# A question about homogeneous distribution

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 $$L^{2}\left(\mathbb{R}^{n}\right)$$.

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 $$-\lambda-n$$.

As a result, the Fourier transform of your $$u$$ is homogeneous with degree $$n-i\tau-n=-i\tau$$ 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 $$1/\vert x-y\vert$$ is not bounded on $$L^2$$, but the Hilbert transform with kernel $$1/(x-y)$$ is bounded on $$L^2$$ (with norm π).

• Thanks a lot ! I already know how to prove it.
– Tau
Mar 12 '20 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 $$f \in L^{2}(\mathbb{R}^{n})$$.

• 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. Mar 12 '20 at 23:10
• " $𝑢$ is homogeneous of degree $-n+i\tau$" indicates that $u$ is a tempered distributions, so fourier transform is meaningful.
– Tau
Mar 13 '20 at 1:57
• 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 $\le -n$. Mar 20 '20 at 17:18