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.
Best Answer
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.