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For $x\in\mathbb{R}^n$ we define the Newton potential as follows:$N(x) = \begin{cases} \frac{\log|x|}{2\pi}, & n=2 \\[10pt] \frac{|x|^{2-n}}{(2-n)\omega_n}, & n>2\end{cases}$

where $\omega_n$ denoted to the volume of the n-ball. Moreover, let $\chi_r$ denote to the characteristic function of the ball $B(0,r)$.

Now my lecture notes say (for $n>2$)

  1. $\chi_r N \in L^1(\mathbb{R}^n)$ (or, more general, $L^p$, where $p<\frac{n}{n-2}$) and

  2. $(1-\chi_r)N \in L^\infty(\mathbb{R}^n)$ (or, more general, $L^p$, where $p>\frac{n}{n-2}$)

Since $N \in L_\mathrm{loc}^1(\mathbb{R}^n)$, it follows immediately that $\chi_r N \in L^1(\mathbb{R}^n)$ for any $r>0$.

How do I see the other claims?

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    Sorry, I wanted to show claims 1 and 2 for n>2 only. So, yes and yes.2012-06-30

1 Answers 1

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We use the following formula, which is true for a radial function $f$ (that is, which depends only of the norm of the argument): $\int_{B(0,r)}f(x)dx=c_n\int_0^r\widetilde f(t)t^{n-1}dt,$ where $c_n$ is a constant I leave to determine, and $\widetilde f$ is such that $f(x)=\widetilde f(|x|)$, where $|\cdot|$ is the euclidian norm.

For $n=2$, $\chi_r N\in L^p$ if and only if $\int_0^rt^{n-1}|\ln t|^pdt $ is finite that is, if and only if $\int_0^1e^{u(n-1)}e^uu^pdu$ is finite, which is always the case.

For $n>2$, $\chi_rN\in L^p$ if and only if $\int_0^1t^{n-1}t^{(2-n)p}dt<\infty$ which is equivalent to $\int_0^1\frac 1{t^{1-(n+(2-n)p)}}dt<\infty$, which occurs if and only if $n+(2-n)p>0$.

We have that \begin{align} (1-\chi_r)N\in L^p&\Leftrightarrow \int_1^{+\infty} t^{n-1}t^{(2-n)p}dt<\infty\\ &\Leftrightarrow \int_1^{+\infty}\frac 1{t^{1-(n+(2-n)p)}}dt<\infty \\ &\Leftrightarrow n+(2-n)p<0. \end{align}

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    For $n=2$, I think the first integral is finite iff $\int_{-\infty}^{\log{r}} e^{u(n-1)} e^u |u|^p du$ is finite what is not clear (at least for me as p>>1). However, the idea for n>2 works fine (here the integral goes from $0$ to $r$ or $r$ to $+\infty$).2012-07-01