Page 1 Continuity and Uniform Continuity 521 May 12, 2010 1. ThroughoutS will denote a subset of the real numbers R andf :S! R will be a real valued function dened on S. The set S may be bounded like S = (0; 5) =fx2 R : 0<x< 5g or innite like S = (0;1) =fx2 R : 0<xg: It may even be all of R. The valuef(x) of the functionf at the pointx2S will be dened by a formula (or formulas). Denition 2. The function f is said to be continuous on S i 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not continuous 1 on S i 9x 0 2S9"> 08> 09x2S jxx 0 j< andjf(x)f(x 0 )j" : Denition 3. The functionf is said to be uniformly continuous on S i 8"> 09> 08x 0 2S8x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not uniformly continuous on S i 9"> 08> 09x 0 2S9x2S jxx 0 j< andjf(x)f(x 0 )j" : 1 For an example of a function which is not continuous see Example 22 below. Page 2 Continuity and Uniform Continuity 521 May 12, 2010 1. ThroughoutS will denote a subset of the real numbers R andf :S! R will be a real valued function dened on S. The set S may be bounded like S = (0; 5) =fx2 R : 0<x< 5g or innite like S = (0;1) =fx2 R : 0<xg: It may even be all of R. The valuef(x) of the functionf at the pointx2S will be dened by a formula (or formulas). Denition 2. The function f is said to be continuous on S i 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not continuous 1 on S i 9x 0 2S9"> 08> 09x2S jxx 0 j< andjf(x)f(x 0 )j" : Denition 3. The functionf is said to be uniformly continuous on S i 8"> 09> 08x 0 2S8x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not uniformly continuous on S i 9"> 08> 09x 0 2S9x2S jxx 0 j< andjf(x)f(x 0 )j" : 1 For an example of a function which is not continuous see Example 22 below. 4. The only dierence between the two denitions is the order of the quan tiers. When you prove f is continuous your proof will have the form Choose x 0 2S. Choose "> 0. Let =(x 0 ;"). Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". The expression for(x 0 ;") can involve bothx 0 and" but must be independent ofx. The order of the quaniers in the denition signals this; in the proof x has not yet been chosen at the point where is dened so the denition of must not involve x. (The represent the proof thatjf(x)f(x 0 )j < " follows from the earlier steps in the proof.) When you prove f is uniformly continuous your proof will have the form Choose "> 0. Let =("). Choose x 0 2S. Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". so the expression for can only involve " and must not involve either x or x 0 . It is obvious that a uniformly continuous function is continuous: if we can nd a which works for all x 0 , we can nd one (the same one) which works for any particular x 0 . We will see below that there are continuous functions which are not uniformly continuous. Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=3. Choose x 0 2 R. Choose x2 R. Assume jxx 0 j<. Then jf(x)f(x 0 )j =j(3x + 7) (3x 0 + 7)j = 3jxx 0 j< 3 =": Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x 2 . Then f is uniformly continuous on S. Proof. Choose " > 0. Let = "=8. Choose x 0 2 S. Choose x2 S. Thus 0<x 0 < 4 and 0<x< 4 so 0<x +x 0 < 8. Assumejxx 0 j<. Then jf(x)f(x 0 )j =jx 2 x 2 0 j = (x +x 0 )jxx 0 j< (4 + 4) =": Page 3 Continuity and Uniform Continuity 521 May 12, 2010 1. ThroughoutS will denote a subset of the real numbers R andf :S! R will be a real valued function dened on S. The set S may be bounded like S = (0; 5) =fx2 R : 0<x< 5g or innite like S = (0;1) =fx2 R : 0<xg: It may even be all of R. The valuef(x) of the functionf at the pointx2S will be dened by a formula (or formulas). Denition 2. The function f is said to be continuous on S i 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not continuous 1 on S i 9x 0 2S9"> 08> 09x2S jxx 0 j< andjf(x)f(x 0 )j" : Denition 3. The functionf is said to be uniformly continuous on S i 8"> 09> 08x 0 2S8x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not uniformly continuous on S i 9"> 08> 09x 0 2S9x2S jxx 0 j< andjf(x)f(x 0 )j" : 1 For an example of a function which is not continuous see Example 22 below. 4. The only dierence between the two denitions is the order of the quan tiers. When you prove f is continuous your proof will have the form Choose x 0 2S. Choose "> 0. Let =(x 0 ;"). Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". The expression for(x 0 ;") can involve bothx 0 and" but must be independent ofx. The order of the quaniers in the denition signals this; in the proof x has not yet been chosen at the point where is dened so the denition of must not involve x. (The represent the proof thatjf(x)f(x 0 )j < " follows from the earlier steps in the proof.) When you prove f is uniformly continuous your proof will have the form Choose "> 0. Let =("). Choose x 0 2S. Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". so the expression for can only involve " and must not involve either x or x 0 . It is obvious that a uniformly continuous function is continuous: if we can nd a which works for all x 0 , we can nd one (the same one) which works for any particular x 0 . We will see below that there are continuous functions which are not uniformly continuous. Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=3. Choose x 0 2 R. Choose x2 R. Assume jxx 0 j<. Then jf(x)f(x 0 )j =j(3x + 7) (3x 0 + 7)j = 3jxx 0 j< 3 =": Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x 2 . Then f is uniformly continuous on S. Proof. Choose " > 0. Let = "=8. Choose x 0 2 S. Choose x2 S. Thus 0<x 0 < 4 and 0<x< 4 so 0<x +x 0 < 8. Assumejxx 0 j<. Then jf(x)f(x 0 )j =jx 2 x 2 0 j = (x +x 0 )jxx 0 j< (4 + 4) =": 7. In both of the preceeding proofs the function f satised an inequality of form jf(x 1 )f(x 2 )jMjx 1 x 2 j (1) for x 1 ;x 2 2S. In Example 5 we had j(3x 1 + 7) (3x 2 + 7)j 3jx 1 x 2 j and in Example 6 we had jx 2 1 x 2 2 j 8jx 1 x 2 j for 0<x 1 ;x 2 < 4. An inequality of form (1) is called a Lipschitz inequality and the constant M is called the corresponding Lipschitz constant. Theorem 8. Iff satises (1) for x 1 ;x 2 2S, thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=M. Choose x 0 2S. Choose x2S. Assume thatjxx 0 j<. Then jf(x)f(x 0 )jMjxx 0 j<M =": 9. The Lipschitz constant depend might depend on the interval. For example, jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j 2ajx 1 x 2 j for 0 < x 1 ;x 2 < a but the function f(x) = x 2 does not satisfy a Lipschitz inequality on the whole interval (0;1) since jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j>Mjx 1 x 2 j if x 1 =M and x 2 =x 1 + 1. In fact, Example 10. The function f(x) =x 2 is continuous but not uniformly con tinuous on the interval S = (0;1). Proof. We show f is continuous on S, i.e. 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jx 2 x 2 0 j<" : Page 4 Continuity and Uniform Continuity 521 May 12, 2010 1. ThroughoutS will denote a subset of the real numbers R andf :S! R will be a real valued function dened on S. The set S may be bounded like S = (0; 5) =fx2 R : 0<x< 5g or innite like S = (0;1) =fx2 R : 0<xg: It may even be all of R. The valuef(x) of the functionf at the pointx2S will be dened by a formula (or formulas). Denition 2. The function f is said to be continuous on S i 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not continuous 1 on S i 9x 0 2S9"> 08> 09x2S jxx 0 j< andjf(x)f(x 0 )j" : Denition 3. The functionf is said to be uniformly continuous on S i 8"> 09> 08x 0 2S8x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not uniformly continuous on S i 9"> 08> 09x 0 2S9x2S jxx 0 j< andjf(x)f(x 0 )j" : 1 For an example of a function which is not continuous see Example 22 below. 4. The only dierence between the two denitions is the order of the quan tiers. When you prove f is continuous your proof will have the form Choose x 0 2S. Choose "> 0. Let =(x 0 ;"). Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". The expression for(x 0 ;") can involve bothx 0 and" but must be independent ofx. The order of the quaniers in the denition signals this; in the proof x has not yet been chosen at the point where is dened so the denition of must not involve x. (The represent the proof thatjf(x)f(x 0 )j < " follows from the earlier steps in the proof.) When you prove f is uniformly continuous your proof will have the form Choose "> 0. Let =("). Choose x 0 2S. Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". so the expression for can only involve " and must not involve either x or x 0 . It is obvious that a uniformly continuous function is continuous: if we can nd a which works for all x 0 , we can nd one (the same one) which works for any particular x 0 . We will see below that there are continuous functions which are not uniformly continuous. Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=3. Choose x 0 2 R. Choose x2 R. Assume jxx 0 j<. Then jf(x)f(x 0 )j =j(3x + 7) (3x 0 + 7)j = 3jxx 0 j< 3 =": Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x 2 . Then f is uniformly continuous on S. Proof. Choose " > 0. Let = "=8. Choose x 0 2 S. Choose x2 S. Thus 0<x 0 < 4 and 0<x< 4 so 0<x +x 0 < 8. Assumejxx 0 j<. Then jf(x)f(x 0 )j =jx 2 x 2 0 j = (x +x 0 )jxx 0 j< (4 + 4) =": 7. In both of the preceeding proofs the function f satised an inequality of form jf(x 1 )f(x 2 )jMjx 1 x 2 j (1) for x 1 ;x 2 2S. In Example 5 we had j(3x 1 + 7) (3x 2 + 7)j 3jx 1 x 2 j and in Example 6 we had jx 2 1 x 2 2 j 8jx 1 x 2 j for 0<x 1 ;x 2 < 4. An inequality of form (1) is called a Lipschitz inequality and the constant M is called the corresponding Lipschitz constant. Theorem 8. Iff satises (1) for x 1 ;x 2 2S, thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=M. Choose x 0 2S. Choose x2S. Assume thatjxx 0 j<. Then jf(x)f(x 0 )jMjxx 0 j<M =": 9. The Lipschitz constant depend might depend on the interval. For example, jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j 2ajx 1 x 2 j for 0 < x 1 ;x 2 < a but the function f(x) = x 2 does not satisfy a Lipschitz inequality on the whole interval (0;1) since jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j>Mjx 1 x 2 j if x 1 =M and x 2 =x 1 + 1. In fact, Example 10. The function f(x) =x 2 is continuous but not uniformly con tinuous on the interval S = (0;1). Proof. We show f is continuous on S, i.e. 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jx 2 x 2 0 j<" : Choose x 0 . Let a = x 0 + 1 and = min(1;"=2a). (Note that depends on x 0 since a does.) Choose x2S. Assumejxx 0 j<. Thenjxx 0 j< 1 so x<x 0 + 1 =a so x;x 0 <a so jx 2 x 2 0 j = (x +x 0 )jxx 0 j 2ajxx 0 j< 2a 2a " 2a =" as required. We show that f is not uniformly continuous on S, i.e. 9"> 08> 09x 0 2S9x2S jxx 0 j< andjx 2 x 2 0 j" : Let " = 1. Choose > 0. Let x 0 = 1= and x =x 0 +=2. Thenjxx 0 j = =2< but jx 2 x 2 0 j = 1 + 2 2 1 2 = 1 + 2 4 > 1 =" as required. (Note that x 0 is large when is small.) 11. According to the Mean Value Theorem from calculus for a dierentiable function f we have f(x 1 )f(x 2 ) =f 0 (c)(x 2 x 1 ): for some c between x 1 and x 2 . (The slope (f(x 1 )f(x 2 ))=(x 1 x 2 ) of the secant line joining the two points (x 1 ;f(x 1 )) and (x 2 ;f(x 2 )) on the graph is the same as the slope f 0 (c) of the tangent point at the intermediate point (c;f(c)).) If x 1 and x 2 lie in some interval S andjf 0 (c)j M for all c2 S we conclude that the Lipschitz inequality (1) holds on S. We don't want to use the Mean Value Theorem without rst proving it, but we certainly can use it to guess an appropriate value of M and then prove the inequality by other means. 12. For example, consider the function f(x) = x 1 dened on the interval S = (a;1) where a> 0. For x 1 ;x 2 2S the Mean Value Theorem says that x 1 1 x 1 2 =c 2 (x 1 x 2 ) where c is between x 1 and x 2 . If x 1 ;x 2 2S then c2 S (as c is between x 1 and x 2 ) and hence c > a so c 2 < a 2 . We can prove the inequality jx 1 1 x 1 2 ja 2 jx 2 x 2 j Page 5 Continuity and Uniform Continuity 521 May 12, 2010 1. ThroughoutS will denote a subset of the real numbers R andf :S! R will be a real valued function dened on S. The set S may be bounded like S = (0; 5) =fx2 R : 0<x< 5g or innite like S = (0;1) =fx2 R : 0<xg: It may even be all of R. The valuef(x) of the functionf at the pointx2S will be dened by a formula (or formulas). Denition 2. The function f is said to be continuous on S i 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not continuous 1 on S i 9x 0 2S9"> 08> 09x2S jxx 0 j< andjf(x)f(x 0 )j" : Denition 3. The functionf is said to be uniformly continuous on S i 8"> 09> 08x 0 2S8x2S jxx 0 j< =) jf(x)f(x 0 )j<" : Hence f is not uniformly continuous on S i 9"> 08> 09x 0 2S9x2S jxx 0 j< andjf(x)f(x 0 )j" : 1 For an example of a function which is not continuous see Example 22 below. 4. The only dierence between the two denitions is the order of the quan tiers. When you prove f is continuous your proof will have the form Choose x 0 2S. Choose "> 0. Let =(x 0 ;"). Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". The expression for(x 0 ;") can involve bothx 0 and" but must be independent ofx. The order of the quaniers in the denition signals this; in the proof x has not yet been chosen at the point where is dened so the denition of must not involve x. (The represent the proof thatjf(x)f(x 0 )j < " follows from the earlier steps in the proof.) When you prove f is uniformly continuous your proof will have the form Choose "> 0. Let =("). Choose x 0 2S. Choose x2S. Assumejxx 0 j<. Thereforejf(x)f(x 0 )j<". so the expression for can only involve " and must not involve either x or x 0 . It is obvious that a uniformly continuous function is continuous: if we can nd a which works for all x 0 , we can nd one (the same one) which works for any particular x 0 . We will see below that there are continuous functions which are not uniformly continuous. Example 5. LetS = R andf(x) = 3x + 7. Thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=3. Choose x 0 2 R. Choose x2 R. Assume jxx 0 j<. Then jf(x)f(x 0 )j =j(3x + 7) (3x 0 + 7)j = 3jxx 0 j< 3 =": Example 6. Let S =fx2 R : 0 < x < 4g and f(x) = x 2 . Then f is uniformly continuous on S. Proof. Choose " > 0. Let = "=8. Choose x 0 2 S. Choose x2 S. Thus 0<x 0 < 4 and 0<x< 4 so 0<x +x 0 < 8. Assumejxx 0 j<. Then jf(x)f(x 0 )j =jx 2 x 2 0 j = (x +x 0 )jxx 0 j< (4 + 4) =": 7. In both of the preceeding proofs the function f satised an inequality of form jf(x 1 )f(x 2 )jMjx 1 x 2 j (1) for x 1 ;x 2 2S. In Example 5 we had j(3x 1 + 7) (3x 2 + 7)j 3jx 1 x 2 j and in Example 6 we had jx 2 1 x 2 2 j 8jx 1 x 2 j for 0<x 1 ;x 2 < 4. An inequality of form (1) is called a Lipschitz inequality and the constant M is called the corresponding Lipschitz constant. Theorem 8. Iff satises (1) for x 1 ;x 2 2S, thenf is uniformly continuous on S. Proof. Choose "> 0. Let ="=M. Choose x 0 2S. Choose x2S. Assume thatjxx 0 j<. Then jf(x)f(x 0 )jMjxx 0 j<M =": 9. The Lipschitz constant depend might depend on the interval. For example, jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j 2ajx 1 x 2 j for 0 < x 1 ;x 2 < a but the function f(x) = x 2 does not satisfy a Lipschitz inequality on the whole interval (0;1) since jx 2 1 x 2 2 j = (x 1 +x 2 )jx 1 x 2 j>Mjx 1 x 2 j if x 1 =M and x 2 =x 1 + 1. In fact, Example 10. The function f(x) =x 2 is continuous but not uniformly con tinuous on the interval S = (0;1). Proof. We show f is continuous on S, i.e. 8x 0 2S8"> 09> 08x2S jxx 0 j< =) jx 2 x 2 0 j<" : Choose x 0 . Let a = x 0 + 1 and = min(1;"=2a). (Note that depends on x 0 since a does.) Choose x2S. Assumejxx 0 j<. Thenjxx 0 j< 1 so x<x 0 + 1 =a so x;x 0 <a so jx 2 x 2 0 j = (x +x 0 )jxx 0 j 2ajxx 0 j< 2a 2a " 2a =" as required. We show that f is not uniformly continuous on S, i.e. 9"> 08> 09x 0 2S9x2S jxx 0 j< andjx 2 x 2 0 j" : Let " = 1. Choose > 0. Let x 0 = 1= and x =x 0 +=2. Thenjxx 0 j = =2< but jx 2 x 2 0 j = 1 + 2 2 1 2 = 1 + 2 4 > 1 =" as required. (Note that x 0 is large when is small.) 11. According to the Mean Value Theorem from calculus for a dierentiable function f we have f(x 1 )f(x 2 ) =f 0 (c)(x 2 x 1 ): for some c between x 1 and x 2 . (The slope (f(x 1 )f(x 2 ))=(x 1 x 2 ) of the secant line joining the two points (x 1 ;f(x 1 )) and (x 2 ;f(x 2 )) on the graph is the same as the slope f 0 (c) of the tangent point at the intermediate point (c;f(c)).) If x 1 and x 2 lie in some interval S andjf 0 (c)j M for all c2 S we conclude that the Lipschitz inequality (1) holds on S. We don't want to use the Mean Value Theorem without rst proving it, but we certainly can use it to guess an appropriate value of M and then prove the inequality by other means. 12. For example, consider the function f(x) = x 1 dened on the interval S = (a;1) where a> 0. For x 1 ;x 2 2S the Mean Value Theorem says that x 1 1 x 1 2 =c 2 (x 1 x 2 ) where c is between x 1 and x 2 . If x 1 ;x 2 2S then c2 S (as c is between x 1 and x 2 ) and hence c > a so c 2 < a 2 . We can prove the inequality jx 1 1 x 1 2 ja 2 jx 2 x 2 j for x 1 ;x 2 a as follows. First a 2 x 1 x 2 since ax 1 and ax 2 . Then jx 1 1 x 1 2 j = jx 1 x 2 j x 1 x 2 jx 1 x 2 j a 2 (2) where we have used the fact that 1 < 1 if 0 < < . It follows that that the function f(x) is uniformly continuous on any interval (a;1) where a > 0. Notice however that the Lipschitz constant M = a 2 depends on the interval. In fact, the function f(x) = x 1 does not satisfy a Lipshitz inequality on the interval (0;1). 13. We can discover a Lipscitz inequality for the square root functionf(x) = p x in much the same way. Consider the function f(x) = p x dened on the interval S = (a;1) where a > 0. For x 1 ;x 2 2 S the Mean Value Theorem says that p x 1 p x 2 = (x 1 x 2 )=(2 p c) where c is between x 1 and x 2 . If x 1 ;x 2 2 S then c 2 S (as c is between x 1 and x 2 ) and hence c > a so (2 p c) 1 < (2 p a) 1 . We can prove the inequality j p x 1 p x 2 j jx 1 x 2 j 2 p a (3) for x 1 ;x 2 a as follows: Divide the equation ( p x 1 p x 2 )( p x 1 + p x 2 ) = (( p x 1 ) 2 ( p x 2 ) 2 ) =x 1 x 2 by ( p x 1 + p x 2 ), take absolute values, and use ( p x 1 + p x 2 ) 2 p a. Again the Lipschitz constant M = (2 p a) 1 depends on the interval and the function does not satisfy a Lipschitz inequality on the interval (0;1). Example 14. The function f(x) = x 1 is continuous but not uniformly continuous on the interval S = (0;1). Proof. We show f is continuous on S, i.e. 8x 0 2S8"> 09> 08x2S jxx 0 j< =) 1 x 1 x 0 <" : Choose x 0 . Let a =x 0 =2 and = min(x 0 a;a 2 "). Choose x2S. Assume jxx 0 j < . Then x 0 xjxx 0 j < x 0 a sox <a so a < x so x;x 0 <a so by (2) 1 x 1 x 0 jx 1 x 2 j a 2 < a 2 a 2 " a 2 ="Read More
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