Home page for accesible maths 4 Continuity vs. discontinuity 4.1 Continuous functions 4.3 Discontinuities. Variation of a theme

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4.2 Uniform continuity

Let us consider the continuous function $f:x\to x^{2}$ . The continuity of $f$ at a certain $x\in\mathbb{R}$ means that for any $\varepsilon>0$ there exists a $\delta>0$ such that $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta$ . Let us pick a good $\delta$ for a given $0<\varepsilon<1$ . If $x=0$ , then $\delta=\sqrt{\varepsilon}$ is a good choice. On the other hand, even if you pick a tiny little $\delta$ , it will not be good for ALL $x$ . Indeed, let $x=\frac{1}{\delta}$ . Then

f(x+\delta)-f(x)=(\frac{1}{\delta}+\delta)^{2}-\frac{1}{\delta^{2}}=2+\delta^{% 2}>2>\varepsilon\,.

This little example shows that if $\varepsilon$ is given, we need to pick smaller and smaller $\delta$ for larger and larger $x$ . The situation is completely different if we have a continuous function $f:[a,b]\to\mathbb{R}$ .

Theorem 4.2.1 (The Uniform Continuity Theorem)

Let $f:[a,b]\to\mathbb{R}$ be a continuous function. Then for any $\varepsilon>0$ there exists a $\delta>0$ such that for ALL $x$ : $|f(x)-f(y)|<\varepsilon$ , if $|x-y|<\delta$ .

Proof: We prove the theorem by contradiction. What does it mean that we cannot find a good $\delta$ for our $\varepsilon>0$ ??? It means that there exists a sequence of numbers $\{x_{n}\}^{\infty}_{n=1}\subset[a,b]$ for which we need smaller and smaller $\delta$ ’s. That is: there exists a sequence $\{y_{n}\}^{\infty}_{n=1}$ such that $|f(x_{n})-f(y_{n})|\geq\varepsilon$ , but $|x_{n}-y_{n}|\to 0$ . Indeed, since $\frac{1}{n}$ is not a good $\delta$ for a certain $x_{n}$ , there exists $y_{n}$ such that $|f(x_{n})-f(y_{n})|\geq\varepsilon$ but $|x_{n}-y_{n}|<\frac{1}{n}.$

Now we use the fact that our function is defined on a bounded interval. By the Bolzano-Weierstrass Theorem, we have a convergent subsequence $\{x_{n_{k}}\}\to x$ . By the Sum Rule, $\{y_{n_{k}}\}\to x$ , as well. Since $f$ is continuous at $x$ , $\lim_{k\to\infty}f(x_{n_{k}})=f(x)$ and $\lim_{k\to\infty}f(y_{n_{k}})=f(x)$ . Hence $|f(x_{n_{k}})-f(y_{n_{k}})|\to 0$ , in contradiction with the assumption that for all $n>1$ , $|f(x_{n})-f(y_{n})|\geq\varepsilon$ . $\square$