Showing $F(f)=\sin(f)$ is Continuous We have a Banach Space $C[0,1]$; consider a function $F:C[0,1]\to C[0,1]$, where $$(F(f))(t):=\sin(f(t))$$ and this is $\forall t\in [0,1]$ Prove F is continuous. I tried showing that F was a contaction, so then in a Banach Space, which is complete, this would insinuate that F is uniformly continuous, which is a stronger condition than I need to show. Maybe I'm approaching the question incorrectly, any tips? Thanks

This is easier than it sounds. Note that $|\sin x-\sin y| = |\int_x^y \cos t dt| \leq |x-y|$ for all $x\leq y \in \mathbb{R}$. In particular for any $f,g \in C[0,1],$ we have $ |\sin f(t)- \sin g(t)| \leq |f(t)-g(t)|$ for all $t \in [0,1]$, so $$ ||F(f)-F(g)|| = \max_{0\leq t \leq 1} |\sin f(t)-\sin g(t)| \leq \max_{0 \leq t \leq 1} | f(t)-g(t)| = ||f-g|| $$ which implies that $F$ is Lipschitz continuous (clearly stronger than continuous).

Thanks to Joonas who pointed out that my original answer only showed continuity at $f=0$.