Reconstruct an operator by its eigenfunctions $e^{-g(p \circ x)}$ - MathOverflow most recent 30 from http://mathoverflow.net2013-06-18T05:53:01Zhttp://mathoverflow.net/feeds/question/90738http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/90738/reconstruct-an-operator-by-its-eigenfunctions-e-gp-circ-xReconstruct an operator by its eigenfunctions $e^{-g(p \circ x)}$Nimza2012-03-09T17:32:48Z2012-03-09T17:46:22Z
<p>Is there some well-known description of an operator $D$ (pseudodifferential of order greater than 0) for which $e^{-g(x_1 p_1,...,x_n p_n)}$, where $g \colon \mathbb{R}^n_+ \to \mathbb{R}_+$ is a homogeneous of order 1 function, will be eigenfunctions for any $(p_1,...,p_n) \in \mathbb{R}^n_+$? I.e.
$$
D e^{-g(p_1 x_1,...,p_n x_n)} = \lambda_p e^{-g(p_1 x_1,...,p_n x_n)}.
$$</p>