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Let $\Delta$ be the Laplacian on $\mathbb R^d$. There are no eigenfunctions of the Laplacian in $L^2(\mathbb R^d)$, but $e^{ik\cdot x}$ is an eigenfunction since $$ \Delta e^{ik\cdot x} = -|k|^2 e^{-ik\cdot x}. $$ The Fourier transform $\mathcal F e^{ik\cdot x}$ can be considered as a tempered distribution and it is the Dirac delta funciton $\delta(\omega - k)$.

I have heard something like the eigenfunctions of the Laplacian are the equal to the distributions/measures supported on spheres $|\omega|=\lambda$ in Fourier space (maybe satisfying certain properties--I'm not sure what the correct statement is). I can see that the Fourier transform of the equation $(\Delta-\lambda)f = 0$ gives $(|\omega|^2-\lambda)\tilde f = 0$, so the distribution should be supported on that above sphere, but I don't think that all distributions supported on the sphere are eigenfunctions on the Laplacian. For example, $\delta'(\omega-k)$. I don't know what the correct statement for these eigenfunctions in Fourier space is or how to prove it.

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    $\begingroup$ If $k \neq 0$ one can prove that $\hat f$ is a distribution of order $0$, hence a measure supported on the sphere. One needs Theorem 2.3.11 in Hormander, The analysis of linear partial differential operators, vol I. $\endgroup$
    – Giorgio Metafune
    Commented 2 days ago

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