with $a_n = 1$, one would have that $\|f-\chi_{A}\|^2 = (1/2 - |A|)^2 + |A|^{1/2}(1 - |A|^{1/2})$. It should be an exercise to obtain a lower bound. It is clearly > 0. – Helge Sep 15 '11 at 17:39
Helge's remark is correct. Since $g=f-\chi$ is not smooth and actually not better than bounded, your norms are all infinite if $a_n$ is not a (say) decreasing sequence. Recall that if $a_n$ is a power of n, your norm is simply a Sobolev norm – Piero D'Ancona Sep 15 '11 at 18:26
For the applications, it is certainly reasonable to assume that the $a_n$ decay rapidly when $n$ is large. I'd be interested in results with whatever assumptions on the $a_n$ seem convenient. – Neil Strickland Sep 15 '11 at 19:12

Your Answer


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

Browse other questions tagged or ask your own question.