Metric Spaces: Basic Properties and Examples

Transcription

1 1 Metrc Spces: Bsc Propertes d Exmples 1.1 NTODUCTON Metrc spce s dspesble termedte course of evoluto of the geerl topologcl spces. Metrc spces re geerlstos of Euclde spce wth ts vector spce structure removed but oly dstce structure reted. t geerlses the de of dstce betwee two pots o the rel le. Wheever we study the theory of fuctos of rel vrble, the oto of dstce betwee two rel umbers tutvely comes. As for exmple, f we sy x, we me tht the bsolute dfferece betwee pressged rel umber d the vlues the vrble x ssumes, pproches zero mthemtcl otto, x 0. f oe keeps md Ctor s geometrc presetto of rel umbers by pots o drected le, the the otto x 0 s vewed s equvlet to mkg dstce betwee two pots x d o the rel le ted to zero. Thus the de of dstce betwee two pots o the rel le plys vtl role formultg the bsc thgs lke lmt, cotuty, dfferetblty, covergece the rel lyss. Let s observe some otble propertes of dstce x betwee two rel umbers x d. We gree to wrte x d(x, ), otto whch we wll crry to more geerlsed dscussos comg up. (P1) d(x, ) x 0 (o-egtvty) d d(x, ) = 0 ff x = 0,.e., ff x = (postve-defteess) (P) d(x, ) x = x d(, x) (symmetry) (P3) d(x, y) x y = x + y x + y d(x, ) + d(, y), ( trgle equlty). We ow geerlse the cocept of dstce to rbtrry o-empty set X, where dstce fucto s defed y wy we lke, the oly costrt beg the smulteous stsfcto of the propertes (P1), (P), (P3) by t. fct, we xomtze the three propertes, vz, o-egtvty d postve-defteess, symmetry d trgle equlty the followg defto. Defto Let X be o-empty set d d: X X be fucto tht stsfes the codtos : (d1) d(x, y) 0 x, y X d d(x, y) = 0 ff x = y (d) d(x, y) = d(y, x) x, y X (d3) d(x, y) d(x, z) + d(z, y) for y x, y, z X. 1

2 METC SPACES AND COMPLEX ANALYSS The fucto d(, ) stsfyg (d1), (d) d (d3) s clled metrc d the structure (X, d) s clled metrc spce. ere we re ot cocered wth the specfc objects (clled pots) of X d ot eve the specfc rule of ssgmet d(, ). Note 1: f oe defes d: X X + {0}, the the o-egtvty property s redudt. Note : Oce we re covced bout the uderlyg metrc d, we express (X, d) by mere X wth the metrc structure mpled. Note 3: The codtos whch d(, ) stsfes just mmc the propertes of the dstce we re ccustomed for rel umbers, d hece these propertes ber sme mes s ther rel-le couterprts. Note 4: The o-egtvty property of metrc s cosequece of ts other propertes s for y x, y X, 0 = d(x, x) d(x, y) + d(y, x) = d(x, y). 1 Note 5: metrc spce (X, d), d(x 1, x ) dx (, x + 1) for y x 1, x,..., x X. t s exteso of trgle equlty d kow s polygol equlty (see exercse 5). y d(x, x 4 5 ) x 4 x 5 d(x, x 3 4 ) d(x, y) d(z, y) 1 5 ) d(x, x x 3 x d(x, z) z 3 ) d(x, x x 1 d(x, x 1 ) x g () Trgle equlty g (b) Polygol equlty ( = 5) Note 6: At lest for hstorcl terest t s very curous tht the motvto of troducg metrc fucto spce evolved from clsscl brchstochroe problem of vrtol clculus. The brchstochroe problem, s we ll kow, dels wth fdg the shpe of smooth curve vertcl ple log whch hevy prtcle should slde uder the cto of grvty so tht t cosumes lest tme trversg from gve pot A to other gve pot B, A beg sted hgher th but ot vertclly bove B. Thus we get rel vlued fucto (tme) defed o the fmly of smooth curves jog two gve pots A d B. f l be specfc curve of ths fmly d t(l) be the correspodg tme of descet, brchstochroe problem ms t mmsg t(l) d hece fd the curve of quckest descet. A(x, y ) 1 1 B(x, y ) 0 0 Curve of quckest descet g mly of smooth curves (prmeter l ) jog A d B

3 METC SPACES: BASC POPETES AND EXAMPLES 3 Bsed o the dfferece the tme of descet oe my be cled to defe dstce betwee two smooth curves jog two gve pots A d B d mke foudto for defg rel-vlued fucto, vz. metrc, hvg spce of curves s ts dom. Exmple Let d: be fucto defed by d(x, y) = x y, x, y. To show tht (, d) s metrc spce. Soluto: Sce x y 0 d x y = 0 ff x = y, (d1) follows. The other propertes lso follow s they re bsclly (P) d (P3). Thus d s metrc o d cosequetly (, d) s metrc spce. As prerequste to the ext exmple, we ow stte d prove Cuchy-Schwrz equlty: Sttemet: f {p 1, p,..., p } d {q 1, q,..., q } be two sets of rel umbers, the G pq Proof: Let µ be y rel umber. Defe f (µ) = ( p + µ q) G p q G =1 f (µ) 0 d s qudrtc µ. f µ 1 d µ be two dstct rel roots of f (µ) = 0, the we my wrte f (µ) = ( p + µ q ) = q ( µ µ )( µ µ ) 1 Thus for y rel t stsfyg µ 1 < t < µ, f (t) becomes egtve cotrdctg the fct f(µ) 0 for ll µ. Thus f (µ) = 0,.e., the qudrtc equto cot hve two dstct rel roots. µ q + µ p q + p =0 ece pq p q G J G J G Altertvely, 0 pq p q d j j j = 1 G G G G + G = j = 1 e 0 j j j j p q + p q p p q q = p qj pj q pq pjq J j = 1 j = 1 j = 1 G = p q p q G j j (q.e.d.) (sce d j re dummy dces of summto)

4 4 METC SPACES AND COMPLEX ANALYSS pq p q G J G J G (q.e.d.) emrk Cuchy-Schwrz equlty my be deemed s exteso of the de of dot product- orm relto b b ecoutered vector lyss. Corollry: ( p + q) p + q G J G U V W 1/ Proof: ( p + q ) = p + q + p q 1/ 1/ p + q + p q + p q p q G G J + G J G L NM 1/ 1/ = p + q G J G O QP 1/ Tkg the postve squre root we get our desred result. 1/ (by C.S. equlty) Exmple 1.1. The -dmesol Euclde spce s metrc spce wth respect to the fucto d:, defed by d(x, y) = ( x y ) = 1 where x (x 1, x,..., x ) d y (y 1, y,..., y ), x, y s belogg to. Soluto: Obvously d(x, y) 0 x, y, d(x, y) = 0 ff ( x y ) U V W 1/ U V W 1/ = 0.e., ff x = y,,...,. ece, x = y ff d(x, y) = 0. Now let x (x 1, x,..., x, y,..., y ) d z (z 1, z,..., z ) be three rbtrry elemets of. Sce x, y, z,,..., d p = x y d q = y z, obvously p + q = (x z ),,...,.

5 METC SPACES: BASC POPETES AND EXAMPLES 5 By the corollry we just proved, 1/ 1/ 1/ U bp + qg V p + q W G G 1/ 1/ U b g V + W G b g G b g 1 1.e., x z x y y z = =.e., d(x, z) d(x, y) + d(y, z) (Trgle equlty) lly, d(x, y) = x y = y x U V W 1/ 1/ b g Gb g 1/ = d(y, x), (symmetry). All these prove tht d(, ) s metrc kow populrly s Euclde Metrc or sometmes Usul Metrc. As by product of ths cse we hve the followg exmple: Exmple The Euclde metrc d: s defed by d(x, y) = ( x1 y1) + ( x y), where x (x 1, x ) d y (y 1, y ). The metrc spce (, d) s clled Euclde ple. The proof s sme letter d sprt s the exmple So o the sme o-empty set X my metrcs c be defed, s result of whch the sme set X s edowed wth dfferet metrc spce structures. The followg exmple s ce llustrto. Exmple Let X be y o-empty set d d s metrc defed over X. Let m be y turl umber so tht we defe d m (x, y) = md(x, y) for y x, y X. We re to show tht d m (, ) s lso metrc. The ew metrc spces {(X, d m )/m = 1,,...} re thus obted from (X, d). Soluto: () d m (x, y) = md(x, y) 0 x, y X Moreover d m (x, y) = 0 ff md(x, y) = 0.e., ff d(x, y) = 0 (sce m s turl umber t our dsposl),.e., ff x = y. () d m (x, y) = d m (y, x) sce d(x, y) = d(y, x) () d m (x, y) md(x, y) m(d(x, z) + d(z, y)) = md(x, z) + md(z, y) d m (x, z) + d m (z, y), for y x, y, z X. ece propertes (d1) (d3) re stsfed by d m (, ). Ths metrc s clled dlto metrc. emrk 1.1. The choce of m beg turl umber hs o specfc dvtge. owever for m > 1, dlto d for 0 < m < 1, cotrcto of dstce occurs. Exmple The set s lso metrc spce wth respect to other metrc defed by d*(x, y) = x y, where x (x 1, x,..., x, y,..., y ), x, y, (1). = 1

6 6 METC SPACES AND COMPLEX ANALYSS Soluto: The codtos (d1) d (d) re strght forwrd s exmple or (d3), let x, y, z where x (x 1, x,..., x, y,..., y ) d z (z 1, z,..., z ); x, y, z,,,...,. urther d*(x, z) = x z = { + } x y + y z x y y z = x y + y z = d*(x, y) + d*(y, z) Thus (, d*) s metrc spce. The metrc d* s clled the rectgulr metrc o. Erler we hve show exmple tht the fucto d 1 defed by d 1 (x, y) = ( x y ) 1 1 ( x y ) + wth x (x 1, x, y ) s dstce fucto. Ag t redly follows from exmple tht d (x, y) = x 1 y 1 + x y s lso dstce fucto o. We re terested scg the dstce fuctos d 1 d d from the geometrcl pot of vew. The metrc d (, ) s kow s Txcb metrc s t mesures the dstce tx would trvel from pot A(x 1, x ) to some other pot B(y 1, y ) f there were o oe wy streets. Txcb metrc or ts geerlsto, vz, rectgulr metrc geometrclly presets the sum of projectos of the stdrd Euclde dstce [c.f. exmple 1.1. d 1.1.3] o the co-ordte xes. D x y C 1 1 (x y ) + (x y ) A(x, x ) 1 B(y, y ) 1 C x y 1 1 D g epresetto of Euclde metrc, Txcb metrc d Chebyshev metrc the bckrop of. The rectgulr metrc s used commucto theory uder the me mmg dstce tht mesures the dscrepcy betwee two dgtl messges. t ws troduced by. mmg (1950). (By dgtl messge of legth we me -compoet colum vector of 0 s d 1 s). The mmg dstce betwee two dgtl messges of sme legth s defed to be the umber of co-ordtes whch they dffer. So f x = (x 1, x,..., x ) T d y = (y 1, y,..., y ) T be y two dgtl messges of legth,.e., x s d y s re oly 0 s d 1 s, ther mmg dstce d (x, y) s gve by x y. f there s sgle dscrepcy betwee the set d

7 METC SPACES: BASC POPETES AND EXAMPLES 7 receved dgtl messges, ther mmg dstce s uty. Thus mmg dstce s metrc o the set of ll dgtl messges of pressged legth. Ag cosder the semcrculr pth wth AB s dmeter. The obvously t wll pss through C. 1 1, D x + y x + y B(y, y ) 1 y 1 A(x, x ) x C(y, x ) 1 g Semcrculr pth jog two pots x1 + y1 x + y Clerly D,, the md-pot of AB, s the cetre of the semcrculr pth d the legth of the semcrculr pth ACB s π DC = π x1 1 G J + G + y x + y y1 x J = π = π d 1 ( x y ) ( x y ) (x, y) Sce d 1 (x, y) s dstce fucto o d π > 1, by exmple 1.1.4, t follows tht d 3 (x, y) = π d 1 (x, y) s lso dstce fucto. Thus we observe tht upo the sme o-empty set, oe c defe more th oe dstce fucto or metrc; t mght be the strght ler dstce or broke-le dstce or eve semcrculr rcul dstce. So wheever we tlk of metrc spce over, we must keep md wht specfc kd of dstce we re thkg. Exmple The set s metrc spce wth respect to the metrc defed by d(x, y) = Mx.{ x y ;,,..., } where x (x 1, x,..., x, y,..., y ), x, y,,,...,. Soluto: Sce x y 0,,...,, Mx. { x y ;,,..., } 0. Ag f x = y, the x = y,,...,. So x y = 0,,..., d hece Mx. { x y ;,,..., } = 0 ;.e., d(x, y) = 0. O the other hd f d(x, y) = 0, the Mx. { x y ;,,..., } = 0

8 8 METC SPACES AND COMPLEX ANALYSS x y = 0,,...,, sce ech x y 0 x = y. Thus d(x, y) = 0 f d oly f x = y. Next let x, y where x (x 1, x,..., x, y,..., y ) wth x, y,,,...,. The d(x, y) = Mx. { x y ; =1,,..., } = Mx. { y x ;,,..., } = d(y, x). lly for x, y, z, where x (x 1, x,..., x, y,..., y ), z (z 1, z,..., z ), d x, y, z,,,...,. d(x, z) = Mx. { x z,,,..., } = Mx. { x y + y z ;,,..., } Mx. { x y + y z ;,,..., } = Mx. { x y ;,,..., } + Mx. { y y ;,,..., } = d(x, y) + d(y, z). Thus d s metrc o d hece (, d) s lso metrc spce. emrk exmple (1.1.6), the Chebyshev metrc presets the mxmum of the projectos of the stdrd Euclde dstce o the co-ordte xes (see fg 1.1.3). The metrc d bove s kow s Chebyshev metrc. Exmple The set of rel umbers s metrc spce wth respect to the metrc defed by d(x, y) = M. {1, x y }, x, y. Soluto: Sce x y 0 x, y, M. {1, x y } 0 Also f x = y, the M. {1, x y } = M. {1, 0} = 0 Ag f M. {1, x y } = 0, the x y = 0 whch mples x = y. Thus d(x, y) = 0, f d oly f x = y. Next d(x, y) = M. {1, x y } = M. {1, y x } = d(y, x). lly let x, y, z. So d(x, z) = M. {1, x z } f M. {1, x z } = 1 the s x z x y + y z, M. {1, x z } = M. {1, ( x y + y z )} M. {1, x y } + M. {1, y z } Ag f M. {1, x z } = x z, the lso M. {1, x z } M. {1, x y } + M. {1, y z } Uder ll crcumstces, d (x, z) d(x, y) + d(y, z) Thus d s metrc o d hece (, d) s metrc spce. emrk f (X, d) be y metrc spce, the t s esy to prove tht d 1 (, ) defed by d 1 (x, y) = M. {1, d (x, y)} x, y X s lso metrc o X. Ths metrc s kow s stdrd bouded metrc o X. fct, correspodg to y metrc d(, ) there lwys exsts metrc d 1 (, ) defed bove. the ext chpter we shll see tht ths metrc spce (X, d 1 ) every subset s bouded. Oe c geerlse the defto of bouded metrc correspodg to d (, ) s d (, ) where d (x, y) = M. {, d (x, y)} x, y X wth > 0 Exmple Let C[, b] be the set of ll rel-vlued cotuous fuctos over [, b]. The C[, b] s metrc spce wth respect to the metrc defed by d(f, g) = sup u [,b] f(u) g(u), f, g C[, b]

9 METC SPACES: BASC POPETES AND EXAMPLES 9 Soluto: Accordg to the defto, d(f, g) 0. urther f f = g the f (u) = g(u) u [, b]. Therefore f (u) g(u) = 0 u [, b] d hece sup f d(f, g) = 0 the sup u [, b] f = g. Thus d(f, g) = 0 f d oly f f = g. Next let f, g C[, b]. The, d(f, g) = sup u [, b] f (u) g(u) = 0. O the other hd f (u) g(u) = 0. Ths mes f (u) g(u) = 0 for ll u [, b] so tht u [, b] f(u) g (u) = sup u [, b] lly let f, g, h C[, b]. The u [, b], f (u) h(u) f (u) g(u) + g(u) h(u) sup u [, b] f (u) g(u) + sup = d(f, g) + d(g, h). u [, b] g(u) f (u) = d(g, f). g(u) h(u) Tkg supremum over [, b], we get sup u [, b] f (u) h(u) d(f, g) + f (g, h) d(f, h) d(f, g) + d(g, h). Thus d stsfes ll the codtos (d1) to (d3) mkg (C[, b], d) metrc spce. Note: Ths metrc d s clled the supmetrc o C[, b]. Exmple (1.1.8), the so clled supmetrc or uform metrc geometrclly presets mxmum potwse seprto betwee two cotuous fuctos f d g defed over [, b]. f(x) g(x) g f x= x=b g epresetto of supmetrc of exmple (1.1.8) Exmple C[, b] s lso metrc spce wth respect to the metrc defed by d*(f, g) = z b f ( u ) g ( u ) du for f, g C[, b]. Soluto: Sce f(u) g(u) s lso cotuous for f, g C[, b], t s tegrble over [, b]. So the defto s megful. Sce f(u) g(u) s o-egtve, d*(f, g) 0 for ll f, g C[, b]. urther f f = g, the f(u) g(u) = 0 u [, b] d cosequetly d*(f, g) z b = 0. Ag f f(u) g(u) du = 0, the sce f(u) g(u) s o-egtve d cotuous o [, b], f(u) g(u) = 0 u [, b] whch mples f = g.

10 10 METC SPACES AND COMPLEX ANALYSS Next let f, g C[ b]. d*(f, g) = zb f u g u du ( ) ( ) = zb gu f u du lly f f, g, h C[, b], the for ll u [, b] f(u) h(u) f(u) g(u) + g(u) h(u) z z b b f(u) h(u) du {f(u) g(u) + g(u) h(u) } du = zb zb ( ) ( ) = d*(g, f). f ( u) g( u) du + g( u) h( u) du d*(f, h) d*(f, g) + d*(g, h). Thus d* s metrc d (C[, b], d*) s metrc spce. Ths metrc d* s clled the tegrl metrc o C[, b]. Exmple (1.1.9), the tergrl metrc represets the bsolute re squeezed betwee two cotuous fuctos f d g over the tervl [, b]. D f C g x= x=b x b g epresetto of tegrl metrc gve exmple (1.1.9) emrk f we defe the tegrl d*(f, g) o [, b], the set of ll -tegrble fuctos over [, b], the d*(f, g) wll ot be metrc o [, b]. fct d*(f, g) = 0 does ot lwys mply f = g. e.g., let, f(x) = x [0, ] d g(x) = for x [0, 1) = 1 for x = 1 The obvously f g but z 0 f( u) g( u) du = 0. = for x (1, ] Exmple Let S be the set of ll sequeces of rel umbers. Let x = { x } d y = { y } be y two members of S. Defe d: S S by 1 x y d(x, y) =, m 1+ x y 1 = m beg y teger greter th 1. Show tht d s metrc o S.