Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $n \geq 2$?

share|improve this question
add comment

1 Answer

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.

share|improve this answer
    
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
1  
@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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.