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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, Bruno

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1 Answer 1

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 $$\sum \frac{1}{n \log n}.$$

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