# Uniform Continuity

(redirected from*Uniformly continuous function*)

## uniform continuity

[′yü·nə‚fȯrm känt·ən′ü·əd·ē]*x*

_{1}) - ƒ(

*x*

_{2})| < ε="" provided="" |="">

*x*

_{1}-

*x*

_{2}| < δ="" for="" any="" pair="">

*x*

_{1},

*x*

_{2}in the set.

*The Great Soviet Encyclopedia*(1979). It might be outdated or ideologically biased.

## Uniform Continuity

an important concept in mathematical analysis. A function *f(x)* is said to be uniformly continuous on a given set if for every ∊ > 0, it is possible to find a number δ = δ(∊) > 0 such that ǀ*f(x*_{1}) - *f*(*x*_{2})ǀ < ∊ for any pair of numbers *x*_{1} and *x*_{2} of the given set satisfying the condition ǀ*x*_{1} - *x*_{2}ǀ < δ (*see*). For example, the function *f(x) = x ^{2}*, is uniformly continuous on the closed interval [0, 1] because if ǀ

*x*

_{1}- x

_{2}ǀ < ∊/2, then ǀ

*f*(

*x*

_{1}) –

*f*(

*x*

_{2})ǀ = ǀ

*x*

_{1}- x

_{2}ǀǀ

*x*

_{1}+ x

_{2}ǀ < ∊ (since necessarily ǀx

_{1}+

*x*

_{2}ǀ ≤ 2 when 0 ≤ x

_{1}≤ 1 and 0 ≤ x

_{2}≤ 1).

According to Cantor’s theorem, a function continuous at every point of a closed interval *[a, b]* is in general uniformly continuous on the interval. This theorem may not hold for an open interval. For example, the function *f(x)* = *1/x* is continuous at every point of the interval 0 < *x* < 1 but is not uniformly continuous on the interval. This can be shown as follows. Suppose ∊ = 1. For any δ > 0 (δ < 1), the numbers *x*_{1} = δ/2 and *x*_{2} = δ satisfy the inequality ǀ*x*_{1} - *x*_{2}ǀ < δ, but ǀ*f*(*x*_{1})- *f*(*x*_{2})ǀ = 1/δ > 1.