22\. Since \(F\) commutes with rotations, the Fourier transform of a radial
function is radial; that is, if \(F \in L^{1}\left(\mathbb{R}^{n}\right)\) and
\(F(x)=f(|x|)\), then \(\widehat{F}(\xi)=g(|\xi|)\), where \(f\) and \(g\) are related
as follows.
a. Let \(J(\xi)=\int_{S} e^{i x \xi} d \sigma(x)\) where \(\sigma\) is surface
measure on the unit sphere \(S\) in \(\mathbb{R}^{n}\) (Theorem 2.49). Then \(J\) is
radial - say, \(J(\xi)=j(|\xi|)\) - and \(g(\rho)=\int_{0}^{\infty} j(2 \pi r
\rho) f(r) r^{n-1} d r .\)
b. \(J\) satisfies \(\sum_{1}^{n} \partial_{k}^{2} J+J=0\).
c. \(j\) satisfies \(\rho j^{\prime \prime}(\rho)+(n-1) j^{\prime}(\rho)+\rho
j(\rho)=0\). (This equation is a variant of Bessel's equation. The function \(j\)
is completely determined by the fact that it is a solution of this equation,
is smooth at \(\rho=0\), and satisfies \(j(0)=\sigma(S)=\) \(2 \pi^{n / 2} /
\Gamma(n / 2)\). In fact, \(j(\rho)=(2 \pi)^{n / 2} \rho^{(2-n) / 2} J_{(n-2) /
2}(\rho)\) where \(J_{\alpha}\) is the Bessel function of the first kind of order
\(\alpha .)\)
d. If \(n=3, j(\rho)=4 \pi \rho^{-1} \sin \rho\). (Set \(f(\rho)=\rho j(\rho)\)
and use (c) to show that \(f^{\prime \prime}+f=0\). Alternatively, use spherical
coordinates to compute the integral defining \(J(0,0, \rho)\) directly.)
(use induction on \(k\) ), and in particular,
$$
h_{k}(x)=\frac{(-1)^{k}}{\left[\pi^{1 / 2} 2^{k} k !\right]^{1 / 2}} e^{x^{2}
/ 2}\left(\frac{d}{d x}\right)^{k} e^{-x^{2}} .
$$
f. Let \(H_{k}(x)=e^{x^{2} / 2} h_{k}(x)\). Then \(H_{k}\) is a polynomial of
degree \(k\), called the \(k\) th normalized Hermite polynomial. The linear span
of \(H_{0}, \ldots, H_{m}\) is the set of all polynomials of degree \(\leq m\).
(The kth Hermite polynomial as usually defined is \(\left[\pi^{1 / 2} 2^{k} k
!\right]^{1 / 2} H_{k}\).)
g. \(\left\\{h_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for
\(L^{2}(\mathbb{R})\). (Suppose \(f \perp h_{k}\) for all \(k\), and let \(g(x)=f(x)
e^{-x^{2} / 2}\). Show that \(\widehat{g}=0\) by expanding \(e^{-2 \pi i \xi \cdot
x}\) in its Maclaurin series and using (f).)
h. Define \(A: L^{2} \rightarrow L^{2}\) by \(A f(x)=(2 \pi)^{1 / 4} f(x \sqrt{2
\pi})\), and define \(\tilde{f}=\) \(A^{-1} \mathcal{F} A f\) for \(f \in L^{2}\).
Then \(A\) is unitary and \(\widetilde{f}(\xi)=(2 \pi)^{-1 / 2} \int f(x) e^{-i
\xi x} d x\). Moreover, \(\widetilde{T f}=-i T(\widetilde{f})\) for \(f \in
\mathcal{S}\), and \(\widetilde{h}_{0}=h_{0}\); hence \(\widetilde{h}_{k}=(-i)^{k}
h_{k}\). Therefore, if \(\phi_{k}=A h_{k}
.\left\\{\phi_{k}\right\\}_{0}^{\infty}\) is an orthonormal basis for \(L^{2}\)
consisting of eigenfunctions for \(\mathcal{F}\); namely,
\(\widehat{\phi}_{k}=(-i)^{k} \phi_{k}\).
