My one-line question would be, what is the Riesz transform of the constant function, identically equal to 1 on $\mathbb{R}^2$?
But more fundamentally, my question stems from some confusion about the definition of the Riesz transform. On $\mathbb{R}^2$, the Riesz transform is given by
$$R_jf(x)=cP.V.\int_{\mathbb{R}^2}\frac{x_j-y_j}{|x-y|^3}f(y)\,dy$$ for some constant $c$, and most sources I've seen (including many textbooks and wikipedia) defines the principal value by the following $$P.V.\int_{\mathbb{R}^2}\frac{x_j-y_j}{|x-y|^3}f(y)\,dy=\lim_{\epsilon\to 0^+}\int_{|x-y|\ge \epsilon}\frac{x_j-y_j}{|x-y|^3}f(y)\,dy.$$ Now if we take this definition, then it's not clear that the Riesz transform of a constant function should be defined since there's not enough decay at infinity. On the other hand, I've seen in some sources (particularly in defining the Hilbert transform) the following "alternative" definition of the principal value $$P.V.\int_{\mathbb{R}^2}\frac{x_j-y_j}{|x-y|^3}f(y)\,dy=\lim_{\epsilon\to 0^+}\int_{\epsilon^{-1}\ge |x-y|\ge \epsilon}\frac{x_j-y_j}{|x-y|^3}f(y)\,dy.$$ Now if we do define the principal value this way, then the Riesz transform of a constant function would simply be 0 because of the cancellation property of the kernel, i.e. the kernel integrated over any circle centered at $x$ vanishes.
So, how is the Riesz transform "truly" defined (as a singular integral; not fourier transforms)?
