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

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Mar 14, 2017 at 13:20	comment	added	Bazin		@Mizar If you consider the two brackets of duality in my comment above, when $f\in L^1$, the second one is a true integral; now using that and the proof that you have for the convergence in the Lorentz space, don't you obtain what you wish?
Mar 13, 2017 at 23:58	comment	added	Mizar		Yes but I am just curious to know whether this agrees with the singular integral definition, i.e. the limit in $L^{1,\infty}$ of the convolution with the truncated kernels.
Mar 13, 2017 at 21:04	comment	added	Bazin		@Mizar Yes because the formula holds true in a weak sense, by the definition of my answer; you have $\mathcal R_jf=\text{Fourier}^{-1}(\tau_j \hat f)$, which means that the bracket of duality $\langle \mathcal R_jf,\phi\rangle$ is equal to $\langle \tau_j\hat f,\text{Fourier}^{-1}\phi\rangle$, for any function $\phi\in \mathscr S(\mathbb R^n)$.
Mar 13, 2017 at 15:24	comment	added	Mizar		Thanks for your reply. I am aware of the fact that you can define the Riesz transform as a tempered distribution. My question basically is: assuming that the singular integral definition produces a function in $L^1$, do the two definition coincide? I slightly edited my question to clarify it.
Mar 13, 2017 at 14:12	history	answered	Bazin	CC BY-SA 3.0