Inverse of laplacian operator I recently read a paper, the author treats 
$$\int_{\mathbb{R}^d}f(y)\cdot \frac{1}{|x-y|^{d-2}}\,dx = (- \Delta)^{-1} f(y)$$
up to a constant in $\mathbb{R}^d$.
I am not familiar with unbounded operator, so my question is: Under what condition can one take the inverse of an unbounded operator like above? Can anyone refer some references? Thanks!
 A: If you associate the Laplacien with a bilinear form $ \langle \Delta u , u \rangle $, then if the boundary terms of $u$ vanish, you have $$\langle \Delta u , u \rangle = \int u \Delta u \; dx = \int \nabla u \cdot \nabla u \; dx$$ Then by Poincare's inequality, there exists some constant $\beta $, such that $$ \langle \Delta u , u \rangle = \int \nabla u \cdot \nabla u \; dx \geq \beta \int  u^2 \; dx.$$ 
But then if $ \Delta : H \rightarrow H$, where $H$ is some function space, then $ \Delta $ is one to one. To see this let $u_1 \neq u_2$, then if $-\Delta u_1 = -\Delta u_2$, then
$$ \langle \Delta (u_1 - u_2) , (u_1 - u_2) \rangle = \langle \Delta u_1 - \Delta u_2 , u_1 - u_2 \rangle = \langle 0, u_1 - u_2 \rangle = 0,$$
but 
$$ \langle \Delta (u_1 - u_2) , (u_1 - u_2) \rangle \geq \int (u_1 - u_2)^2 \; dx, $$
since $u_1 \neq u_2$. Thus $\Delta$ is one to one, and hence has a well defined inverse. 
So to answer your question, you can invert the operator if it is bounded below by the square of the norm of it's arguments. I think this is called the ellipticity condition.  
A: To make sense of this sort of problem, it's best to work with Distributions also known as generalized functions. The sort of solution you gave above is sometimes called  Greens function or a fundamental solution.
See for example Friedlander--Joshi Theory of Distributions.
