CONTINUOUS FUNCTIONS
Section 5.6 Monotone and Inverse Functions
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 (In) of subintervals with the length ofInequal toðbaÞ=2n such thatIndoes not have ad-fine partition.
(c) Letj2 \1n¼1Inand letp2Nbe such thatðbaÞ=2p<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.
In this section, we will be concerned with monotone functions that are defined on an interval IR. We will discuss increasing functions explicitly, but it is clear that there are corresponding results for decreasing functions. These results can either be obtained directly from the results for increasing functions or proved by similar arguments.
Monotone functions are not necessarily continuous. For example, if fðxÞ:¼0 for x2 ½0;1 and fðxÞ:¼1 for x2 ð1;2, then f is increasing on [0, 2], but fails to be continuous atx¼1. However, the next result shows that a monotone function always has both one-sided limits (see Definition 4.3.1) inRat every point that is not an endpoint of its domain.
5.6.1 Theorem Let IRbe an interval and let f :I!Rbe increasing on I. Suppose that c 2I is not an endpoint of I. Then
(i) lim
x!cf ¼supffðxÞ:x2I; x<cg, (ii) lim
x!cþf ¼infffðxÞ:x2I;x>cg.
Proof. (i) First note that if x2I and x < c, then fðxÞ fðcÞ. Hence the set fðxÞ:x2I;x<c
f g, which is nonvoid sincecis not an endpoint ofI, is bounded above byf(c). Thus the indicated supremum exists; we denote it byL. Ife>0 is given, thenLe is not an upper bound of this set. Hence there exists ye2I; ye<c such that Le<fðyeÞ L.
Sincefis increasing, we deduce that ifde:¼cyeand if 0<cy<de, thenye<
y<cso that
Le<fðyeÞ fðyÞ L:
ThereforejfðyÞ Lj<ewhen 0<cy<de. Sincee>0 is arbitrary we infer that (i) holds.
The proof of (ii) is similar. Q.E.D.
The next result gives criteria for the continuity of an increasing functionfat a pointc that is not an endpoint of the interval on which fis defined.
5.6.2 Corollary Let IRbe an interval and let f :I!Rbe increasing on I. Suppose that c2I is not an endpoint of I. Then the following statements are equivalent. (a) f is continuous at c.
(b) lim
x!cf ¼fðcÞ ¼lim
x!cþf.
(c) supffðxÞ:x2I;x<cg ¼fðcÞ ¼infffðxÞ:x2I;x>cg.
This follows easily from Theorems 4.3.3 and 5.6.1. We leave the details to the reader.
LetIbe an interval and letf :I!Rbe an increasing function. Ifais the left endpoint of I, it is an exercise to show thatfis continuous ata if and only if
fðaÞ ¼infffðxÞ:x2I;a<xg or if and only if fðaÞ ¼lim
x!aþf. Similar conditions apply at a right endpoint, and for decreasing functions.
Iff :I!Ris increasing onIand ifcis not an endpoint ofI, we define thejump off at c to be jfðcÞ:¼lim
x!cþf lim
x!cf. (See Figure 5.6.1.) It follows from Theorem 5.6.1 that
jfðcÞ ¼infffðxÞ:x2I;x>cg supffðxÞ:x2I; x<cg
for an increasing function. If the left endpointaofIbelongs toI, we define thejump offat ato bejfðaÞ:¼lim
x!aþf fðaÞ. If the right endpointbofIbelongs toI, we define thejump of fatb to bejfðbÞ:¼fðbÞ lim
x!bf.
5.6.3 Theorem Let IRbe an interval and let f :I!Rbe increasing on I. If c2I, then f is continuous at c if and only if jfðcÞ ¼0.
Proof. Ifcis not an endpoint, this follows immediately from Corollary 5.6.2. Ifc2Iis the left endpoint of I, then f is continuous at cif and only if fðcÞ ¼ lim
x!cþf, which is equivalent tojfðcÞ ¼0. Similar remarks apply to the case of a right endpoint. Q.E.D.
We now show that there can be at most a countable set of points at which a monotone function is discontinuous.
5.6.4 Theorem Let IRbe an interval and let f :I!Rbe monotone on I. Then the set of points DI at which f is discontinuous is a countable set.
Proof. We shall suppose thatf is increasing onI. It follows from Theorem 5.6.3 that D¼x2I:jfðxÞ 6¼0
. We shall consider the case thatI:¼ ½a;bis a closed bounded interval, leaving the case of an arbitrary interval to the reader.
We first note that since f is increasing, then jfðcÞ 0 for all c2I. Moreover, if ax1< xn b, then (why?) we have
ð1Þ fðaÞ fðaÞ þjfðx1Þ þ þjfðxnÞ fðbÞ;
whence it follows that
jfðx1Þ þ þjfðxnÞ fðbÞ fðaÞ:
(See Figure 5.6.2.) Consequently there can be at most k points in I¼ ½a;b where jfðxÞ ðfðbÞ fðaÞÞ=k. We conclude that there is at most one point x2I where jfðxÞ ¼fðbÞ fðaÞ; there are at most two points in I where jfðxÞ ðfðbÞ fðaÞÞ=2;
at most three points inIwherejfðxÞ ðfðbÞ fðaÞÞ=3, and so on. Therefore there is at Figure 5.6.1 The jump offatc
5.6 MONOTONE AND INVERSE FUNCTIONS 155
most a countable set of points x wherejfðxÞ>0. But since every point in Dmust be included in this set, we deduce thatDis a countable set. Q.E.D.
Theorem 5.6.4 has some useful applications. For example, it was seen in Exercise 5.2.12 that ifh:R!R satisfies the identity
ð2Þ hðxþyÞ ¼hðxÞ þhðyÞ for all x;y2R;
and ifhis continuous at a single pointx0, thenhis continuous ateverypoint ofR. Thus, ifh is a monotone function satisfying (2), thenhmust be continuous onR. [It follows from this thath(x)¼Cxfor allx2R, where C:¼hð1Þ.]
Inverse Functions
We shall now consider the existence of inverses for functions that are continuous on an intervalIR. We recall (see Section 1.1) that a functionf :I!Rhas an inverse function if and only iffis injective (¼one-one); that is,x;y2Iandx6¼yimply thatfðxÞ 6¼fðyÞ. We note that a strictly monotone function is injective and so has an inverse. In the next theorem, we show that if f :I!R is a strictly monotone continuous function, then f has an inverse functiongonJ:¼fðIÞthat is strictly monotone and continuous onJ. In particular, iffis strictly increasing then so isg, and iffis strictly decreasing then so isg. 5.6.5 Continuous Inverse Theorem Let IR be an interval and let f :I!R be strictly monotone and continuous on I. Then the function g inverse to f is strictly monotone and continuous on J:¼fðIÞ.
Proof. We consider the case thatfis strictly increasing, leaving the case thatfis strictly decreasing to the reader.
Sincefis continuous andIis an interval, it follows from the Preservation of Intervals Theorem 5.3.10 thatJ:¼fðIÞis an interval. Moreover, sincefis strictly increasing onI, it is injective onI; therefore the functiong:J!R inverse tof exists. We claim thatgis
Figure 5.6.2 jfðx1Þ þ þjfðxnÞ fðbÞ fðaÞ
strictly increasing. Indeed, ify1;y22Jwithy1<y2, theny1¼fðx1Þandy2¼fðx2Þfor some x1;x22I. We must have x1<x2; otherwise x1x2, which implies that y1¼fðx1Þ fðx2Þ ¼y2, contrary to the hypothesis that y1<y2. Therefore we have gðy1Þ ¼x1<x2¼gðy2Þ. Sincey1andy2are arbitrary elements ofJ withy1<y2, we conclude thatgis strictly increasing on J.
It remains to show thatgis continuous onJ. However, this is a consequence of the fact thatgðJÞ ¼Iis an interval. Indeed, ifgis discontinuous at a pointc2J, then the jump ofg at c is nonzero so that lim
y!cg< lim
y!cþg. If we choose any number x6¼gðcÞ satisfying
x!clim g<x< lim
x!cþg, then x has the property that x6¼gðyÞ for any y2J. (See Figure 5.6.3.) Hence x2= I, which contradicts the fact that I is an interval. Therefore
we conclude thatg is continuous onJ. Q.E.D.
Thenth Root Function
We will apply the Continuous Inverse Theorem 5.6.5 to thenth power function. We need to distinguish two cases: (i)n even, and (ii)nodd.
(i)n even. In order to obtain a function that is strictly monotone, we restrict our attention to the intervalI:¼ ½0;1Þ. Thus, letfðxÞ:¼xnforx2I. (See Figure 5.6.4.) We have seen (in Exercise 2.1.23) that if 0x<y, thenfðxÞ ¼xn<yn¼fðyÞ; thereforefis strictly increasing onI. Moreover, it follows from Example 5.2.3(a) thatfis continuous on I. Therefore, by the Preservation of Intervals Theorem 5.3.10, J:¼fðIÞ is an interval.
We will show thatJ¼ ½0;1Þ. Lety0 be arbitrary; by the Archimedean Property, there existsk2Nsuch that 0y<k. Since
fð0Þ ¼0y<kkn¼fðkÞ;
it follows from Bolzano’s Intermediate Value Theorem 5.3.7 thaty2J. Since y0 is arbitrary, we deduce thatJ¼ ½0;1Þ.
We conclude from the Continuous Inverse Theorem 5.6.5 that the functiongthat is inverse tofðxÞ ¼xnonI¼ ½0;1Þis strictly increasing and continuous onJ¼ ½0;1Þ. We usually write
gðxÞ ¼x1=n or gðxÞ ¼ ffiffiffi x pn
Figure 5.6.3 gðyÞ 6¼xfory2J
5.6 MONOTONE AND INVERSE FUNCTIONS 157
forx0 (neven), and callx1=n¼ ffiffiffi x pn
thenth root ofx0 (neven). The functiongis called the nth root function(neven). (See Figure 5.6.5.)
Sinceg is inverse tofwe have
g fð ðxÞÞ ¼x and f gðxÞð Þ ¼x for all x2 ½0;1Þ:
We can write these equations in the following form:
ðxnÞ1=n¼x and ðx1=nÞn¼x for allx2 ½0;1Þandneven.
(ii)nodd. In this case we letFðxÞ:¼xnfor allx2R; by 5.2.3(a),Fis continuous onR. We leave it to the reader to show thatFis strictly increasing onRand thatFðRÞ ¼R. (See Figure 5.6.6.)
It follows from the Continuous Inverse Theorem 5.6.5 that the function G that is inverse toFðxÞ ¼xn for x2R, is strictly increasing and continuous onR. We usually write
GðxÞ ¼x1=n or GðxÞ ¼ ffiffiffi x pn
forx2R;nodd;
and callx1=nthenth root ofx2R. The functionGis called thenth root function(nodd).
(See Figure 5.6.7.) Here we have xn
ð Þ1=n¼x and x1=n n
¼x for allx2R andnodd.
Figure 5.6.4 Graph of
fðxÞ ¼xn ðx0;nevenÞ Figure 5.6.5 Graph of gðxÞ ¼x1=n ðx0;nevenÞ
Figure 5.6.6 Graph of F(x)¼xn (x2R,nodd)
Figure 5.6.7 Graph of G(x)¼x1=n (x2R,nodd)
Rational Powers
Now that thenth root functions have been defined forn2N, it is easy to define rational powers.
5.6.6 Definition (i) Ifm,n2N andx0, we definexm=n:¼x1=n m . (ii) Ifm,n2N andx>0, we definexm=n:¼x1=n m
.
Hence we have definedxrwhenris a rational number andx>0. The graphs ofx7!xr depend on whetherr>1,r¼1, 0<r<1,r¼0, orr<0. (See Figure 5.6.8.) Since a rational numberr2Q can be written in the formr¼m=nwithm2Z,n2N, in many ways, it should be shown that Definition 5.6.6 is not ambiguous. That is, ifr¼m=n¼p=q withm,p2Zandn,q2Nand ifx>0, thenx1=n m
¼x1=q p
. We leave it as an exercise to the reader to establish this relation.
5.6.7 Theorem If m2Z,n2N,and x>0, then xm=n¼ð Þxm 1=n.
Proof. Ifx>0 andm,n2Z, then (xm)n¼xmn¼(xn)m. Now lety:¼xm=n¼x1=n m
>0 so that yn¼x1=n m n
¼x1=n n m
¼xm. Therefore it follows that y¼ð Þxm 1=n.
Q.E.D.
The reader should also show, as an exercise, that ifx>0 andr,s2Q, then xrxs¼xrþs¼xsxr and ð Þxr s¼xrs¼ð Þxs r:
Figure 5.6.8 Graphs ofx!xr ðx0Þ
5.6 MONOTONE AND INVERSE FUNCTIONS 159
Exercises for Section 5.6
1. IfI:¼[a,b] is an interval andf:I!Ris an increasing function, then the pointa[respectively, b] is an absolute minimum [respectively, maximum] point forfonI. Iffis strictly increasing, thenais the only absolute minimum point forfonI.
2. Iffandgare increasing functions on an intervalIR, show thatfþgis an increasing function onI. Iffis also strictly increasing onI, thenfþgis strictly increasing onI.
3. Show that bothf(x) :¼xandg(x) :¼x1 are strictly increasing onI:¼[0, 1], but that their productf gis not increasing onI.
4. Show that iffandgare positive increasing functions on an intervalI, then their productfgis increasing onI.
5. Show that ifI:¼[a,b] and f :I!Ris increasing onI, thenfis continuous ataif and only if f(a)¼inf{f(x) :x2(a,b]}.
6. LetIR be an interval and letf:I!R be increasing onI. Suppose thatc2I is not an endpoint ofI. Show thatfis continuous atcif and only if there exists a sequence (xn) inIsuch thatxn<cforn¼1, 3, 5, . . . ;xn>cforn¼2, 4, 6, . . . ; and such thatc¼lim(xn) and f(c)¼lim (f(xn)).
7. LetIR be an interval and let f:I!R be increasing onI. If cis not an endpoint ofI, show that the jumpjf(c) offatcis given by inf{f(y)f(x) :x<c<y,x,y2I}.
8. Let f, g be strictly increasing on an interval IR and let f(x) > g(x) for all x 2 I. If y2f Ið Þ \g Ið Þ, show thatf1ð Þy <g1ð Þ. [Hint: First interpret this statement geometrically.]y 9. LetI:¼[0, 1] and letf :I!Rbe defined byf(x) :¼xforxrational, andf(x) :¼1xforx irrational. Show thatfis injective onIand thatf(f(x))¼xfor allx2I. (Hencefis its own inverse function!) Show thatfis continuous only at the pointx¼12.
10. LetI:¼[a,b] and letf :I!Rbe continuous onI. Iffhas an absolute maximum [respectively, minimum] at an interior pointcofI, show thatfis not injective onI.
11. Letf(x) :¼xforx2[0, 1], andf(x) :¼1þxforx2(1, 2]. Show thatfandf1are strictly increasing. Arefandf1continuous at every point?
12. Letf:½0;1 !Rbe a continuous function that does not take on any of its values twice and with f(0)<f(1). Show thatfis strictly increasing on [0, 1].
13. Leth:½0;1 !R be a function that takes on each of its values exactly twice. Show thath cannot be continuous at every point. [Hint: If c1 < c2 are the points where h attains its supremum, show thatc1¼0,c2¼1. Now examine the points wherehattains its infimum.]
14. Letx2R,x>0. Show that ifm,p2Z,n,q2N, andmq¼np, thenx1=n m
¼x1=q p . 15. Ifx2R,x>0, and ifr,s2Q, show thatxrxs¼xrþs¼xsxrandð Þxrs¼xrs¼ð Þxsr.