The functionfis said to beuniformly continuousonDif for everynumberε >0there is aδ >0such that wheneverx,y∈Dandx−y< δ,thenf(x)−f(y)< ε.The main difference between continuity and uniform continuity is thatfor continuity,δcan depend onx0as well asε, whereas in uniform conti-nuityδdepends only onε. Thus any uniformly continuous function is alsocontinuous. An example of a function that is continuous but not uniformlycontinuous is given in Exercise 2 in this supplement. The distinction be-tween the notions of continuity and uniform continuity can be rephrased:For a functionfthat is continuous but not uniformly continuous,δcannotbe chosen independently of the point of the domain (thex0in the defini-tion). The definition of uniform continuity states explicitly that once youare given anε >0, aδcan be found independent of any point ofD.Recall from§3.3 that a setD⊂Rnisboundedif there exists a numberM >0such thatx≤Mfor allx∈D. A set isclosedif it contains allits boundary points. Thus a set is bounded if it can be strictly containedin some (large) ball. The next theorem states that under some conditionsa continuous function is actually uniformly continuous.Theorem. The Uniform Continuity Principle.Every function thatis continuous on a closed and bounded setDinRnis uniformly continuousonD.The proof of this theorem will take us too far afield;1however, we canprove a special case of it, which is, in fact, suﬃcient for many situationsrelevant to this text.1The proof can be found in texts on mathematical analysis. See, for example, J.Marsden and M. Hoffman,Elementary Classical Analysis, 2nd ed., Freeman, New York,1993, or W. Rudin,Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, NewYork, 1976.
725 Double and Triple IntegralsProof of a special case.Let us assume thatD= [a, b] is a closed intervalon the line, thatf:D→Ris continuous, thatdf/dxexists on the openinterval (a, b), and thatdf/dxis bounded (that is, there is a constantC >0such that|df(x)/dx| ≤Cfor allxin (a, b)). To show that these conditionsimplyfis uniformly continuous, we use the mean value theorem as follows:Letε >0 be given and letxandylie onD. Then by the mean valuetheorem,f(x)−f(y) =f(c)(x−y)for somecbetweenxandy. By the assumed boundedness of the derivative,|f(x)−f(y)| ≤C|x−y|.Letδ=ε/C. If|x−y|< δ, then|f(x)−f(y)|< CεC=ε.Thusfis uniformly continuous. (Note thatδdepends on neitherxnory,which is a crucial part of the definition.)This proof also works for regions inRnthat are convex; that is, for anytwo pointsx,yinD, the line segmentc(t) =tx+ (1−t)y,0≤t≤1,joining them also lies inD. We assumefis differentiable (on an open setcontainingD) and that∇f(x)≤Cfor a constantC. Then the meanvalue theorem applied to the functionh(t) =f(c(t)) gives a pointt0suchthath(1)−h(0) = [h(t0)][1−0]orf(x)−f(y)−h(t0) =∇f(c(t0))·c(t0) =∇f(c(t0))·(x−y)by the chain rule. Thus by the Cauchy-Schwartz inequality,