Skip to main content
5 events
when toggle format what by license comment
May 15, 2018 at 8:28 comment added Robert Israel I was considering $\widehat{f}(x) = \sum_{k=-\infty}^\infty f(k) e^{ikx}$ which extends from $[0,2\pi)$ to an entire function on $\mathbb C$. Of course the two points of view are equivalent.
May 14, 2018 at 20:49 comment added Jochen Glueck @RobertIsrael: Good point! I can't resist, though, to point out the following two things: (i) The dual group of $\mathbb{Z}$ is the complex unit circle $\mathbb{T}$, so I guess you'd rather consider the set $\{z \in \mathbb{T}: \hat f(z) = t\}$; (ii) The Fourier transform $\hat f$ is a rational function which might have a pole at $0$. (Or did I missunderstand you and you would like to consider the function $\hat f(x) := \sum_{k=-\infty}^\infty f(k) e^{ikx}$ rather than $\hat f(z) := \sum_{k=-\infty}^\infty f(k)z^k$)?
May 14, 2018 at 20:14 comment added Robert Israel ... and this is essentially the only example on $\mathbb Z$. That is, if $f$ has finite support in $\mathbb Z$, $\widehat{f}$ extends to an entire function, so if $\{x \in \mathbb R:\; \widehat{f}(x) = t\}$ has positive measure it must be all of $\mathbb R$, and $T_f$ is a constant multiple of the identity.
May 14, 2018 at 18:16 history edited Jochen Glueck CC BY-SA 4.0
added 120 characters in body
May 14, 2018 at 17:23 history answered Jochen Glueck CC BY-SA 4.0