4
$\begingroup$

I am having trouble proving / disproving the question in the title.

That is, let $f : \mathbb{R}^3 \to \mathbb{R}$ be a real-valued Schwartz function. Then, I wonder if \begin{equation} F(x):=\int_{\mathbb{R}^3} \frac{f(y)}{\lvert x-y \rvert} d^3y \end{equation} is also a Schwartz function.

At least it seems clear from the property of convolution that $F(x)$ is smooth. However, I cannot figure out decay properties of $F(x)$. Could anyone please help me?

$\endgroup$
1
  • 1
    $\begingroup$ You could use the substitution $y-x = y'$ to get $$\int \frac{f(x+y')}{|y'|}d^3y'$$ to make things easier $\endgroup$ Jul 15, 2023 at 23:54

3 Answers 3

7
$\begingroup$

$F$ is not necessarily Schwartz. If it was, then its Fourier transform will also be Schwartz. Now since $F = f * |\cdot|^{-1}$, and the Fourier transform of $|\cdot|^{-1}$ as a tempered distribution is $c_3 |\xi|^{-2}$ (where $c_3$ is a dimensional constant), we see that $\hat{F} = c_3\hat f |\cdot|^{-2}$ as convolution corresponds to multiplication on the Fourier side. We could design $\hat f $ to be a smooth compactly supported bump function equal to $1$ on $|\xi| \le 1$ and consequently, $$ \sup_{\xi \in \mathbb{R^3}} c_3|\xi|^{-2} |\hat f(\xi)| = +\infty, $$ i.e. $\hat F$ is not Schwartz.

$\endgroup$
5
$\begingroup$

The Fourier transform is an isomorphism in the Schwartz space $S$. Note that the $\frac{1}{|x|}$ is proportional to the Green's function of the 3D laplacian which implies:

\begin{equation} \int_{\mathbb{R}^3} \frac{f(y)}{\lvert x-y \rvert} d^3y \in S \iff \frac{1}{|k|^2}\hat{f}(k)\in S \end{equation} This will clearly not be true in general as it requires $\hat{f}(0)=0$. In fact, your condition is true iff $f$ is the Laplacian of a Schwartz function.

$\endgroup$
2
  • $\begingroup$ Thank you. How is it linked to the requirement of being a Laplacian of another Schwartz function? $\endgroup$
    – Keith
    Jul 16, 2023 at 0:19
  • 3
    $\begingroup$ -$\frac{1}{4\pi|x|}$ is the Green's function of the 3D Laplacian. This is another way of saying that that $G(y)=-\frac{1}{4\pi}F(y)=-\int_{\mathbb{R}^3} \frac{f(y)}{4\pi\lvert x-y \rvert} d^3y$ is a particular solution to the partial differential equation $\nabla^2G = f$. Your condition of $F$ being Schwartz is clearly equivalent to $G$ being Schwartz. $\endgroup$
    – Andrew
    Jul 16, 2023 at 0:32
3
$\begingroup$

Another way of seeing that $F$ is in general not a Schwartz function (avoiding tempered distributions), is to consider your favourite non-trivial nonnegative Schwartz function $f$ (for example $f(x)=e^{-\vert x\vert^2}$). Then $F$ is not even in $L^1$ and thus not Schwartz. Indeed, we have

$$\int_{\mathbb{R}^3} \vert F(x)\vert dx =\int_{\mathbb{R}^3} \vert \int_{\mathbb{R}^3} \frac{f(y)}{\vert x-y\vert} dy \vert dx = \Vert f\Vert_1 \int_{\mathbb{R}^3} \frac{1}{\vert x\vert} dx =\infty.$$

Note that this simplification comes at the price of not getting a characterization when $F$ is a Schwartz function. This basic proof only allows to construct some counterexamples.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .