Let $$ X=C([0,1]) $$ $$ f: [0,1] \rightarrow {R} $$ with supremum metric:
$$ d(f,g)=\sup_{x \in [0,1]}|f(x)-g(x)| $$ where $$ f,g \in X $$ Check if $$F: X \rightarrow X $$ $$ F(f)=\arctan{(f(x))} $$ is continous
My attempt: I want to say that this function fulfils Lipschitz contidion. I have: $$ d(F(f), G(g))= \sup_{x \in [0,1]}|\arctan{f(x)}-\arctan{g(x)}|$$ This is the point where I am stuck. I wanted to say that $$ \sup_{x \in [0,1]}|\arctan{f(x)}-\arctan{g(x)}|\leqslant \sup_{x \in [0,1]}|f(x)-g(x)|$$ Then it would be everything, but I'm not sure that this is true. Can you help me with this?
Actually,
Use Lagrange inequality giving :
$$ \exists c\in]0,\frac{\pi}{2}[, \forall(p,q)\in]0,\frac{\pi}{2}[ , (\arctan(p) -\arctan(q)) \leq |p-q|*\arctan'(c) $$