Let $(X, d)$ be a compact metric space, and $\mathscr{F}$ an equicontinuous family of functions from $X$ to itself. Suppose that $g: X → R$ is continuous.
Show that the family $\mathscr{G} = \{g ◦ f: f \in \mathscr{F} \subset C(X, R)\}$ is equicontinuous and that $\overline{\mathscr{G}}$ is compact.
I have finished the proof of the equicontinuous and because $X$ is compact, it is actually uniformly equicontinuous. But I'm stuck in the proof of $\overline{\mathscr{G}}$ is compact. I try to apply the Arzela-Ascoli theorem and prove that $\overline{\mathscr{G}}$ is uniformly bounded and uniformly equicontinuous and so it is compact. But I'm stuck. I also try that assume $\overline{\mathscr{G}}$ is not compact and apply the Arzela-Ascoli theorem show that $\mathscr{G}$ is not uniformly equicontinuous so get the contradiction. But I'm also stuck because I don't know how to prove $\overline{\mathscr{G}}$ is bounded.
Can anyone give me some hints?
For every $f \in \mathscr{F}$ we have $$(g \circ f)(X) = g(f(X)) \subseteq g(X)\,.$$ Now $X$ is compact and $g$ continuous, hence $g(X)$ is a compact subset of $\mathbb{R}$, thus there is a $c > 0$ such that $g(X) \subseteq [-c,c]$. This implies $$\lvert g(f(x))\rvert \leqslant c$$ for all $f \in \mathscr{F}$ and $x \in X$. This global weak inequality is preserved under pointwise limits (a fortiori under uniform limits), hence $$\lvert h(x)\rvert \leqslant c$$ holds for all $h \in \overline{\mathscr{G}}$ and all $x \in X$.