Let assume $ X= \Omega \times (0,T) $ where $\Omega \subset \mathbb{R} $ is a bounded open set and $T>0$.
Now define the operator $ \mathcal{A} : C^{\sigma, \sigma/2}(X) \to C^{\sigma, \sigma/2}(X) $ in the following way:
$$ \mathcal{A}(\phi)(t, x)= \\ \nonumber A_0(x) \mathcal{B}(\phi)(t, x, 0)+ \int_0^t a(s, x)b(s, x) | \phi(s, x)| \mathcal{B}(\phi)(t,x,s) ds $$
such that
$$ \mathcal{B}(\phi)(t,x,s)=\chi_{[s, T]}(t) \, \exp\Big(-\int_s^t \frac{a(\zeta,x) b(\zeta, x)}{c(\zeta, x)}\vert \phi(\zeta, x)\vert \: d\zeta\Big) \\ \nonumber \times \exp \Big(-\int_s^t [d(\zeta, x)+e(\zeta, x)+f(\zeta, x) \chi_{\omega}]\: d\zeta \Big).$$
where
$$ a(t,x) , b(t,x) , c(t,x), d(t,x), e(t,x), f(t,x) \in C^{\sigma, \sigma/2} (\overline{\Omega} \times (0, T)),$$
$\omega \subset \Omega$ and $A_0 \in C^{\sigma}(\overline{\Omega}).$
In solving a problem related to my thesis, for proving existence of a local solution by classical Contraction Mapping Theorem, I need to prove that the above operator is lipschitz with the standard norm of $C^{\sigma, \sigma/2}$, but I can't succeed.
Can someone help me to know whether this opeator can be lipschitz or not, in order to use other way to find a local solution instead of classical Contraction Mapping Theorem.
Thank you.
