A continuous function on a closed interval is uniformly continuous

The notion of uniform continuity is a ``stronger'' version of (simple) continuity. If a function is uniformly continuous, it is continuous, but the converse does not generally hold (that is, a continuous function may not be uniformly continuous). However, if we restrict a continuous function on a closed interval, it is always uniformly continuous.

Definition (Uniform continuity)

The function $f(x)$ on an interval $I$ is said to be uniformly continuous on $I$ if it satisfies the following condition.

• For any $\epsilon >0$, there exists $\delta >0$, such that, for all $x,y\in I$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon$.

  In a logical form, this condition is expressed as 

$$\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,\mathrm{\forall }x,y\in I(|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon ).$$

Remark. Compare the above condition for uniform continuity with the condition for the continuous function on $I$:

$$\mathrm{\forall }y\in I,\mathrm{\forall }\epsilon >0,\mathrm{\exists }\delta >0,\mathrm{\forall }x\in I(|x-y|<\delta ⟹|f(x)-f(y)|<\epsilon ).$$

What's the difference? In the uniform continuity, $\delta$ does not depend on $x$ or $y$, whereas in the (ordinary) continuity on an interval, $\delta$ may depend on $y$. The latter means that we may need to choose different values of $\delta$ depending on where we are in the interval $I$. In uniform continuity, on the other hand, we can choose a constant $\delta$ irrespective of where we are in the interval $I$. In this sense, uniform continuity imposes a more stringent condition on the function. □

Example. The function $f(x)=x^2$ is continuous on the entire $\mathbb{R}$, but it is not uniformly continuous on $\mathbb{R}$. To see this, let $\epsilon =2$. If $f(x)$ were uniformly continuous, we could choose some $\delta$ such that for all $x,y\in \mathbb{R}$, $|x-y|<\delta$ implies $|x^2-y^2|<2$. Now, let us pick a sufficiently large natural number $n$ such that $\frac{1}{n}<\delta$ (which is always possible), and let $x=n+\frac{1}{n}$ and $y=n$. Then $|x-y|=\frac{1}{n}<\delta$, but $|x^2-y^2|=2+\frac{1}{n^2}>2$, which is a contradiction. □

Example. Consider again the function $f(x)=x^2$, but this time we restrict its domain to some finite closed interval $[a,b]$. Then, this function is uniformly continuous on $[a,b]$. Let $c=max\{|a|,|b|\}$. For any $\epsilon >0$, let $\delta =\frac{\epsilon }{2c}$. Then for all $x,y\in [a,b]$, if $|x-y|<\delta$, noting $|x|,|y|\le c$,

$$|x^2-y^2|=|x-y||x+y|<\delta \cdot |x+y|\le \delta \cdot (|x|+|y|)\le \delta \cdot 2c=\epsilon .$$ □

More generally, we have the following theorem.

Theorem

A continuous function on a closed interval is uniformly continuous.

Proof. We prove this by contradiction. Suppose the function $f(x)$ that is continuous on the closed interval $[a,b]$ is not uniformly continuous. Then, the following holds:

• There exists some $\epsilon >0$ such that, for all $\delta >0$, there exists $x,y\in [a,b]$ such that $|x-y|<\delta$ and $|f(x)-f(y)|\ge \epsilon$.

Now for each $n\in \mathbb{N}$, let $\delta =\frac{1}{n}$, and there exist $x_n,y_n\in [a,b]$ such that

$$|x_n-y_n|<\frac{1}{n}\text{ and }|f(x_n)-f(y_n)|\ge \epsilon .$$

Thus we can define sequences $\{x_n\}$ and $\{y_n\}$ bounded on $[a,b]$.

  By the Bolzano-Weierstrass theorem, $\{x_n\}$ contains a subsequence $\{x_{n_k}\}$ that converges to some $\alpha \in [a,b]$. Using these indices $n_1,n_2,\cdots$ in $\{x_{n_k}\}$, we can define a subsequence $\{y_{n_k}\}$ of $\{y_n\}$. But $\{y_{n_k}\}$ contains yet another subsequence that converges to some $\beta \in [a,b]$. By reindexing, let us denote this converging subsequence by $\{y_{n_k}\}$. After this reindexing, $\{x_{n_k}\}$ still converges to $\alpha$ (Theorem: a subsequence of a converging sequence converges to the same value.). Then $x_{n_k}-y_{n_k}$ converges to $\alpha -\beta$. However, for any $k$, by assumption, we have $-\frac{1}{n_k}<x_{n_k}-y_{n_k}<\frac{1}{n_k}$ and $n_k\to \mathrm{\infty }$ as $k\to \mathrm{\infty }$ so that $\alpha -\beta =0$ or $\alpha =\beta$. On the one hand, $f(x)$ is continuous so that $\underset{k\to \mathrm{\infty }}{lim}f(x_{n_k})=\underset{k\to \mathrm{\infty }}{lim}f(y_{n_k})=f(\alpha )$. On the other hand, for any $k$, $|f(x_{n_k})-f(y_{n_k})|\ge \epsilon$ so that $\underset{k\to \mathrm{\infty }}{lim}|f(x_{n_k})-f(y_{n_k})|\ge \epsilon$. Hence we have

$$0=|f(\alpha )-f(\alpha )|=\underset{k\to \mathrm{\infty }}{lim}|f(x_{n_k})-f(y_{n_k})|\ge \epsilon >0$$

which is a contradiction. ■