2
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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 $f\in \mathcal{F}$.

$\endgroup$
0
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.