The function f is said to be uniformly continuous on D if for every number \u03b5

# The function f is said to be uniformly continuous on

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The function f is said to be uniformly continuous on D if for every number ε > 0 there is a δ > 0 such that whenever x , y D and x y < δ , then f ( x ) f ( y ) < ε . The main difference between continuity and uniform continuity is that for continuity, δ can depend on x 0 as well as ε , whereas in uniform conti- nuity δ depends only on ε . Thus any uniformly continuous function is also continuous. An example of a function that is continuous but not uniformly continuous is given in Exercise 2 in this supplement. The distinction be- tween the notions of continuity and uniform continuity can be rephrased: For a function f that is continuous but not uniformly continuous, δ cannot be chosen independently of the point of the domain (the x 0 in the defini- tion). The definition of uniform continuity states explicitly that once you are given an ε > 0, a δ can be found independent of any point of D . Recall from § 3.3 that a set D R n is bounded if there exists a number M > 0 such that x M for all x D . A set is closed if it contains all its boundary points . Thus a set is bounded if it can be strictly contained in some (large) ball. The next theorem states that under some conditions a continuous function is actually uniformly continuous. Theorem. The Uniform Continuity Principle. Every function that is continuous on a closed and bounded set D in R n is uniformly continuous on D . The proof of this theorem will take us too far afield; 1 however, we can prove a special case of it, which is, in fact, suﬃcient for many situations relevant to this text. 1 The 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, New York, 1976.
72 5 Double and Triple Integrals Proof of a special case. Let us assume that D = [ a, b ] is a closed interval on the line, that f : D R is continuous, that df/dx exists on the open interval ( a, b ), and that df/dx is bounded (that is, there is a constant C > 0 such that | df ( x ) /dx | ≤ C for all x in ( a, b )). To show that these conditions imply f is uniformly continuous, we use the mean value theorem as follows: Let ε > 0 be given and let x and y lie on D . Then by the mean value theorem, f ( x ) f ( y ) = f ( c )( x y ) for some c between x and y . By the assumed boundedness of the derivative, | f ( x ) f ( y ) | ≤ C | x y | . Let δ = ε/C . If | x y | < δ , then | f ( x ) f ( y ) | < C ε C = ε. Thus f is uniformly continuous. (Note that δ depends on neither x nor y , which is a crucial part of the definition.) This proof also works for regions in R n that are convex; that is, for any two points x , y in D , the line segment c ( t ) = t x + (1 t ) y , 0 t 1, joining them also lies in D . We assume f is differentiable (on an open set containing D ) and that f ( x ) C for a constant C . Then the mean value theorem applied to the function h ( t ) = f ( c ( t )) gives a point t 0 such that h (1) h (0) = [ h ( t 0 )][1 0] or f ( x ) f ( y ) h ( t 0 ) = f ( c ( t 0 )) · c ( t 0 ) = f ( c ( t 0 )) · ( x y ) by the chain rule. Thus by the Cauchy-Schwartz inequality,