Let $f: \mathbb{R}\longrightarrow \mathbb{R}$ be a locally Lipschitz continuous function. Meaning that: $$ \forall M>0,\quad \exists L_M>0, \quad \max\{{x},{y}\}\leq M \Longrightarrow \vert f(x)-f(y)\vert \leq L_M \vert x-y\vert. $$ Consider the following mollifier $$ \forall x\geq 0, \forall k>0,\quad \eta_k(x)=\begin{cases}1\quad \text{if}\quad x\leq k; \\ 0\quad \text{if}\quad x \geq 2k. \end{cases} $$ Then the Nemytskii operator \begin{equation} \begin{split} F: &L^2(\Omega) \longrightarrow L^2(\Omega)\\ &u\mapsto F(\eta_k(\Vert u \Vert_{L^2(\Omega)}) u) \end{split} \end{equation} where $\Omega \subset \mathbb{R}^n$ is a bounded open set with regular boundary, and $$ F(u)(x)=f(u(x)), \quad \forall x\in \Omega. $$ A published article states that the operator $F$ is globally Lipschitz?
However, I have tried to distinguish the cases. For the first case, I let $k>0,$ then $$\max\{\Vert u \Vert_{L^2(\Omega)},\Vert v \Vert_{L^2(\Omega)}\} \leq k \Longrightarrow \Vert F(u)-F(v) \Vert_{L^2(\Omega)}^2=\int_\Omega \vert f(u(x))-f(v(x))\vert^2 dx.$$ I can't estimate the right hand side by using just the Local Lipschitzity of $f$ since I don't have $$ \max \{ \vert u (x) \vert,\vert v(x) \vert\} \leq k.$$
Am I approaching this in the wrong way?