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https://www.sciencedirect.com/science/article/pii/S0079816908602779
This chapter discusses the Fourier transforms of distributions with compact support and Paley-Wiener theorem. This chapter considers a continuous function f with compact support in R n . The chapter mentions that the Fourier transform of a continuous function with compact support can be extended to the complex space C n , as an entire analytic function of exponential type.
https://en.wikipedia.org/wiki/Paley%E2%80%93Wiener_theorem
Generally, the Fourier transform can be defined for any tempered distribution; moreover, any distribution of compact support v is a tempered distribution. If v is a distribution of compact support and f is an infinitely differentiable function, the expression = (↦ ()) is well defined.
https://mathoverflow.net/questions/29991/fourier-transforms-of-compactly-supported-functions
Then $\hat{f}$ should remain unchanged when convolved with the Fourier transform of $\chi$, but since $\hat{f}$ lives on the reals you would like to think that this convolution would partially fill up some open set connected to the boundary of the original support of $\hat{f}$ and thus enlarge the support of $\hat{f}$.
https://www3.canisius.edu/~kahngb/research/papqfourier.pdf
Using the Haar measure on G, it is then possible to define the Fourier transform of a continuous function having compact support f ∈ C c (G) (or, even a Schwartz function),obtaining fˆ ∈ C 0 (Gˆ).
https://math.stackexchange.com/questions/154454/a-function-and-its-fourier-transform-cannot-both-be-compactly-supported
A funtion and its fourier transformation cannot both be compactly supported unless f=0 1 If a function is compactly supported, then its Fourier series converge?
https://www.researchgate.net/post/Can_we_say_that_if_Fourier_transform_of_f_has_compact_support_the_Fourier_transform_of_Tf_where_T_is_any_bounded_linear_operator_on_Hilbert_space
The Fourier transform maps convolution to the usual product, so the Fourier transform of T u (f) will be the product of Fourier transforms of f and u, so the compactness of the support remains....
http://math.uchicago.edu/~may/REU2013/REUPapers/Hill.pdf
the basic properties of the Fourier transform and show that a function and its Fourier transform cannot both have compact support. From there we prove the Fourier inversion theorem and use this to prove the classical uncertainty principle which shows that the spread of a function and its Fourier transform are inversely proportional. Finally, we
https://en.wikipedia.org/wiki/Pontryagin_duality
In particular, the Fourier transform is an isometry from the complex-valued continuous functions of compact support on G to the -functions on ^ (using the -norm with respect to μ for functions on G and the -norm with respect to ν for functions on ^).
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