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So, if $x\in \mathbb{R}$, then the fourier coefficients of $|x|$ decay like $k^{-2}$. I had a homework problem that asked us to show that the fourier coefficients of $|x|^b$ decay like $k^{-1-b}$.

My question is about doing this in $\mathbb{R}^n$. My friend and I were talking about it, and I think that if $x\in\mathbb{R}^n$, the higher dimensional fourier series should decay like $k^{-2n}$ and then if $0<a<n$ the fourier coefficients of $|x|^{a}$ should be like $k^{-n^2-a}$.

Does this seem right? Is there a way to show this?

johndavis1187
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2 Answers2

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This question does broach an important idea! Yes, the tempered distribution given by the meromorphic (distribution-valued) continuation of $|x|^{-s}$ on $\mathbb R^n$ is a (determinable... apply it to a Gaussian and use properties of the Gamma function) constant multiple of $|x|^{-(n-s)}$. This is provable in the range where $|x|^{-s}$ is locally integrable essentially by the definition, and then ... by whatever semi-magical invocations one prefers... "by analytic continuation" for all values of $s\in\mathbb C$.

(My own choice of "semi-magic" here would use the Snake Lemma to prove existence and uniqueness of an extension to the whole space of Schwartz functions of the distribution $|x|^{-s}$ on the subspace of Schwartz functions vanishing to infinite order at $0$...)

paul garrett
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  • Thanks! I think I get it, but I am still slightly confused; I think it is because I don't know enough facts. So, what you are saying is that the fourier transform of $|x|^{n-a}$ is a tempered distribution and if $\phi$ vanishes with infinite order at $0$, then the action of $\mathcal{F}|x|^{n-a}$ is given by the integral $\int |x|^{-n-a} \phi$. Then this implies that $\mathcal{F}|x|^{n-a}$ is actually the function $|x|^{-n-a}$ because the scwartz functions that vanish at $0$ are "dense" (in an appropriate meaning of the word) in $\mathcal{S}$? Its the extension that I am now confused about. – johndavis1187 Sep 10 '15 at 01:30
  • No, the extesion argument cannot be by density, because the vanishing conditions are $0$ are closed conditions. Rather, characterize $|x|^{-s}$ as rotation-invariant eigenfunctions for $T=x{d\over dx}$ in the one-dimensional case, or for a suitable analogous operator $T$ in general. The short exact sequence $0\to\mathcal S_o\to \mathcal S \to \hbox{germs at 0}\to 0$ dualizes to a short exact sequence (using Hahn-Banach), then a short exact sequence of complexes of the form $0\to A\to A\to 0$ mapping by $T$, and then by the snake lemma to ... – paul garrett Sep 10 '15 at 13:06
  • ... a six-term long exact sequence. The classification of rotation-invariant distributions at $0$ (via Taylor-Maclaurin) for generic $s$ makes terms vanish, giving existence and uniqueness of the extension. This is not the most elementary way to do this, but it does simply make use of standard things... Questions about extensions really are substantially homological, I think. – paul garrett Sep 10 '15 at 13:08
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I'm new here, so my apologies if I am doing something wrong since this isn't a full answer, but I can't leave a comment.

I've been thinking about something similar. First, let $k\in\mathbb{Z}^n$ and assume that $|k_1|=||k||_{\infty}$ (the other cases are similar.) Then $|k_1|\simeq ||k||_{1}$. So, integrate by parts $n$ times in the $x_1$ variable. This is why this isn't a proper answer: I don't know of a closed form for this $n$ fold derivative. But, it should be something like $|x|^{-a}$ (or maybe something slightly different, by the exponent should be something that is integrable locally.) Now, you essentially want to compute the fourier coefficient of $|x|^{-a}$. Now you integrate by parts $k$ more times until $k+1+a >n$ but $k+a<n$. Now you want to do a"fractional" integration by parts to get the rest of the decay. (You can see an example of that here: https://mathoverflow.net/questions/129830/integration-by-parts-for-the-fractional-laplacian).

Again, I know there are some gaps above, but this is the right idea and I don't have any reputation so I can't just leave this in the comments.

edit: change $\infty$ to $1$. edit: I changed the wrong one.

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