# Diagonal part of Functional Derivative

Given an analytic function $$f$$, define $$$$J[G]=\mathrm{Tr}\log\big(1-if(G)\big)$$$$ with $$G=G(x,y)$$ a function of two (real) variables and the logarithm defined in a convolutional sense as $$$$\log\big(1-if(G)\big)(x,y)=-if\big(G(x,y)\big)-\frac12\int\text{d}z\ if\big(G(x,z)\big)if\big(G(z,y)\big)+\dots\ .$$$$ Is it true that \begin{align} \frac{\delta J[G]}{\delta G(x,y)}=&-if'\big(G(x,y)\big)\big[1-if(G)\big]^{-1}(y,x)\\\\ =&-if'\big(G(x,y)\big)\Big(\delta(x-y)+\int\text{d}z\ \big[1-if(G)\big]^{-1}(y,z)if(z,x)\Big)\ ? \end{align} I am having trouble evaluating the above expression for a function $$f$$ depending only on the diagonal elements of $$G$$, say $$$$f\big(G(x,y)\big)=e^{G(x,x)}\ .$$$$ It is then $$$$f'\big(G(x,y)\big)=\delta(x-y)e^{G(x,x)}$$$$ right? Such that $$$$\frac{\delta J[G]}{\delta G(x,y)}=-i\delta(x-y)^2e^{G(x,x)}+\dots\ .$$$$ In the end I would like to decompose the functional derivative of $$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.

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