Is $|x|^{-r}$ tempered distribution?

Question

The Schwartz space, $S(\mathbb R): = \{f\in C^{\infty}(\mathbb R): \sup_{x\in \mathbb R} |(1+|x|)^{\alpha} D^{\beta}f(x)|< \infty , \forall \alpha, \beta \in \mathbb N \cup \{0\} \}$

$\|f\|_{(\alpha, \beta)}:= \sup_{x\in \mathbb R} |(1+|x|)^{\alpha} D^{\beta}f(x)|$

Put $g(x)= \frac{1}{|x|^{r}}, (r>0, x\in \mathbb R).$

My Question is: Can we expect to find constant $C$ and $\alpha$ such that $$|\int_{\mathbb R} g(x) \phi (x) dx| \leq C \|\phi \|_ {(\alpha, \beta)}$$ for all $\phi \in \mathcal{S}$ for some $r >0$?

Answer 1

A soft way could be observing that $\frac d{dx} x|x|^{-r} = (1-r)|x|^{-r}$. The left hand side is the derivative of a tempered distribution, so it is a tempered distribution.

Answer 2

It does if and only if $r<1$. To show that it does if $r<1$, it suffices to observe that

$$ \int_\mathbb{R} g \phi dx = \int_{[-a,a]} g\phi dx + \int_{|x|>a}g\phi dx. $$

On the first term, you can use Holder inequality, with $g\in L^p$ and $\phi\in L^q$, and $p$ chosen in such a way that $|x|^{-pr}$ is still integrable ($p$ has to be small enough). Since $[-a,a]$ is bounded, you can then bound the $L^q$ norm of $\phi$ with the $L^\infty$ norm. The second term is slightly harder, since, although we can still use Holder (this time $p$ has to be large enough), we can't bound the $L^q$ norm of $\phi$ with the $L^\infty$ norm. However, we can say that on $|x|>a$, $\phi\leq \frac{\phi(a)}{|x|^{2-r}}$ (this must happen after a certain $x=a$, since $\phi$ is a rapidly decreasing function). Then

$$ \left|\int_{|x|>a} g\phi dx \right|\leq |\phi(a)| \int_{|x|>a}\frac{1}{x^2}dx\leq \|\phi\|_\mathcal{S}\frac{1}{a} $$

Put the two integrals together and you've shown that if $r<1$, $|x|^{-r}$ is a tempered distribution.

To show that if $r\geq1$ then $g$ is not in $\mathcal{S}'$, simply consider $\phi_\varepsilon$ being a smoothed version of $\chi_{[-1,1]}$, such that $\phi_\varepsilon = 1$ in $[-1,1]$ and $\phi_\varepsilon=0$ outside $[-2,2]$ and $\phi_\varepsilon\geq 0$ in $\mathbb{R}$ (you can get it using mollifiers). Then,

$$ \int_\mathbb{R} g\phi_\varepsilon dx \geq \int_{[-1,1]}g dx $$ and this is clearly unbounded.

Edit: a comment on the fact that on $|x|>a$, $\phi\leq \frac{\phi(a)}{|x|^{2-r}}$. Let me restate it a little differently. WLOG, we can assume that $\sup |\phi(x)|=1$. If not, just factor $\max |\phi(x)|$ out from every integral. Then note that $1\leq \|\phi\|_\mathcal{S}$. Then I claim that, for $|x|>a$, $|\phi(x)|\leq |x|^{2-r}$. Why is this true? Well, since $\phi$ is rapidly decreasing, we have that $|x^s| \frac{|\phi|}{|x|^{2-r}}$ is bounded $\forall s>2-r$. Assume that there are infintely many point $x_n$ (with $x_n\to\infty$) such that $|\phi(x_n)|>\frac{1}{|x_n|^{2-r}}$. Well, then, at those points,

$$ |{x_n}|^s\frac{|\phi(x_n)|}{|x_n|^{2-r}}>|x_n|^s. $$ But this is not bounded for $s>0$, which contradicts the fact that $\phi\in \mathcal{S}$.

Now the question is: how can I find $a$? Well, you can't, unless you have the expression for $\phi$. But do you need it? No. Notice that, in the estimate of the second integral, you end up with $\|\phi\|_\mathcal{S}\frac{1}{a}$. Notice that $a\geq 1$, since $\sup |\phi|=1$ and $1/|x|^{2-r}>1$ for $x<1$. Well, then you can simply replace $1/a$ with $1$ in the estimate.