Logarithmic potential Given a continuous, compactly supported function $f$ on $R^2$, it is known that the logarithmic potential of $f$, that is
$$
U_{f}(x):=-\frac{1}{2\pi}\int\log|x-y|f(y)dy
$$
has the following decay at infinity
$$
U_{f}(x)=-\frac{M}{2\pi}\log|x|+O(1/|x|)\quad as\ |x|\rightarrow\infty.
$$
My question is rather simple: can this result be extended when we consider $f\in L^{1}\cap L^{\infty}$.
Thank you very much for any suggestion you can give me.
 A: No, I don't think so. Consider the function 
$$f(x) = \frac{1}{(|x|^2 +1)(\log (|x|+2))^2}$$
Then this function is clearly in $L^\infty(\mathbb{R}^2)$. That this is in $L^1(\mathbb{R}^2)$ follows if you consider its integration in polar coordinates, and recall that $$\sum_{n \geq 2} \frac{1}{n (\log n)^2}$$ is a convergent sum. However, the integral defined by the potential is a divergent integral as $|y| \rightarrow \infty$, and is more comparable to the sum $$\sum \frac{1}{n \log n}.$$
A: For the behavior of the logarithmic potential
$$U_{\mu}(x)=-\int\log{|x-y|}d\mu(y),$$
at infinity, where $\mu$ is a positive measure of finite mass, there are two cases :
1) the measure $\mu$ does not integrate the logarithm at infinity, that is 
$$\int\log(1+|y|)d\mu(y)=\infty.$$
This is in some sense an extreme case since, then, the potential $U_{\mu}(x)$ is just the constant function equal to $-\infty$, everywhere on $\mathbb{C}$. The example given in the answer by Ray Yang belongs to this case.
2)  the measure $\mu$ does integrate the logarithm at infinity, that is 
$$\int\log(1+|y|)d\mu(y)<\infty.$$
In this case, the behavior at infinity, that holds for compactly supported measures, remains almost true in the following sense :
$$
\lim_{|x|\to\infty,~x\in\mathbb{C}\setminus E}U_{\mu}(x)+\mu(\mathbb{C})\log|x|=0,
$$
where $E$ is a set thin at infinity, which means that its inverse $E^{*}$ (e.g. by the map $T:z\to1/z$) is thin at 0, where a set is thin at 0 if 0 is a regular point of its complement for the Dirichlet problem. For instance, the real line is not thin at infinity, as one may show that for a set $E$ which is thin at a point, there are arbitrarily small circles centered at that point which do not intersect $E$.
By inversion, the previous limit becomes
$$
\lim_{|x|\to0,~x\in\mathbb{C}\setminus E^{*}}U_{T_{*}\mu}(x)=U_{T_{*}\mu}(0),
$$
where $T_{*}\mu$ is the image measure of $\mu$ by $T$.
The above limit says that the potential $U_{T_{*}\mu}$ is continuous at 0 with respect to the fine topology, where a set is a neighborhood of a point $z$ if and only if its complement is thin at $z$.
