Introduction and definition

Uniform continuity is a property on functions that is similar to but stronger than continuity. The usefulness of the concept is mainly due to the fact that it turns out that any continuous function on a compact set is actually uniformly continuous; in particular this is used to prove that continuous functions are Riemann integrable.


A function f : I → R (where I ⊂ R is an interval) is uniformly continuous if given ε > 0 there exists δ > 0 such that

|f(x) - f(y)| < ε for all x, y ∈ I with |x - y| < δ

Note that the definition is almost the same as for a function to be continuous. The only difference is that for continuity we allow the value of δ to depend on x, while for uniform continuity we require that the same δ works for all x ∈ I, so that the rate at which f(y) approaches f(x) as y approaches x is in some sense uniform.

In order to get a picture of what uniform continuity means we can prove the following simple result.


If f is differentiable on I and its derivative is bounded then f is uniformly continuous.


Let M be an upper bound for |f'(x)|. Then by the mean value theorem

|f(x) - f(y)| < M |x - y| for x, y ∈ I

Therefore given ε > 0 we may take δ = ε/M in the definition of uniform continuity.

This does of course only give us a sufficient condition for a function to be uniformly continuous but not a necessary one. Still it gives us a clue about what kind of function we would expect to be non-uniformly continuous.


f(x) = 1/x is not uniformly continuous on (0, ∞) since |f(x) - f(2x)| → ∞ as x → 0.
f(x) = x2 is not uniformly continuous on R, since for any δ > 0 |f(x+δ) - f(x)| → ∞ as x → ∞.
f(x) = cos(πx2) is not uniformly continuous on R since |f(sqrt(n+1)) - f(sqrt(n))| = 2 for all n ∈ N, but |sqrt(n+1) - sqrt(n)| → 0 as n → ∞.

Important results


Any continuous function on a closed and bounded interval I is uniformly continuous.


Suppose that there is a function f : I → R that is continuous but not uniformly continuous.
Since f is not uniformly continuous there is ε > 0 and two sequences xn, yn in I such that

|f(xn) - f(yn)| > ε for all n

but |xn - yn| → 0 as n → ∞.

Since I is closed and bounded the Bolzano-Weierstrass theorem asserts that xn has a convergent subseqence. Call the limit of this subsequence x.
Since f is continuous at x there is δ > 0 such that

|f(x) - f(y)| < ε/2 for |x - y| < δ

By definition of convergence we have that there is an N such that

|xN - yN| < δ/2
|xN - x| < δ/2

Hence by the triangle inequality

|xN - yN| ≤ |xN - x| + |yN - x| < ε

which contradicts |xn - yn| > ε for all n.
Thus any continuous function on I is uniformly continuous.

Together with the following result this gives a very important corollary.


Any uniformly continuous function f on [a, b] is Riemann integrable on [a, b].


Given any ε > 0 we wish to show that there is a δ such that for any divison of [a, b] with norm < δ the upper and lower sums in the definition of the Riemann integral differ by at most ε.
Since f is uniformly continuous there is a δ such that

|(f(x) - f(y)| < ε/(b-a) for |x - y| < δ

This δ will obviously do.


Any continuous function on [a, b] is Riemann integrable.

This corollary is obviously a fundamental result in one dimensional analysis.


The concept of uniform continuity can be naturally generalised to a property of functions between metric spaces. Our main theorem still holds (the proof is the same as before but with different notation) and is often useful.


Let (X, d), (Y, e) be metric spaces. A function f : X → Y is said to be uniformly continuous if given ε > 0 there is δ > 0 such that

e(f(x), f(y)) < ε for d(x, y) < δ


Let f : X → Y be continuous and suppose that (X, d) is compact. Then f is uniformly continuous.

Log in or register to write something here or to contact authors.