# Equicontinuous family composed with a continuous function

I have an equicontinuous family of functions $\mathcal{F}$, each $f:X\rightarrow X$ in the family is a function on a compact metric space $(X,d)$. I also have another continuous function $g:X \rightarrow \mathbb{R}$. I can use the composition $g\circ f$ to form a new family of functions $\mathcal{G} = \{g\circ f: f \in \mathcal{F}\}$, I wish to show that this new family is also equicontinuous.

My approach is as follows

• $\mathcal{F}$ has a $\delta$ satisfying the definition of continuity for all $f \in\mathcal{F}$ (Since $\mathcal{F}$ is equicontinuous).
• $g$ has a $\delta^{\prime}$ satisfying the definition of continuity.
• Use these facts to show that there exists some combination of the deltas above which gives us equicontinuity on $\mathcal{G}$.

However, I am stuck on the third step, in particular, the details of the combination of the deltas which allows us to form the equicontinuity of the composition.

Thanks

For $\epsilon > 0$, we let $\delta_f(\epsilon) > 0$ be defined such that $d(x_1, x_2) < \delta_f(\epsilon)$ implies $d(f(x_1), f(x_2)) < \epsilon$ for all $f\in \mathcal{F}$ and $\delta_g(\epsilon) > 0$ be defined such that $d(x_1, x_2) < \delta_g(\epsilon)$ implies $\lvert g(x_1)-g(x_2)\rvert < \epsilon$. Then, for $\epsilon > 0$, if $d(x_1, x_2) < \delta_f(\delta_g(\epsilon))$, we have that $d(f(x_1), f(x_2)) < \delta_g(\epsilon)$, or $\lvert g(f(x_1))-g(f(x_2))\rvert < \epsilon$ for all $f\in \mathcal{F}$.

For $X$ compact metric, the equicontinuity condition is "uniform" (see my remarks on EQ1 and EQ2 in my note here, if necessary), so

$$\forall \varepsilon>0 : \exists \delta>0: \forall f \in \mathcal{F}: \forall x,x' \in X: d(x,x') < \delta \implies d(f(x), f(x')) < \varepsilon$$

And your $g: X \to \mathbb{R}$ is uniformly continuous, as $X$ is compact, so

$$\forall \varepsilon > 0: \exists \delta>0: \forall x,x' < \delta : d(x,x') < \delta \implies |g(x)- g(x')| < \varepsilon$$

So start with $\varepsilon$, get $\delta$ from the second, and apply that $\delta$ as the $\varepsilon$ for the equicontinuiy I gave. Done.

• Im not following your explanation at the bottom, start with which $\epsilon$? and $\delta$ from the second argument? – jamesmartini May 23 '18 at 4:56