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I'm searching for an inequality in the form $$\forall s>2,\quad\forall u\in H^{s-2}(\mathbb{R}^{2}),\quad\|\Delta^{-1}u\|_{H^{s}(\mathbb{R}^{2})}\lesssim\|u\|_{H^{s-2}(\mathbb{R}^{2})}$$ where $H^s$ is the nonhomogeneous Sobolev space defined by $$H^{s}(\mathbb{R}^{2})=\big\{u\in\mathcal{S}'(\mathbb{R}^{2})\quad\mbox{s.t.}\quad\|u\|_{H^{s}(\mathbb{R}^{2})}:=\left(\int_{\mathbb{R}^{2}}(1+|\xi|^2)^s|\widehat{u}(\xi)|^{2}d\xi\right)^{\frac{1}{2}}<+\infty\big\}$$ and where $\Delta^{-1}$ has symbol $-|\xi|^{-2}$. Such an inequality in homogeneous Sobolev spaces $\dot{H}^s$ (without the addition by $1$ in the norm) seems trivial, but I do not manage to prove the above one.

Such an estimate means in particular that the operator $\Delta^{-1}$ is bounded in $H^s.$ In bounded domains, there exists an elliptic regularity result for the Dirichlet problem, but the proof strongly uses the fact that the domain is bounded (at least in one direction), which is not the case here since we work in the full space.

For me this result may be false, but I do no manage to disprove it (with for instance dilation arguments) and it would be great if it reveals true..

Thank you for your help.

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    $\begingroup$ No. What you would like to have is$$\int(1+\xi^2)^s|\xi|^4|\hat u|^2\,d\xi\,\ge\,c\cdot\int(1+\xi^2)^{s+2}|\hat u|^2\,d\xi$$with some $c>0$. This turns out to be false when you choose bump functions for $\hat u$ with smaller and smaller support around zero. $\endgroup$
    – amsmath
    Sep 4, 2021 at 15:30
  • $\begingroup$ Yeah, if though about it and I think that if you use Banach isomorphism theorem (or open mapping theorem if you prefer), you see that this operator can't be bounded since the Laplacian isn't bounded. $\endgroup$
    – Maticawa
    Sep 5, 2021 at 16:05
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    $\begingroup$ No. That's not the reason. The reason is that the Laplacian is not bounded from below. $\endgroup$
    – amsmath
    Sep 5, 2021 at 17:47
  • $\begingroup$ Ok I understand your argument. Thank you for that. Can you tell me where does mine (with isomorphism Banach theorem) fails ? $\endgroup$
    – Maticawa
    Sep 6, 2021 at 20:19
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    $\begingroup$ To invoke the inverse mapping theorem you would have to show that your Laplacian is injective and has a closed image. The latter fails. $\endgroup$
    – amsmath
    Sep 6, 2021 at 23:04

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