= Solution
For the <maximum-entropy regularization functional>, differentiation of the scalar integrand gives $(s\log s-s)'=\log s$. Thus the <Fréchet derivative> at a positive reference function $y$ is $E'(y)=\log y$, and the stated differentiability result gives $\partial E(y)=\{\log y\}$. Consequently,
$$
\boxed{D_E(x,y)=\int_\Omega\left[x\log\frac{x}{y}-x+y\right]dt.}
$$
This is the <generalized Kullback–Leibler divergence>. When both functions integrate to the same mass, the last two terms cancel after integration; for <probability density functions> it becomes the usual <Kullback-Leibler divergence>. For two positive functions, adding the reverse distance gives
$$
\boxed{D_E^{\mathrm{sym}}(x,y)=\int_\Omega(x-y)(\log x-\log y)\,dt.}
$$
A sufficient setting for the differentiability argument is $L^\infty(\Omega)$ with $y$ bounded away from zero. For nonnegative $x$, use $0\log(0/y)=0$ wherever appropriate. Positivity alone in $L^2$ does not automatically provide an open domain on which the <Fréchet derivative> exists.
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