Diagonal part of Functional Derivative

Question

Given an analytic function $f$, define \begin{equation} J[G]=\mathrm{Tr}\log\big(1-if(G)\big) \end{equation} with $G=G(x,y)$ a function of two (real) variables and the logarithm defined in a convolutional sense as \begin{equation} \log\big(1-if(G)\big)(x,y)=-if\big(G(x,y)\big)-\frac12\int\text{d}z\ if\big(G(x,z)\big)if\big(G(z,y)\big)+\dots\ . \end{equation} Is it true that \begin{align} \frac{\delta J[G]}{\delta G(x,y)}=&-if'\big(G(x,y)\big)\big[1-if(G)\big]^{-1}(y,x)\\\\ =&-if'\big(G(x,y)\big)\Big(\delta(x-y)+\int\text{d}z\ \big[1-if(G)\big]^{-1}(y,z)if(z,x)\Big)\ ? \end{align} I am having trouble evaluating the above expression for a function $f$ depending only on the diagonal elements of $G$, say \begin{equation} f\big(G(x,y)\big)=e^{G(x,x)}\ . \end{equation} It is then \begin{equation} f'\big(G(x,y)\big)=\delta(x-y)e^{G(x,x)} \end{equation} right? Such that \begin{equation} \frac{\delta J[G]}{\delta G(x,y)}=-i\delta(x-y)^2e^{G(x,x)}+\dots\ . \end{equation} In the end I would like to decompose the functional derivative of $J$ into a diagonal part $\sim\delta(x-y)$ and an off-diagonal part, is my calculation correct? I encounter the dreaded $\delta(x-y)^2$ which I am not sure how to work with here... Any help is greatly appreciated.

Answer 1

After revisiting the problem I found that when \begin{equation} \frac{\delta f(x',y')}{\delta G(x,y)}=\frac12\big(\delta(x'-x)\delta(x'-y)+\delta(y'-x)\delta(y'-y)\big)A(x',y')+\delta(x'-x)\delta(y'-y)B(x',y')\ , \end{equation} as is the case for example with \begin{equation} f(x,y)=\exp\big\{\frac12\big(G(x,x)+G(y,y)\big)\big\}\ , \end{equation} the derivative evaluates to \begin{equation} \frac{\delta J[G]}{\delta G(x,y)}=-i\delta(x-y)\Big[\big(A*\big[1-if\big]^{-1}\big)(x,x)+B(x,x)\Big]+B(x,y)\big(\big[1-if\big]^{-1}*f\big)(y,x)\ . \end{equation}