Let $X = C([0,1],\mathbb{R})$ be the set of continuous functions on $[0,1]$ equipped with the sup-norm $d(f,g) := \sup\limits_{x\in [0,1]} \{|f(x)-g(x)|\}$ for each $f,g \in X$.
Define a function $F: X \rightarrow X$ by $F(g)(x) := \int_{0}^x \cos(\frac{g(t)}{2}) dt$.
Prove that for each $g \in X$ we have that $F(g) \in X$, i.e. $F(g)(x)$ is continuous.
Can I just do this: $F(g)'(x) = \cos(\frac{g(t)}{2})$, and $|\cos(\frac{g(t)}{2})| \leq 1$ then since the derivative is bounded, from the Mean Value Theorem, $F(g)(x)$ is Lipschitz continuous hence uniformly continuous?
Strange way to prove continuity for such a function... If you can prove that $$|F(f)-F(g)|\leq K\|f-g\|_X,$$ for some constant $K$, then you are done.
Hint
There is $C$ s.t. for all $a,b\in\mathbb R$, $$|\cos(a)-\cos(b)|\leq C|a-b|.$$