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Can someone help me the question below? Is there a positive-valued compactly supported function $f$ such that the Fourier transform ${{f}^{\operatorname{ft}}}\left( t \right)=\int_{-\infty }^{\infty }{f\left( x \right){{e}^{itx}}dx}, t\in\mathbb{R}$, and derivative of the Fourier transform, $\frac{d}{dt}{{f}^{\operatorname{ft}}}\left( t \right)$, decay with exponential rates at infinity ? Thank so much for helping.

Cao
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1 Answers1

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There is a close relationship between smoothness of a function and the decay rates of its Fourier transform (and vice versa). Roughly speaking:

Some number of derivatives of $f$ existing implies some power law decay of the Fourier transform; an extra derivative speeds up the decay by a power.

Smoothness of $f$ (infinitely differentiable) gives decay of $\hat{f}$ faster than any power.

Extensibility of $f$ to an analytic function in a strip $|Im(z)|<a$ gives exponential ($e^{-at}$) decay.

Extension to an entire function of exponential type gives a Fourier transform of compact support.

See Paley-Wiener theorem for more precise statements.

In your case, you won't be able to find a compactly supported function with exponentially decaying Fourier transform, because that would require analyticity in some strip, which is clearly impossible.

Holographer
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