3
$\begingroup$

Suppose that $K=[-N, N]$ is some compact subset of $\mathbb R$, for some $N>2.$

Can we expect to choose $h$ such that $h=1$ on $K$ and the support of the Fourier transform of $\widehat{h}$ contained in $[-2/N, 2/N]$ and $\int_{\mathbb R} h(x) dx \leq N$? if so how?

My attempts: Take $\widehat{f}= \chi_{[-M, M]}^{\vee}g$ where the support of $g$ lies in $[-2/N, 2/N]$. Then by taking the inverse Fourier transform we get $f= \chi_{[-M, M]}\ast g^{\vee}$. Now how to choose $g$ and $M$ so that $f=1$ on $[-N, N]$? (My approach could be wrong as well?)

$\endgroup$

1 Answer 1

8
$\begingroup$

No. It is already impossible for $h$ to be "band-limited" (i.e. with $\widehat h$ of compact support) and constant on an interval. Indeed by the Fourier inversion formula a band-limited function is analytic, so if $h=1$ on an interval then $h=1$ on all of ${\mathbf R}$, whence $h$ does not have a Fourier transform at all (and at any rate can't have $\int_{\mathbf R} h < \infty$ even if you allow $\widehat 1$ to be a "delta function").

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .