Continuity and Uniform Continuity

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)|< ].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.

2 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. (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.

3 We will see below that there are continuous functionswhich are not uniformly (x) = 3x+ 7. 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. An inequality of form (1) is called aLipschitz inequalityand the constantMis called the correspondingLipschitz (1) forx1,x2 S, thenfis uniformly >0.

4 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, ). 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| ].

5 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).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.

6 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. 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.

7 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. 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|<.

8 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 ].Let = 1. Choose >0. Letx0= min( ,1) andx=x0/2. Then|x x0|=x0/2 /2< but 1x 1x0 = 1x0/2 1x0 =1x0 1 = as functionf(x) = xis uniformly continuous on the setS= (0, ).Remark example shows that a function can be uniformly contin-uous on a set even though it does not satisfy a Lipschitz inequality on thatset, the method of Theorem 8 is not the only method for proving afunction uniformly continuous. The proof we give will use the following choosing >0 we specify two numbersaandbwhich will depend on . These numbers will satisfy 0< a < b.

9 We will choose so that (amongother things) < b a. Then after we choosex,x0 Sand assume that|x x0|< we will be able to conclude that either bothx0andxare lessthanbor both are greater thana. We will choosebso small that xand x0are within of zero forx,x0< b. We will use a Lipschitz inequality tohandle the case wherex,x0> a. We give the details of this proof after somepreliminary lemmas. The only properties of that square root function thatwe will use are that xis defined forx 0 and satisfies x 0,( x)2=x, x2= square root function is increasing:0 a < b= a < 0 a < b. If b athenb= ( b)2 ( a)2=acontradictinga < b. Hence a < 18. ab= a bfora,b ( a b)2= ( a)2( b)2=abso a b= ( a b)2= < b. Then for any two numbersxandyat least oneof the four alternatives(i)x < b&y < b, (ii)a x&a y,(iii)x < a&b y, (iv)y < a&b one of the three alternativesx < a,a x < b,b xholdsand exactly one of the three alternativesy < a,a y < b,b are thus nine cases which we can arrange in a table:x < a a x < b b xy < a(i)(i)(iv)a y < b(i)(i),(ii)(ii)b y(iii)(ii)(ii)In each entry of the table we have indicated the alternative (i iv) whichholds in the corresponding we prove what is claimed in Example 15, viz.

10 That the squareroot function is uniformly continuous on the positive real numbers, >0 >0 x,x0>0[|x x0|< = | x x0|< ].Choose >0. Let = min( 2/2, 2 2). Choosex,x0>0. Assume|x x0|< . Reada= 2/2 andb= 2in Lemma 19: we need consider only four cases:(i)x < 2&x0< 2,(ii) 2/2 x& 2/2 x0,(iii)x < 2/2 & 2 x0, (iv)x0< 2/2 & 2 (iii) and (iv) contradict the assumption that|x x0|< 2/2 =b aso we need only consider cases (i) and (ii). In case (i) we have x < and x0< by Lemma 17 so| x x0| max( x, x0)< . In case (ii) weuse the inequality (3) and get| x x0| |x x0|2 2/2=|x x0| 2 < 2 .7 Remark course, min( 2/2, 2 2) = 2/2. The more complicatedformula is given to help the reader understand how the proof was can also prove that the square root function is uniformlycontinuous by taking = 2and using the inequality a+b a+ b.