Let $\Omega\subset\mathbb{R}^n$ be a bounded open set . Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that $\displaystyle\lim_{s\to\pm\infty}\frac{f(s)}{s}=f_{\pm}$ with $f_{\pm}$ finite . Show that the mapping $u\to f(u)$ defined by $f(u)(x)=f(u(x))$ is continuous from $L^2(\Omega)$ into itself .
It is clear that $f$ is satisfying a growth condition $|f(x)|\leq M_1|x|$ for all $|x|>R>0$ , also by uniform continuity $|f(x)|\leq M_2$ for all $|x|\leq R$ . Now $$|f(u)(x)-f(v)(x)|=|f(u(x))-f(v(x))|\leq|f(u(x))-u(x)|+|u(x)-v(x)|+|v(x)-f(v(x))|$$ for all $u,v\in L^2(\Omega)$ . Afterwards how to estimate the RHS with only $|u(x)-v(x)|$ ? I suspect this has something to do with fixed point theorem but I am not getting the correct idea to apply it . Any help is appreciated .
