I'm trying to prove the following result, Let $f:I\rightarrow[c,d]$ be bounded, let $g:[c,d]\rightarrow \mathbb{R}$ be Lipschitz with constant C. Then $$\sup_{x\in I}(g\circ f)-\inf_{x\in I}(g\circ f)\le C(\sup_{x\in I}(f)-\inf_{x\in I}(f))$$ My attempt was quite straight forward:
Since $g$ is Lipschitz we have: $$|g(f(x))-g(f(y))|\leq C|f(x)-f(y)|\forall x,y\in I\leq C(\sup_{x\in I}(f)-\inf_{x\in I}(f))$$ Which is close to the result but not quite close enough...
We notice that since $ g $ is Lipschitz we have: $$|g(f(x))-g(f(y))|\leq C|f(x)-f(y)| \leq C|\sup_{x\in I}f(x)-\inf_{x\in I}f(x)| \; \forall x,y \in I $$ Then we construct two sequences, $\{x_n \}$, $\{y_n \}$ such that $(g\circ f)(x_n) \rightarrow \sup_{x\in I}(g\circ f)(x) $ as $n\to \infty $ and $(g\circ f)(y_n) \rightarrow \inf_{x\in I}(g\circ f)(x) $ as $n\to \infty $.
We have: $$|g(f(x_n))-g(f(y_n))|\leq C|\sup_{x\in I}f(x)-\inf_{x\in I}f(x)| \; \forall n\in\mathbb{N} $$ Taking the limit as $n\to\infty$ on both sides gives: $$|\sup_{x\in I}(g\circ f)(x)-\inf_{x\in I}(g\circ f)(x)|\leq C|\sup_{x\in I}f(x)-\inf_{x\in I}f(x)|$$
Which is what we wanted.