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Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of smooth functions $C^{\infty}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in \mathscr{S}(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $C^{\infty}(\Omega)$. This seems to imply that the functions $\psi_{p} \in \mathscr{S}(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $C^{\infty}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

ADD: Okay, after some thoughts maybe the situation is simpler. Because $f$ is bounded and continuous on $[0,\infty)$ and the spectrum $\sigma(-\Delta) \subset [0,\infty)$, then $f(-\Delta)$ makes sense as a bounded linear operator on the whole $L^{2}(\Omega)$ and, by (\ref{1}), I can claim that $A = f(-\Delta)$?

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of smooth functions $C^{\infty}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in C^\infty(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $C^{\infty}(\Omega)$. This seems to imply that the functions $\psi_{p} \in C^\infty(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $C^{\infty}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

ADD: Okay, after some thoughts maybe the situation is simpler. Because $f$ is bounded and continuous on $[0,\infty)$ and the spectrum $\sigma(-\Delta) \subset [0,\infty)$, then $f(-\Delta)$ makes sense as a bounded linear operator on the whole $L^{2}(\Omega)$ and, by (\ref{1}), I can claim that $A = f(-\Delta)$?

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of rapidly decreasing functions $\mathscr{S}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in \mathscr{S}(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $\mathscr{S}(\Omega)$. This seems to imply that the functions $\psi_{p} \in \mathscr{S}(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $\mathscr{S}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

ADD: Okay, after some thoughts maybe the situation is simpler. Because $f$ is bounded and continuous on $[0,\infty)$ and the spectrum $\sigma(-\Delta) \subset [0,\infty)$, then $f(-\Delta)$ makes sense as a bounded linear operator on the whole $L^{2}(\Omega)$ and, by (\ref{1}), I can claim that $A = f(-\Delta)$?

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of smooth functions $C^{\infty}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in \mathscr{S}(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $C^{\infty}(\Omega)$. This seems to imply that the functions $\psi_{p} \in \mathscr{S}(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $C^{\infty}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

ADD: Okay, after some thoughts maybe the situation is simpler. Because $f$ is bounded and continuous on $[0,\infty)$ and the spectrum $\sigma(-\Delta) \subset [0,\infty)$, then $f(-\Delta)$ makes sense as a bounded linear operator on the whole $L^{2}(\Omega)$ and, by (\ref{1}), I can claim that $A = f(-\Delta)$?

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of rapidly decreasing functions $\mathscr{S}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in \mathscr{S}(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $\mathscr{S}(\Omega)$. This seems to imply that the functions $\psi_{p} \in \mathscr{S}(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $\mathscr{S}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

Let $L \ge 1$ and consider a finite box $\Omega = [0,L]^{d} \subset \mathbb{R}^{d}$. The set of functions: $$\psi_{p}(x) = \frac{1}{L^{d/2}}e^{i\langle p,x\rangle} \quad p\in \frac{2\pi}{L}\mathbb{Z}^{d}$$ form a countable orthonormal basis for $L^{2}(\Omega)$.

The Laplacian operator $\Delta$ is well-defined on the space of rapidly decreasing functions $\mathscr{S}(\Omega) \subset L^{2}(\Omega)$. Thus, on this space, we can consider the operator $f(-\Delta)$ where, for $a>0$ and $b < 0$, $f$ is given by: $$f(x) = \frac{1}{e^{a(x-b)}-1}$$

Suppose $A$ is a bounded positive (hence self-adjoint) compact operator on $L^{2}(\Omega)$. Suppose further that, for every $\varphi, \psi \in \mathscr{S}(\Omega)$, we have the following identity: $$\langle \varphi, A\psi\rangle = \langle \varphi, f(-\Delta)\psi\rangle \tag{1}\label{1}$$

In simple words, my question is: can we establish a connection between these two operators? Let me explain what I am trying to do. The functions $\psi_{p}$ are eigenstates of $f(-\Delta)$ and they form an ONB for $L^{2}(\Omega)$. I know that $A$ is compact, so by the Hilbert-Schmidt Theorem it has a representation: $$A = \sum_{n\in \mathbb{N}}\lambda_{n}\langle \phi_{n},\cdot\rangle \phi_{n}$$ where $\lambda_{n}$ are the eigenvalues of $A$ and $\phi_{n}$ its corresponding eigenvectors.

Question 1: I wonder if I can use this representation in terms of these $\psi_{p}$ functions, since these are eigenstates of $f(-\Delta)$. In other words, by (\ref{1}), $A = f(-\Delta)$ on $\mathscr{S}(\Omega)$. This seems to imply that the functions $\psi_{p} \in \mathscr{S}(\Omega)$, $p \in \frac{2\pi}{L}\mathbb{Z}^{d}$, are eigenvectors of $A$. Does this mean that I can represent $A$ as: $$A = \sum_{p\in \frac{2\pi}{L}\mathbb{Z}^{d}}\lambda_{p}\langle \lambda_{p}\psi_{p},\cdot\rangle\psi_{p}$$ where $\lambda_{p} = f(-p)$, since $A\psi_{p} = f(-\Delta)\psi_{p} = f(-p)\psi_{p}$?

Question 2: Can we use some density argument to extend $f(-\Delta)$ to $L^{2}(\Omega)$ by means of (\ref{1}), using the fact that $\mathscr{S}(\Omega)$ is dense in $L^{2}(\Omega)$? This would, in particular, imply $A = f(-\Delta)$, and the answer to Question 1 would be positive.

ADD: Okay, after some thoughts maybe the situation is simpler. Because $f$ is bounded and continuous on $[0,\infty)$ and the spectrum $\sigma(-\Delta) \subset [0,\infty)$, then $f(-\Delta)$ makes sense as a bounded linear operator on the whole $L^{2}(\Omega)$ and, by (\ref{1}), I can claim that $A = f(-\Delta)$?