A distribution in $\mathscr{S}^{\prime}\left(\mathbb{R}^{n}\right)$ is called homogeneous of degree $\gamma \in \mathbb{C}$ if for all $\lambda>0$ and for all $\varphi \in \mathscr{S}\left(\mathbb{R}^{n}\right),$ we have $$ \left\langle u, \delta^{\lambda} \varphi\right\rangle=\lambda^{n\gamma}\langle u, \varphi\rangle. $$ where $\delta^{\lambda} \varphi(x)=\varphi(\lambda x)$. Now suppose that $u \in C^{\infty}\left(\mathbb{R}^{n} \backslash\{0\}\right)$ is homogeneous of degree $n+i \tau, \tau \in \mathbb{R} .$ How to prove that the operator given by convolution with $u$ maps $L^{2}\left(\mathbb{R}^{n}\right)$ to $L^{2}\left(\mathbb{R}^{n}\right)$.
2 Answers
Let $u$ be a smooth function on $\mathbb R^n\backslash\{0\}$ homogeneous with degree $\lambda$ (on $\mathbb R^n\backslash\{0\}$). If $\lambda$ is not an integer $\le n$, then $u$ can be uniquely extended to a tempered distribution homogeneous with degree $\lambda$. Moreover, the Fourier transform of an homogeneous distribution with degree $\lambda$ is an homogeneous distribution of degree $\lambdan$.
As a result, the Fourier transform of your $u$ is homogeneous with degree $ni\taun=i\tau$ when $\tau\in \mathbb R^*$, so is in one dimension a linear combination of $\xi_\pm^{i\tau}$ which is thus bounded, proving the sought $L^2$ boundedness.
If $\tau =0$, then $u$ is homogeneous of degree $1$ in one dimension. You need $u$ to be odd for the $L^2$ boundedness to hold. Take for instance (still in one dimension) $u(x)=1/\vert x\vert$, obviously homogeneous with degree $1$ and smooth on $\mathbb R^*$. The singular integral with kernel $1/\vert xy\vert$ is not bounded on $L^2$, but the Hilbert transform with kernel $1/(xy)$ is bounded on $L^2$ (with norm π).

$\begingroup$ Thanks a lot ! I already know how to prove it. $\endgroup$– TauMar 12, 2020 at 14:40
By the Proposition 2.4.8. of L. Grafakos(GTM249), we have $\hat{u}\in C^{\infty}\left(\mathbb{R}^{n} \backslash\{0\}\right)$. It follows from $u$ is homogeneous of degree $n+i\tau$ that $\hat{u}$ is a homogeneous distribution of degree $i\tau$.Then, $$\hat{u}(\xi)=\xi^{i\tau}\hat{u}(\frac{\xi}{\xi}),$$ which implies that $\hat{u}(\xi)=\sup_{x=1}\hat{u}(x)<\infty$, i.e. $\hat{u}\in L^{\infty}(\mathbb{R}^{n})$. Thus, $$\u*f\_{L^{2}}=\\hat{u}\hat{f}\_{L^{2}}\leq \\hat{u}\_{L^{\infty}}\\hat{f}\_{L^{2}}=\\hat{u}\_{L^{\infty}}\f\_{L^{2}}$$ for all $f \in L^{2}(\mathbb{R}^{n})$.

$\begingroup$ The meaning of $\hat u$ is not clear, take for instance $u(x)=1/\vert x\vert$ and see the constraints in my answer above. $\endgroup$– BazinMar 12, 2020 at 23:10

$\begingroup$ " $𝑢$ is homogeneous of degree $n+i\tau$" indicates that $u$ is a tempered distributions, so fourier transform is meaningful. $\endgroup$– TauMar 13, 2020 at 1:57

$\begingroup$ Well, the extension of an homogenous functions on $\mathbb R^n\backslash\{0\}$ to an homogeneous distribution on $\mathbb R^n$ could be impossible for integer indices $\le n$. $\endgroup$– BazinMar 20, 2020 at 17:18