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Sep 15, 2011 at 19:12 comment added Neil Strickland 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.
Sep 15, 2011 at 18:26 comment added Piero D'Ancona 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
Sep 15, 2011 at 17:39 comment added Helge 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.
Sep 15, 2011 at 16:22 comment added Neil Strickland @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.
Sep 15, 2011 at 16:15 comment added Helge 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).
Sep 15, 2011 at 15:05 history asked Neil Strickland CC BY-SA 3.0