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In this answer @icurays1 states:

In this same vein, Fourier analysis leads to an extremely powerful theory of smoothness, because of the correspondence between differentiability and decay of the Fourier coefficients.

Similarly, in this section of a presentation dealing with the decomposition of functions using spherical harmonics, the presenter states, "[...] spectral smoothness corresponds to spatial decay[...]"

My question is two-fold. First, does this correspondence go both ways (i.e. does spatial decay imply spectral smoothness)? Second, if the converse holds, what is the minimal condition on a function on $\mathbb{R}^n$ for the Fourier transform to exist and be continuous everywhere in Fourier space (differentiability is not needed for my purposes)? For example, is the minimal condition that the function has a finite $L^2$ norm? A finite second moment?

Sean Lake
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    Well, the Riemann-Lebesgue lemma asserts that $L^1$ functions have continuous Fourier transforms going to $0$ at infinity. Since (Plancherel-extended) FT is an isomorphism of $L^2$ to itself, $L^2$ does not imply continuity of the FT. At a somewhat subtler level, there are Sobolev spaces, Sobolev imbedding, and so on. What kind of thing were you wanting? – paul garrett Jan 27 '23 at 18:32
  • @paulgarrett Honestly, I'm not sure. I have an application that requires the $n$-dimensional Fourier transform of a function to be continuous, so I wanted to know what condition I need to impose on the function to guarantee that. $L^1$ might be sufficient, since I'm imposing that for other reasons, anyway. – Sean Lake Jan 27 '23 at 21:10
  • The reason I mentioned $L^2$ is because that's a natural additional thing to impose in the context I'm working in. – Sean Lake Jan 27 '23 at 21:12
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    Well, for $L^2$-type conditions: for example, if $f$ itself is in various weighted $L^2$ (or other) spaces, then its Fourier transform is (by definition...) in a corresponding Sobolev space... and Sobolev imbedding theorems give (for example) the continuity of that function. Would this sort of thing work for you? – paul garrett Jan 27 '23 at 23:00
  • I think it would. If you could make it an answer with links to good reading on the Sobolev embedding theorem, I'll mark it as the accepted answer. – Sean Lake Jan 27 '23 at 23:01
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    Ok! :) I'll write something tomorrow... :) – paul garrett Jan 27 '23 at 23:03

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Apart from the Riemann-Lebesgue lemma, which asserts that $f\in L^1(\mathbb R^n$ implies that $\widehat{f}$ is continuous (and goes to $0$ at infinity), there are also $L^2$-style conditions coming from Sobolev theory.

One basic "Sobolev imbedding theorem" is that for $f$ on $\mathbb R^n$ with $|\widehat{f}|^2\cdot (1+|x|^2)^{\frac{n}{2}+\varepsilon}$ in $L^2(\mathbb R^n)$, $f$ is continuous. Reversing the roles gives an $L^2$-style condition for continuity of $\widehat{f}$.

Such basic Sobolev stuff occurs in many places: Folland's TATA notes on PDE, also Folland's fuller PDE book, Evans' PDE book, Taylor's PDE book(s) I, ... and probably Google-able. My own course notes in both Real Analysis and Functional Analysis do several of these basic things, as I'm sure many others' course notes do, as well:

http://www.math.umn.edu/~garrett/m/real/ and http://www.math.umn.edu/~garrett/m/fun/

Sean Lake
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paul garrett
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