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On $L^2(\mathbb{R}^n)$ consider the operator $(-\Delta+z)^{-1}$, for $\Delta$ being the Laplacian and $z\in\mathbb{C}$, on $\mathcal{D}(\Delta^{-1})$.

How can one show that for $\phi\in \mathcal{D}(\Delta^{-1})$ the following holds ? :

$$(-\Delta+z)^{-1} \phi = \mathcal{F}^{-1}\{(\vert k\vert^2 + z)^{-1} \;\hat \phi(k)\},$$

where $\mathcal{F}$ (shorthand $\hat{}$) denotes the Fourier transform.

I was reading about the functional calculus of the Laplacian using spectral theory, but I am not sure how to properly use it for its inverse, maybe this is not even the right tool to approach this.

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  • $\begingroup$ Functional calculus is it. The Fourier transform in this case is a concrete example of a unitary transformation that "diagonalizes" a self-adjoint operator. $\endgroup$ – Michael Nov 11 '13 at 3:56
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What does it mean to say $(-\Delta +zI)^{-1}f=g$? If such an inverse exists, then this equation is equivalent to $f=(-\Delta + zI)g$, which you can easily Fourier transform to obtain $(|s|^{2}+z)\hat{g}(s)=\hat{f}(s)$, or $\hat{g}(s)=\frac{\hat{f}(s)}{|s|^{2}+z}$ where $s$ is a complex $N$-vector ($N$ is the dimension of the space.) So it works as you would think it should, provided the inverse exists.

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