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Continuity and Uniform Continuity

521

May 12, 2010

1.ThroughoutSwill denote a subset of the real numbersRandf:S!R

will be a real valued function dened onS. The setSmay be bounded like

S= (0;5) =fx2R: 0< x <5g

or innite like

S= (0;1) =fx2R: 0< xg:

It may even be all ofR. The valuef(x) of the functionfat the pointx2S will be dened by a formula (or formulas). Denition 2.The functionfis said to becontinuous onSi

8x02S8" >09 >08x2S

jxx0j< =) jf(x)f(x0)j< "

Hencefis not continuous1onSi

9x02S9" >08 >09x2S

jxx0j< andjf(x)f(x0)j " Denition 3.The functionfis said to beuniformly continuous onSi

8" >09 >08x02S8x2S

jxx0j< =) jf(x)f(x0)j< "

Hencefis not uniformly continuous onSi

9" >08 >09x02S9x2S

jxx0j< andjf(x)f(x0)j " :1 For an example of a function which isnotcontinuous see Example 22 below. 1

4.The only dierence between the two denitions is the order of the quan-

tiers. When you provefis continuous your proof will have the form

Choosex02S. Choose" >0. Let=(x0;"). Choosex2S.

Assumejxx0j< .Thereforejf(x)f(x0)j< ".

The expression for(x0;") can involve bothx0and"but must be independent ofx. The order of the quaniers in the denition signals this; in the proofx has not yet been chosen at the point whereis dened so the denition of must not involvex. (Therepresent the proof thatjf(x)f(x0)j< " follows from the earlier steps in the proof.) When you provefis uniformly continuous your proof will have the form

Choose" >0. Let=("). Choosex02S. Choosex2S.

Assumejxx0j< .Thereforejf(x)f(x0)j< ".

so the expression forcan only involve"and must not involve eitherxor x 0. It is obvious that a uniformly continuous function is continuous: if we can nd awhich works for allx0, we can nd one (the same one) which works for any particularx0. We will see below that there are continuous functions which are not uniformly continuous. Example 5.LetS=Randf(x) = 3x+7. Thenfis uniformly continuous onS. Proof.Choose" >0. Let="=3. Choosex02R. Choosex2R. Assume jxx0j< . Then jf(x)f(x0)j=j(3x+ 7)(3x0+ 7)j= 3jxx0j<3=":Example 6.LetS=fx2R: 0< x <4gandf(x) =x2. Thenfis uniformly continuous onS. Proof.Choose" >0. Let="=8. Choosex02S. Choosex2S. Thus

0< x0<4 and 0< x <4 so 0< x+x0<8. Assumejxx0j< . Then

jf(x)f(x0)j=jx2x20j= (x+x0)jxx0j<(4 + 4)=":2

7.In both of the preceeding proofs the functionfsatised an inequality of

form jf(x1)f(x2)j Mjx1x2j(1) forx1;x22S. In Example 5 we had j(3x1+ 7)(3x2+ 7)j 3jx1x2j and in Example 6 we had jx21x22j 8jx1x2j for 0< x1;x2<4. An inequality of form (1) is called aLipschitz inequality and the constantMis called the correspondingLipschitz constant. Theorem 8.Iffsatises (1) forx1;x22S, thenfis uniformly continuous onS. Proof.Choose" >0. Let="=M. Choosex02S. Choosex2S. Assume thatjxx0j< . Then jf(x)f(x0)j Mjxx0j< M=":9.The Lipschitz constant depend might depend on the interval. For example, jx21x22j= (x1+x2)jx1x2j 2ajx1x2j for 0< x1;x2< abut the functionf(x) =x2does not satisfy a Lipschitz inequality on the whole interval (0;1) since jx21x22j= (x1+x2)jx1x2j> Mjx1x2j ifx1=Mandx2=x1+ 1. In fact, Example 10.The functionf(x) =x2is continuous but not uniformly con- tinuous on the intervalS= (0;1).

Proof.We showfis continuous onS, i.e.

8x02S8" >09 >08x2S

jxx0j< =) jx2x20j< " 3 Choosex0. Leta=x0+ 1 and= min(1;"=2a). (Note thatdepends on x

0sinceadoes.) Choosex2S. Assumejxx0j< . Thenjxx0j<1 so

x < x

0+ 1 =asox;x0< aso

jx2x20j= (x+x0)jxx0j 2ajxx0j<2a2a"2a=" as required. We show thatfis not uniformly continuous onS, i.e.

9" >08 >09x02S9x2S

jxx0j< andjx2x20j " Let"= 1. Choose >0. Letx0= 1=andx=x0+=2. Thenjxx0j= =2< but jx2x20j= 1 +2 2 1

2= 1 +24

>1 ="

as required. (Note thatx0is large whenis small.)11.According to the Mean Value Theorem from calculus for a dierentiable

functionfwe have f(x1)f(x2) =f0(c)(x2x1): forsomecbetweenx1andx2. (The slope (f(x1)f(x2))=(x1x2) of the secant line joining the two points (x1;f(x1)) and (x2;f(x2)) on the graph is the same as the slopef0(c) of the tangent point at the intermediate point (c;f(c)).) Ifx1andx2lie in some intervalSandjf0(c)j Mforallc2S we conclude that the Lipschitz inequality (1) holds onS. 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 ofMand then prove the inequality by other means.

12.For example, consider the functionf(x) =x1dened on the interval

S= (a;1) wherea >0. Forx1;x22Sthe Mean Value Theorem says that x

11x12=c2(x1x2) wherecis betweenx1andx2. Ifx1;x22Sthen

c2S(ascis betweenx1andx2) and hencec > asoc2< a2. We can prove the inequality jx11x12j a2jx2x2j 4 forx1;x2aas follows. Firsta2x1x2sinceax1andax2. Then jx11x12j=jx1x2jx

1x2jx1x2ja

2(2) where we have used the fact that1< 1if 0< < . It follows that that the functionf(x) is uniformly continuous on any interval (a;1) where a >0. Notice however that the Lipschitz constantM=a2depends on the interval. In fact, the functionf(x) =x1doesnotsatisfy a Lipshitz inequality on the interval (0;1).

13.We can discover a Lipscitz inequality for the square root functionf(x) =pxin much the same way. Consider the functionf(x) =pxdened on the

intervalS= (a;1) wherea >0. Forx1;x22Sthe Mean Value Theorem says thatpx 1px

2= (x1x2)=(2pc) wherecis betweenx1andx2. If

x

1;x22Sthenc2S(ascis betweenx1andx2) and hencec > aso

(2pc)1<(2pa)1. We can prove the inequality j px 1px

2j jx1x2j2

pa (3) forx1;x2aas follows: Divide the equation px 1px 2)(px 1+px

2) = ((px

1)2(px

2)2) =x1x2

by ( px 1+px

2), take absolute values, and use (px

1+px

2)2pa. Again the

Lipschitz constantM= (2pa)1depends on the interval and the function doesnotsatisfy a Lipschitz inequality on the interval (0;1). Example 14.The functionf(x) =x1is continuous but not uniformly continuous on the intervalS= (0;1).

Proof.We showfis continuous onS, i.e.

8x02S8" >09 >08x2S

jxx0j< =)1x 1x 0 Choosex0. Leta=x0=2 and= min(x0a;a2"). Choosex2S. Assume jxx0j< . Thenx0x jxx0j< x0asox 0< aso by (2) 1x 1x 0 jx1x2ja 208 >09x02S9x2S jxx0j< and1x 1x 0 Let"= 1. Choose >0. Letx0= min(;1) andx=x0=2. Thenjxx0j= x

0=2=2< but

1x 1x 0 =1x 0=21x 0 =1x 01 =" as required.Example 15.The functionf(x) =pxis uniformly continuous on the set

S= (0;1).

Remark 16.This example shows that a function can be uniformly contin- uous on a set even though it does not satisfy a Lipschitz inequality on that set, i.e. the method of Theorem 8 is not the only method for proving a function uniformly continuous. The proof we give will use the following idea. After choosing" >0 we specify two numbersaandbwhich will depend on ". These numbers will satisfy 0< a < b. We will chooseso that (among other things) < ba. Then after we choosex;x02Sand assume that jxx0j< we will be able to conclude that either bothx0andxare less thanbor both are greater thana. We will choosebso small thatpxandpx

0are within"of zero forx;x0< b. We will use a Lipschitz inequality to

handle the case wherex;x0> a. We give the details of this proof after some preliminary lemmas. The only properties of that square root function that we will use are thatpxis dened forx0 and satises px0;(px)2=x;px 2=x:

Lemma 17.The square root function is increasing:

0a < b=)pa <

pb:

Proof.Assume 0a < b. Ifpbpathenb= (pb)2(pa)2=a

contradictinga < b. Hencepa < pb.Lemma 18. pab=pa pbfora;b0. 6

Proof.(pa

pb)2= (pa)2(pb)2=absopa pb=q( pa pb)2=pab.Lemma 19.Assumea < b. Then for any two numbersxandyat least one of the four alternatives (i)x < b&y < b, (ii)ax&ay, (iii)x < a&by, (iv)y < a&bx. Proof.Exactly one of the three alternativesx < a,ax < b,bxholds and exactly one of the three alternativesy < a,ay < b,byholds. There are thus nine cases which we can arrange in a table:x < a ax < b bxy < a(i) (i) (iv) ay < b(i) (i);(ii) (ii) by(iii) (ii) (ii) In each entry of the table we have indicated the alternative (iiv) which

holds in the corresponding case.Proof.Now we prove what is claimed in Example 15, viz. that the square

root function is uniformly continuous on the positive real numbers, i.e.

8" >09 >08x;x0>0

jxx0j< =) jpxpx 0j< " Choose" >0. Let= min("2=2;p2"2). Choosex;x0>0. Assumejxx0j< . Reada="2=2 andb="2in Lemma 19: we need consider only four cases: (i)x < "2&x0< "2, (ii)"2=2x&"2=2x0, (iii)x < "2=2 &"2x0, (iv)x0< "2=2 &"2x. Cases (iii) and (iv) contradict the assumption thatjxx0j< "2=2 =ba so we need only consider cases (i) and (ii). In case (i) we havepx < "andpx

0< "by Lemma 17 sojpxpx

0j max(px;

px

0)< ". In case (ii) we

use the inequality (3) and get j pxpx

0j jxx0j2

p"

2=2=jxx0jp2" Remark 20.Of course, min("2=2;p2"2) ="2=2. The more complicated formula is given to help the reader understand how the proof was discovered. Remark 21.We can also prove that the square root function is uniformly continuous by taking="2and using the inequality pa+bpa+pb: (To prove the inequality square both sides.) Ifjxx0j< "2then either x

0x < x0+"2(which impliespx

0px < px

0+") or elsexx0< x+"2

(which impliespxpx

0

0j< ". This

proof is simpler than the one given but is harder to nd since it uses a clever inequality which the reader might not think of.

Example 22.The functionfdened by

f(x) =1 ifx >0

0 ifx0

is not continuous.

Proof.In other words,

9x02R9" >08 >09x2R

jxx0j< andjf(x)f(x0)j " Letx0= 0. (No other value ofx0will work here.) Let"= 1. Choose >0.

Letx==2. Thenjxx0j==2< butjf(x)f(x0)j=j10j= 1".8

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