Usual example of tempered distribution

Question

I have read in my class notes that $\int_{\mathbb{R^n}}\frac{1}{|x|^p}\phi(x)dx$ defines clearly a tempered distribution when $0<p<n$. It doesn't seem so easy to me and I have tried to prove it:

My attempt

If $\{\phi_n\}$ is a sequence of tempered distributions which converges to zero function: $$\int_{\mathbb{R^n}}\frac{1}{|x|^p}\phi(x)dx=\int_{B(0,1)}\frac{1}{|x|^p}\phi(x)dx+\int_{\mathbb{R^n}\backslash B(0,1)}\frac{1}{|x|^p}\phi(x)dx.$$

We have that $\int_{B(0,1)}\frac{1}{|x|^p}\phi(x)dx\leq{\sup|\phi(x)|}\int_{B(0,1)}\frac{1}{|x|^p}dx\to 0$.

On the other hand $\int_{\mathbb{R^n}\backslash B(0,1)}\frac{1}{|x|^p}\phi(x)dx=\int_{\mathbb{R^n}\backslash B(0,1)}|x|^n\frac{1}{|x|^{p+n}}\phi(x)dx\leq{\sup\{|x|^n\phi(x)\}}\int_{\mathbb{R^n}\backslash B(0,1)}\frac{1}{|x|^{p+n}}dx.$

How can we conclude that this second term converges to zero?

This is the way that I have followed. Please, if someone knows a better way, tell me.

Answer 1

You are almost done. First note that you are missing the subscript $n$ on all of your functions. Then as $\phi_n$ is an element of the Schwartz space $\textbf{S}(\mathbb{R}^n)$ you get $$\lim_{n\to\infty}\sup_{x\in\mathbb{R}^n}||x|^n\phi_n(x)|=0,$$ by the convergence of $\phi_n$ to $0$ in $\textbf{S}(\mathbb{R}^n)$. The last integral exists by $p+n>n$ and therefore the expression coverges to $0$.

Answer 2

You don't want to index your test functions by $n$, because you're already using $n$ for the dimension of the space $\mathbb R^n$.

The point is that if $\phi_k \to 0$ as test functions, $|x^m \phi_k(x)| \to 0$ uniformly for all positive integers $m$. Also use the fact that $\int_{{\mathbb R}^n \backslash B(0,1)} |x|^{-n-1}\; dx < \infty$.