CONTINUOUS FUNCTIONS
Section 5.3 Continuous Functions on Intervals
4. Letx7!vxbdenote the greatest integer function (see Exercise 5.1.4). Determine the points of continuity of the functionfðxÞ:¼xvxb;x2R.
5. Letgbe defined onRbygð1Þ:¼0, andgðxÞ:¼2 ifx6¼1, and letfðxÞ:¼xþ1 for allx2R. Show that lim
x!0gf6¼ ðgfÞð0Þ. Why doesn’t this contradict Theorem 5.2.6?
6. Letf,gbe defined onRand letc2R. Suppose that lim
x!cf ¼band thatgis continuous atb. Show that lim
x!cgf¼gðbÞ. (Compare this result with Theorem 5.2.7 and the preceding exercise.)
7. Give an example of a functionf:½0;1 !Rthat is discontinuous at every point of [0, 1] but such thatjfjis continuous on [0, 1].
8. Letf,gbe continuous fromRtoR, and suppose thatfðrÞ ¼gðrÞfor all rational numbersr. Is it true thatfðxÞ ¼gðxÞfor allx2R?
9. Leth:R!Rbe continuous onR satisfyinghðm=2nÞ ¼0 for allm2Z;n2N. Show that hðxÞ ¼0 for allx2R.
10. Letf:R!Rbe continuous onR, and letP:¼ fx2R:fðxÞ>0g. Ifc2P, show that there exists a neighborhoodVdðcÞ P.
11. Iffandgare continuous onR, letS:¼ fx2R:fðxÞ gðxÞg. IfðsnÞ Sand limðsnÞ ¼s, show thats2S.
12. A functionf:R!Ris said to beadditiveiffðxþyÞ ¼fðxÞ þfðyÞfor allx,yinR. Prove that if f is continuous at some point x0, then it is continuous at every point of R. (See Exercise 4.2.12.)
13. Suppose thatfis a continuous additive function onR. Ifc:¼fð1Þ, show that we havefðxÞ ¼cx for allx2R. [Hint: First show that ifris a rational number, thenfðrÞ ¼cr.]
14. Letg:R!R satisfy the relationgðxþyÞ ¼gðxÞgðyÞfor allx,yinR. Show that ifgis continuous atx¼0, thengis continuous at every point ofR. Also if we havegðaÞ ¼0 for some a2R, thengðxÞ ¼0 for allx2R.
15. Letf;g:R!Rbe continuous at a pointc, and lethðxÞ:¼supffðxÞ;gðxÞgforx2R. Show that hðxÞ ¼12ðfðxÞ þgðxÞÞ þ12jfðxÞ gðxÞj for all x2R. Use this to show that h is continuous atc.
bounded. In the next theorem, however, we show that continuous functions on a certain type of interval are necessarily bounded.
5.3.2 Boundedness Theoremy Let I:¼ ½a;b be a closed bounded interval and let f:I!R be continuous on I. Then f is bounded on I.
Proof. Suppose thatfis not bounded onI. Then, for anyn2Nthere is a numberxn2I such thatjfðxnÞj >n. SinceIis bounded, the sequenceX:¼ ðxnÞis bounded. Therefore, the Bolzano-Weierstrass Theorem 3.4.8 implies that there is a subsequenceX0¼ð Þxnr ofX that converges to a numberx. SinceIis closed and the elements ofX0belong toI, it follows from Theorem 3.2.6 thatx2I. Thenfis continuous atx, so thatðfðxnrÞÞconverges tofðxÞ. We then conclude from Theorem 3.2.2 that the convergent sequence ðf xð Þnr Þ must be bounded. But this is a contradiction since
f xð Þnr
j j>nrr for r2N:
Therefore the supposition that the continuous function f is not bounded on the closed
bounded intervalIleads to a contradiction. Q.E.D.
To show that each hypothesis of the Boundedness Theorem is needed, we can construct examples that show the conclusion fails if any one of the hypotheses is relaxed.
(i) The interval must be bounded. The functionfðxÞ:¼xforxin the unbounded, closed intervalA:¼ ½0;1Þis continuous but not bounded onA.
(ii) The interval must be closed. The functiongðxÞ:¼1=x forx in the half-open intervalB:¼ ð0;1is continuous but not bounded onB.
(iii) The function must be continuous. The functionhdefined on the closed interval C:¼ ½0;1byhðxÞ:¼1=xforx2 ð0;1andhð0Þ:¼1 is discontinuous and unbounded onC.
The Maximum-Minimum Theorem
5.3.3 Definition LetARand letf :A!R. We say thatf has an absolute maximum onAif there is a pointx 2A such that
fðx Þ fðxÞ for all x2A:
We say thatf has an absolute minimumonA if there is a pointx 2Asuch that fðx Þ fðxÞ for all x2A:
We say thatx is anabsolute maximum point for f on A, and that x is an absolute minimum point forfonA, if they exist.
We note that a continuous function on a setAdoes not necessarily have an absolute maximum or an absolute minimum on the set. For example,fðxÞ:¼1=xhas neither an absolute maximum nor an absolute minimum on the setA:¼ ð0;1Þ. (See Figure 5.3.1.) There can be no absolute maximum forfonAsincefis not bounded above onA, and there is no point at which f attains the value 0¼infffðxÞ:x2Ag. The same function has
yThis theorem, as well as 5.3.4, is true for an arbitrary closed bounded set. For these developments, see Sections 11.2 and 11.3.
5.3 CONTINUOUS FUNCTIONS ON INTERVALS 135
neither an absolute maximum nor an absolute minimum when it is restricted to the set (0, 1), while it has both an absolute maximum and an absolute minimum when it is restricted to the set [1, 2]. In addition, fðxÞ ¼1=x has an absolute maximum but no absolute minimum when restricted to the set [1, 1), but no absolute maximum and no absolute minimum when restricted to the set (1,1).
It is readily seen that if a function has an absolute maximum point, then this point is not necessarily uniquely determined. For example, the function gðxÞ:¼x2 defined for x2A:¼ ½1;þ1has the two pointsx¼ 1 giving the absolute maximum onA, and the single pointx ¼0 yielding its absolute minimum onA. (See Figure 5.3.2.) To pick an extreme example, the constant functionhðxÞ:¼1 forx2Ris such thatevery pointofRis both an absolute maximum and an absolute minimum point for h.
5.3.4 Maximum-Minimum Theorem Let I:¼ ½a;bbe a closed bounded interval and let f :I!R be continuous on I. Then f has an absolute maximum and an absolute minimum on I.
Proof. Consider the nonempty setfðIÞ:¼ ffðxÞ:x2Igof values offonI. In Theorem 5.3.2 it was established thatf(I) is a bounded subset ofR. Lets :¼sup fðIÞands :¼inf fðIÞ. We claim that there exist pointsx andx inIsuch thats ¼fðx Þands ¼fðx Þ. We will establish the existence of the pointx , leaving the proof of the existence ofx to the reader.
Sinces ¼supfðIÞ, ifn2N, then the numbers 1=nis not an upper bound of the setf(I). Consequently there exists a numberxn2I such that
ð1Þ s 1
n<fðxnÞ s for all n2N:
Since I is bounded, the sequence X:¼ ðxnÞ is bounded. Therefore, by the Bolzano- Weierstrass Theorem 3.4.8, there is a subsequenceX0¼ ðxnrÞofXthat converges to some numberx. Since the elements ofX0belong toI¼ ½a;b, it follows from Theorem 3.2.6 that x 2I. Thereforefis continuous atx so that limðf xð Þnr Þ ¼fðx Þ. Since it follows from (1) that
s 1
nr<f xð Þ nr s for all r2N;
we conclude from the Squeeze Theorem 3.2.7 that limðf xð Þnr Þ ¼s . Therefore we have fðx Þ ¼limðfðxnrÞÞ ¼s ¼supfðIÞ:
We conclude thatx is an absolute maximum point offonI. Q.E.D.
Figure 5.3.1 The function
fðxÞ ¼1=x ðx>0Þ Figure 5.3.2 The function gðxÞ ¼x2 ðjxj 1Þ
The next result is the theoretical basis for locating roots of a continuous function by means of sign changes of the function. The proof also provides an algorithm, known as the Bisection Method, for the calculation of roots to a specified degree of accuracy and can be readily programmed for a computer. It is a standard tool for finding solutions of equations of the formf(x)¼0, wherefis a continuous function. An alternative proof of the theorem is indicated in Exercise 5.3.11.
5.3.5 Location of Roots Theorem Let I¼ ½a;band let f :I!Rbe continuous on I.If fðaÞ<0<fðbÞ, or if fðaÞ>0>fðbÞ, then there exists a number c2 ða;bÞsuch that fðcÞ ¼0.
Proof. We assume that fðaÞ<0<fðbÞ. We will generate a sequence of intervals by successive bisections. LetI1:¼ ½a1;b1, wherea1:¼a;b1:¼b, and letp1be the midpoint p1:¼12ða1þb1Þ. Iffðp1Þ ¼0, we takec:¼p1and we are done. Iffðp1Þ 6¼0, then either fðp1Þ>0 orfðp1Þ<0. Iffðp1Þ>0, then we set a2 :¼a1;b2:¼p1, while if fðp1Þ<0, then we seta2:¼p1;b2 :¼b1. In either case, we letI2:¼ ½a2;b2; then we haveI2I1and fða2Þ<0;fðb2Þ>0.
We continue the bisection process. Suppose that the intervals I1;I2;. . .;Ik have been obtained by successive bisection in the same manner. Then we havefðakÞ<0 and fðbkÞ>0, and we setpk:¼12ðakþbkÞ. IffðpkÞ ¼0, we takec:¼pk and we are done.
If fðpkÞ>0, we set akþ1:¼ak;bkþ1:¼pk, while if fðpkÞ<0, we set akþ1 :¼ pk;bkþ1:¼bk. In either case, we let Ikþ1:¼ ½akþ1;bkþ1; then Ikþ1Ik and fðakþ1Þ<0;fðbkþ1Þ>0.
If the process terminates by locating a pointpnsuch thatfðpnÞ ¼0, then we are done.
If the process does not terminate, then we obtain a nested sequence of closed bounded intervals In:¼ ½an;bnsuch that for everyn2N we have
fðanÞ<0 and fðbnÞ>0:
Furthermore, since the intervals are obtained by repeated bisection, the length of Inis equal to bnan ¼ ðbaÞ=2n1. It follows from the Nested Intervals Property 2.5.2 that there exists a point c that belongs to In for all n2N. Since ancbn for all n2Nand limðbnanÞ ¼0, it follows that limðanÞ ¼c¼limðbnÞ. Sincefis continuous atc, we have
limðfðanÞÞ ¼fðcÞ ¼limðfðbnÞÞ:
The fact thatfðanÞ<0 for alln2Nimplies thatfðcÞ ¼limðfðanÞÞ 0. Also, the fact thatfðbnÞ>0 for alln2Nimplies thatfðcÞ ¼limðfðbnÞÞ 0. Thus, we conclude that fðcÞ ¼0. Consequently,cis a root of f. Q.E.D.
The following example illustrates how the Bisection Method for finding roots is applied in a systematic fashion.
5.3.6 Example The equation fðxÞ ¼xex2¼0 has a root c in the interval [0, 1], becausefis continuous on this interval andfð0Þ ¼ 2<0 andfð1Þ ¼e2>0. Using a calculator we construct the following table, where the sign off(pn) determines the interval at the next step. The far right column is an upper bound on the error whenpnis used to approximate the rootc, because we have
jpncj 12ðbnanÞ ¼1=2n:
5.3 CONTINUOUS FUNCTIONS ON INTERVALS 137
We will find an approximationpnwith error less than 102.
n an bn pn f(pn) 12ðbnanÞ
1 0 1 .5 1.176 .5
2 .5 1 .75 .412 .25
3 .75 1 .875 þ.099 .125
4 .75 .875 .8125 .169 .0625
5 .8125 .875 .84375 .0382 .03125
6 .84375 .875 .859375 þ.0296 .015625
7 .84375 .859375 .8515625 — .0078125
We have stopped atn¼7, obtainingcp7 ¼:8515625 with error less than .0078125.
This is the first step in which the error is less than 102. The decimal place values of p7 past the second place cannot be taken seriously, but we can conclude that
:843<c< :860. &
Bolzano’s Theorem
The next result is a generalization of the Location of Roots Theorem. It assures us that a continuous function on an interval takes on (at least once) any number that lies between two of its values.
5.3.7 Bolzano’s Intermediate Value Theorem Let I be an interval and let f :I!Rbe continuous on I. If a;b2I and if k2Rsatisfies fðaÞ<k<fðbÞ, then there exists a point c2I between a and b such that fðcÞ ¼k.
Proof. Suppose thata<band letgðxÞ:¼fðxÞ k; thengðaÞ<0<gðbÞ. By the Location of Roots Theorem 5.3.5 there exists a pointcwitha<c<bsuch that 0¼gðcÞ ¼fðcÞ k. ThereforefðcÞ ¼k.
Ifb<a, lethðxÞ:¼kfðxÞso thathðbÞ<0<hðaÞ. Therefore there exists a pointc withb<c<asuch that 0¼hðcÞ ¼kfðcÞ, whence fðcÞ ¼k. Q.E.D.
5.3.8 Corollary Let I¼ ½a;b be a closed, bounded interval and let f :I!R be continuous on I. If k2R is any number satisfying
inffðIÞ ksupfðIÞ;
then there exists a number c2I such that fðcÞ ¼k.
Proof. It follows from the Maximum-Minimum Theorem 5.3.4 that there are pointsc andc inIsuch that
inffðIÞ ¼fðc Þ kfðcÞ ¼ supfðIÞ:
The conclusion now follows from Bolzano’s Theorem 5.3.7. Q.E.D.
The next theorem summarizes the main results of this section. It states that the image of a closed bounded interval under a continuous function is also a closed bounded interval.
The endpoints of theimage interval are the absolute minimum and absolute maximum values of the function, and the statement that all values between the absolute minimum and the absolute maximum values belong to the image is a way of describing Bolzano’s Intermediate Value Theorem.
5.3.9 Theorem Let I be a closed bounded interval and let f:I!Rbe continuous on I.
Then the set fðIÞ:¼ ffðxÞ:x2Ig is a closed bounded interval.
Proof. If we let m:¼inffðIÞ andM :¼supfðIÞ, then we know from the Maximum- Minimum Theorem 5.3.4 thatmandMbelong tof(I). Moreover, we havefðIÞ ½m;M. If kis any element of [m,M], then it follows from the preceding corollary that there exists a point c2I such that k¼fðcÞ. Hence, k2fðIÞ and we conclude that ½m;M fðIÞ.
Therefore,f(I) is the interval [m,M]. Q.E.D.
Warning IfI:¼ ½a;bis an interval andf :I!R is continuous onI, we have proved thatf(I) is the interval [m,M]. We havenotproved (and it is not always true) thatf(I) is the
interval½fðaÞ;fðbÞ. (See Figure 5.3.3.) &
The preceding theorem is a ‘‘preservation’’ theorem in the sense that it states that the continuous image of a closed bounded interval is a set of the same type. The next theorem extends this result to general intervals. However, it should be noted that although the continuous image of an interval is shown to be an interval, it isnottrue that the image interval necessarily has thesame formas the domain interval. For example, the continuous image of an open interval need not be an open interval, and the continuous image of an unbounded closed interval need not be a closed interval. Indeed, iffðxÞ:¼1=ðx2þ1Þfor x2R, then f is continuous on R [see Example 5.2.3(b)]. It is easy to see that if I1:¼ ð1;1Þ, then fðI1Þ ¼12;1
, which is not an open interval. Also, if I2:¼ ½0;1Þ, thenfðI2Þ ¼ ð0;1, which is not a closed interval. (See Figure 5.3.4.)
Figure 5.3.3 fðIÞ ¼ ½m;M
Figure 5.3.4 Graph offðxÞ ¼1=ðx2þ1Þ ðx2RÞ
5.3 CONTINUOUS FUNCTIONS ON INTERVALS 139
To prove the Preservation of Intervals Theorem 5.3.10, we will use Theorem 2.5.1 characterizing intervals.
5.3.10 Preservation of Intervals Theorem Let I be an interval and let f :I!R be continuous on I. Then the set f(I)is an interval.
Proof. Let a;b2fðIÞwith a<b; then there exist points a;b2I such thata¼fðaÞ and b¼fðbÞ. Further, it follows from Bolzano’s Intermediate Value Theorem 5.3.7 that if k2 ða;bÞ then there exists a number c2I with k¼fðcÞ 2fðIÞ. Therefore
½a;b fðIÞ, showing that f(I) possesses property (1) of Theorem 2.5.1. Therefore
f(I) is an interval. Q.E.D.
Exercises for Section 5.3
1. LetI:¼ ½a;band letf :I!Rbe a continuous function such thatfðxÞ>0 for eachxinI.
Prove that there exists a numbera>0 such thatfðxÞ afor allx2I.
2. LetI:¼ ½a;band letf:I!Randg:I!Rbe continuous functions onI. Show that the set E:¼ fx2I:fðxÞ ¼gðxÞghas the property that ifðxnÞ Eandxn!x0, thenx02E. 3. LetI:¼ ½a;band letf:I!Rbe a continuous function onIsuch that for eachxinIthere
existsyinIsuch thatjfðyÞj 12jfðxÞj. Prove there exists a pointcinIsuch thatfðcÞ ¼0.
4. Show that every polynomial of odd degree with real coefficients has at least one real root.
5. Show that the polynomialpðxÞ:¼x4þ7x39 has at least two real roots. Use a calculator to locate these roots to within two decimal places.
6. Letfbe continuous on the interval [0, 1] toRand such thatfð0Þ ¼fð1Þ. Prove that there exists a pointcin [0,12] such thatfðcÞ ¼f c þ12
. [Hint: ConsidergðxÞ ¼fðxÞ f x þ12
.] Conclude that there are, at any time, antipodal points on the earth’s equator that have the same temperature.
7. Show that the equationx¼cosx has a solution in the interval½0;p=2. Use the Bisection Method and a calculator to find an approximate solution of this equation, with error less than 103.
8. Show that the functionfðxÞ:¼2 lnxþpffiffiffix2 has root in the interval [1, 2], Use the Bisection Method and a calculator to find the root with error less than 102.
9. (a) The functionfðxÞ:¼ ðx1Þðx2Þðx3Þðx4Þðx5Þhas five roots in the interval [0, 7]. If the Bisection Method is applied on this interval, which of the roots is located?
(b) Same question forgðxÞ:¼ ðx2Þðx3Þðx4Þðx5Þðx6Þon the interval [0, 7].
10. If the Bisection Method is used on an interval of length 1 to findpnwith errorjpncj<105, determine the least value ofnthat will assure this accuracy.
11. Let I:¼ ½a;b, letf :I!R be continuous on I, and assume that fðaÞ<0;fðbÞ>0. Let W:¼ fx2I:fðxÞ<0g, and letw:¼supW. Prove thatfðwÞ ¼0. (This provides an alter- native proof of Theorem 5.3.5.)
12. LetI:¼ ½0;p=2and letf:I!Rbe defined byfðxÞ:¼supfx2;cosxgforx2I. Show there exists an absolute minimum pointx02IforfonI. Show thatx0is a solution to the equation cosx¼x2.
13. Suppose thatf:R!Ris continuous onRand that lim
x!1f¼0 and lim
x!1f¼0. Prove thatfis bounded onRand attains either a maximum or minimum onR. Give an example to show that both a maximum and a minimum need not be attained.
14. Letf:R!Rbe continuous onRand letb2R. Show that ifx02Ris such thatfðx0Þ<b, then there exists ad-neighborhoodUofx0such thatfðxÞ<bfor allx2U.
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.