I am struggling with the following concept in a rigorous sense. I am writing a paper where a Heaviside function arises and I need to compute the Fréchet derivative. Since I am a math researcher, my question is more on formalism than correctness, as I know a way to make everything correct without calling the below a "Fréchet derivative."
Consider the operator $G:C_0^1(\mathbb{R})\to C_0^1(\mathbb{R})$ given by $G(\psi)(z)=\int_\mathbb{R} J(x,z)H(\psi(x))\,\mathrm{d}x,$ with $\psi'\neq 0$ (say $>0$) when $\psi=0.$ Formally, the notation of a Fréchet derivative makes sense by saying $$ G'(\psi)(h)(z)=\int_\mathbb{R}J(x,z)\delta(\psi(x))h(x)\,\mathrm{d}x. $$ Since $\psi'>0$ when $\psi=0,$ there's an interval $(-\epsilon,\epsilon)$ such that $\psi$ is increasing when $\psi\in(-\epsilon,\epsilon).$ Then we can rewrite above as $$ G'(\psi)(h)(z)=\int_{-\epsilon}^{\epsilon} \frac{J(\psi^{-1}(x),z)h(\psi^{-1}(x))}{\psi'(\psi^{-1}(x))}\delta(x)\,\mathrm{d}x=\frac{J(\psi^{-1}(0),z)h(\psi^{-1}(0))}{\psi'(\psi^{-1}(0))}. $$ Assume that the space is given the usual norm $||\psi||=||\psi||_{\infty} + ||\psi'||_{\infty}$ to make it a Banach space. Clearly, this output is a legitimate function, but is it called a "Fréchet derivative"? I've tried arriving at this using the definition of a Fréchet derivative. i.e. the unique linear operator $A$ s .t. $$ G'(\psi)(h)(z)=\lim_{||h||\to 0}\frac{||G(\psi+h)-G(\psi)-Ah||}{||h||}=0. $$ However, I am unable to show that this holds other than by intuition. Is there a way to show this or is it really just a distributional derivative formalism and I can't really call it a Frechet derivative?