1.2 Continuity and convergence of functions

Continuity for functions

Definition 1 A function $f:(X,d_x)\to (Y,d_y)$ is continuous at a point $x_0\in X$ if: for any $ϵ>0$ there exists a $\delta >0$ such that if $d_x(x_0,x_1)<\delta$ , then $d_y(f\left(x_0\right),f\left(x_1\right))<ϵ$ .

Definition 2 A function: $f:(X,d_x)\to (Y,d_y)$ is uniformly continuous if: for every $ϵ>0$ there exists a $\delta >0$ such that for all $x_0\in X$ : if $d_x(x_0,x_1)<\delta$ , then $d_y(f\left(x_0\right),f\left(x_1\right))<ϵ$ .

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 $f:(C\left[T\right],d_{\infty })\to (\mathbb{R},d_0)$ 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: $d_{\infty }(x_1\left(t\right),x_2\left(t\right))<\delta$ , where
  $$d_{\infty }(x_1\left(t\right),x_2\left(t\right))=\underset{t\in T}{sup}|x_1\left(t\right)-x_2\left(t\right)|.$$
  In other words, $su{p}_{t\in T}|x_1\left(t\right)-x_2\left(t\right)|<\delta$ .
• Next, we see that $d_0(f\left(x_1\right),f\left(x_2\right))=|x_1\left(t_0\right)-x_2\left(t_0\right)|$ .
• 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 $d_0(f\left(x_1\right),f\left(x_2\right))<\delta$ , then
  $$d_0(f\left(x_1\right),f\left(x_2\right))\le d_{\infty }(x_1\left(t\right),x_2\left(t\right))<\delta =ϵ.$$

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 $\left\{f_n\right\}$ , $f_n:(X,d_x)\to (Y,d_y)$ converges pointwise to $f:(X,d_x)\to (Y,d_y)$ if: for each $x\in X$ , the sequence of values $\left\{f_n\left(x\right)\right\}$ converges to $f\left(x\right)$ in $(Y,d_y)$ .

Definition 4 The sequence of functions $\left\{f_n\right\}$ , $f_n:(X,d_x)\to (Y,d_y)$ converges uniformly to $f:(X,d_x)\to (Y,d_y)$ if: for each $ϵ>0$ , there exists $n_0\in {\mathbb{Z}}^{+}$ such that if $n\ge n_0$ , then $d_y(f\left(x\right),f_n\left(x\right))<ϵ$ 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 $x_n\left(t\right):([0,1],d_0)\to ([0,1],d_0)$ 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 $\{x_1\left(t_0\right),x_2\left(t_0\right),x_3\left(t_0\right),x_4\left(t_0\right),...\}$ is converging to 0: Denote $a_n=x_n\left(t_0\right)=\frac{t_0}{n}$ , and notice that $d_0(a_n,0)=|a_n-0|=\frac{t_0}{n}$ . 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 $t\in [0,1]$ , 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 $t_0\in [0,1]$ , and so the sequence $\left\{x_n\right\}$ also converges uniformly to the zero function.