Fourier transforms of homogeneous functions [closed]

Question

Compute Fourier transforms of homogeneous functions of the form, $$ \frac{1}{|x|^{n+d}}P_d(x) $$ where $P_d$ is a homogenous harmonic polynomial of degree $d$ in $n+1$ variables.

Answer 1

Your function is, with $P_d$ homogeneous harmonic polynomial of degree $d$ in $n$ variables, $$ u(x)=\frac{P_d(x)}{\vert x\vert^{n+d}}. \tag{1} $$ This is an homogeneous distribution of degree $-n$ on $\mathbb R^n\backslash\{0\}$. The first thing to do is to extend that distribution to a distribution on $\mathbb R^n$. Let us first check for a test function $\phi\in C_c^\infty(\mathbb R^n\backslash\{0\})$, the absolutely converging integral \begin{multline} \langle T, \phi\rangle= \int_{\mathbb R^n}\frac{P_d(x)}{\vert x\vert^{n+d}}\phi(x) dx =\int_0^{+\infty} r^{n-1}\int_{\mathbb S^{n-1}} \frac{P_d(r\omega)}{r^{n+d}}\phi(r\omega)d\sigma(\omega) dr \\= \int_{0}^{+\infty}\int_{\mathbb S^{n-1}} {P_d(\omega)}\phi(r\omega)d\sigma(\omega)\frac{dr}r \end{multline}

$\mathbf {\text{Claim:}}$ Let $\psi$ be a function in the Schwartz class of $\mathbb R$ and let $H=\mathbf 1_{\mathbb R^+}$. We have $$\int_0^{1}\frac{\psi(r)-\psi(0)}{r} dr +\int_1^{+\infty}\frac{\psi(r)}{r} dr = \langle \frac{d}{dr}\bigl( H(r) \ln r \bigr),\psi(r)\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) }. $$ Indeed we have, \begin{align} \langle \frac{d}{dr}\bigl( H(r) \ln r \bigr),\psi(r)\rangle_{\mathscr S'(\mathbb R),\mathscr S(\mathbb R) } &=-\int_0^{+\infty}\psi'(r) \ln r\ dr =\lim_{\epsilon \rightarrow 0_+} -\int_\epsilon^{+\infty} \psi'(r) \ln r\ dr, \\ &\text{and}\hskip25pt \\ -\int_\epsilon^{+\infty} \psi'(r) \ln r\ dr&=\bigl[\psi(r) \ln r\bigr]^{\epsilon}_{+\infty} +\int_{\epsilon}^{+\infty}\psi(r) r^{-1} dr \\&=\psi(\epsilon) \ln \epsilon +\int_{\epsilon}^{+\infty} \psi(r) r^{-1} dr \\ &= \int_{1}^{+\infty}\psi(r) r^{-1} dr +\int_{\epsilon}^{1}\psi(r) r^{-1} dr -\int_{\epsilon}^{1}\psi(\epsilon)r^{-1} dr \\ &= \int_{1}^{+\infty}\psi(r) r^{-1} dr +\int_{\epsilon}^{1} \bigl[\psi(r) -\psi(\epsilon)\bigr]r^{-1} dr, \end{align} proving the claim.$\ \mathbf{\square}$

We may thus extend $T$ as a temperate distribution $\tilde T$ on $\mathbb R^n$ with the definition for $\phi\in\mathscr S(\mathbb R^n )$, $$ \langle \tilde T, \phi\rangle_{\mathscr S'(\mathbb R^n),\mathscr S(\mathbb R^n) } = \int_{\mathbb S^{n-1}} {P_d(\omega)} \langle \frac{d}{dr}\bigl( H(r) \ln r \bigr),\phi(r\omega)\rangle_{\mathscr S'(\mathbb R_r),\mathscr S(\mathbb R_r) } d\sigma(\omega). \tag{2}$$ However the distribution $\tilde T$ fails to be homogeneous on $\mathbb R^n$, as it is seen when you calculate $ \langle \tilde T(\lambda x), \phi(x)\rangle. $ Note that an homogeneous distribution of degree $\mu$ is temperate and its Fourier transform is also homogeneous with degree $-\mu-n$. That does not mean that you cannot calculate the Fourier transform of $\tilde T$. In particular Theorem 7.1.18 in the ALPDO of Lars Hörmander (Grundlehren 256) shows that the Fourier transform of ${\tilde T}$ is smooth outside of the origin. It is also possible to calculate directly $$ \langle \widehat{\tilde T}, \phi\rangle=\langle \tilde T, \hat \phi\rangle, $$ when $\hat \phi(0)=\int \phi(x) dx =0$, but in general you will need to use the Fourier transform of the one-dimensional distribution $\frac{d}{dr}\bigl( H(r) \ln r \bigr)$.