Question about the action of the operator $(-\Delta+i)^{-1}$ It is well-known that minus the Laplace operator $-\Delta:\mathcal{D}(-\Delta)\to L^{2}(\mathbb{R}^{3})$ is self-sdjoint on the domain $\mathcal{D}(-\Delta)=H^{2}(\mathbb{R}^{3})$ (Sobolev space). Therefore, the spectrum of $-\Delta$ is contained in $\mathbb{R}$, which means that $i$ is in the resolvent and hence the inverse of $(-\Delta+i):\mathcal{D}(-\Delta)\to L^{2}(\mathbb{R}^{3})$ exists and
$$(-\Delta+i)^{-1}\in\mathcal{B}(L^{2}(\mathbb{R}^{3}))$$
where $\mathcal{B}(\mathcal{H})$ denotes the set of bounded, linear operators on the Hilbert space $\mathcal{H}$.
Now, in a lecture I have seen the following formula:
$$\frac{1}{-\Delta+i}f(x)=\int_{\mathbb{R}^{3}}\frac{\hat{f}(k)}{4\pi^{2}k^{2}+i}e^{2\pi ik\cdot x}\,\mathrm{d}^{3}k$$
for some $f\in L^{2}(\mathbb{R}^{3})$ and where $\hat{f}$ denotes the Fourier transform $f$. I have two questions about this:
(1) Is $\frac{1}{-\Delta+i}$ here just a notation for the inverse operator $(-\Delta+i)^{-1}$? I guess that this is the case, but just to get sure, because when talking about real functions for example, the symbols $1/f$ and $f^{-1}$ mean completely different things.
(2) How is the above formula derived?
Thanks a lot!
 A: (1) Yes, $(-\Delta +i)^{-1}$ refers to the inverse operator relative to the operation of composition.
There is an issue regarding what is the inverse of a
function, since one may either be talking about the  pointwise multiplication or the composition of functions,  hence
the two somewhat well established notations $1/f$ and $f^{-1}$.  For operators, the pointwise multiplication makes no
sense so people often use these notations indistinctly
(2) By a simple integration by parts argument,  one can show that for every $f$ in the Schwartz space,
$$
  \widehat{\frac {\partial f}{\partial \xi _i}}(\xi ) = 2\pi  i \xi _i \widehat f(\xi ), \quad \forall \xi \in  \widehat {\mathbb R^3} = \mathbb R^3.
  $$
Thus,  denoting by $\mathfrak F$ the Fourier transform,  we have that
$$
  \mathfrak F \circ  \frac{\partial} {\partial \xi _i} = 2\pi  i M_{i} \circ  \mathfrak F,
  \tag 1
  $$
where $M_i$ is the multiplication operator
$$
  M_i(f)|_\xi  = \xi _if(\xi ).
  $$
Another way to state (1) is to say that
$$
  \mathfrak F  \frac{\partial} {\partial \xi _i}  \mathfrak F^{-1}= 2\pi  i M_{i},
  $$
whence
$$
  \mathfrak F  \Delta    \mathfrak F^{-1}=
  \sum_{i=1}^3   \mathfrak F  \frac{\partial^2} {\partial \xi _i^2}  \mathfrak F^{-1} =
  -4\pi ^2  \sum_{i=1}^3   M_i^2 =   -4\pi ^2 M,
  $$
where $M$ is now the multiplication operator
$$
  M(f)|_\xi  = |\xi |^2f(\xi ).
  $$
Since an operator ($\Delta $) and its conjugate ($-4\pi ^2 M$) are essentially a mirror image of each other, all you need to do
is compute $(4\pi ^2 M+i)^{-1}$, and this is now much easier since you only have to deal with multiplication operators!
