a simple proof of sharkovsky's theorem

8/13/2019 A Simple Proof of Sharkovsky's Theorem

1/6

A Simple Proof of Sharkovsky's TheoremAuthor(s): Bau-Sen DuSource: The American Mathematical Monthly, Vol. 111, No. 7 (Aug. - Sep., 2004), pp. 595-599Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/4145161.

Accessed: 31/05/2013 18:13

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at.

http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of

content in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new forms

of scholarship. For more information about JSTOR, please contact [email protected].

.

Mathematical Association of Americais collaborating with JSTOR to digitize, preserve and extend access to

The American Mathematical Monthly.

http://www.jstor.org

This content downloaded from 147.65.167.156 on Fri, 31 May 2013 18:13:49 PMAll use subject to JSTOR Terms and Conditions
http://www.jstor.org/action/showPublisher?publisherCode=maahttp://www.jstor.org/stable/4145161?origin=JSTOR-pdfhttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/stable/4145161?origin=JSTOR-pdfhttp://www.jstor.org/action/showPublisher?publisherCode=maa


2/6

HerevsinTo s the vertex ine at To, hequantity iven byvsin To=

V1 2 cos ZToTicos LToT2cos LToT3 co2 1 - cos2 TT2 - cos2 LTo3.It is nothard o prove hisformulaandwe leave it to the reader.From hepowerseriesexpansions f thevarious erms n (7), we extractts localversion i.e., thewell-known osine lawforEuclideanetrahedra):

A2 A A + A - 2A1A2CO ToT3-2A A3cos LToT2 2A2A3COSLToT1.(8)For an octant etrahedroni.e., for a tetrahedronn whichall dihedralanglesat Toarerightangles) hisreduces o thePythagoreanheoremn three-space6]:A2 = A2 A2 A.A 2 A 3 .

Ina similar ashion,onecanprove hat n the casec-fan octant phericaletrahe-dron t is true hatA' < A + A2 + A .

Infact,it reduces by settingAo/2 = w, A,/2 = x, A2/2 = y, andA3/2 = z) to thefollowing:when0 < x, y, z, w < r, theconditionscos w = cos x cosy cos z + sin x siny sinz,

x + y + z > w, y + z + w > x, x + y + w > z, andx + z + w > y implythatw2


3/6


4/6

Proof Let P = {xi : 1 3 and odd, thenf has periodic points of all evenperiods. Furthermore,f has a periodic point of leastperiod nfor each integer n with n > m + 1.Proof Let P = {xi : 1 < i < m}, with xl < x2 < ... < Xm, be a period-m orbit of f.Let x, = max{x E P : x < f (x)}. Then x,+l < f (xs) and f (xs+) < x,, so f has afixed point z in [xs, xs+ ]. Since m is odd, for some integer t such that 1 < t < m - 1andt : s thepointsf (xt) andf(x,+l) lie onopposite idesof z. Thusf ([xt,xt+1]) D[xs,xs+l]. Forsimplicity,we assume hatx, < x,. Ifxs+1< x,, theproof s similar.Letq be thesmallestpositive nteger uch thatfq (x,) < xt. Then2 < q < m - 1.Firstassumethat m = 3. Without oss of generality,we assumethatf(x1) = x2,f (x2) = x3, and f(x3) = x1. Let Jo = [x1, x2] and J1 = [x2, x3]. For any n > 2, wecanapplyLemma4 to thecycle Jo0JJ1 .. J1Jo of lengthn to obtaina period-npoint.Accordingly,f f has a period-3point, henf hasperiodicpointsof allperiods.Nowassumethatm > 3. Since q is the smallestpositive integersuch that fq(x,) < xt,xt+1 < f'(xs) whenever1 < i < q - 1. Ifxt+l fq9- (Xs) < xs, Lemma4 applies othecycle

[xfq-(X,)][fq-(Xs),][fq-I(',), ][X,,q-1(X,)]and establishes he existence of a period-3pointof f. If fq-I(xs) = Xs+l,we canapplyLemma4 to thecycle

[z,xs+l][X,,t+ ][x,,Xs+1][z,s+1]to obtaina period-3pointof f.Weproceedassuminghatxs+1 < fq-I(x,). Ifk = min{1 < i < q - 1 : fq- (xs)


5/6

[u, v][z, w][u, v] and, for every even integer n > 4, to the cycle[u,v]([z,w][v, ])(n-2)/2[z,][u, ]

(here ([z, w][v, z])(n-2)/2 represents (n - 2)/2 copies of [z, w][v, z]) of length n, weconclude that f has periodic points of all even periods. On the other hand, let Ji =[z : f'(x,)] for i = 0, 1, .., q - 1, where [a : b] denotes the closed interval with aand b as endpoints.Forany n > m + 1, we appealto Lemma4 for the cycle of length nJo J1 -J_-1 Jq-1 [Xt, Xt+l]J . . JJo, where J = [x,, x,s+ ], to confirm the existence ofa period-n point. 04. A PROOF OF SHARKOVSKY'S THEOREM. We now combine (a), (b), (c),and Lemma 1 to prove Sharkovsky's theorem.Theorem 6 (Sharkovsky). Assume that f : I - I is a continuous map. If f hasa period-m point, then f also has a period-n point precisely when m -< n in theSharkovskyordering.Proof By (b) and (c), we have 3 - 3 and odd and if k > 2,then by Lemma 1(1) f2k-I has period-(2

-m) points. It follows from what we havejustproved that f2k-I has period-(2 - (m + 2)) points and period-(22 . 3) points. In viewof Lemma 1(2), f has period-(2k . (m + 2)) points and period-(2k+1 - 3) points. Fur-

thermore, because f has period-(2k ?m) points, f2k has period-m points. By (b), f2khas period-2 points as long as 2 > m, so by Lemma 1(2) f has period-(2k+n) pointsfor all integers n such that 2 > m. Finally, if f has period-2' points for some inte-ger i > 2, then f2i-2 has period-4 points. As a result of (a), f2i-2 has period-2 points,ensuring that f has period-2i-' points. This proves the sufficiency of Sharkovsky'stheorem.

For the converse, it suffices to assume that I = [0, 1]. Let T(x) = 1 - 12x - 11be the tent map on I. Then for any k > 1 the equation Tk x) = x has exactly 2kdistinct solutions in I. It follows that T has finitely many period-k orbits. Amongthese period-k orbits, let Pkbe one with the smallest diameter max Pk- min Pk. Forany x in I, let Tk(x) = min Pk if T(x) < min Pk, Tk(x) = max Pkif T(x) > max Pk,and Tk(x) = T(x) if min Pk < T(x) < max Pk. It is then easy to see that Tkhasexactly one period-k orbit (i.e., Pk) and no period-j orbit for any j with j -< k inthe Sharkovskyordering(see also [1, pp. 32-34]). Now let Q3 be any period-3 orbit ofT of minimal diameter. Then [min Q3, max Q3] contains finitely many period-6 orbitsof T. If Q6 is one of smallest diameter,then [min Q6, max Q61contains finitely manyperiod-12 orbits of T. We choose one, say Q 2, of minimal diameterand continue theprocess inductively.Let qo = sup{min Q2n.3 : n > 0) andq, = inf{max Q2n.3 : n > 0}.Let T,(x) = qo if T(x) < qo, T,(x) = ql if T(x) > q1, and T,(x) = T(x) if qo 0, buthas no periodic points of any other periods. This establishes the otherdirection in Sharkovsky'stheorem. U598 ? THEMATHEMATICALSSOCIATIONFAMERICA[Monthly 111

http://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsp


6/6

Remark. Ourmethodcan alsobe used to prove hat, f f has a periodicpointof oddperiodm > 1butno periodicpointsof oddperiodstrictlybetween1 andm, thenanyperiodicorbitof oddperiodm mustbe a Stefanorbit cf. [4]).ACKNOWLEDGMENTS. I would like to thankM. Misiurewicz,A. N. Sharkovsky,and the referee for manyconstructivesuggestionsthatled to improvements n this note.

REFERENCES1. L. Alsed&,J. Llibre, and M. Misiurewicz, CombinatorialDynamics and Entropyin Dimension One,2nd ed., WorldScientific, Singapore,2000.2. R. Barton and K. Bums, A simple special case of Sharkovskii'stheorem, this MONTHLY07 (2000)932-933.3. L. S. Block and W. A. Coppel,Dynamicsin OneDimension,LectureNotes in Math.,no. 1513, Springer-

Verlag,Berlin, 1992.4. K. Burns,A note about Sharkovskii's heorem,2003 (preprint).5. M. Misiurewicz,Remarkson Sharkovsky's heorem,this MONTHLY04 (1997) 846-847.6. A. N. Sharkovsky,Coexistence of cycles of a continuous map of a line into itself, Ukrain. Mat. Zh. 16(1964) 61-71; English translation, nternat.J. Bifur ChaosAppl.Sci. Engrg.5 (1995) 1263-1273.7. P. Stefan, A theorem of Sarkovskiion the existence of periodic orbitsof continuousendomorphismsofthe real line, Comm.Math.Phys. 54 (1977) 237-248.8. P. D. Straffin,Jr.,Periodicpoints of continuousfunctions,Math.Mag. 51 (1978) 99-105.Instituteof Mathematics,AcademiaSinica, Taipei11529, Taiwanmabsdu@ccvax. inica.edu. w

The Hairy Ball Theoremvia Sperner'sLemmaTylerJarvis and JamesTanton

1. INTRODUCTION.It is well knownthatanycontinuousangentvectorfieldonthesphereS2 must,at somelocation,be zero.This result s knownas theHairyBallTheoremor it can belooselyinterpreteds follows:

It is impossible to comb all the hairs of a fuzzy ball so that (i) each hair liestangentto the surface of the ball and (ii) the angles of the hairs varycontinuouslyover the surface of the ball. (By this we mean that the angle between two hairs atpositions p and q, say, can be made arbitrarilysmall by choosing q sufficientlyclose to p.) Any attemptto accomplish thisfeat mustproduce a cowlick.We areassuming hateverypointof the ball'ssurface proutsa hair. t is a surpriseto learn that this topologicalresult,like Brouwer's amousfixedpointtheorem[2,pp.21-24], alsofollows fromanapplication f Sperner'semma.

2. SPERNER'SLEMMA. In 1928 EmanuelSpernerpresenteda simple,yet sur-prisinglypowerful,combinatorialemmaabouttriangles 5]. We workwith a slightgeneralizationf hisoriginal esult:August-September 2004] NOTES 599

http://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsphttp://www.jstor.org/page/info/about/policies/terms.jsp

a simple proof of sharkovsky's theorem

Documents