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Consider a function $f \in L^2(\mathbb{R}^n)$ such that $f$ is radial. My question is, is the Fourier transform $\hat{f}(\xi)$ automatically radial (I can see it is even in each variable $x_i$), or we need some conditions on $f$? Thanks for your help.

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    $\begingroup$ I think so and you might be able to prove it by showing it is invariant under the action of a special orthogonal matrix. It should just be a change of variables at that point. $\endgroup$ Jun 3 '15 at 17:59
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Preliminaries: i) $f$ is radial iff $f\circ T = f$ for every orthogonal transformation $T$ on ${R}^n.$ ii) An orthogonal transformation $T$ preserves the inner product: $\langle Tx,Ty \rangle = \langle x,y \rangle$ for all $x,y \in \mathbb {R}^n.$ iii) If $T$ is orthogonal, then $|\det J_T|=1,$ where $J_T$ is the Jacobian matrix of $T.$

So suppose $f\in L^1(\mathbb {R}^n)$ and $f$ is radial. Fix an orthogonal transformation $T.$ Then

$$\hat {f} (Tx) = \int_{\mathbb {R}^n} f(t) e^{-i\langle Tx,t \rangle}\,dt = \int_{\mathbb {R}^n} f(Ts) e^{-i\langle Tx,Ts \rangle}\,ds= \int_{\mathbb {R}^n} f(s) e^{-i\langle x,s \rangle}\,ds = \hat f (x).$$

Thus $\hat f$ is radial as desired.

That was for $L^1,$ but the question was about $L^2$ as @AdamHughes reminded me. See the comments below to get the result for $L^2.$

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  • $\begingroup$ In theory you want to say that this is true for the more basic functions in $L^1\cap L^2$, since the Fourier integral doesn't define the transform on all of $L^2$ and extend by density. Alternatively use the improper integral formula. $\endgroup$ Jun 5 '15 at 2:36
  • $\begingroup$ @zhw how you deduce that the $L^2$ Fourier transform of an $L^2$ radial function function,is radial ? According to this answer ,can it be done by approximation? $\endgroup$ Aug 8 '19 at 13:43
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    $\begingroup$ @MariosGretsas Yes, if $f\in L^2$ then $f_R = f\cdot \chi_{|x|\le R}\in L^1.$ Apply the $L^1$ result to $f_R$ and then take the limit as $R\to \infty.$ (The last limit exists in $L^2$.) $\endgroup$
    – zhw.
    Aug 8 '19 at 16:15
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    $\begingroup$ @AdamHughes Sorry, I didn't see your comment until now, four years later. You were quite right. $\endgroup$
    – zhw.
    Aug 8 '19 at 16:16
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It is: you can decompose the integral as follows: write $x= r n$, where $n$ is a unit vector, and then $$ \int_{\mathbb{R}^n} e^{i\xi \cdot x} f(\lvert x \rvert) d^n x = \int_{r=0}^{\infty} f(r) \left( \int_{|n|=1} e^{i(r \xi) \cdot n} \, dn \right) r^{n-1} \, dr, $$ by considering the volume as composed of spherical shells.

Now, the inner integral is in fact only a function of $r\xi$. You can take my word for it, or evaluate it: setting $\rho=\lvert \xi \rvert$, then we can write $\xi \cdot x = r\rho \cos{\theta} $, where $0 \leqslant \theta \leqslant \pi$, and then integrate over the remaining angles on the unit sphere. This gives $$ S_{n-2} \int_0^{\pi} e^{i r \rho \cos{\theta}} \sin^{n-2}{\theta} d\theta, $$ where $S_{n-2}$ is the area of the $(n-2)$-dimensional sphere. The remaining integral can be done in terms of Bessel functions; I have a note on it here (last section), although it's not very detailed.

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It indeed will be radial. It is called a Hankel transform:

enter image description here

enter image description here

Where $J_{\nu}$ is a bessel function of the first kind such that $\nu>-1/2$. For it to match fourier transform you use $J_0$ in 2 dimensional radial functions. For three dimensional radial functions, you use spherical bessel functions instead. https://en.wikipedia.org/wiki/Hankel_transform

There is a nice table on wiki. It is quite difficult to do this transform.

If you are interested in how fourier transform connects to Hankel transform, the 4th part derives it (quite short): http://math.arizona.edu/~faris/methodsweb/hankel.pdf

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