Uniform Continuity

Definition Let $f$ be defined on a set $S$. Then $f$ is continuous on $S$ (pointwise) if for every $c \in S$ and every $\varepsilon > 0$ there is $\delta > 0$ such that

Here $\delta$ may depend on both $\varepsilon$ and $c$. The function $f$ is uniformly continuous on $S$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that

Here $\delta$ depends only on $\varepsilon$. Every uniformly continuous function is continuous, but not the other way around.

Definition To prove that $f$ is uniformly continuous on $S$, start from an arbitrary $\varepsilon > 0$ and produce a single $\delta > 0$, written in terms of $\varepsilon$ only, such that $\lvert f(x) - f(y) \rvert < \varepsilon$ for all $x, y \in S$ with $\lvert x - y \rvert < \delta$. A common route is a bound $\lvert f(x) - f(y) \rvert \le K \lvert x - y \rvert$ valid on all of $S$ for one constant $K > 0$, which allows $\delta = \frac{\varepsilon}{K}$.

Definition If $f$ is continuous on a closed bounded interval $[a, b]$, then $f$ is uniformly continuous on $[a, b]$. The proof uses the completeness of the real numbers. Here is a sketch. If $f$ were not uniformly continuous, there would be some $\varepsilon > 0$ and, for each $n$, points $x_n, y_n$ in $[a, b]$ with $\lvert x_n - y_n \rvert < \frac{1}{n}$ but $\lvert f(x_n) - f(y_n) \rvert \ge \varepsilon$. The numbers $x_n$ form a bounded sequence, and the Bolzano-Weierstrass theorem, which comes back with sequences later in the course, says that some subsequence $x_{n_k}$ converges to a number $c$. This $c$ lies in $[a, b]$ because the interval is closed. The matching $y_{n_k}$ converge to the same $c$. Continuity at $c$ then gives $f(x_{n_k}) \to f(c)$ and $f(y_{n_k}) \to f(c)$, so $\lvert f(x_{n_k}) - f(y_{n_k}) \rvert \to 0$, which contradicts $\lvert f(x_{n_k}) - f(y_{n_k}) \rvert \ge \varepsilon$.

Definition On an interval that is unbounded or not closed, a continuous function may fail to be uniformly continuous. To show that $f$ is not uniformly continuous on $S$, find one $\varepsilon > 0$ such that for every $\delta > 0$ there are $x, y \in S$ with $\lvert x - y \rvert < \delta$ and $\lvert f(x) - f(y) \rvert \ge \varepsilon$. On a bounded open interval $(a, b)$, a continuous $f$ is uniformly continuous exactly when it has a continuous extension to $[a, b]$, that is, when $\lim_{x \to a^+} f(x)$ and $\lim_{x \to b^-} f(x)$ both exist as finite numbers. If the limits exist, the extension is continuous on $[a, b]$, so it is uniformly continuous there by the theorem above, and so is $f$ on the smaller set $(a, b)$. In the other direction, uniform continuity keeps the values of $f$ at any two points close to $a$ close to each other, and by completeness this forces the limit at $a$ to exist, and likewise at $b$.

Definition Let $f$ be continuous on $[a, b]$. By uniform continuity, for every $\varepsilon > 0$ there is $\delta > 0$ with the following property. If $[a, b]$ is cut into subintervals each of length less than $\delta$, then on every subinterval the largest and smallest values of $f$ differ by less than $\varepsilon$. For such a cut, the upper sum, built from the largest value of $f$ on each subinterval, and the lower sum, built from the smallest, differ by less than $\varepsilon (b - a)$, which can be made as small as we like. This is the key step in proving that every continuous function on $[a, b]$ is integrable, a fact used throughout the study of integrals.