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?

  • $\begingroup$ 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 May 12 '16 at 22:25
  • $\begingroup$ @Ali yes, this is also what I thought, see the linked question $\endgroup$ – Leopold May 12 '16 at 22:47

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

  • $\begingroup$ 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 May 12 '16 at 23:35
  • $\begingroup$ Yet another reference: Chapter 4 of Folland's Introduction to Partial Differential Equations. $\endgroup$ – Fan Zheng May 13 '16 at 1:51

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 $(1-\Delta)u=f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.