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Let $G_n$ be the following distribuitions for $n\geq3$ (for $n=2$ it is just a function) in $\mathbb{R^n}$ (the fundamental solutions of the Laplace equation in $\mathbb{R^n}$ ):

$$G_n=\left\{\begin{matrix} \frac{1}{2 \pi} \cdot \log \left \| x \right \|, \; \; \; n=2 \\ \frac{\left \| x \right \|^{2-n}}{(2-n) \sigma_{n-1}}, \; \; \; n \geq 3 \end{matrix}\right.$$

Where $\sigma_{n-1}$ is the surface area of the unit radius $n$-sphere.

If f $\in L^1(\mathbb{R^n})$ for $n\geq 3$ and $g(y)=f(y) \cdot \log(\left \| y \right \|) \in L^1(\mathbb{R^2})$ for $n=2$ can I get any hint as to how to show that $U=G_n \ast f$ is locally integrable in $\mathbb{R^n}$ for any $n \geq 2$?

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At first, you should clarify that $G_n(x) = C_n ||x||^{2-n}$ is the fundamental solution of the Laplace equation. Then $G_n\ast f$ is a continuous function of $x$, so definitely locally integrable.

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Thanks for the reply. I didn't clarify at first that the $G_n$ are the fundamental solutions of the Laplace equation because i didn't think it would matter. I don't see, however, why $G_n \ast f$ should be continuous as I can't even be sure that it makes sense, seeing as f is just in $L^1(\mathbb{R}^n)$ – grizzlyjoker Jun 3 '13 at 12:01
Or $g(y)=f(y) \cdot \log(\left \| y \right \|) \in L^1(\mathbb{R^2})$ for n=2 – grizzlyjoker Jun 3 '13 at 12:09
@grizzlyjoker So you mean $G_n$ as you have stated or as I have stated? – Vobo Jun 3 '13 at 12:54
As you have stated, I am thinking about the $G_n$'s as regular distribuitions – grizzlyjoker Jun 3 '13 at 13:12
@grizzlyjoker Then please edit the question accordingly. I will edit my answer to explain the case $n=2$ in more details. – Vobo Jun 3 '13 at 13:44

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