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According to the answers in the the following questions: How to prove the spectrum of the Laplace operator? and What is spectrum for Laplacian in $\mathbb{R}^n$ , the spectrum of the Laplace operator $\Delta :H^2(\mathbb{R}^2)\subset L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ is in fact $\sigma(\Delta)=(-\infty,0].$ However, I was not able to find a discussion on the eigenvalues of $\Delta$.

The set of eigenvalues $\sigma_p(\Delta)$ (also called point spectrum) is known to be contained in $\sigma(\Delta)$ and one can have $\sigma_p(\Delta)\subsetneq \sigma(\Delta)$. Indeed, by taking the Fourier transform $\mathcal{F}:L^2(\mathbb{R}^2)\to L^2(\mathbb{R}^2)$ of the eigenvalue problem one has $$\Delta u(x) = \lambda u(x),\;\;\forall x\in \Bbb R^2 \;\;\;\overset{\mathcal F}{\longrightarrow}\;\;\;\;-4\pi^2|\xi|^2\hat u(\xi) =\lambda \hat u(\xi), \;\;\;(\text{a.e})\;\xi\in \Bbb R^2,$$ and this can only be satisfied by $\hat u=u=0$. This means that the only eigenvalue-eigenvector pair in this setting is $(\lambda,u)=(0,0)$ . Also, the same argument applies when $\Delta$ is seen as $\Delta:W^{m,p}(\mathbb{R}^2)\subset L^p(\mathbb{R}^2)\to L^p(\mathbb{R}^2)$ with $p\in [1,2)$ and $\mathcal F:L^p(\mathbb{R}^2)\to L^{p^*}(\mathbb{R}^2)$ with $1/p+1/p^*=1$.

Question 1. What happens when $p>2$ and the Fourier transform becomes distribution valued, so that the above elementary argument cannot be applied directly?

It seems if $u\in C^2(\mathbb{R}^2)$ is in fact an eigenvalue of $\Delta$, then it cannot be in $L^p(\mathbb{R}^2)$ for any $p\in [1,2]$. Depending on the answer to Q1, this might also hold for $p>2$. In any case, it seems that the $L^p$ framework is not suitable for this problem.

Question 2. On what space(s) could one define the domain of $\Delta$ to obtain non-trivial eigenvalues?

Edit. The crossed out sentence should be replaced by: "The point spectrum $\sigma_p(\Delta)$ is therefore empty."

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The point spectrum coincides with the spectrum minus 0 if $p>2n/(n-1)$ and it is empty in the remaining cases ($n$ is the dimension). This is proved in G. Talenti: "Spectrum of the Laplace operator acting in $L^p(R^n)$", Indam, Symposia Mathematica vol VII, Academic Press 1971.

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    $\begingroup$ The full spectrum is in any case $(-\infty, 0]$, for every $p$. The inclusion in the half line follows from semigroup theory, since the heat semigroup is bounded holomorphic of angle $\pi/2$, the reverse inclusion by showing that negative numbers are approximate eigenvalues. $\endgroup$
    – Giorgio Metafune
    Commented Jan 29, 2022 at 10:37
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    $\begingroup$ @JochenGlueck Jochen you are right, the argument works for the spectrum minus 0. Of course, there is a way of avoiding semigroups. For $l\lambda$ not a negative number (and not 0), write formally the inverse through the Fourier transform and then use Mikhlin to check the boundedness. If $\lambda<0$ one shows it is an approximate eigenvlaue (the functions $e^{iax}$ should be an approximate eigenvector). For the point spectrum, one first show that if there is an eigenfuntions, then there is a radial one, by averaging, and then ends with a Bessel equation whose asymptotic is known. $\endgroup$
    – Giorgio Metafune
    Commented Jan 29, 2022 at 12:26
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    $\begingroup$ @GiorgioMetafune: Thanks a lot for your reply, and for the explanation of how to construct the eigenfunctions! Quite embarrassingly for me, I really wasn't aware (before reading your post) that $\Delta$ has any point spectrum on $L^p$ for sufficiently large $p$) $\endgroup$
    – Jochen Glueck
    Commented Jan 29, 2022 at 12:31
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    $\begingroup$ @JochenGlueck Since you are interested I go on. If $\Delta u+k^2 u=0$ ($K \neq 0$) and $u$ is radial, then setting $u(r)=r^{(1-n)/2}w(r)$ you get a Bessel equation whose solutions at infinity oscillate lile $\sin$. Then the asympotics for $v$ at inifinity is like $r^{(1-n)/2}$ which is in $L^p$ for $p>2n(n-1)$. $\endgroup$
    – Giorgio Metafune
    Commented Jan 29, 2022 at 12:52
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    $\begingroup$ I checked better the book. Some papers are in italian but this one is in English. If you need I can scan and send a copy via mail. Then you should write to me via mail to [email protected] $\endgroup$
    – Giorgio Metafune
    Commented Jan 29, 2022 at 18:02

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