Let $A$ denotes a self-adjoint operator such that $Tr(A)< \infty$. Let $t\geq 0$. How to show the following \begin{equation} \int_{0}^{\infty}Tr(e^{-tA})< \infty\,. \end{equation} Some facts \begin{equation} Tr(A) =\sum_j<\phi_j|A|\phi_j>\,, \end{equation} where $\{ |\phi_j>\}_j$ is any orthogonal basis (it could be the set of the eigen functions of A). Moreover, \begin{equation} A= \sum_{\lambda\in\sigma(A)} \lambda E_{\lambda}\,, \end{equation} where $\sigma(A)$ is the spectrum of operator $A$ and \begin{equation} E_{\lambda} := \sum_{\lambda_j=\lambda}<\phi_j,\cdot> |\phi_j> \end{equation} is the orthogonal projection on to the eigen space of $\lambda$.
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It seems I need to have $Tr(A^{-1}) < \infty$ to get the desired result, which make sense if I am not mistaken. – Aban Oct 07 '23 at 14:35
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My idea is to use the formula written here to express \begin{equation} e^{-tA} = \sum_j e^{-t\lambda_j} E_{\lambda_j}\,, \end{equation} which implies \begin{equation} Tr(e^{-tA}) = \sum_j e^{-t\lambda_j}\,. \end{equation} Hence, \begin{equation} \int_0^{\infty} e^{-tA} dt = \sum_j\int_0^{\infty} e^{-t\lambda_j} dt = \sum_j\frac{1}{\lambda_j}\,. \end{equation}
Aban
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