Uniform continuity

In mathematical analysis, a function "f"("x") is called uniformly continuous if, roughly speaking, small changes in the input "x" effect small changes in the output "f"("x") ("continuity"), and furthermore the "size" of the changes in "f"("x") depends only on the size of the changes in "x" but not on "x" itself ("uniformity").

Continuity itself is a "local" property of a function—that is, a function "f" is continuous, or not, at a particular point, and when we speak of a function being continuous on an interval, we mean only that it is continuous at each point of the interval. In contrast, uniform continuity is a "global" property of a function.

Uniform continuity, unlike continuity, is meaningless in an arbitrary topological space, since it relies on comparing sizes of open sets at distant points of a space. Instead, uniform continuity can be defined in a metric space where such comparisons are possible, or in a uniform space.

Definition

Given metric spaces $(X,\; d\_1)$ and $(Y,\; d\_2)$, a function $f\; :\; X\; o\; Y$ is called uniformly continuous if for every real number $epsilon\; >\; 0$ there exists $delta\; >0$ such that for every $x,y\; in\; X$ with

If $X$ and $Y$ are subsets of the real numbers, $d\_1$ and $d\_2$ can be the standard Euclidean norm, $|cdot|$, yielding the definition: for all $epsilon\; >\; 0$ there exists a $delta>0$ such that for all $x,y\; in\; X$, implies

Properties

It is an immediate consequence of the definitions that every uniformly continuous function is continuous.

The converse does not hold. Consider for instance the function $f:\; R\; ightarrow\; R,\; x\; mapsto\; x^2$. Let $epsilon$ be any positive real number. Then uniform continuity requires the existence of a positive number $delta$ such that for all $x\_1,\; x\_2$ with , we have . But for any positive number $delta$, we have $f(x+delta)-f(x)\; =\; 2xdelta\; +\; delta^2\; =\; (delta)(2x+delta)$, and for all sufficiently large "x" this quantity is greater than $epsilon$.

If $X\; subset\; R$, $f:\; X\; ightarrow\; R$ is uniformly continuous and $S\; subset\; X$ is bounded, then $f(S)$ is a bounded subset of "R". In particular, the function $x\; mapsto\; frac\{1\}\{x\}$ from (0,1) to R is continuous but not uniformly continuous.

More generally, the image of a totally bounded subset under a uniformly continuous function is totally bounded. Beware that the image of a bounded subset of an arbitrary metric space under a uniformly continuous function need not be bounded. For instance, consider the identity function from the integers endowed with the discrete metric to the integers endowed with the usual Euclidean metric.

The Heine-Cantor theorem asserts that if "X" is compact, then every continuous "f" : "X" → "Y" is uniformly continuous.

In particular, we have the following important uniform continuity theorem: if a function is continuous at every point of a closed bounded interval, it is uniformly continuous on that interval. This theorem is to be compared to the Intermediate value theorem and the Extreme value theorem: all three are interval theorems, i.e., theorems which give important properties satisfied by any continuous function defined on a closed, bounded interval. The uniform continuity theorem gets short shrift in most contemporary calculus texts, perhaps because of the difficulty of giving a description of uniform continuity which is less formal than the epsilon-delta definition (epsilon-delta definitions being themselves omitted or given minimal emphasis in most contemporary texts). This is unfortunate, since the Darboux integrability of continuous functions follows almost immediately from the uniform continuity theorem and is otherwise not so easy to establish.

Every Lipschitz continuous map between two metric spaces is uniformly continuous.

Relations with the extension problem

Let "X" be a metric space, "S" a subset of "X", and $f:\; S\; ightarrow\; R$ a continuous function. When can "f" be extended to a continuous function on all of "X"?

If "S" is closed in "X", the answer is given by the Tietze extension theorem: always. So it is necessary and sufficient to extend "f" to the closure of "S" in "X": that is, we may assume without loss of generality that "S" is dense in "X", and this has the further pleasant consequence that if the extension exists, it is unique.

Let us suppose moreover that "X" is complete, so that "X" is the completion of "S". Then a continuous function $f:\; S\; ightarrow\; R$ extends to all of "X" if and only if "f" is Cauchy-continuous, i. e., the image under "f" of a Cauchy sequence remains Cauchy. (In general, Cauchy continuity is necessary and sufficient for extension of "f" to the completion of "X", so is a priori stronger than extendability to "X".)

It is easy to see that every uniformly continuous function is Cauchy-continuous and thus extends to "X". However, for functions defined on unbounded spaces like "R", uniform continuity is a rather strong condition, and it is desirable to have a weaker condition from which to deduce extendability.

For example, suppose "a > 1" is a real number. At the precalculus level, the function $f:\; x\; mapsto\; a^x$ can be given a precise definition only for rational values of "x" (assuming the existence of qth roots of positive real numbers, an application of the Intermediate Value Theorem). One would like to extend "f" to a function defined on all of "R". The identity $f(x+delta)-f(x)\; =\; a^x(a^\{delta\}-1)$

shows that "f" is not uniformly continuous on all of "Q"; however for any bounded interval "I" the restriction of "f" to $Q\; cap\; I$ is uniformly continuous, hence Cauchy-continuous, hence f extends to a continuous function on "I". But since this holds for every "I", there is then a unique extension of "f" to a continuous function on all of "R".

More generally, a continuous function $f:\; S\; ightarrow\; R$ whose restriction to every bounded subset of "S" is uniformly continuous is extendable to "X", and the converse holds if "X" is locally compact.

Generalization to topological vector spaces

In the special case of two topological vector spaces $V$ and $W$, the notion of uniform continuity of a map $f:V\; o\; W$ becomes : for any neighborhood $B$ of zero in $W$, there exists a neighborhood $A$ of zero in $V$ such that $v\_1-v\_2in\; A$ implies $f(v\_1)-f(v\_2)in\; B.$

Generalization to uniform spaces

Just as the most natural and general setting for continuity is topological spaces, the most natural and general setting for the study of "uniform" continuity are the uniform spaces.A function "f" : "X" → "Y" between uniform space is called "uniformly continuous" if for every entourage "V" in "Y" there exists an entourage "U" in "X" such that for every ("x"₁, "x"₂) in "U" we have ("f"("x"₁), "f"("x"₂)) in "V".

In this setting, it is also true that uniformly continuous maps transform Cauchy sequences into Cauchy sequences and that continuous maps on compact uniform spaces are automatically uniformly continuous.

References

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*citation|first=John L.|last=Kelley|title=General topology|year=1955|publisher=Springer-Verlag|series=Graduate Texts in Mathematics|isbn=0-387-80125-6
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*Citation | last1=Rudin | first1=Walter | author1-link=Walter Rudin | title=Principles of Mathematical Analysis | publisher=McGraw-Hill | location=New York | isbn=978-0-07-054235-8 | year=1976