Does convolution with $(1+|x|)^{-n}$ define an operator $L^p(\mathbb R^n) \to L^p(\mathbb R^n)$

Question

Suppose that $f : \mathbb R^n \to \mathbb R^n$ is a locally integrable function. I am interested in the integral

$$ x \to \int_{\mathbb R^n} ( 1 + |y| )^{-n} f(x-y) \;dy $$

If the decay of the integrand's first factor where a little bit faster, say, $( 1 + |y| )^{-n-\epsilon}$, then the integral would be the convolution of $f$ with an integrable function. By Young's convolution inequality, this would define a bounded mapping $L^p(\mathbb R^n) \to L^p(\mathbb R^n)$.

Unfortunately, $( 1 + |x| )^{-n}$ is not integrable. However, I am wondering the integral still converges if $f \in L^p(\mathbb R^n)$ for some $1 < p < \infty$. Clearly, $p = \infty$ can already be excluded, but if $p < \infty$, there is hope that some decay of $f$ leads to convergence of the integral.

I am aware of some mapping theorems for Lorentz spaces, and I think that convolution with $( 1 + |x| )^{-n}$ defines a mapping between Lorentz spaces. Unfortunately, the results in the literature do not seem to provide a mapping between Lebesgue spaces.

Question: Is there a sufficiently strong result stated in the literature to make the integral converge, and does it define a mapping between $L^p$ spaces? Do you have a suggestion how to modify an existing weaker result?

Answer 1

No. E.g., let $n=1$ and $$f(u)=\frac{1(u>0)}{(1+u)^{1/2}\ln(2+u)}$$ for real $u$. Then, letting $I_f(x)$ denote the integral in question and letting $x\to\infty$, we have $$I_f(x)\ge\int_0^x\frac{dy}{1+y}\,f(x)=f(x)\ln(1+x) \sim\frac{1}{(1+x)^{1/2}},$$ so that $I_f\notin L^2$ whereas $f\in L^2$. $\quad\Box$