If $\mu$ has no atoms, then $\frac{x^2}{x^2+y^2} *\mu$ is continuous

Question

Let $\mu$ be a probability measure on $\mathbb{R}^2$, having no atomic points.

How to prove that $\frac{x^2}{x^2+y^2} *\mu$ is a continuous function on $\mathbb{R}^2$?

Here $f * \mu(z)=\int_{\mathbb{R}^2} f(z-w)d\mu(w)$, where $z=(x,y)$.

Answer 1

As your function $f(x,y):=\frac{x^2}{x^2+y^2}$ is continuous and bounded belongs to $L_1(\mu)$, and so there is some $g\in C_c(\mathbb{R}^2)$ such that $\|f-g\|_1 <\epsilon $ (as $C_c$ is dense in any $L_p$ space of any Euclidean space equiped with a Radon measure). Then its easy to see that $$ |f*\mu(z_1)-f*\mu(z_0)|\leqslant 2\epsilon +\int_{\mathbb{R}^2}|g(z_1-w)-g(z_0-w)|\mathop{}\!d \mu(w) $$ As the support of $g$ is compact then finally from the definition of compactness we can see that $g(z_1-\cdot )\to g(z_0-\cdot )$ uniformly as $z_1\to z_0$, so we conclude that $f*\mu$ is continuous. (Note that we don't need that $\mu$ will be atomless.)

Answer 2

I now think it is rather easy, actually:

Set $f(x,y)=\frac{x^2}{x^2+y^2}$. Suppose that $z_n \to z_0$. Then the functions $$ f_n(w):=f(z_n-w) \to f_0(w):=f(z_0-w), $$ for every $w$ except $w=z_0$.

Thus $f_n $ converges pointwise on $\mathbb{R}^2 \setminus \{z_0\}$ to $f_0$.

Since $\mu$ have no atomic points, $\mu(\{z_0\})=0$, so $f_n \to f_0$ pointwise $\mu$-a.e on $\mathbb{R}^2$.

Since $f \le 1$ and $\mu$ is a probability measure, the Lebesgue dominated convergence theorem implies that $$ \int f_n(w)d\mu(w) \to \int f_0d\mu(w), $$ which implies the claim.

The same result would be true for any bounded function, which has a finite number of discontinuities.