Neither operator has an eigenfunction on ${\mathbb R}^n$. 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, thus can be diagonalized.

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 $$t\Delta u=\sum_{m=0}^\infty\frac1m(1-\lambda)^mu=t\left(\log\frac1\lambda\right) u.$$

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 $$t\Delta u=\sum_{m=0}^\infty\frac1m(1-\lambda)^mu=t\left(\log\frac1\lambda\right) u.$$

Neither operator has an eigenfunction on ${\mathbb R}^n$. 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, thus can be diagonalized.

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

Neither operator has an eigenfunction on ${\mathbb R}^n$. 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, thus can be diagonalized.

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 $$t\Delta u=\sum_{m=0}^\infty\frac1m(1-\lambda)^mu=t\left(\log\frac1\lambda\right) u.$$