If $\mathcal R_j f\in L^1$ then $\widehat{\mathcal R_j f}=-i\frac{\xi_j}{|\xi|}\widehat{f}(\xi)$

Question

For any $f\in L^1(\mathbb{R}^n)$ and $1\le j\le n$, recall that the Riesz transform $\mathcal{R}_jf\in L^{1,\infty}(\mathbb{R}^n)$ is given by $$ \mathcal{R}_jf:=c_n\lim_{\epsilon\to 0}\left(\frac{x_j}{|x|^{n+1}}1_{\mathbb{R}^n\setminus B_\epsilon(0)}(x)\right)*f, $$ for some irrelevant dimensional constant $c_n$. It is well-known that the above limit holds in the $L^{1,\infty}$-topology.

Assume now that, for some $f\in L^1(\mathbb{R}^n)$, we have $\mathcal{R}_j f\in L^1(\mathbb{R}^n)$. Can we conclude that the usual formula $\widehat{\mathcal{R}_j f}=-i\frac{\xi_j}{|\xi|}\widehat{f}(\xi)$ holds for this $f$?

This question is not trivial since it is not clear whether the above limit holds in the $L^1$-topology.

Moreover, given $f\in L^1(\mathbb{R}^n)$, is the following equivalence $$\mathcal{R}_j f\in L^1(\mathbb{R}^n) \ \Longleftrightarrow\ \mathcal{F}^{-1}\left(-i\frac{\xi_j}{|\xi|}\widehat{f}(\xi)\right)\in L^1(\mathbb{R}^n)$$ true? (The second formula is meant in the sense of tempered distributions)

Answer 1

The answer to the first question is yes, but some precisions in the definitions are needed. The operator $\mathcal R_j$ is indeed the convolution with an homogeneous distribution $T_j$ (homogeneous implies tempered) with homogeneity $-n$. The best way to define $T_j$, to avoid the business with $\epsilon$ in your definition is to say that it is the inverse Fourier transform of $\tau_j=\xi_j/\vert\xi\vert$, which is $L^\infty(\mathbb R^n)$, thus a tempered distribution. The homogeneity with degree 0 of $\tau_j$ implies homogeneity of $T_j$ with degree $-n$ (a general fact for the Fourier transform of homogeneous distributions). For $f$ continuous with compact support, you recover your formula and if $f\in L^1(\mathbb R^n)$, $\hat f$ is (uniformly) continuous going to zero at infinity (let us say belongs to $C^0_{(0)}$). As a result, the product $\hat f(\xi)\xi_j/\vert\xi\vert$ belongs to $L^\infty(\mathbb R^n)\subset \mathscr S'(\mathbb R^n)$ and thus the inverse Fourier transform makes sense.

The second question does not make obvious sense since the Fourier multiplier $\xi_j/\vert\xi\vert$ does not belong to the space $\mathscr O_M$, the space of multipliers of $\mathscr S(\mathbb R^n)$: the multiplication by $\xi_j/\vert\xi\vert$ does NOT send $\mathscr S(\mathbb R^n)$ into itself. However I believe that the claimed result would hold true if you assume $\mathcal R_j f$ in $L^1$ and $\hat f$ in $L^\infty$.