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We know that the operator $A=\Delta$ with domain $D(A)=\{u\in W^{2, 2}(\Omega): u=0 \ \ \text{on } \partial\Omega\}$ (say $\Omega$ is a bounded nice domain) has eigenvalues $\lambda_1>\lambda_2\ge \cdots \lambda_n\ge$ with corresponding eigenvectors $\{\phi_n\}_{n=1}^\infty$ such that $\{\phi_n\}_{n=1}^\infty$ can be normalized to be an orthonormal basis of $L^2(\Omega)$.

Questions: if $A$ is not symmetric. Say $A=\Delta + a\partial_x$, where $a\in C(\bar\Omega)$. Can the eigenvectors of $A$ still form an orthonormal basis of $L^2(\Omega)$? My guess is if $(\lambda-A)^{-1}$ is normal, then we can apply the spectral theory for compact normal operators. But is $(\lambda-A)^{-1}$ normal for large $\lambda$? Thanks!!!

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  • $\begingroup$ I believe that this convection diffusion operator is a good examples of a non-normal operator for which the spectral theory is more complicated. See the first couple of pages of the introduction of this survey: arxiv.org/pdf/1002.4844.pdf $\endgroup$ Commented Jan 25, 2022 at 13:48
  • $\begingroup$ See also this related post mathoverflow.net/questions/409944/… $\endgroup$ Commented Jan 25, 2022 at 16:06
  • $\begingroup$ Thanks for the comments! I think I found a counter example: $\Delta (U\phi)/U=\lambda \phi$ $\endgroup$
    – Y Wu
    Commented Jan 25, 2022 at 17:55

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