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Uniform continuity andgrowth of real continuousfunctionsSmail DjebaliPublished online: 11 Nov 2010.

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Uniform continuity and growth of real continuous functions

SMAIL DJEBALI

Department of Mathematics, E cole Normale Supe rieure, B.P. 92 Kouba, 16050 Algiers,Algeria; e-mail: djebali@hotmail.com

(Received 10 May 1999)

The purpose of this paper is to understand whether there exists any linkbetween the uniform continuity of a real function de®ned on an unboundedinterval and its growth at in®nity. The primary objective is to present someresults from teaching experience which help in the comprehension of thisnotion and yield some classroom techniques. It is well known that a uniformlycontinuous function has a monomial growth; it will be proved that there doesnot exist another growth of positive order. After introducing three kinds ofgrowth, some results are recalled in connection with the behaviour near in®nityof a uniformly continuous function. Using a series of counterexamples, it isshown that the uniform continuity of a function cannot be described by itsasymptotic behaviour near in®nity. Finally, some useful properties of theaveraging convergence are reviewed, and how this is related to uniformcontinuity is investigated.

1. Introduction

The notion of uniform continuity in real analysis turns out to be an epistemo-logical obstacle. The concept of continuity is often transmitted to students andpupils through the intuitive representation of an unbroken curve; but what aboutuniform continuity? Is there any global geometrical apprehension of this concept?

For many students of elementary mathematical analysis, the geometricalinterpretation of the uniform continuity of real functions is not assimilated aswell as some teachers would like. Indeed, in most classical mathematical textbooks,the relation between the behaviour at in®nity of a function and its uniformcontinuity over some unbounded real interval is not made precise.

Moreover, it should be emphasized that books often deal with functions de®nedon bounded intervals or merely discuss elementary illustrative examples with nodistinction made between continuity and uniform continuity. Note that, from ahistorical point of view, the notion of uniform continuity had ®rst been introducedfor functions de®ned on bounded intervals by Heine (1828±1881), a student ofWeierstrass (1815±1897) (see [1], p. 657).

Nevertheless, we know that very simple criteria from classical analysis mayoften enable us to assert whether a function de®ned over the real line is uniformlycontinuous or not. For instance, a glance at the graph of a function may sometimesexplain quickly why it is not uniformly continuous. Furthermore, it seems to beworthwhile to gather and to summarize some known as well as less known resultsabout this concept.

int. j. math. educ. sci. technol., 2001, vol. 32, no. 5, 677±689

International Journal of Mathematical Education in Science and TechnologyISSN 0020±739X print/ISSN 1464±5211 online # 2001 Taylor & Francis Ltd

http://www.tandf.co.uk/journalsDOI: 10.1080/00207390110053757

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So, the aim of this work is to carry through a self-contained investigation of thesubject in order to provide the teacher with a survey and to help the student tograsp better the aformentioned relation. In particular, it is hoped that the resultsand the conclusion of this contribution plug a gap by providing some pedagogicalmeans of presentation and a somewhat new approach in the teaching of theimportant concept of uniform continuity.

In what follows, I ˆ ‰a; ‡1†…a > 1† will denote an interval of ‡: ˆ ‰0; ‡1†while ‡

¤ : ˆ …0; ‡1† and ¤ stands for the set of non-negative integers; f : I ¡! isassumed to be a continuous function. Let us consider the open stripE ˆ …¡1; 1† £ and the sets

E0 ˆ E { f…1; y† 2 2 : y < 0g and E1 ˆ E0 { f…1; 0†g

2. De®nitionsThe function f is said to have a monomial growth if

lim supx!‡1

j f…x†jx

< ‡1 …1†

We say that f admits a growth of order ¬ in terms of the variable x and of order ­ withrespect to logarithm x if there exists a pair …¬; ­ † 2 E0 such that

lim supx!‡1

j f…x†jx¬ ln­ x

< ‡1 …2†

A growth is of Bernstein type when the function f satis®es the following condition:9 k > 0, 9 x0 2 I, 9 h: I¡! Š0; ‡1‰ such that

‡1

a

ds

h…s† ˆ …0; ‡1† and j f …x†j 4 kh…x†; 8 x 5 x0 …3†

Hereafter, we will need to de®ne the following function

g¬;­ …x† ˆ x¬ ln­ …x†; for x 2 I …4†

f is said to be uniformly continuous if it satis®es

8 " > 0; 9 ¯ > 0; 8…x; y† 2 I2; …jx ¡ yj 4 ¯ ) j f…x† ¡ f …y†j 4 "†

Equivalently, we have the following sequential criterion (see, for instance, [2],Theorem 11.1): f is uniformly continuous if and only if for any two sequences…xn†n2N ; …yn†n2N 2 I such that limn!1…xn ¡ yn† ˆ 0; we have limn!1… f…xn†¡f…yn† ˆ 0: From pedagogical experiences, this characterization turns out to bee� cient in practice to show either uniform continuity or nonuniform continuity; itwill be frequently used in this paper.

3. Remarks

(a) Condition (2) is read f…x† ˆx!‡1 O g¬;­ …x† in which the notation O standsfor big o (see [2], p. 391).

(b) As easily seen, (1) is a particular case of (2) though …¬; ­ † 6ˆ …1; 0†:(c) One may extend (2) to some other growth conditions such as

x¬ ln­ …ln …. . . ln …x† . . .†† or x¬ ln­ 1 …x†: ln­ 2 …ln …x††: ln­ 3 …ln …ln …x††† . . .

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In fact, the following estimates are straightforward

x¬ 4 x­ ; 8 x 2 I and 8 ¬ 2Š…0; ­ ŠŠ

9 x0 2 I; ln­ …ln …x†† 4 ln® …x†; 8 x 5 x0 and 8 …­ ; ®† 2 £ ‡¤

(d) (3) is a growth condition of integral type. It may include Condition (2);in particular, every Lipschitz continuous function satis®es (3) withh…s† ˆ s:

4. Preliminaries

We ®rst assert some auxiliary results we need for developing our subsequentresults.

Lemma 4.1. Every Lipschitz function on I is uniformly continuous.

Lemma 4.2. f is uniformly continuous on I whenever limx!‡1 f…x† exists.

Lemma 4.3. Every uniformly continuous function f admits a monomialgrowth.

Lemma 4.4. The function g¬;­ de®ned in (4) is uniformly continuous if andonly if …¬; ­ † 2 E1:

Lemma 4.5. Let …¬; ­ † 2 E1; then the function g¬;­ has a Bernstein typegrowth.

The ®rst lemma is obvious; Lemmas 4.2 and 4.3 are rather classical (see forinstance [3, probleÁ me 4, p. 48]); but for the sake of completeness, we give here asuccinct proof of each one.

Proof of Lemma 4.2. Assume limx!‡1 f …x† ˆ ` and consider some " > 0; thenthere exists x0 2 I such that …x 5 x0 ) j f …x† ¡ `j 4 "=2†¢ We infer that for any…x; y† 2 ‰x0; ‡1Š2; j f …x† ¡ f …y†j 4 ": Now, if …x; y† 2 ‰a; x0 ‡ 1Š2; the inequalityj f …x† ¡ f …y†j 4 " is again valid by virtue of Heine’s Theorem ([2], Theorem11.5); then the claim of the lemma follows.

Proof of Lemma 4.3. f being uniformly continuous, there exists some ¯ > 0such that

8 …x; y† 2 I2; …jx ¡ yj 4 ¯ ) j f …x† ¡ f…y†j 4 1† …5†

Let x 2 I and n ˆ ‰x=¯Š where ‰tŠ stands for the integer part of the real t; thenn¯ 4 x < …n ‡ 1†¯, that is

0 4 x ¡ n¯ < ¯ …6†

Writing f…x† ˆ n¡1kˆ0‰ f…x ¡ k¯† ¡ f…x ¡ …k ‡ 1†¯†Š ‡ f …x ¡ n¯†; we get, owing to

(5), (6) j f …x† 4 n ‡ j f…x ¡ n¯†j 4 x=¯ ‡ sup0 4 t 4 ¯ j f…t†j; and the lemmafollows.

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Proof of Lemma 4.4. First assume …¬; ­ † 2 E0; then limx!‡1 g 0…x† ˆ 0 sothat g¬;­ is a lipschitz continuous function whence uniformly continuous byLemma 4.1 (the case …¬; ­ † ˆ …1; 0† is obvious). Conversely, if …¬; ­ † 62 E1, then

limx!‡1 …g¬;­ …x†=x† ˆ ‡1

hence g¬;­ is, by Lemma 4.3, nonuniformly continuous.

Proof of Lemma 4.5. Note that for any …¬; ­ † 2 Š ¡ 1; 1‰ £ ; there exist® 2Š¬; 1‰ and x0 2 I such that for any x 5 x0; x¬ ln­ …x† 4 x®; therefore

‡1

a

dx

g¬;­ …x† 5‡1

a

dx

x®ˆ ‡1

The case ¬ ˆ 1; ­ 4 0 is treated in a similar manner with the help of the inequalityx ln­ …x† 4 x ln­ …a†; 8 x 2 I:

5. Uniform continuity and asymptotic behaviour

Throughout this section, some results related to the behaviour near in®nity of auniformly continuous function are summed up. We present a classi®cation whichtakes into account the value of limx!‡1 … f …x†=x†. The ®rst one is a consequence ofLemma 4.3.

Proposition 5.1. If limx!‡1 … f…x†=x† ˆ ‡1; then f is not uniformly contin-uous on I:

Proposition 5.2. If limx!‡1 … f…x†=x† ˆ 0; then apart from the case in whichlimx!‡1 f…x† ˆ `; the uniform continuity of f is not a priori determined.

Proof. The case limx!‡1 f …x† ˆ ` is Lemma 4.2. Otherwise, three situationsmay occur; in each one, we give a counterexample showing that the uniformcontinuity cannot be foreseen.

. The case where limx!‡1 j f…x†j† ˆ ‡1:

Both functions xp

and xp

exp …sin …x†† satisfy this condition; the ®rst one isuniformly continuous (Appendix A.2) while the second one is not; to be convinced,one may use the sequences xn ˆ 2n4º and yn ˆ xn ‡ 1=n. Then, doing a two ordersdevelopment of the sine and the exponential functions, we obtain

f…yn† ¡ f…xn† ˆ xnp yn

xn

exp sin1

n¡ 1

ˆ n2 2ºp 1

n‡

1

2

1

n

2

‡1

n2"

1

n

ˆ 2ºp

n ‡ 1

2‡ "

1

n

with limn!‡1 " 1=n… † ˆ 0: Therefore, limn!‡1 f …yn† ¡ f …xn†… † ˆ ‡1 whilelimn!‡1 yn ¡ xn… † ˆ 0:

. The case where limx!‡1 f…x† does not exist and f is bounded:

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Consider the functions f …x† ˆ sin …x† and f …x† ˆ sin …x2†; the ®rst one isuniformly continuous while the second one is not (A.1 in the Appendix).

. The case where limx!‡1 f…x† does not exist and f is not bounded:

The function f …x† ˆ xp

sin … xp † is uniformly convergent but f …x† ˆ x

psin …x†

is not (A.1 in the Appendix).

Remark 5.3.(a) In Proposition 5.2, the case ` 6ˆ 0 obviously goes back to the case ` ˆ 0:(b) The function x

pexp …sin …x†† provides an example of a nonuniformly

continuous function bounded by two nontrivial uniformly continuousfunctions (e x

pand e¡1 x

p) growing at the same speed.

6. Uniform continuity and growth at 1This section deals with the crucial point of our analysis, namely the relation

that may exist between the uniform continuity of a function and its growth atin®nity. To this end, we begin by introducing two assumptions

9 ` 5 0; 9 …¬; ­ † 2 E0 such that limx!‡1

f …x† ¡ `x¬ ln­ x ˆ 0 …7†

9 ` 5 0; 9 …¬; ­ † 2 E0 such that limx!‡1

f…x†x¬ ln­ x

ˆ ` …8†

Remark 6.1. Assumptions (7) and (8) are read respectivelyf …x† ¡ lg¬;­ …x† ˆx!‡1

o…1† and f ¹x!‡1`g¬;­ where g¬;­ has been introduced in (4).

It is meaningful to ®rst compare these hypotheses with De®nitions (1)±(3).

Proposition 6.2. Let us consider the statements:

(a) f satis®es (7).(b) f satis®es (7).(c) f satis®es (2).(d) limx!‡1 … f…x†=x† ˆ 0:(e) f satis®es (1).(f) f satis®es (3).

Then, the following implications hold true

…a† )̂ …b† () …c† )̂ …d† )̂ …e† )̂ … f †

In addition, none of the converse implications hold true.

Proof. Assume (a) and let ` 5 0; …¬; ­ † 2 E0 satisfy (7). When ¬ > 0 or

…¬ ˆ 0 and ­ > 0†, limx!‡1 x¬ ln­ …x† ˆ ‡1 so that ` and …¬; ­ † also satisfy(8). If ¬ < 0 or (¬ ˆ 0 and ­ < 0), (6.1) implies limx!‡1 f …x† ˆ 0; in thiscase, any pair …¬ 0; ­ 0† 2 E0 \ … ‡

¤ £ † satis®es (8) with ` ˆ 0; the case

¬ ˆ ­ ˆ 0 is obvious; (b) is then proved.. The implication …b† ) …c† is straightforward. Conversely, assume (c); then

there exist k > 0; x0 2 I and …¬; ­ † 2 E0 such that j f …x†j 4 kx¬ ln­ …x†; for

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any x 5 x0: It follows that f veri®es (8) with ` ˆ 0 together with some pair…®; ¯† 2 E0 such that either ¬ < ® < 1 (this occurs when ¬ < 1), or ® ˆ 1 and0 < ¯ (whenever ¬ ˆ 1 and ­ < 0); (b) follows.

. The implication …e† ) … f † is checked with h…s† ˆ s; the remaining implica-tions are easily proved.

. At last, we list some counterexamples showing that the converse implicationsdo not hold true.

(1) f…x† ˆ xp

sin …x† satis®es (b) not (a).(2) f…x† ˆ x ln¡¯…ln …x†† with some ¯ > 0 satis®es (d) not (c).(3) f…x† ˆ x satis®es (e) not (d).(4) f…x† ˆ x ln …x† satis®es (f) not (e).

It remains to clarify the connection existing between the uniform continuityand the statements occuring in Proposition 5.2. This is given by

Proposition 6.3. (a) A function f is uniformly continuous on I wheneverAssumption (7) is satis®ed.

(b) Assumption (8) does not imply the uniform continuity of the function f on I.In particular, none of Statements (c)±(f) can imply the uniform con-tinuity.

(c) None of Statements (a)±(d) is a necessary condition of uniform continuity.

Proof. Writing f …x† ˆ … f …x† ¡ `x¬ ln­ x† ‡ `x¬ ln­ x, part (a) follows straight-forwardly from Lemmas 4.2 and 4.4.

(b) f…x† ˆ xp

sin …x† satis®es (8) with ` ˆ 0 for any …¬; ­ † 2 ‰ 12 ; 1† £ ‡ while it

is not uniformly continuous (A.1 in the Appendix).(c) The identity mapping f ˆ IjR is uniformly continuous without satisfying

(d) and hence (a)±(c).

Remark 6.4.(a) Note that Assumption (7) does not imply a Lipschitz condition; this can be

checked by studying the function

f …x† ˆ x¬ ln­ x ‡ sin …x2†x

p

with some …¬; ­ † 2 E0; so, we cannot insert the Lipschitz Property betweenStatements (a)±(f) though this condition implies (e). Moreover, the LipschitzCondition does not imply (a) as shows the identity mapping f…x† ˆ x:

(b) The function f de®ned by f…x† ˆ x sin …1=x† is uniformly continuous(Appendix A.1) without being Lipschitz continuous on I; this can be checkedby considering the sequences

xn ˆ2

…2n ‡ 1†ºand yn ˆ

2

…2n ¡ 1†º

(c) Theorem 6.3 shows that Assumption (7) is not very strong with respect tothe uniform continuity concept. To ®nd a weaker hypothesis, that is a sharperresult, one may introduce an intermediary assumption between (7) and (8)which ensures the uniform continuity; this is an open question. Nevertheless,

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we will make another attempt in the subsequent section by recalling a new kind ofgrowth.

7. Uniform continuity and averaging convergence

Up to now, we have been interested essentially in polynomial or logarithmictype growths. In the following, we will show that Bernstein type growth (see (3)) isclosely related to what is commonly called the averaging convergence; further-more, the latter is known to be more concerned with the uniform continuityconcept. First, we recall

De®nition 7.1. (a) f is said to be averaging convergent if the improper integral‡1a

f…x† dx converges.(b) If the integral

‡1a

j f …x†j dx converges, we say that f is absolutely averagingconvergent.

Immediately, we have

Proposition 7.2. A function f admits a Bernstein type growth whenever it isabsolutely averaging convergent.

The proof relies on the following lemma which may easily be deduced from theCauchy±Schwartz inequality applied to the functions j f j and 1= j f j¢

Lemma 7.3. Let f : ‰a; bŠ 7¡! be a function such that f as well as 1=f areRiemann integrable; in addition, assume j f j > 0 over ‰a; bŠ: Then, the followingestimate holds true

b

a

j f…x†j dx ¢b

a

1

j f …x†jdx 5 …b ¡ a†2

Subsequently, writing

b

a

1

j f …x†j dx 5…b ¡ a†2

b

a

j f …x†j dx

and taking b large enough yields the proof of Proposition 7.2.

Remark 7.4.(a) The function x ,! x¬ with j¬j < 1 shows that the converse of Proposition 7.2

does not hold true.(b) At a ®rst glance, the following question arises: Is it possible to insert the

averaging convergence between Statements (a)±(f) in Proposition 6.2; more pre-cisely either between (d) and (e) or between (e) and (f)? Unfortunately, the answeris `no’; indeed, the function x ,! ln …x† satis®es Assumption (d) but is not averagingconvergent.

The last result in connection with the uniform continuity is given by

Theorem 7.5. Assume f to be averaging convergent; then the followingstatements hold true

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(a) If, in addition f is supposed to be a C2…I† class function such that‡1a

f 00…x† dx converges, then f and f 0 are uniformly continuous.(b) When f is monotonic, it is uniformly continuous and grows better than 1=x¢(c) If f does not possess a limit at ‡1; then it is not uniformly continuous.

Proof. (a) We begin by writing f 0…x† ˆ f 0…a† ‡ x

af 00…t† dt: Since the integral

‡1a

f 00…x† dx converges, limx!‡1 f 0…x† exists; we infer that f 0 is bounded (anduniformly continuous); Therefore, f is a Lipschitz function and we conclude withLemma 4.1.

(b) Let f be a nonincreasing function. Then

xf…2x† 42x

x

f …t† dt 4 xf…x† 4 2x

x=2

f …t† dt …9†

The Cauchy Criterion yields

limx!‡1

2x

x

f…t† dt ˆ limx!‡1

x

x=2

f…t† dt ˆ 0

Then (9) implies limx!‡1 xf…x† ˆ 0; whence the boundedness of f : Since f ismonotonic, it follows that limx!‡1 f…x† ˆ ` and ` ˆ 0: Lemma 4.2 then implies theuniform continuity of f ; moreover, there exists ·xx > 0 such that for any x 5 ·xx,f…x† 4 1=x; in case of an increasing function, the same argument holds for thefunction …¡f†:

(c) Arguing by contradiction, suppose f is uniformly continuous; then

8 " > 0; 9 ¬ > 0; 8 …x; y† 2 I2: jx ¡ yj 4 ¬ ) j f…x† ¡ f…y†j 4"

2…10†

By the Cauchy Criterion, we have

9 x0 2 I:x‡¬

x

f …t† dt 4 ¬"

2; 8 x 5 x0 …11†

Since f is continuous on I, we can make use of the First Mean Theorem to assertthat

8 x 5 x0; 9 ~xx 2 ‰x; x ‡ ¬Š: f …~xx† ˆ 1

¬

x‡¬

x

f …t† dt

With (11), we then derive the bound

j f …~xx†j 4"

2…12†

Putting y ˆ ~xx in (10) leads to the estimate

j f…x† ¡ f…~xx†j 4"

2…13†

Both (12) and (13) give in turn

8 " > 0; j f …x†j 4 j f …~xx†j ‡ j f…x† ¡ f…~xx†j 4"

2‡ "

2ˆ "; 8 x 5 x0

that is limx!‡1 f …x† ˆ 0 contradicting our assumption. The proof of Theorem 7.5is now complete.

Combining Lemma 4.2 with part (c) of the theorem yields the following

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Corollary 7.6. Let f be an averaging convergent function. Then, f is uni-formly continuous if and only if limx!‡1 f …x† exists (and equals 0).

Remark 7.7.(a) The uniform continuity does not imply the averaging convergence of a

function f even if limx!‡1 f …x† ˆ 0; as a counterexample, consider the functiong¬;­ de®ned in (4). Then, Bertrand’s integral

‡1a

g¬;­ …x† dx converges if and only if…¬ < ¡1† _ …¬ ˆ ¡1 ^ ­ < ¡1† [4, Lemma I, p. 278]. So, Lemma 4.4 both withsome ¡1 < ¬ < 0 lead to the required result.

(b) Now, consider the function x ,! x¬ sin x with some ¡1 4 ¬ < 0; thenlimx!‡1 f…x† ˆ 0, f is uniformly continuous (Appendix A.1), f is averagingconvergent but not absolutely averaging convergent [4, p. 280].

8. Applications

Remarks 8.1. To illustrate the scope of Theorem 7.5, we mention thefollowing examples:

(a) The function f1…x† ˆ xk sin …x† is averaging convergent for any k 2 …0; 1†:Since limx!‡1 f1…x† does not exist, it is not uniformly continuous on I. It can bechecked that it is still averaging convergent for k < 0 and in this caselimx!‡1 f1…x† ˆ 0; then f1 is uniformly continuous on I.

(b) For the same reason, both functions f2…x† ˆ x¬ sin …x2† and f3…x† ˆx¬ cos …x2† are not uniformly continuous for each ¬ 2 ‰0; 1†:

(c) More generally, the function f4…x† ˆ x¬ sin …x­ † is averaging convergent forvalues ­ > 1 and ¬ < ­ ¡ 1; hence it is not uniformly continuous whenever…¬; ­ † 2 ‰0; ­ ¡ 1† £ …1; ‡1†:

(d) f4 is a Lipschitz continuous function for any pair of real numbers…¬; ­ † satisfying ¬ ‡ ­ 4 1; now limx!‡1 f …x† does not exist if ever ¬ 5 0: There-fore, the integral

‡1a

x¬ sin …x­ † dx diverges for all values of parameters…¬; ­ † 2 ‡ £ …¡1; 1 ¡ ¬Š:

We point out that it is possible to get more useful applications of the analysisdiscussed above. Hereafter, we present without proof some of them as supple-mentary statements.

Proposition 8.2.(1) Let f be an absolutely averaging convergent function such that

‡1a

f 2…x† dxdiverges. Then f is not uniformly continuous.(Hint: Argue by contradiction).

(2) Let f be a C1…I† class function, absolutely averaging convergent, such that‡1a

f 02…x† dx converges. Then f is uniformly continuous, hence limx!‡1 f …x† ˆ 0:(Hint: Use 8x 2 I; 2j f 0…x†j j f …x†j 4 f 0…x†2 ‡ j f …x†j†:

(3) Let f be a C2…I† class function such that the integrals‡1a

f 2…x† dx and‡1a

f 002…x† dx are convergent. Then

. f is uniformly continuous.

. ‡1a

f 02…x† dx converges.. limx!‡1 f…x† ˆ 0:

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(Hint: Use 8x 2 I;x

aj f 0j2…t† dt ˆ f…t†f 0…t†‰ Šxa¡ x

af …t†f 00…t† dt; and ff 00…x† 4

12…j f j2…x† ‡ j f 00j2…x††:

Remark 8.3. From Proposition 8.2 (1) and (2), we have, for any positive and

averaging convergent function f 2 C1…I†, the following implications:

‡1

a

f 02…x† dx converges ) f (u.c.) )‡1

a

f 2…x† dx converges

9. Summary and concluding remarks

(a) We are now in position to sum up the results obtained in the preceding

sections; in the following diagram, the notations (Lip.), (u.c.), (a.c.) and (a.a.c.)

stand respectively for the Lipschitz Property, uniform continuity, averaging

convergence and absolute averaging convergence while (a)±(f) refer to the state-ments in Proposition 6.2. We have the implications:

…b† () …c† )̂ …d† )̂ …e† )̂ … f †* * *

…a† )̂ (u.c.) (̂ (a.c.) (̂ (a.a.c.)*

(Lip.)

Note that the implication ……u:c:† (̂ …a:c:†† holds true either if f is monotonic or

limx!‡1 f…x† ˆ 0: For the values of the parameters …¬ < ¡1† or …¬ ˆ 1 and

­ 4 ¡ 1†, the implication …c† ) …a:a:c:† holds true (see Remark 7.7 (a)). Moreover,Remark 7.7 shows that both implications …u:c:† ) …a:c:† and …u:c:† ‡ …a:c:† )…a:a:c:† do not hold true.

(b) The analysis given in this paper leads to the following conclusion:

In spite of its very close relation with the behaviour near in®nity of a function,

the concept of uniform continuity can hardly be predicted by some classical

properties of growth. Improving or contradicting this observation is, as far as

the author knows, an open as well as an interesting question. Actually, uniform

continuity, contrary to the general properties of growth, is a global concept which

preserves the metrical properties of sets.(c) The author is not acquainted with similar works dealing with the subject

covered by this paper. Further research may thus be undertaken to discover a

su� cient condition of growth type, weaker than Assertion (7), which guarantees

uniform continuity. It is also desirable that more specialized work on mathematical

education investigate thoroughly various facets of this question.

(d) This investigation shows that, when presenting the notion of uniformcontinuity in analysis, teachers should pay attention to the description of the

asymptotic behaviour near in®nity rather than con®ning themselves to its abstract

aspects. The de®nition stating that f is uniformly continuous if lim¯!0‡ !¯… f† ˆ 0,

where !¯… f† ˆ supjx¡yj 4 ¯ j f …x† ¡ f…y†j stands for oscillation, is far from all con-

crete interpretations. In addition, both examples and counterexamples provided

herein may be useful for lesson preparation.

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Appendix

In this appendix, we are concerned with the analysis of the uniform continuityof the function x ,! x¬ sin®…x­ †; for …¬; ­ ; ®† 2 3; it has been used many times inthis work. Special cases of this function are treated in classical textbooks (see forinstance [2, p. 285]).

A.1. Analysis of the case ® ˆ 1We seek to prove that

f is uniformly continuous , …¬ < 0† or …¬ ‡ ­ 4 1†

(a) The condition is su� cient:. If ¬ < 0; f is uniformly continuous since limx!‡1 f …x† ˆ 0:. A simple computation yields

f 0…x† ˆ ¬x¬¡1 sin …x­ † ‡ ­ x¬x­ ¡1 cos …x­ †

which we rewrite as

f 0…x† ˆ x¬‡­ ¡1…¬ sin …x­ †=x­ ‡ ­ cos …x­ ††

If ¬ ‡ ­ 4 1; f 0 is bounded so f is again uniformly continuous.

(b) The condition is necessary:Assume ¬ ‡ ­ > 1 and ¬ 5 0:. The case ­ > 1 and ¬ ˆ 0 is treated in Remark 8.1 (c).. If ­ 4 0, then f …x† ˆ x¬‡­ sin …x­ †=x­ goes to in®nity when x ! ‡1; the

same arguing holds for the ratio

f…x†x

ˆ x¬‡­ ¡1 sin …x­ †=x­

From Proposition 5.1, we deduce that f is not uniformly continuous.. We now prove the nonuniform continuity in case of positive real constants ¬

and ­ satisfying ¬ ‡ ­ > 1: To this end, we consider the sequences

xn ˆ …2npº†1=­ and yn ˆ 2npº ‡ 1

nl

1=­

in which p 2 ¤ and l 2 ‡¤ need to be chosen judiciously later on. From

de®nition, limn!‡1 xn ˆ ‡1 and f …xn† ˆ 0: Moreover, for large values of theinteger n; the following asymptotic equivalences hold true

yn ¹ …2npº†1=­ 1 ‡1

2­ ºnp‡`and yn ¡ xn ¹

1

­…2º†…1=­ †¡1:n…p=­ †¡p¡`

Hence, limn!‡1…yn ¡ xn† ˆ 0 whenever …p=­ † ¡ p ¡ l < 0: However,

f …yn† ˆ n…¬p=­ †¡` 2º ‡ 1

n`‡p

¬=­

n` sin1

n`

Therefore, limn!‡1 f…yn† 6ˆ 0; whenever ¬…p=­ † ¡ ` 5 0: Consequently, the func-tion f cannot be uniformly continuous if the following two conditions are ful®lled

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p1

­¡ 1 < `

p¬

­5 `

Now, two cases need to be examined separately:

? ­ 5 1: The ®rst constraint is satisfactorily settled so that it remains to pick upan integer p 5 `…­ =¬†.

? 0 < ­ < 1: We must select an integer p satisfying simultaneously

p 5 `­

¬and p <

`­

1 ¡ ­

which is possible only if`­

1 ¡ ­¡ `

­

¬> 1

that is

` >¬…1 ¡ ­ †

­ …¬ ‡ ­ ¡ 1†Archimedes’ axiom ensures the existence of such a real number and the proof is

complete.

A.2. Analysis of the case ­ ˆ 1 and ® 5 0We are going to prove the following equivalence

f is uniformly continuous () …¬ 4 0† or …¬ 4 1 and ® ˆ 0†

The condition ® 5 0 assures that f is well de®ned near in®nity.(a) The condition is su� cient:. If ® ˆ 0; we know, from Lemma 4.4, that f is uniformly continuous if and

only if ¬ 4 1:. As for the extreme case ¬ ˆ 0; we have got the uniform continuity of the

function f ; indeed, either ® 5 1 and f satis®es a Lipschitz condition;otherwise f may be viewed as the composition of two uniformly continuousfunctions: the sine function and the function x ,! x® … with 0 4 ® < 1†.

. In the case ¬ < 0; limx!‡1 f …x† ˆ 0 so that the uniform continuity followsvacuously.

(b) The condition is necessary:Assume ¬; ® > 0 and consider the sequences xn ˆ 2npº and yn ˆ xn ‡ …1=n†

with some p 2 ¤: Then, limn!‡1…yn ¡ xn† ˆ 0: In addition, there is some realconstant C such that, for n large enough

f…xn† ˆ 0 and f…yn† ˆ2npº ‡

1

n

¬

n®n® sin® 1

n¹ C ¢ np¬¡®

By selecting p 5 ®=¬ we infer that limn!‡1 f…yn† 6ˆ 0: As a consequence, f is notuniformly continuous. The case ¬ > 1 and ® ˆ 0 is in Lemma 4.4.

A.3. A last pointWe leave to the reader the study of the uniform continuity of the more general

function x ,! x¬ sin®…x­ †; x 2 I, in term of the parameters …¬; ­ ; ®† 2 2 £ ‡:

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References[1] Katz, Victor J., 1993, A History of Mathematics. An Introduction (Harper Collins

College Publishers, New York).[2] Fisher, E., 1983, Intermediate Real Analysis (undergraduate Texts in Mathematics;

New York: Springer Verlag).[3] DieudonnEÂ , J., 1968, EÂ leÂments d’Analyse; 1. Fondements de l’Analyse Moderne (Paris:

Gauthiers Villars).[4] Ramis, E., Deschamps, C., and Odoux, J., 1995, Cours de MatheÂmatiques SpeÂciales; 3.

Topologie et EÂ leÂments d’Analyse (Paris: Masson).

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