This is really a math question in which I will expose to the world my apparent lack of expertise with Greens functions, but it has appeared in a physics context so I guess it might be useful to somebody else. I am studying the $O(N)$ non-linear sigma model in two dimensional Euclidean space. The generating functional for correlation functions involving $\sigma^a$ fields is $$ Z[J^a]=\int d\alpha\,d\sigma\,\exp\bigg(\int d^2x\bigg[-\frac{1}{2}\partial_{\mu}\sigma^a\partial^{\mu}\sigma^a+\frac{1}{2}i\alpha\bigg(\sigma^a\sigma^a-\frac{N}{t_0}\bigg)\bigg]+\int d^2xJ^a\sigma^a\bigg) $$ Upon integrating the $\sigma^a$ fields, which is fairly easy as the functional integral is just a Gaussian we arrive at $$ Z[J^a]=\int d\alpha\exp(\frac{-N}{2} tr\log(-\partial^2-i\alpha)-\frac{i}{2}\frac{N}{t_0}\int d^2x\alpha+\frac{1}{2}\int d^2x\,d^2yJ^a(x)\Delta(x-y)J^a(y)) $$ and $\Delta(x-y)$ is $$ (-\partial_x^2-i\alpha(x))\Delta(x-y)=\delta(x-y) $$ The Green function of the linear operator $(-\partial_x^2-i\alpha(x))$. This object is the subject of my question. In the next line ( I am reading the book Instantons and Large N by Marcos Mariño https://www.cambridge.org/core/books/instantons-and-large-n/B031949424E5E22418E376A5CE73C615 chapter 6) the author claims that $$ \Delta^{-1}(x-y)=(-\partial_x^2-i\alpha(x))\delta(x-y) $$ I wanna figure out in which sense this object above is the inverse of $\Delta$. To simplify notation I will call $L\equiv (-\partial_x^2-i\alpha(x))$. Thus, by definition $$ L\Delta(x-y)=\delta(x-y) $$ and also according to the author $$ \Delta^{-1}(x-y)=L\delta(x-y) $$ Thus we have that $$ L^2\Delta(x-y)=L\delta(x-y)=\Delta^{-1}(x-y) $$ I expect that multiplying $\Delta$ and its alleged inverse we get the Dirac delta somehow, so I try $$ \Delta^{-1}(x-y)\Delta(y-z)=(L^2\Delta(x-y))\Delta(y-z) $$ and I do not know how to proceed from here. At least not in a clean way. I could try $$ =(L\delta(x-y))\Delta(y-z) $$ and now integrate in the $y$ variable $$ \int d^2y\,\Delta^{-1}(x-y)\Delta(y-z)=\int d^2y(L\delta(x-y))\Delta(y-z) $$ and very very naively use the Dirac delta above to integrate over the $y$ variable $$ =L\Delta(x-z)=\delta(x-z) $$ which is the sought after relation. Nevertheless, the last step is very distasteful to me because when I integrated over the delta I still had a linear operator acting on it. How can this argument be made rigorous(er at least)?
2 Answers
What's going on here is $\Delta(x;y)=\Delta(x-y)$ is the integration kernel of a linear operator, \begin{align} L^{-1}f(x)= \int\Delta(x-y)\,f(y)\operatorname{d}y. \end{align} Likewise, the operator $L$ has an integration kernel \begin{align} Lf(x)=\int\Delta^{-1}(x-y)\,f(y)\operatorname{d}y. \end{align}
The relationship between and operator in function space and integration kernel is analogous to the relationship between a linear operator on a finite dimensional vector space and the matrix elements of a representation of that operator in a particular basis.
So, we say that $LL^{-1}$ gives us the identity operator $I$, by definition of the operator inverse. We then say that the integration kernel represents the operator in a particular basis, and we denote representation with an arrow; so \begin{align} MM^{-1}&\rightarrow \delta_{ij} \ \mathrm{just\ as}\\ LL^{-1}&\rightarrow \delta(x-y), \end{align} where $M$ is an operator on a finite dimensional space.
For example, you can show using integration by parts with $x\in(a,b)$ that the ordinary derivative is represented by \begin{align} f'(x) &= \int_a^b \delta(x-y)\, f'(y)\operatorname{d}y \\ &= \left.f(y)\delta(x-y)\right|_{y=a}^b-\int_a^b \left[\frac{\partial}{\partial y}\delta(x-y)\right] f(y)\operatorname{d}y\\ &=\int_a^b \left[\frac{\partial}{\partial x}\delta(x-y)\right] f(y)\operatorname{d}y \\ &=\int_a^b \delta'(x-y)\, f(y)\operatorname{d}y\\ \frac{\operatorname{d}}{\operatorname{d}x} &\rightarrow \delta'(x-y)=-\delta'(y-x). \end{align}
It is in this sense that \begin{align} L&\rightarrow \Delta^{-1}(x-y)\ \mathrm{and}\\ L^{-1}&\rightarrow \Delta(x-y), \end{align} where the defining property of $\Delta^{-1}(x-y)$ is $$\int \Delta^{-1}(x-z)\,\Delta(z-y)\operatorname{d}z=\delta(x-y),$$ which is a representation of the definition of an operator inverse.
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To streamline @SeanELake 's impeccable argument, you only need to recall these operators are associative, and the definition of the inverse, $$\int dz ~\Delta(x-z)\,\Delta^{-1}(z-y) =\delta(x-y).$$
You then merely operate your definition on $\Delta^{-1}$, $$ \int dy ~L(x)\Delta(x-y) \Delta^{-1}(y-z)=\int dy ~\delta(x-y) \Delta^{-1}(y-z)~. $$ The r.h.side is just $\Delta^{-1}(x-y)$, but grouping the last two factors in the l.h.side into the definition of the inverse, it reduces to $L(x)\delta(x-z)$, that is, $$ L\delta(x-z)=\Delta^{-1}(x-z). $$
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