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Let $\Delta$ be the usual Laplacian on $\mathbb R^n$, $\Delta=-\sum_{i=1}^n\frac{\partial^2}{\partial x_i^2}$. Consider the heat operator $H_t=e^{-t\Delta}$. Is there an eigenfunction of $H_t$ which is not an eigenfunction of $\Delta$?

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Neither operator has an eigenfunction in $L^2({\mathbb R}^n)$. But if you replace ${\mathbb R}^n$ by a bounded domain $\Omega$ with a smooth boundary, you may consider the Heat equation with the Dirichlet boundary condition $u=0$ on $\partial\Omega$. Then $e^{-\Delta}$ and $\Delta^{-1}$ are compact and self-adjoint on $L^2(\Omega)$, thus can be diagonalized. In addition $e^{-\Delta}$ is a contraction in $L^2(\Omega)$.

To see that every eigenfunction of $e^{-t\Delta}$ is an eigenfunction of $\Delta$, you may use the formula $$t\Delta=\sum_{m=1}^\infty\frac1m(I-e^{-t\Delta})^m.$$ This is valid over the domain $D(\Delta)$. If $e^{-t\Delta}u=\lambda u$, then you obtain $$t\Delta u=\sum_{m=0}^\infty\frac1m(1-\lambda)^mu=t\left(\log\frac1\lambda\right) u.$$

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Do you mean $\Delta$ rather than $\delta$ in the second paragraph? – Jon May 28 '12 at 13:07
Shouldn't that sum go from $m=1$ and add up to $-\textrm{ln}(\lambda)$? The eigenvectors of $\Delta$ and $e^{t\Delta}$ should hopefully match, but not necessarily so the corresponding eigenvalues. – Emilio Pisanty May 28 '12 at 13:38
(terrific answer, otherwise!) – Emilio Pisanty May 28 '12 at 13:38
I apologize for the typos and I edit ... – Denis Serre May 28 '12 at 15:12
I am not talking about $L^2$-eigenfunctions. There are $L^\infty$-eigenfunctions of $\Delta$ on $\mathbb R^n$: $x↦e^{−i(x_1y_1+⋯+x_ny_n)}$. There are eigenfunctions in some other $L^p$ spaces depending on $n$. Any eigenfunction of $\Delta$ on $\mathbb R^n$ is clearly an eigenfunction of $e^{-t\Delta}$. I am asking if there is an eigenfunction of $e^{-t\Delta}$ on $\mathbb R^n$ for a fixed $t>0$ which is not an eigenfunction of $\Delta$. – spr May 29 '12 at 6:48

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