Continuity and Gauges y

CONTINUOUS FUNCTIONS

Section 5.5 Continuity and Gauges y

15. Letfandgbe Lipschitz functions onA.

(a) Show that the sumfþgis also a Lipschitz function onA.

(b) Show that iffandgare bounded onA, then the productfgis a Lipschitz function onA. (c) Give an example of a Lipschitz functionfon [0,1) such that its squaref²isnota Lipschitz

function.

16. A function is calledabsolutely continuouson an intervalIif for anye>0 there exists ad>0 such that for any pair-wise disjoint subintervals ½xk;y_k;k¼1;2;. . .;n, of I such that Pjxky_kj<dwe haveP

f xð Þ k f yð Þ_k

j j<e. Show that iffsatisfies a Lipschitz condition

onI, thenfis absolutely continuous onI.

5.5.3 Lemma If a partitionP_ of I:¼ ½a;bisd-fine and x2I, then there exists a tag ti

inP_ such thatjxt_ij dðt_iÞ.

Proof. Ifx2I, there exists a subinterval ½x_i1;x_ifromP_ that containsx. SinceP_ is d-fine, then

ð2Þ t_idðt_iÞ x_i1xx_it_iþdðt_iÞ;

whence it follows that jxt_ij dðt_iÞ. Q.E.D.

In the theory of Riemann integration, we will use gauges d that are constant functions to control the fineness of the partition; in the theory of the generalized Riemann integral, the use of nonconstant gauges is essential. But nonconstant gauge functions arise quite naturally in connection with continuous functions. For, letf :I! R be continuous on I and let e>0 be given. Then, for each point t2I there existsd_eðtÞ>0 such that ifjxtj<d_eðtÞandx2I, thenjfðxÞ fðtÞj <e. Since d_e is defined and is strictly positive on I, the function d_e is a gauge on I. Later in this section, we will use the relations between gauges and continuity to give alternative proofs of the fundamental properties of continuous functions discussed in Sections 5.3 and 5.4.

5.5.4 Examples (a) Ifdandgare gauges onI:¼ ½a;band if 0<dðxÞ gðxÞfor all x2I, then every partitionP_ that isd-fine is alsog-fine. This follows immediately from the inequalities

tigðtiÞ tidðtiÞ and tiþdðtiÞ tiþgðtiÞ which imply that

t_i2½t_idðt_iÞ;t_iþdðt_iÞ ½t_igðt_iÞ;t_iþgðt_iÞ for i¼1;. . .;n: (b) Ifd1 andd2 are gauges onI:¼ ½a;band if

dðxÞ:¼minfd1ðxÞ; d2ðxÞg for all x2I;

thendis also a gauge onI. Moreover, sincedðxÞ d1ðxÞ, then everyd-fine partition isd1- fine. Similarly, everyd-fine partition is alsod2-fine.

(c) Suppose thatd is defined onI:¼ ½0;1by dðxÞ:¼

1

10 if x¼0;

1

2x if 0<x1: (

Thendis a gauge on [0, 1]. If 0<t1, then½tdðtÞ;tþdðtÞ ¼¹₂t;³₂t

, which does not contain the point 0. Thus, ifP_ is ad-fine partition ofI, then the only subinterval inP_ that contains 0 must have the point 0 as its tag.

Figure 5.5.1 Inclusion (1)

(d) Let g be defined onI:¼ ½0; 1by

gðxÞ:¼

1

10 if x¼0 orx¼1;

1

2x if 0<x¹₂;

1

2ð1xÞ if ¹₂<x<1: 8>

><

>>

:

Theng is a gauge on I, and it is an exercise to show that the subintervals in anyg-fine partition that contain the points 0 or 1 must have these points as tags.

&

Existence ofd-Fine Partitions

In view of the above examples, it is not obvious that an arbitrary gaugedadmits ad-fine partition. We now use the Supremum Property of R to establish the existence ofd-fine partitions. In the exercises, we will sketch a proof based on the Nested Intervals Theorem 2.5.2.

5.5.5 Theorem Ifdis a gauge defined on the interval[a,b], then there exists ad-fine partition of [a, b].

Proof. LetEdenote the set of all pointsx2 ½a;bsuch that there exists ad-fine partition of the subinterval [a, x]. The setE is not empty, since the pair ([a,x],a) is a d-fine partition of the interval [a,x] whenx2½a; aþdðaÞandxb. SinceE ½a;b, the set Eis also bounded. Letu:¼supEso thata<ub. We will show thatu2Eand that u¼b.

We claim that u2E. Since udðuÞ<u¼supE, there exists v2E such that udðuÞ<v<u. Let P_1 be a d-fine partition of [a, v] and let P_2:¼P_1[ ½v;ð u;uÞ. ThenP_2 is ad-fine partition of [a,u], so thatu2E.

Ifu<b, letw2 ½a;bbe such thatu<w<uþdðuÞ. IfQ_1 is ad-fine partition of [a,u], we let Q_2:¼Q_1[ ½u;ð w; uÞ. Then Q_2 is a d-fine partition of [a, w], whence w2E. But this contradicts the supposition that u is an upper bound of E. Therefore

u ¼b. Q.E.D.

Some Applications

Following R. A. Gordon (see hisMonthlyarticle in the References), we will now show that some of the major theorems in the two preceding sections can be proved by using gauges.

Alternate Proof of Theorem 5.3.2: Boundedness Theorem. Sincefis continuous onI, then for each t2I there exists dðtÞ>0 such that if x2I and jxtj dðtÞ, then jfðxÞ fðtÞj 1. Thusdis a gauge on I. LetfðI_i;t_iÞgⁿ_i¼1 be ad-fine partition ofIand letK:¼maxfjfðt_iÞj:i¼1;. . .;ng. By Lemma 5.5.3, given anyx2Ithere existsiwith jxt_ij dðt_iÞ, whence

jfðxÞj jfðxÞ fðtiÞj þ jfðtiÞj 1þK:

Sincex2I is arbitrary, thenfis bounded by 1þKonI. Q.E.D.

5.5 CONTINUITY AND GAUGES 151

Alternate Proof of Theorem 5.3.4: Maximum-Minimum Theorem. We will prove the existence ofx . LetM:¼supffðxÞ:x2Igand suppose thatf(x)<Mfor allx2I. Sincef is continuous onI, for eacht2Ithere existsdðtÞ>0 such that ifx2Iandjxtj dðtÞ, thenfðxÞ<¹₂ðMþfðtÞÞ. Thusdis a gauge onI, and iffðIi;tiÞgⁿ_i¼1is ad-fine partition ofI, we let

M~:¼¹₂maxfMþfðt1Þ;. . .;MþfðtnÞg:

By Lemma 5.5.3, given anyx2I, there existsiwithjxt_ij dðt_iÞ, whence fðxÞ<¹₂ðMþfðtiÞÞ M~:

Sincex2Iis arbitrary, thenM~ð<MÞis an upper bound forfonI, contrary to the definition

of Mas the supremum off. Q.E.D.

Alternate Proof of Theorem 5.3.5: Location of Roots Theorem. We assume thatfðtÞ 6¼0 for allt2I. Sincefis continuous att, Exercise 5.1.7 implies that there existsdðtÞ>0 such that ifx2Iandjxtj dðtÞ, thenf(x)<0 iff(t)<0, andf(x)>0 iff(t)>0. Thendis a gauge onIand we letfðI_i;t_iÞgⁿ_i¼1be ad-fine partition. Note that for eachi, eitherf(x)<0 for allx2 ½x_i1;x_iorf(x)>0 for all suchx. Sincefðx0Þ ¼fðaÞ<0, this implies that f(x1) < 0, which in turn implies that f(x2) < 0. Continuing in this way, we have fðbÞ ¼fðx_nÞ<0, contrary to the hypothesis thatf(b)>0. Q.E.D.

Alternate Proof of Theorem 5.4.3: Uniform Continuity Theorem. Lete>0 be given.

Sincefis continuous att2I, there existsdðtÞ>0 such that ifx2Iandjxtj 2dðtÞ, thenjfðxÞ fðtÞj ¹₂e. Thus dis a gauge on I. IffðIi;tiÞgⁿ_i¼1 is ad-fine partition ofI, letd_e:¼minfdðt1Þ;. . .;dðtnÞg. Now suppose thatx;u2Iandjxuj d_e, and choosei withjxtij dðtiÞ. Since

jutij juxj þ jxtij d_eþdðtiÞ 2dðtiÞ;

then it follows that

jfðxÞ fðuÞj jfðxÞ fðtiÞj þ jfðtiÞ fðuÞj ¹₂eþ¹₂e¼e:

Therefore,fis uniformly continuous onI. Q.E.D.

Exercises for Section 5.5

1. Letdbe the gauge on [0, 1] defined bydð0Þ:¼¹₄anddðtÞ:¼¹₂tfort2 ð0;1. (a) Show thatP_1¼ 0; ¹₄

;0

; ¹₄; ¹₂

; ¹₂

; ¹₂;1

; ³₄

isd-fine.

(b) Show thatP_2¼ 0; ¹₄

;0

; ¹₄; ¹₂

; ¹₂

; ¹₂;1

; ³₅

is notd-fine.

2. Suppose thatd1is the gauge defined byd1ð0Þ:¼¹₄;d1ðtÞ:¼³₄tfort2 ð0;1. Are the partitions given in Exercise 1d1-fine? Note thatdðtÞ d1ðtÞfor allt2 ½0;1.

3. Suppose thatd2 is the gauge defined byd2ð0Þ:¼₁₀¹ and d2ðtÞ:¼₁₀⁹t fort2 ð0;1. Are the partitions given in Exercise 1d2-fine?

4. Letgbe the gauge in Example 5.5.4(d).

(a) Ift2 ð0;¹₂show that½tgðtÞ;tþgðtÞ ¼¹₂t;³₂t ð0;³₄. (b) Ift2 ð¹₂;1Þshow that½tgðtÞ;tþgðtÞ ð¹₄;1Þ.

5. Leta<c<band letdbe a gauge on [a,b]. IfP_⁰is ad-fine partition of [a,c] and ifP_⁰⁰is ad-fine partition of [c,b], show thatP_⁰[P_⁰⁰is ad-fine partition of [a,b] havingcas a partition point.

6. Leta<c<band letd⁰andd⁰⁰be gauges on [a,c] and [c,b], respectively. Ifdis defined on [a,b] by

dðtÞ:¼

d⁰ðtÞ if t2 ½a;cÞ;

minfd⁰ðcÞ;d⁰⁰ðcÞg if t¼c;

d⁰⁰ðtÞ if t2 ðc;b;

8>

><

>>

:

thendis a gauge on [a,b]. Moreover, ifP_⁰is ad⁰-fine partition of [a,c] andP_⁰⁰ is ad⁰⁰-fine partition of [c,b], thenP_⁰[P_⁰⁰is a tagged partition of [a,b] havingcas a partition point.

Explain whyP_⁰[P_⁰⁰may not bed-fine. Give an example.

7. Letd⁰andd⁰⁰ be as in the preceding exercise and letd be defined by

dðtÞ:¼

mind⁰ðtÞ; ¹₂ðctÞ

if t2 ½a;cÞ;

minfd⁰ðcÞ;d⁰⁰ðcÞg if t¼c;

mind⁰⁰ðtÞ; ¹₂ðtcÞ

if t2 ðc;b:

8>

><

>>

:

Show thatd is a gauge on [a,b] and that everyd -fine partitionP_of [a,b] havingcas a partition point gives rise to ad⁰-fine partitionP_⁰of [a,c] and ad⁰⁰-fine partitionP_⁰⁰of [c,b] such that P ¼_ P_⁰[P_⁰⁰.

8. Letdbe a gauge onI:¼ ½a;band suppose thatI does nothave ad-fine partition.

(a) Letc:¼¹₂ðaþbÞ. Show that at least one of the intervals [a,c] and [c,b] does not have a d-fine partition.

(b) Construct a nested sequence (I_n) of subintervals with the length ofI_nequal toðbaÞ=2ⁿ such thatIndoes not have ad-fine partition.

(c) Letj2 \¹_n¼1Inand letp2Nbe such thatðbaÞ=2^p<dðjÞ. Show thatIp½jdðjÞ;

jþdðjÞ, so the pairIp;j

is ad-fine partition ofIp.

9. LetI:¼ ½a;band letf:I!Rbe a (not necessarily continuous) function. We say that fis

‘‘locally bounded’’ at c2I if there existsdðcÞ>0 such thatfis bounded on I\½cdðcÞ;

cþdðcÞ. Prove that iffis locally bounded at every point ofI, thenfis bounded onI.

10. LetI:¼ ½a;bandf:I!R. We say thatfis ‘‘locally increasing’’ atc2Iif there existsdðcÞ>

0 such thatfis increasing onI\½cdðcÞ;cþdðcÞ. Prove that iffis locally increasing at every point ofI, thenfis increasing onI.