Riesz transform does not preserve continuity I've read somewhere that the Riesz operator $R_j$ defined by $$R_j f(t) := c(n) \, \text{pv} \int_{\mathbb{R}^n} \frac{x_j}{|x|^{n+1}} f(t-x) \, dx$$doesn't preserve the continuity, but I can't produce a counterexample (actually not even when $n=1$). My idea was to take a continuous function $f$ (maybe with compact support) such that its transformed explodes at $0$, but I couldn't find such $f$. Do you have any hint on how to construct such function?
Thanks in advance!
 A: Let $g:[0,\infty) \to (-\infty,0]$ be continuous and compactly supported, with $g(0) = 0$ and $g(x) = 1/\ln x, 0< x < 1/2.$ On $\mathbb R^n,$ define
$$f(x) = (x_1/|x|)g(|x|), x \ne 0, \,f(0) = 0.$$
Then $f$ is continuous with compact support on $\mathbb R^n.$ Let's get at $R_1f(0):$
$$\int_{\epsilon<|x|<\infty} \frac{x_1}{|x|^{n+1}}f(-x)\,dx =  \int_{\epsilon<|x|<\infty} \frac{-x_1^2}{|x|^{n+2}}g(x)\,dx \ge \int_{\epsilon<|x|<1/2} \frac{-x_1^2}{|x|^{n+2}\ln |x|}\,dx. $$
Let' reduce the domain of integration to $U_\epsilon = \{\epsilon <|x|<1/2\} \cap \{x_1>|x|/2\}.$ The last integral above is then greater than
$$\tag 1\int_{U_\epsilon}(1/4) \frac{-1}{|x|^{n}\ln |x|}\,dx.$$
Note that $U_\epsilon$ can be written as the set of points $r\zeta,$ where $0<r<1/2$ and $\zeta \in \kappa,$ where $\kappa$ is an open "spherical cap" on the unit sphere centered at $(1,0,\dots,0).$ Using the polar coordinates formula* in $\mathbb R^n,$ we see the integral in $(1)$ equals
$$\tag 2 \sigma(\kappa)\int_\epsilon^{1/2} \frac{dr}{r\ln r},$$
where $\sigma$ is $(n-1)$-dimensional area measure on the unit sphere. The limit of $(2)$ is $\infty$ as $\epsilon \to 0^+$ and thus we've shown $R_1f(0) = \infty.$
*This formula can be found in Rudin's Real and Complex Analysis.
