Denote by $\mathbb{E}(g)$ the solution of the PDE $$\Delta v(x,y) = 0 \quad\text{in $\Omega$}$$ $$v(x,0) = g(x) \quad\text{on $\partial\Omega$}.$$ Let $X=L^1(\partial\Omega)$. Let $h(t)$ be a differentiable function and define $A(t)\colon D(A) \subset X \to X$ by $$ A(t;v) = -\text{trace}\left(\frac{\partial \mathbb{E}(v^m)}{\partial y}\right)h(t),$$ $$D(A(t)) = D(A)= \{v \in L^\infty(\partial\Omega) \mid A(v) \in L^1(\partial\Omega), |{v}|_{L^\infty(\partial\Omega)} \leq |{f}|_{L^\infty(\partial\Omega)}\}.$$
It follows that $A(t)$ is accretive, that is, that $J_\lambda(A(t)) := (I+\lambda A(t))^{-1}$ is a contraction: $$|{J_{\lambda}(A) g_1-J_{\lambda}(A) g_2}|_{L^1(\partial\Omega)} \leq |{ g_1 - g_2}|_{L^1(\partial\Omega)}.$$
I want to show that there exists a continuous $F\colon [0,T] \to X$ and a monotone increasing function $L\colon [0,\infty) \to [0,\infty)$ such that $$ |{J_\lambda(A(t)) g - J_\lambda(A(s)) g}|_{X} \leq \lambda |{F(t) - F(s)}|_{X}L(|{ g}|_{X})\tag{1}$$ for $0 < \lambda$, $ \leq t, s \leq T$ and $ g \in \overline{D(A)}$.
The following is my attempt:
Let $u_t = J_{\lambda}(A(t)) g$ and $u_s$ similarly. So with $w=u^m$, $$\Delta_\Omega w_t = 0$$ $$\frac{\partial w_t}{\partial y} = \frac{ u_t - g}{\lambda h(t)} $$ and the weak formulation is $$\int_{\Omega}\nabla_\Omega w_t \nabla_\Omega \varphi = \int_{\partial\Omega}\varphi\left(\frac{ g- u_t }{\lambda h(t)}\right)$$ and subtracting one from the other: $$ \lambda\int_{\Omega}\nabla_\Omega ( w_t- w_s) \nabla_\Omega \varphi = \int_{\partial\Omega}\varphi\left(\frac{1}{h(t)}-\frac{1}{h(s)}\right)( g- u_s) - \varphi\left(\frac{ u_t- u_s}{h(t)}\right). $$ Now I pick $ \varphi = p_\delta( w_t - w_s)$ where $p_\delta$ is a monotone approximation of the sign function. Then we find $$ \lambda\int_{\Omega}p_\delta'( w_t - w_s)|\nabla_\Omega ( w_t- w_s)|^2 + \int_{\partial\Omega}p_{\delta}( w_t - w_s)\left(\frac{ u_t- u_s}{h(t)}\right) = \int_{\partial\Omega}p_{\delta}( w_t - w_s)\left(\frac{1}{h(t)}-\frac{1}{h(s)}\right)( g - u_s)$$ Throwing away the first term on the left (safe to assume it is positive) and passing to the limit: $$ \int_{\partial\Omega}\text{sign}( w_t - w_s)\left(\frac{ u_t- u_s}{h(t)}\right) = \int_{\partial\Omega}\text{sign}( w_t - w_s)\left(\frac{1}{h(t)}-\frac{1}{h(s)}\right)( g - u_s)\\ \leq \left|\frac{1}{h(t)}-\frac{1}{h(s)}\right|\int_{\partial\Omega}| g| + \left|\frac{1}{h(t)}-\frac{1}{h(s)}\right|\int_{\partial\Omega}| u_s| $$ Now note that $J_{\lambda}(A(s))0 = 0$, so $|{ u_s}|{} = |{J_{\lambda}(A(s)) g - J_{\lambda}(A(s))0}|{} \leq |{ g}|{}$ by contractivity. So the above becomes $$ \int_{\partial\Omega}| u_t- u_s| \leq 2h(t)\left|\frac{1}{h(t)}-\frac{1}{h(s)}\right||{g}|_{X} $$
This is as close as I get. I can probably bound the $h(t)$ from above so forget about that. The issue is that it obviously does not satisfy the desired (1). How do I get the $\lambda$ in there?!
Edit: The reason I need this condition is to obtain existence of PDE with the time-dependent operator $A(t)$ I defined above. This condition is from the paper by Crandall and Pazy I think. Maybe there is a different work that gives the same result with an easier condition to check on the time dependence?
