CONTINUOUS FUNCTIONS
Section 5.4 Uniform Continuity
15. Examine which open [respectively, closed] intervals are mapped byfðxÞ:¼x2forx2Ronto open [respectively, closed] intervals.
16. Examine the mapping of open [respectively, closed] intervals under the functions gðxÞ:¼ 1=ðx2þ1ÞandhðxÞ:¼x3forx2R.
17. Iff:½0;1 !Ris continuous and has only rational [respectively, irrational] values, mustfbe constant? Prove your assertion.
18. LetI:¼ ½a;band letf :I!Rbe a (not necessarily continuous) function with the property that for everyx2I, the functionfis bounded on a neighborhood Vdxð Þx ofx (in the sense of Definition 4.2.1). Prove thatfis bounded onI.
19. LetJ:¼ ða;bÞand letg:J!Rbe a continuous function with the property that for everyx2J, the functiongis bounded on a neighborhood VdxðxÞ ofx. Show by example thatg is not necessarily bounded onJ.
x;u2A. We note that the value ofdðe;uÞgiven in (2) certainly depends on the pointu2A. If we wish to considerall u2A, formula (2) does not lead to one valuedðeÞ>0 that will
‘‘work’’ simultaneously for allu>0, since inffdðe;uÞ:u>0g ¼0.
In fact, there is no way of choosing one value ofdthat will ‘‘work’’ for allu>0 for the function gðxÞ ¼1=x. The situation is exhibited graphically in Figures 5.4.1 and 5.4.2 where, for a givene-neighborhood Ve 1
2
about 12¼fð2ÞandVeð2Þabout 2¼f 12 , the corresponding maximum values ofdare seen to be considerably different. Asutends to 0, the permissible values ofdtend to 0.
5.4.1 Definition LetAR and letf :A!R. We say thatfisuniformly continuous onAif for eache>0 there is adðeÞ>0 such that ifx;u2Aare any numbers satisfying jxuj<dðeÞ, thenjfðxÞ fðuÞj<e.
It is clear that iffis uniformly continuous onA, then it is continuous at every point ofA. In general, however, the converse does not hold, as is shown by the functiongðxÞ ¼1=xon the setA:¼ fx2R:x>0g.
It is useful to formulate a condition equivalent to saying that f is not uniformly continuous onA. We give such criteria in the next result, leaving the proof to the reader as an exercise.
5.4.2 Nonuniform Continuity Criteria Let ARand let f :A!R.Then the follow- ing statements are equivalent:
(i) f is not uniformly continuous on A.
(ii) There exists ane0>0such that for everyd>0there are points xd;udin A such that jxdudj<dand jfðxdÞ fðudÞj e0.
(iii) There exists ane0>0and two sequences(xn)and (un)in A such thatlimðxn unÞ ¼0and jfðxnÞ fðunÞj e0 for all n2N.
We can apply this result to show that gðxÞ:¼1=x is not uniformly continuous
on A:¼ fx2R:x>0g. For, if xn:¼1=n and un:¼1=ðnþ1Þ, then we have
limðxnunÞ ¼0, butjgðxnÞ gðunÞj ¼1 for alln2N.
We now present an important result that assures that a continuous function on a closed bounded intervalIis uniformly continuous onI. Other proofs of this theorem are given in Sections 5.5 and 11.3.
Figure 5.4.1 gðxÞ ¼1=x ðx>0Þ Figure 5.4.2 gðxÞ ¼1=x ðx>0Þ
5.4.3 Uniform Continuity Theorem Let I be a closed bounded interval and let f :I!R be continuous on I. Then f is uniformly continuous on I.
Proof. Iffis not uniformly continuous on Ithen, by the preceding result, there exists e0>0 and two sequences (xn) and (un) inIsuch thatjxnunj<1=nandjfðxnÞ fðunÞj e0 for all n2N. Since I is bounded, the sequence (xn) is bounded; by the Bolzano- Weierstrass Theorem 3.4.8 there is a subsequenceð Þxnk of (xn) that converges to an element z. Since I is closed, the limit z belongs to I, by Theorem 3.2.6. It is clear that the corresponding subsequence ð Þunk also converges toz, since
unkz
j j junkxnkj þjxnkzj:
Now iffis continuous at the pointz, then both of the sequencesðf xð Þnk Þandðf uð Þnk Þ must converge tofðzÞ. But this is not possible since
jfðxnÞ fðunÞj e0
for alln2N. Thus the hypothesis thatfis not uniformly continuous on the closed bounded interval I implies that f is not continuous at some point z2I. Consequently, if f is continuous at every point ofI, then fis uniformly continuous onI. Q.E.D.
Lipschitz Functions
If a uniformly continuous function is given on a set that is not a closed bounded interval, then it is sometimes difficult to establish its uniform continuity. However, there is a condition that frequently occurs that is sufficient to guarantee uniform continuity. It is named after Rudolf Lipschitz (1832–1903) who was a student of Dirichlet and who worked extensively in differential equations and Riemannian geometry.
5.4.4 Definition LetARand letf :A!R. If there exists a constantK>0 such that ð4Þ jfðxÞ fðuÞj Kjxuj
for all x;u2A, then f is said to be a Lipschitz function (or to satisfy a Lipschitz condition) onA.
The condition (4) that a functionf :I!Ron an intervalIis a Lipschitz function can be interpreted geometrically as follows. If we write the condition as
fðxÞ fðuÞ xu
K; x;u2I;x6¼u;
then the quantity inside the absolute values is the slope of a line segment joining the points ðx;fðxÞÞandðu;fðuÞÞ. Thus a functionfsatisfies a Lipschitz condition if and only if the slopes of all line segments joining two points on the graph ofy¼fðxÞoverIare bounded by some number K.
5.4.5 Theorem If f :A!Ris a Lipschitz function,then f is uniformly continuous on A. Proof. If condition (4) is satisfied, then givene>0, we can taked:¼e=K. If x;u2A satisfyjxuj<d, then
jfðxÞ fðuÞj<K e K¼e:
Thereforef is uniformly continuous onA. Q.E.D.
5.4 UNIFORM CONTINUITY 143
5.4.6 Examples (a) IffðxÞ:¼x2 onA:¼ ½0;b, whereb >0, then jfðxÞ fðuÞj ¼ jxþujjxuj 2bjxuj
for allx,uin [0,b]. Thusfsatisfies (4) withK:¼2bonA, and thereforefis uniformly continuous onA. Of course, sincefis continuous andAis a closed bounded interval, this can also be deduced from the Uniform Continuity Theorem. (Note thatfdoesnotsatisfy a Lipschitz condition on the interval½0;1Þ.)
(b) Not every uniformly continuous function is a Lipschitz function.
LetgðxÞ:¼ ffiffiffi px
forxin the closed bounded intervalI:¼ ½0;2. Sincegis continuous onI, it follows from the Uniform Continuity Theorem 5.4.3 thatgis uniformly continuous onI. However, there is no numberK>0 such thatjgðxÞj Kjxjfor allx2I. (Why not?) Therefore,g is not a Lipschitz function onI.
(c) The Uniform Continuity Theorem and Theorem 5.4.5 can sometimes be combined to establish the uniform continuity of a function on a set.
We considergðxÞ:¼ ffiffiffi px
on the setA:¼ ½0;1Þ. The uniform continuity ofgon the interval I:¼ ½0;2 follows from the Uniform Continuity Theorem as noted in (b). If J:¼ ½1;1Þ, then if both x,uare inJ, we have
gðxÞ gðuÞ j j ¼ ffiffiffi
px ffiffiffi
pu
¼ jxffiffiffiuj px
þ ffiffiffi
p u 12jxuj:
Thusgis a Lipschitz function onJwith constantK¼12, and hence by Theorem 5.4.5,gis uniformly continuous on [1, 1). Since A¼I[J, it follows [by taking dðeÞ:¼ inf 1f ;dIðeÞ;dJðeÞg] that g is uniformly continuous on A. We leave the details to the
reader. &
The Continuous Extension Theorem
We have seen examples of functions that are continuous but not uniformly continuous on open intervals; for example, the functionfðxÞ ¼1=xon the interval (0, 1). On the other hand, by the Uniform Continuity Theorem, a function that is continuous on a closed bounded interval is always uniformly continuous. So the question arises: Under what conditions is a function uniformly continuous on a boundedopen interval? The answer reveals the strength of uniform continuity, for it will be shown that a function on (a,b) is uniformly continuous if and only if it can be defined at the endpoints to produce a function that is continuous on the closed interval. We first establish a result that is of interest in itself.
5.4.7 Theorem If f :A!Ris uniformly continuous on a subset A ofRandif xð Þn is a Cauchy sequence in A, thenðfðxnÞÞis a Cauchy sequence inR.
Proof. Let (xn) be a Cauchy sequence inA, and lete>0 be given. First choosed>0 such that if x, u in A satisfy jxuj <d, then jfðxÞ fðuÞj <e. Since (xn) is a Cauchy sequence, there existsHðdÞsuch thatjxnxmj <dfor alln;m>HðdÞ. By the choice ofd, this implies that forn;m>HðdÞ, we havejfðxnÞ fðxmÞj <e. Therefore the sequence
ðfðxnÞÞis a Cauchy sequence. Q.E.D.
The preceding result gives us an alternative way of seeing thatfðxÞ:¼1=x is not uniformly continuous on (0, 1). We note that the sequence given byxn:¼1=nin (0, 1) is a Cauchy sequence, but the image sequence, wherefðxnÞ ¼n, is not a Cauchy sequence.
5.4.8 Continuous Extension Theorem A function f is uniformly continuous on the interval(a,b)if and only if it can be defined at the endpoints a and b such that the ex- tended function is continuous on[a,b].
Proof. ð(ÞThis direction is trivial.
ð)ÞSupposefis uniformly continuous on (a,b). We shall show how to extendftoa; the argument forbis similar. This is done by showing that lim
x!c fðxÞ ¼Lexists, and this is accomplished by using the sequential criterion for limits. If (xn) is a sequence in (a,b) with limðxnÞ ¼a, then it is a Cauchy sequence, and by the preceding theorem, the sequence ðfðxnÞÞis also a Cauchy sequence, and so is convergent by Theorem 3.5.5. Thus the limit limðfðxnÞÞ ¼L exists. If (un) is any other sequence in (a,b) that converges toa, then limðunxnÞ ¼aa¼0, so by the uniform continuity offwe have
limðfðunÞÞ ¼ limðfðunÞ fðxnÞÞ þlimðfðxnÞÞ
¼0þL¼L:
Since we get the same value L for every sequence converging to a, we infer from the sequential criterion for limits that f has limit L at a. If we define fðaÞ:¼L, then f is continuous ata. The same argument applies tob, so we conclude thatfhas a continuous
extension to the interval [a,b]. Q.E.D.
Since the limit offðxÞ:¼sinð1=xÞat 0 does not exist, we infer from the Continuous Extension Theorem that the function is not uniformly continuous onð0;bfor anyb>0.
On the other hand, since lim
x!0xsinð1=xÞ ¼0 exists, the function gðxÞ:¼xsinð1=xÞ is
uniformly continuous onð0;bfor allb>0.
Approximationy
In many applications it is important to be able to approximate continuous functions by functions of an elementary nature. Although there are a variety of definitions that can be used to make the word ‘‘approximate’’ more precise, one of the most natural (as well as one of the most important) is to require that, at every point of the given domain, the approximating function shall not differ from the given function by more than the preassigned error.
5.4.9 Definition A functions:½a;b !Ris called astep functionif [a,b] is the union of a finite number of nonoverlapping intervalsI1; I2;. . .;Insuch thatsis constant on each interval, that is,s xð Þ ¼ckfor allx2Ik; k¼1; 2;. . .;n.
Thus a step function has only a finite number of distinct values.
For example, the functions:½2;4 !R defined by
s xð Þ:¼
0; 2x<1; 1; 1x0;
1
2; 0<x<12; 3; 12x<1; 2; 1x3; 2; 3<x4; 8>
>>
>>
><
>>
>>
>>
: is a step function. (See Figure 5.4.3.)
yThe rest of this section can be omitted on a first reading of this chapter.
5.4 UNIFORM CONTINUITY 145
We will now show that a continuous function on a closed bounded intervalIcan be approximated arbitrarily closely by step functions.
5.4.10 Theorem Let I be a closed bounded interval and let f :I!Rbe continuous on I. Ife>0,then there exists a step function se:I!Rsuch that f xj ð Þ seð Þx j<efor all x2I.
Proof. Since (by the Uniform Continuity Theorem 5.4.3) the function f is uniformly continuous, it follows that givene>0 there is a numberdð Þe >0 such that ifx;y2Iand
xy
j j<dð Þe , thenjf xð Þ f yð Þj<e. LetI:¼½a;band letm2Nbe sufficiently large so thath:¼ðbaÞ=m<dð Þe . We now divideI¼[a,b] intomdisjoint intervals of lengthh; namely,I1 :¼½a;aþh, andIk:¼ðaþðk1Þh; aþkhÞfork¼2, . . . ,m. Since the length of each subintervalIkish<dð Þe , the difference between any two values offinIkis less thane. We now define
ð5Þ seð Þx :¼f að þkhÞ for x2Ik; k¼1;. . .;m;
so thatseis constant on each intervalIk. (In fact the value ofseonIkis the value offat the right endpoint of Ik. See Figure 5.4.4.) Consequently ifx2Ik, then
f xð Þ seð Þx
j j ¼jf xð Þ f að þkhÞj<e:
Therefore we have jf xð Þ seð Þx j<efor allx2I. Q.E.D.
Figure 5.4.3 Graph ofy¼s(x)
Figure 5.4.4 Approximation by step functions
Note that the proof of the preceding theorem establishes somewhat more than was announced in the statement of the theorem. In fact, we have proved the following, more precise, assertion.
5.4.11 Corollary Let I:¼ ½a;b be a closed bounded interval and let f :I!R be continuous on I. Ife>0,there exists a natural number m such that if we divide I into m disjoint intervals Ik having length h:¼ðbaÞ=m, then the step function se defined in equation(5) satisfies f xj ð Þ seð Þx j<efor all x2I.
Step functions are extremely elementary in character, but they are not continuous (except in trivial cases). Since it is often desirable to approximate continuous functions by elementary continuous functions, we now shall show that we can approximate continuous functions by continuous piecewise linear functions.
5.4.12 Definition LetI:¼ ½a; bbe an interval. Then a functiong:I!Ris said to be piecewise linearonIifIis the union of a finite number of disjoint intervalsI1;. . .;Im, such that the restriction ofgto each intervalIkis a linear function.
Remark It is evident that in order for a piecewise linear functiongto be continuous on I, the line segments that form the graph of g must meet at the endpoints of adjacent subintervalsIk;Ikþ1ðk¼1;. . .; m1Þ.
5.4.13 Theorem Let I be a closed bounded interval and let f :I!Rbe continuous on I.
If e>0, then there exists a continuous piecewise linear function ge:I!R such that f xð Þ geð Þx
j j<efor all x2I.
Proof. Sincefis uniformly continuous onI:¼ ½a;b, there is a numberdð Þe >0 such that ifx,y2I andjxyj<dð Þe , thenjf xð Þ f yð Þj<e. Letm2Nbe sufficiently large so thath:¼ðbaÞ=m<dð Þe . DivideI¼ ½a; bintomdisjoint intervals of lengthh; namely, letI1¼½a; aþh, and letIk¼ðaþðk1Þh;aþkhfork¼2;. . .;m. On each interval Ikwe definegeto be the linear function joining the points
aþðk1Þh; f að þðk1ÞhÞ
ð Þ and ðaþkh; f að þkhÞÞ:
Thengeis a continuous piecewise linear function onI. Since, forx2Ikthe valuef(x) is withineoff að þðk1ÞhÞandf að þkhÞ, it is an exercise to show thatjf xð Þ geð Þx j<e for allx2Ik; therefore this inequality holds for all x2I. (See Figure 5.4.5.) Q.E.D.
Figure 5.4.5 Approximation by piecewise linear function
5.4 UNIFORM CONTINUITY 147
We shall close this section by stating the important theorem of Weierstrass con- cerning the approximation of continuous functions by polynomial functions. As would be expected, in order to obtain an approximation within an arbitrarily preassigned e>0, we must be prepared to use polynomials of arbitrarily high degree.
5.4.14 Weierstrass Approximation Theorem Let I¼ ½a;b and let f :I!R be a continuous function. Ife>0is given, then there exists a polynomial function pesuch that
f xð Þ peð Þx
j j<efor all x2I.
There are a number of proofs of this result. Unfortunately, all of them are rather intricate, or employ results that are not yet at our disposal. (A proof can be found in Bartle, ERA, pp. 169–172, which is listed in the References.)
Exercises for Section 5.4
1. Show that the functionf xð Þ:¼1=xis uniformly continuous on the setA:¼½a;1Þ, whereais a positive constant.
2. Show that the functionf xð Þ:¼1=x2is uniformly continuous onA:¼½1;1Þ, but that it is not uniformly continuous onB:¼ð0;1Þ.
3. Use the Nonuniform Continuity Criterion 5.4.2 to show that the following functions are not uniformly continuous on the given sets.
(a) f xð Þ:¼x2; A:¼½0;1Þ. (b) g xð Þ:¼sin 1=xð Þ; B:¼ð0;1Þ.
4. Show that the functionf xð Þ:¼1=ð1þx2Þforx2R is uniformly continuous onR. 5. Show that iffandgare uniformly continuous on a subsetAof R, thenfþgis uniformly
continuous onA.
6. Show that iffandgare uniformly continuous onARand if they arebothbounded onA, then their productfgis uniformly continuous onA.
7. Iff xð Þ:¼xandg xð Þ:¼sinx, show that bothfandgare uniformly continuous onR, but that their productfgis not uniformly continuous onR.
8. Prove that iffandgare each uniformly continuous onR, then the composite functionfgis uniformly continuous onR.
9. If fis uniformly continuous onAR, andjf xð Þj k>0 for all x2A, show that 1=fis uniformly continuous onA.
10. Prove that iffis uniformly continuous on a bounded subsetAofR, thenfis bounded onA.
11. Ifg xð Þ:¼ ffiffiffi px
forx2½0;1, show that there does not exist a constantKsuch thatjg xð Þj K xj jfor allx2½0;1. Conclude that the uniformly continuousgis not a Lipschitz function on [0, 1].
12. Show that iffis continuous on [0,1) and uniformly continuous on [a,1) for some positive constanta, thenfis uniformly continuous on [0,1).
13. LetARand suppose thatf:A!Rhas the following property: for eache>0 there exists a functionge:A!Rsuch thatgeis uniformly continuous onAandjf xð Þ geð Þx j<efor all x2A. Prove thatfis uniformly continuous onA.
14. A functionf :R!R is said to beperiodiconR if there exists a numberp>0 such that f xð þpÞ ¼f xð Þfor allx2R. Prove that a continuous periodic function onRis bounded and uniformly continuous onR.
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;yk;k¼1;2;. . .;n, of I such that Pjxkykj<dwe haveP
f xð Þ k f yð Þk
j j<e. Show that iffsatisfies a Lipschitz condition
onI, thenfis absolutely continuous onI.