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Fact 1. The function $1/|\xi|^{s}$ locally integrable (in the unit ball) if and only if $s<n$.

Fact 2. If $f\in L^1(\mathbb{R}^n)$, then $\widehat{f}\in\mathcal{C}^{\infty}(\mathbb{R}^n)$ (bounded).

Let $H^{2}:=\left\{u\in L^2(\mathbb{R}^n): \mathcal{F}^{-1}\left((1+|\xi|^2)^2\widehat{f}(\xi)\right)\in L^2(\mathbb{R}^n)\right\}$.

Question 1. For any $f\in L^2$, the equation $-\Delta u=f$ has a unique solution in the Sobolev space $H^{2}$?

My attempt: By fact 1, the function $1/|\xi|^2$ is locally integrable in the unit ball. Then, $1/|\xi|^2\in L^1(\mathbb{R}^n)$ when $2<n$. or not?. The candidate solution is $u_f=\mathcal{F}^{-1}\left( \frac{1}{|\xi|^2}\widehat{f}\right)$. By convolution theorem, $u_f=\mathcal{F}^{-1}\left(\frac{1}{|\xi|^2}\right)*f$. By fact 1, $\mathcal{F}^{-1}\left(\frac{1}{|\xi|^2}\right)$ is bounded, then $u_f\in L^2$. Moreover, \begin{align} \mathcal{F}^{-1}\left((1+|\xi|^2)\widehat{u_f}(\xi)\right)&=\mathcal{F}^{-1}\left( \frac{(1+|\xi|^2)}{|\xi|^2}\widehat{f}(\xi)\right)\\ &=\mathcal{F}^{-1}\left(\frac{1}{|\xi|^2}\widehat{f}\right)+\mathcal{F}^{-1}\widehat{f} \\ &=\mathcal{F}^{-1}\left(\frac{1}{|\xi|^2}\widehat{f}\right)+f\\ &=u_f+f \end{align} and $u_f+f\in L^2(\mathbb{R}^n)$ because $u_f, f\in L^2(\mathbb{R}^n)$

This is right? Thanks you!

eraldcoil
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    Sadly, $H^2(\mathbb{R}^n)$ is not the good space for solving the Poisson equation on the whole space.

    An idea to show this: assume by contraction the inverse laplace operator is bounded from $\mathrm{L}^2$ to $H^2$. In particular, it maps $\mathrm{L}^2$ into itself. But by a dilation argument, you can make the norm of the solution as small as possible and therefore the $\mathrm{L}^2$ of your solution must be $0$...

    I did a post about solving the poisson equation here : https://math.stackexchange.com/questions/4738465/properties-of-the-inverse-laplacian-operator/4740654#4740654

     – ToGle Mar 13 '24 at 00:59
  • Thanks for the reply. But then there isn't a space where there is a unique solution for such an equation? (any space similar to some Sobolev for example?) – eraldcoil Mar 13 '24 at 13:01
  • See the post I linked : the answer is homogeneous Sobolev spaces. The counter example I told you says "ok, we need to give up our desire to control the $\mathrm{L}^2$ norm of our solution". In such a context the only thing you can expect is to control exactly what shows up in the equation : the second order derivatives of your solution, nothing less, nothing more. – ToGle Mar 13 '24 at 14:21
  • In particular, you can never expect, in general, that that your data $f$ and your solution $u$ actually lie in the same space. – ToGle Mar 13 '24 at 14:26
  • I have a doubt. If $f\in L^2$, the equation $(1-\Delta)u=f$ have a unique solution in the Sobolev space $H^{2}$ right?. If $f\in L^2$, the equation $-\Delta u=f$ have a unique solution in $\left{u\in L^2: \mathcal{F}^{-1}(|x|^2\widehat{u})\in L^2\right}$ (homogeneous Sobolev)? – eraldcoil Mar 14 '24 at 18:38
  • Not really. The two spaces you wrote are the same spaces, by Fourier-Plancherel : $$\mathrm{H}^2={,u\in\mathcal{S}',|, (1+|\xi|^2)\mathcal{F}u(\xi)\in\mathrm{L}^2} = {,u\in\mathrm{L}^2,|, (1+|\xi|^2)\mathcal{F}u(\xi)\in\mathrm{L}^2} = {,u\in\mathrm{L}^2,|, |\xi|^2\mathcal{F}u(\xi)\in\mathrm{L}^2}.$$ The homogeneous space is defined by $$\dot{\mathrm{H}}^2={,u\in\mathcal{S}'/\mathcal{P},|, |\xi|^2\mathcal{F}u(\xi)\in\mathrm{L}^2} \nsubseteq \mathrm{L}^2.$$ This last space is the one s.t. $-\Delta,:,\dot{\mathrm{H}}^2\rightarrow \mathrm{L}^2$ is an isomorphism. – ToGle Mar 14 '24 at 20:55
  • However, you got it right for the first part of your comment, we have an isomorphism $\mathrm{I}-\Delta ,:, \mathrm{H}^2\longrightarrow \mathrm{L}^2$. – ToGle Mar 14 '24 at 21:02
  • But, if I can identify $u\in H^2$ with $T_u$ (tempered distribution) then $-\Delta: H^2\to L^2$ is an isomorphism? or is there no way? – eraldcoil Mar 19 '24 at 05:26
  • No way, by the argument in my first comment. This is false. – ToGle Mar 19 '24 at 18:17

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