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Given the equation $(1-\Delta)u=f$ for $f \in S(\mathbb{R}^n)$ (rapidly decreasing functions) we get by taking the Fourier transform that

$u = \left(\frac{1}{2\pi}\right)^{\frac{n}{2}}\mathcal{F}^{-1}\left(\frac{1}{1+|.|^2} \right)*f$

This is all well-defined in the sense of $S'(\mathbb{R}^n).$

Obviously, for $n \le 3$ the function $\frac{1}{1+|.|^2}$ is in $L^2$ so it is meaningful to write

$u$ as $u(x) = \int_{\mathbb{R}^n } K(x-y)f(y) dy$ for some integral kernel defined via the inverse Fourier transform, but is there also a way to make sense out of this integral kernel representation for $n>3$?- Or if not, how can I see that there is no such kernel in higher dimensions?

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  • $\begingroup$ Fourier transform is bounded from $L^p$ to $L^q$ where $1<p \leq 2$ and $q$ is the holder conjugate. This map is not necessarily surjective though for $p<2$. This then becomes a question of whether or not one can find an appropriate $L^p$ function with the fourier transform being your example... $\endgroup$
    – Ali
    Commented May 12, 2016 at 22:25
  • $\begingroup$ @Ali yes, this is also what I thought, see the linked question $\endgroup$
    – Leopold
    Commented May 12, 2016 at 22:47

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This kernel is called the Bessel potential. It is smooth away from $0$, and in your case, this kernel is locally integrable.

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  • $\begingroup$ For another reference, Bessel potentials are treated in detail in Sec V.3 in Stein's Singular Integrals and Differentiability Properties of Functions (PUP, 1970). $\endgroup$
    – Igor Khavkine
    Commented May 12, 2016 at 23:35
  • $\begingroup$ Yet another reference: Chapter 4 of Folland's Introduction to Partial Differential Equations. $\endgroup$
    – Fan Zheng
    Commented May 13, 2016 at 1:51
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Alternatively, the self-adjoint operator $-\Delta$ on the domain $H^2\subseteq L^2$ has spectrum $[0,\infty)$ in any dimension. Since $-1$ is not in the spectrum, $(-\Delta+1)^{-1}$ is bounded on $L^2$ and thus for any $f\in L^2$, there is a unique $u\in H^2$ such that $(1-\Delta)u=f$.

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