Math 0450 Honors intro to analysis Spring, 2009 Notes 17 1 Uniform continuity Read rst: 5.4 Here are some examples of continuous functions, some of which we have done before. 1. A = (0; 1]; f : A R given by f (x) = 1 . ! x Proof. To prove that f is continuous at c (0; 1], suppose that " > 0, and 2 2 let = min c ; c " : If x c < , then rst of all, x > c and so 0 < 1 < 2 . 2 2 j j 2 x c Hence, n o 1 1 c x 1 1 2 1 c2" = = c x < = ": x c xc x c j j c c 2 2. f : R R given by f (x) = x2: ! Proof. Observe that x2 c2 = x c (x + c) : j j j j " If c = 0; let = p": If c = 0; let = min c ; 2 c If c = 0 and x c < , 6 j j j j j j then x < p" and x2 c2 = x2 < ", Ifnc = 0; theno j j j j j j 6 " x2 c2 = x + c x c < 2 c = ": j j j j j j 2 c j j 3. A = [1; ); f (x) = 1 . 1 x Proof. This makes use of item 1 above. Suppose that c A. From item 1, 2 2 we see that if " > 0, and x c < min c ; c " , then 1 1 < ". Hence, j j 2 2 x c 2 let = " : If x c < , and x and c aren in A, theno x c < min c ; c " ; so 2 j j j j 2 2 1 1 < ": n o x c 1 1 4. A = [1; 2] ; f (x) = x . " Proof. Any which works in #3 clearly works here. So if " > 0, let = 2 . This brings out the point that if f is uniformly continuous on a set A, and B A, then f is also uniformly continuous on B. Now we point out an important di erence between examples 1 and 2 and examples 3 and 4. In 1 and 2, the depends on c . In 3 and 4 it does not. De nition 1 Suppose f : A R. We say that f is \uniformly continuous" on A if for each " > 0 there is a >!0 such that if c and x are in A, and x c < , then f (x) f (c) < ". j j j j In this de nition it is very important that is chosen before c, so that does not depend on c. Looking at the examples above, we see that: 1 1. x is not uniformly continuous on (0; 1] 2. x2 is not uniformly continuous on [0; ) 1 3. 1 is uniformly continuous on [1; ) x 1 4. x2 is uniformly continuous on [1; 2]. One further example: Proposition 2 px is uniformly continuous on [0; ). 1 Proof. Suppose that " > 0. Let "1 = min "; 1 . We consider two cases: (i) 2 2 2 "1 "1 f "1 g 0 c < 9 and (ii) 9 c: In case (i), let = 9 . If x c < and x 0 then "2 "2 j j 0 x < c + 1 2 1 : Hence, 9 9 2 1 px pc px + pc < "1 + "1 < "1 ": r9 3 In case (ii), observe that px + pc x c x c x c px pc = px pc = j j j j: px + pc px + pc pc " =3 1 2 "1 Hence, we choose = 3 ". 2 "1 "1 If we let (") = min 9 ; 3 " , then this works in both cases, and shows that f is uniformly continuous onn [0; o). 1 Now we have an important theorem Theorem 3 If A = [a; b] is a closed bounded interval, and f : A R is continuous on A, then f is uniformly continuous on A. ! Proof. If not, then there is an " > 0 such that for each > 0; there is a pair c; x with x c < and f (x) f (c) ". In particular, this is true if = 1 f g j j j j n where n . So we obtain two sequences, (cn) [a; b] and (xn) [a; b] ; such that 2 N1 xn cn < and f (cn) f (xn) ". But by the Bolzano-Weierstrass theorem, j j n j j (cn) must have a convergent subsequence (cnk ) : Suppose that (cnk ) converges to c: 1 Since xn cn < , it must be the case that (xn ) c as well. But f is j k k j nk k ! continuous at c; so that lim (f (xnk )) = f (c) and lim (f (cnk )) = f (c). In other words, lim (f (xn ) f (c)) = 0 k lim (f (cn ) f (c)) = 0: k Since f (cn ) f (xn ) = f (cn ) f (c) + f (c) f (xn ) f (cn ) f (c) + f (xn ) f (c) ; j k k j j k k j j k j j k j it follows that lim (f (cnk ) f (xnk )) = 0: Since f (cn) f (xn) " for all n, this is a contradiction. j j Once again, the Bolzano-Weierstrass theorem plays a key role. Example 4 Now we can easily show that px is uniformly continuous. It was shown in Theorem 5.2.5 that px is continuous at each c 0. In particular, it is continuous on [0; 2] ; and so uniformly continuous. So for each " > 0; there is a 1 independent of c [0; 2] such that if x c < , then px pc < ". Let = min 1; "; 1 : If c 12and x c < , thenj (asj in (iii) in ourj earlier j discussion of px) f g j j x c px pc = x c < ": px + pc j j (We could chose the intervals [0 ; 1] and [1; ), but this leads to unnecessary com- plication when c is at or close to 1.) Hence,1 for any c 0; if x c < , then px pc < ". j j j j 3 Lipschitz continuity De nition 5 A function f : A R is said to be \Lipschitz continuous" on A if there is an L R such that for each! x A and y A; f (x) f (y) L x y : 2 2 2 j j j j It is easily shown that if f is Lipschitz continuous on A; then f is uniformly " continuous on A. Just choose = L : " f (x) f (c) L x y < L = ": j j j j L As an example we have f (x) = x on R. Even though R is unbounded, f is uniformly continuous on R. f is Lipschitz continuous on R; with L = 1: This shows that if A is unbounded, then f can be unbounded and still uniformly continuous. The function x2 is an easy example of a function which is continuous, but not uniformly continuous, on R. If we jump ahead, and assume we know about derivatives, we can see a rela- tion between f 0 (x) and Lipschitz continuity. Eventually we will prove that if f is di erentiable on [a; b], then there is a c [a; b] such that 2 f (b) f (a) = f 0 (c)(b a) : Hence, if the function f 0 is bounded on [a; b], and L is an upper bound for f 0 (x) on [a; b] ; then j j f (b) f (a) L b a j j j j 2 Example 6 f (x) = x on I = [0; 100]. Since f 0 (x) = 2x 200 on I; we see that x2 is Lipschitz continuous, and so uniformlyj continuous,j j onj I. 2 Example 7 f (x) = x = [0; ) = x R : x 0 : The derivative f 0 (x) = 2x is unbounded on on [0; ) , and1f isf not2 Lipschitz continuousg there. We saw earlier that x2 is also not uniformly1 continuous on [0; ). 1 1 Example 8 f (x) = px on I = (0; ). Now f 0 (x) = , which is unbounded on 1 2px I. Hence, f is not Lipschitz continuous on I. However, we saw above that px is uniformly continuous on (0; ). 1 4 1.1 Approximation of functions Often it is desirable to approximate a function f by a simpler function g: Among the so-called \simpler" functions which may be used are the \piecewise constant" functions and the \piecewise linear" functions. To de ne these, we rst de ne a \partition" of an interval, something which will be very useful later on. De nition 9 Suppose that I = [a; b] is a closed bounded interval. Then a \parti- tion" of I is a nite set of points x0; x1; ::; xn with a = x0 < x1 < < xn = b: f g De nition 10 A function g is said to be \piecewise constant" on an interval [a; b] if there is a partition x0; :::; xn such that g is constant on (xi; xi+1) and the values at the partition pointsf are the limitsg from one side or the other. Such functions are also called \`step" functions. The following theorem says that a continuous function on a closed bounded interval can be approximated by step functions. Theorem 11 Suppose that f :[a; b] R and f is continuous in [a; b]. Then for ! each " > 0 there is a piecewise linear function g" such that f (x) g" (x) < " j j for all x [a; b]. 2 The proof is in the text, and relies on the uniform continuity of f. De nition 12 A function g is said to be \piecewise linear"' if there is a partition x0; :::; xn such that g is a linear function (ax + b) on (xi; xi+1), and the values at thef partitiong points are the limits from one side or the other.