# 1.2 Continuity and convergence of functions

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Definitions of continuity and convergence for functions defined on metric spaces.

## Continuity for functions

Definition 1 A function $f:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ is continuous at a point ${x}_{0}\in X$ if: for any $ϵ>0$ there exists a $\delta >0$ such that if ${d}_{x}\left({x}_{0},{x}_{1}\right)<\delta$ , then ${d}_{y}\left(f\left({x}_{0}\right),f\left({x}_{1}\right)\right)<ϵ$ .

Definition 2 A function: $f:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ is uniformly continuous if: for every $ϵ>0$ there exists a $\delta >0$ such that for all ${x}_{0}\in X$ : if ${d}_{x}\left({x}_{0},{x}_{1}\right)<\delta$ , then ${d}_{y}\left(f\left({x}_{0}\right),f\left({x}_{1}\right)\right)<ϵ$ .

The difference between these definitions is that for a function to be simply continuous at a point, one need only find a $\delta$ for the given input ${x}_{0}$ , while for a function to be uniformly continuous, one needs to find for a given $ϵ$ a single value of $\delta$ that works in the definition for every point over which the function is defined.

Example 1 Consider the function $f:\left(C\left[T\right],{d}_{\infty }\right)\to \left(\mathbb{R},{d}_{0}\right)$ defined as $f\left(x\right)=x\left({t}_{0}\right)$ . In words, the input to $f$ is a continuous function over $T$ , and the output from $f$ is the value of the input function evaluated at $t={t}_{0}$ . We ask the question: Is $f$ continuous at some function input ${x}_{1}\left(t\right)$ ?

• Designate a pair of functions ${x}_{1}\left(t\right),{x}_{2}\left(t\right)$ such that: ${d}_{\infty }\left({x}_{1}\left(t\right),{x}_{2}\left(t\right)\right)<\delta$ , where
${d}_{\infty }\left({x}_{1}\left(t\right),{x}_{2}\left(t\right)\right)=\underset{t\in T}{sup}|{x}_{1}\left(t\right)-{x}_{2}\left(t\right)|.$
In other words, $su{p}_{t\in T}|{x}_{1}\left(t\right)-{x}_{2}\left(t\right)|<\delta$ .
• Next, we see that ${d}_{0}\left(f\left({x}_{1}\right),f\left({x}_{2}\right)\right)=|{x}_{1}\left({t}_{0}\right)-{x}_{2}\left({t}_{0}\right)|$ .
• Because $|{x}_{1}\left({t}_{0}\right)-{x}_{2}\left({t}_{0}\right)|\le {sup}_{t\in T}|{x}_{1}\left(t\right)-{x}_{2}\left(t\right)|$ , by the definition of the supremum, we can simply select $\delta =ϵ$ to get that if ${d}_{0}\left(f\left({x}_{1}\right),f\left({x}_{2}\right)\right)<\delta$ , then
${d}_{0}\left(f\left({x}_{1}\right),f\left({x}_{2}\right)\right)\le {d}_{\infty }\left({x}_{1}\left(t\right),{x}_{2}\left(t\right)\right)<\delta =ϵ.$

This shows the continuity of $f\left(x\right)$ at ${x}_{1}\left(t\right)$ . However, because the selection of $\delta$ did not depend on the value of ${x}_{1}$ , we have also shown that the function $f$ is uniformly continuous.

## Convergence of functions

Definition 3 The sequence of functions $\left\{{f}_{n}\right\}$ , ${f}_{n}:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ converges pointwise to $f:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ if: for each $x\in X$ , the sequence of values $\left\{{f}_{n}\left(x\right)\right\}$ converges to $f\left(x\right)$ in $\left(Y,{d}_{y}\right)$ .

Definition 4 The sequence of functions $\left\{{f}_{n}\right\}$ , ${f}_{n}:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ converges uniformly to $f:\left(X,{d}_{x}\right)\to \left(Y,{d}_{y}\right)$ if: for each $ϵ>0$ , there exists ${n}_{0}\in {\mathbb{Z}}^{+}$ such that if $n\ge {n}_{0}$ , then ${d}_{y}\left(f\left(x\right),{f}_{n}\left(x\right)\right)<ϵ$ for all $x\in X$ .

The difference between these definitions is that for uniform convergence, there must exist a single value of ${n}_{0}$ that works in the definition of continuity for all possible values of $x\in X$ .

Example 2 Consider the sequence of functions ${x}_{n}\left(t\right):\left(\left[0,1\right],{d}_{0}\right)\to \left(\left[0,1\right],{d}_{0}\right)$ given by ${x}_{n}\left(t\right)=\frac{t}{n}$ . One naturally suspects that ${x}_{n}\left(t\right)$ may be converging to the zero-valued function. We can show this formally as follows:

• Pick some ${t}_{0}$ , and check if $\left\{{x}_{1}\left({t}_{0}\right),{x}_{2}\left({t}_{0}\right),{x}_{3}\left({t}_{0}\right),{x}_{4}\left({t}_{0}\right),...\right\}$ is converging to 0: Denote ${a}_{n}={x}_{n}\left({t}_{0}\right)=\frac{{t}_{0}}{n}$ , and notice that ${d}_{0}\left({a}_{n},0\right)=|{a}_{n}-0|=\frac{{t}_{0}}{n}$ . So, to pick an ${n}_{0}$ such that ${a}_{n}<ϵ$ if $n\ge {n}_{0}$ , one only needs to note that since $\frac{{t}_{0}}{n}\le \frac{{t}_{0}}{{n}_{0}}$ , it then suffices for $\frac{{t}_{0}}{{n}_{0}}<ϵ$ , or in other words ${n}_{0}>\frac{{t}_{0}}{ϵ}$ . So this sequence converges pointwise to 0.
• Additionally, because the range of possible inputs to $x\left(t\right)$ is $t\in \left[0,1\right]$ , we could select ${n}_{0}>\frac{1}{ϵ}$ . Because 1 is the maximum value for ${t}_{0}$ , ${n}_{0}>\frac{1}{ϵ}>\frac{{t}_{0}}{ϵ}$ will work for all values of ${t}_{0}\in \left[0,1\right]$ , and so the sequence $\left\{{x}_{n}\right\}$ also converges uniformly to the zero function.

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