Transcription of Continuity and Uniform Continuity - University of …
1 Continuity and Uniform Continuity521 May 12, denote a subset of the real numbersRandf:S Rwill be a real valued function defined onS. The setSmay be bounded likeS= (0,5) ={x R: 0< x <5}or infinite likeS= (0, ) ={x R: 0< x}.It may even be all ofR. The valuef(x) of the functionfat the pointx Swill be defined by a formula (or formulas).Definition functionfis said to becontinuous onSiff x0 S >0 >0 x S[|x x0|< = |f(x) f(x0)|< ].Hencefis not continuous1onSiff x0 S >0 >0 x S[|x x0|< and|f(x) f(x0)| ].Definition functionfis said to beuniformly continuous onSiff >0 >0 x0 S x S[|x x0|< = |f(x) f(x0)|< ].
2 Hencefis not uniformly continuous onSiff >0 >0 x0 S x S[|x x0|< and|f(x) f(x0)| ].1 For an example of a function which isnotcontinuous see Example 22 only difference between the two definitions is the order of the quan-tifiers. When you provefis continuous your proof will have the formChoosex0 S. Choose >0. Let = (x0, ). Choosex |x x0|< . Therefore|f(x) f(x0)|< .The expression for (x0, ) can involve bothx0and but must be independentofx. The order of the quanifiers in the definition signals this; in the proofxhas not yet been chosen at the point where is defined so the definition of must not involvex.
3 (The represent the proof that|f(x) f(x0)|< follows from the earlier steps in the proof.) When you provefis uniformlycontinuous your proof will have the formChoose >0. Let = ( ). Choosex0 S. Choosex |x x0|< . Therefore|f(x) f(x0)|< .so the expression for can only involve and must not involve is obvious that a uniformly continuous function is continuous: if we canfind a which works for allx0, we can find one (the same one) which worksfor any particularx0. We will see below that there are continuous functionswhich are not uniformly (x) = 3x+ 7.
4 Thenfis uniformly >0. Let = /3. Choosex0 R. Choosex R. Assume|x x0|< . Then|f(x) f(x0)|=|(3x+ 7) (3x0+ 7)|= 3|x x0|<3 = .Example {x R: 0< x <4}andf(x) =x2. Thenfisuniformly continuous >0. Let = /8. Choosex0 S. Choosex S. Thus0< x0<4 and 0< x <4 so 0< x+x0<8. Assume|x x0|< . Then|f(x) f(x0)|=|x2 x20|= (x+x0)|x x0|<(4 + 4) = . both of the preceeding proofs the functionfsatisfied an inequality ofform|f(x1) f(x2)| M|x1 x2|(1)forx1,x2 S. In Example 5 we had|(3x1+ 7) (3x2+ 7)| 3|x1 x2|and in Example 6 we had|x21 x22| 8|x1 x2|for 0< x1,x2<4.
5 An inequality of form (1) is called aLipschitz inequalityand the constantMis called the correspondingLipschitz (1) forx1,x2 S, thenfis uniformly >0. Let = /M. Choosex0 S. Choosex S. Assumethat|x x0|< . Then|f(x) f(x0)| M|x x0|< M = . Lipschitz constant depend might depend on the interval. For example,|x21 x22|= (x1+x2)|x1 x2| 2a|x1 x2|for 0< x1,x2< abut the functionf(x) =x2does not satisfy a Lipschitzinequality on the whole interval (0, ) since|x21 x22|= (x1+x2)|x1 x2|> M|x1 x2|ifx1=Mandx2=x1+ 1. In fact,Example functionf(x) =x2is continuous but not uniformly con-tinuous on the intervalS= (0, ).
6 Showfis continuous onS, x0 S >0 >0 x S[|x x0|< = |x2 x20|< ].3 Choosex0. Leta=x0+ 1 and = min(1, /2a). (Note that depends onx0sinceadoes.) Choosex S. Assume|x x0|< . Then|x x0|<1 sox < x0+ 1 =asox,x0< aso|x2 x20|= (x+x0)|x x0| 2a|x x0|<2a 2a 2a= as show thatfis not uniformly continuous onS, >0 >0 x0 S x S[|x x0|< and|x2 x20| ].Let = 1. Choose >0. Letx0= 1/ andx=x0+ /2. Then|x x0|= /2< but|x2 x20|= (1 + 2)2 (1 )2 = 1 + 24>1 = as required. (Note thatx0is large when is small.) to the Mean value theorem from calculus for a differentiablefunctionfwe havef(x1) f(x2) =f (c)(x2 x1).
7 Forsomecbetweenx1andx2. (The slope (f(x1) f(x2))/(x1 x2) of thesecant line joining the two points (x1,f(x1)) and (x2,f(x2)) on the graph isthe same as the slopef (c) of the tangent point at the intermediate point(c,f(c)).) Ifx1andx2lie in some intervalSand|f (c)| Mforallc Swe conclude that the Lipschitz inequality (1) holds onS. We don t want touse the Mean value theorem without first proving it, but we certainly canuse it to guess an appropriate value ofMand then prove the inequality byother example, consider the functionf(x) =x 1defined on the intervalS= (a, ) wherea >0.
8 Forx1,x2 Sthe Mean value theorem says thatx 11 x 12= c 2(x1 x2) wherecis betweenx1andx2. Ifx1,x2 Sthenc S(ascis betweenx1andx2) and hencec > asoc 2< a 2. We canprove the inequality|x 11 x 12| a 2|x2 x2|4forx1,x2 aas follows. Firsta2 x1x2sincea x1anda x2. Then|x 11 x 12|=|x1 x2|x1x2 |x1 x2|a2(2)where we have used the fact that 1< 1if 0< < . It follows thatthat the functionf(x) is uniformly continuous on any interval (a, ) wherea >0. Notice however that the Lipschitz constantM=a 2depends onthe interval.
9 In fact, the functionf(x) =x 1doesnotsatisfy a Lipshitzinequality on the interval (0, ). can discover a Lipscitz inequality for the square root functionf(x) = xin much the same way. Consider the functionf(x) = xdefined on theintervalS= (a, ) wherea >0. Forx1,x2 Sthe Mean value Theoremsays that x1 x2= (x1 x2)/(2 c) wherecis betweenx1andx2. Ifx1,x2 Sthenc S(ascis betweenx1andx2) and hencec > aso(2 c) 1<(2 a) 1. We can prove the inequality| x1 x2| |x1 x2|2 a(3)forx1,x2 aas follows: Divide the equation( x1 x2)( x1+ x2) = (( x1)2 ( x2)2) =x1 x2by ( x1+ x2), take absolute values, and use ( x1+ x2) 2 a.
10 Again theLipschitz constantM= (2 a) 1depends on the interval and the functiondoesnotsatisfy a Lipschitz inequality on the interval (0, ).Example functionf(x) =x 1is continuous but not uniformlycontinuous on the intervalS= (0, ). showfis continuous onS, x0 S >0 >0 x S[|x x0|< = 1x 1x0 < ].Choosex0. Leta=x0/2 and = min(x0 a,a2 ). Choosex S. Assume|x x0|< . Thenx0 x |x x0|< x0 aso x < asoa < xsox,x0< aso by (2) 1x 1x0 |x1 x2|a2< a2 a2 a2= 5as show thatfis not uniformly continuous onS, >0 >0 x0 S x S[|x x0|< and 1x 1x0 ].