# If $f : \mathbb{R}^3 \to \mathbb{R}$ is Schwartz, is $F(x):=\int_{\mathbb{R}^3} \frac{f(y)}{\lvert x-y \rvert} d^3y$ also Schwartz?

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 $$$$F(x):=\int_{\mathbb{R}^3} \frac{f(y)}{\lvert x-y \rvert} d^3y$$$$ 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?

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

$$F$$ 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.

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:

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

• Thank you. How is it linked to the requirement of being a Laplacian of another Schwartz function? Jul 16, 2023 at 0:19
• -$\frac{1}{4\pi|x|}$ 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. Jul 16, 2023 at 0:32

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.