Is my proof of the following identity correct?
Let the functional $F[\Psi(\xi);x,y]$ be the inverse of the functional $G[\Psi(\xi); x, y]$ in the following sense:
$\displaystyle \int dz F[\Psi(\xi);x,z] G[\Psi(\xi); z, y] = \delta(x- y), \quad \forall \Psi(\xi) \qquad (1)$
Why is the following identity true:
$\displaystyle \frac{\delta F[\Psi(\zeta);x,y]}{\delta \Psi(z)} = -\int d\xi d\eta F[\Psi(\zeta);x,\xi]\frac{\delta G[\Psi(\zeta); \xi, \eta]}{\delta \Psi(z)}F[\Psi(\zeta);\eta,y] \qquad (2)$
Proof:
Changing the dummy variables in (1) we have:
$\displaystyle \int d\xi F[\Psi(\zeta);x,\xi] G[\Psi(\zeta); \xi, \eta] = \delta(x - \eta)$
Taking the variation of the previous equation and multiplying by $F[\Psi(\zeta);\eta,y]$ we obtain:
$\displaystyle \int d\xi \Bigg\{ \frac{\delta F[\Psi(\zeta);x,\xi]}{\delta \Psi(z)}G[\Psi(\zeta); \xi, \eta]F[\Psi(\zeta);\eta,y] \\\displaystyle+ \frac{\delta G[\Psi(\zeta); \xi, \eta]}{\delta \Psi(z)}F[\Psi(\zeta);x,\xi]F[\Psi(\zeta);\eta,y]\Bigg\} = 0$
which integrated with respect to $\eta$ yields:
$\displaystyle \int d\xi \frac{\delta F[\Psi(\zeta);x,\xi]}{\delta \Psi(z)}\delta(\xi - y) + \int\int d\xi d\eta F[\Psi(\zeta);x,\xi]\frac{\delta G[\Psi(\zeta); \xi, \eta]}{\delta \Psi(z)}F[\Psi(\zeta);\eta,y] = 0$
where we have made use of (1) once again. Using the sifting property of the delta funcion leads to the desired result.
The first identity can be written as $FG=I,$ where $I$ is the identity kernel. We also assume that $GF=I.$ Taking derivatives of both sides gives $(FG)' = 0,$ i.e $F'G+FG'=0,$ leading to $F'G=-FG'.$ Multiply both sides with $F$ from the right to get $F'GF=-FG'F,$ i.e. $F'=-FG'F$ since $GF=I.$