---- > [!definition] Definition. ([[equicontinuous]]) > Let $X$ be a [[topological space]] and let $(Y,d)$ be a [[metric space]]. Let $F \subset C(X,Y):=\text{Hom}_{\mathsf{Top}}(X,Y)$. We call the collection $F$ of [[continuous]] functions **equicontinuous under $d$ at $x_{0} \in X$** if for all $\varepsilon>0$ there exists a [[neighborhood]] $U \ni x_{0}$ such that $d\big( f(x), f(x_{0}) \big)<\varepsilon $for all $f \in F$ and $x \in U$. We call $F$ **equicontinuous under $d$** if it is equicontinuous at all $x_{0} \in X$. > Note that equicontinuity depends on the specific metric $d$ rather than merely on the [[topological space|topology]] of $Y$. ^definition > [!equivalence] Equivalent definition when $X$ is a metric space. > When $X$ is also a [[metric space]], equicontinuity of $F$ at $x_{0} \in X$ is equivalent to the following: for all $\varepsilon>0$, there exists $\delta=\delta(\varepsilon, x_{0})>0$ such that $d\big( x, x_{0} \big)< \delta \implies d\big( f(x), f(x_{0}) \big)< \varepsilon$ for all $f \in F$. If there exists a single $\delta=\delta(\varepsilon)$ that works for all $x_{0}$, then $f$ is said to be **uniformly equicontinuous**. The situation for metric spaces is thus as follows: > > > > | $\delta$ may depend on: | $\varepsilon$ | $f$ | $x_{0}$ | > | -------------------------------------------- | ------------- | --------- | --------- | > | [[continuous\|Pointwise continuity]] | $\bullet$ | $\bullet$ | $\bullet$ | > | [[uniformly continuous\|Uniform continuity]] | $\bullet$ | $\bullet$ | | > | [[equicontinuous\|Pointwise Equicontinuity]] | $\bullet$ | | $\bullet$ | > | [[equicontinuous\|Uniform equicontinuity]] | $\bullet$ | | | > > > [!basicexample] > [[totally bounded]] under [[uniform metric]] $\implies$ equicontinuos ([[totally bounded under uniform metric implies equicontinuous|see here]]) ^basic-example - $X$ a [[norm|normed space]]. For a collection $\mathcal{F} \subset X^{*}$ of [[linear functional|linear functionals]], $\mathcal{F}$ bounded (in operator norm) implies $\mathcal{F}$ [[equicontinuous]]. Indeed suppose there exists $K>0$ such that $\|f\|<K$ for all $f \in \mathcal{F}$. We claim $\delta:=\frac{\varepsilon}{K}$ suffices to show unform equicontinuity of $\mathcal{F}$. Indeed, for any $x_{0} \in X$, $\|x-x_{0}\|<\delta$ implies $|f(x)-f(x_{0})|=|f(x-x_{0})|\leq \underbrace{ \|f\| }_{ < K } \underbrace{ \|x-x_{0}\| }_{ < \frac{\varepsilon}{K} }<\varepsilon.$ Linearity was crucial here! ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```