Simultaneous approximation of polynomials

Transcription

1 Smultaneous approxmaton of polynomals Anre Kupavsk * János Pach Abstract Let P enote the famly of all polynomals of egree at most n one varable x, wth real coeffcents. A sequence of postve numbers x 1 x 2... s calle P -controllng f there exst y 1, y 2,... R such that for every polynomal p P there exsts an nex wth p(x ) y 1. We settle an problem of Maka an Pach (1983) by showng that x 1 x 2... s P -controllng f an only f =1 1 x s vergent. The proof s base on a statement about coverng the Euclean space wth translates of slabs, whch s relate to Tarsk s plank problem. 1 Introucton Let F be a class of real functons R R. We say that a sequence of postve numbers x 1, x 2,... s F-controllng f there exst reals y 1, y 2,... wth the property that for every f F, one can fn an wth f(x ) y 1. In other wors, a sequence x 1, x 2,... s F-controllng f we can fn y 1, y 2,... R such that the ponts p 1 = (x 1, y 1 ), p 2 = (x 2, y 2 ),... R 2 smultaneously approxmate all functons n F, n the sense that the graph of every member f F gets (vertcally) closer than 1 to at least one pont p. In ths paper, we aress the followng queston rase n [11]. Gven a class of functons F, how sparse a F-controllng sequence can be? A smlar queston, motvate by a problem of László Fejes Tóth [5], was stue n [4]. Let P enote the class of polynomals R R of egree at most. It was shown by Maka an Pach [11] that f a sequence of postve numbers x 1 x 2... s P -controllng, then the nfnte seres s vergent. They conjecture that ths conton s also x 1 x 2 suffcent for a sequence x 1 x 2... to be P -controllng (see Conjecture 3.2.B n [11]). The am of ths note s to prove ths statement. * EPFL, Lausanne an MIPT, Moscow. Supporte n part by the grant N of the Russan Founaton for Basc Research. E-mal: kupavsk@ya.ru. EPFL, Lausanne an Rény Insttute, Buapest. Supporte by Hungaran Scence Founaton EuroGIGA Grant OTKA NN , by Swss Natonal Scence Founaton Grants an E-mal: pach@cms.nyu.eu. 1

2 Theorem 1. Let be a postve nteger an x 1 x 2... be a monotone ncreasng nfnte sequence of postve numbers. The sequence x 1, x 2,... s P -controllng f an only 1 f =. x 1 x 2 x 3 We also generalze ths result to other fntely generate functon classes. Gven + 1 real functons, f 0, f 1,..., f : R + R +, let L = L(f 0,..., f ) enote the set of all functons that can be obtane as lnear combnatons of them wth real coeffcents. That s, Here R + stans for the set of postve reals. L = {a 0 f a f : a 0,..., a R}. Theorem 2. Let 1 be an nteger, x 0 > 0, ε > 0, an let f 0, f 1,..., f : R + R + be real functons that are monotone ncreasng for x x 0 an boune over every boune subnterval of R +. Assume that the functons F j (x) = f j (x)/(f (x)) 1 ε (j = 0,..., 1) are monotone ecreasng for x x 0 an ten to 0 as x. An ncreasng sequence of postve numbers x 1 x 2... s L(f 0,..., f )-controllng f an only f =1 1 =. f (x ) Obvously, the functons f (x) = x ( = 0, 1,..., )) meet the above requrements, so that Theorem 2 mples Theorem 1. For the proof of Theorem 1, we wll rephrase the queston as a coverng problem for slabs. A slab (sometmes calle plank or strp) s the set of ponts S lyng between two parallel hyperplanes n R. The stance w between these two hyperplanes s calle the wth of the slab. We can wrte S as S = {x R : b w 2 v, x b + w 2 }, for some unt vector v an real number b. We say that a sequence of slabs S 1, S 2,... permts a translatve coverng of a subset R f there are sutable translates S of S ( = 1, 2,...) such that =1S = R. As t was shown n [11], Theorem 1 (an, n fact, Theorem 2, too) woul easly follow from Conjecture 1. ([11], [3]) Let be a postve nteger. A sequence of slabs n R wth wths w 1, w 2,... permts a translatve coverng of R f an only f =1 w =. The fact that ths conton s necessary follows, for example, from Tarsk s result [12] whch states that the total wth of any system of slabs the unon of whch covers a sk of unt ameter s at least 1. Tarsk s plank problem, whether ths statement remans true n hgher mensons, remane open for almost twenty years. In 1950, Bang [1, 2] answere ths queston n the affrmatve. For = 2, Conjecture 1 was prove by Maka an Pach [11] an, accorng to [6], nepenently, by Erős an Straus (unpublshe). (See [7, 8] for 2

3 some refnements.) For 3, the problem s open. Groemer [6] prove that any sequence of slabs n R wth wths w 1, w 2,... satsfyng =1 w +1 2 = permts a translatve coverng of R. Recently, the authors of the present note [9] have come close to settlng Conjecture 1 by replacng Groemer s suffcent conton wth the weaker assumpton lm sup n w 1 + w w n log(1/w n ) In partcular, any sequence of slabs of wths 1, 1, 1,... permts a translatve coverng of 2 3 space. To establsh Theorem 1, t s enough to verfy Conjecture 1 for specal sequences of slabs, whose normal vectors le on a moment curve. We wll o precsely ths n Secton 2, by explorng the natural orerng of these vectors. In Secton 3, we generalze our arguments to establsh Theorem 2. The last secton contans a few conclung remarks. > 0. Fgure 1: Controllng polynomals of egree at most. 2 Proof of Theorem 1 We only have to prove the f part of the theorem. 3

4 Let x 1 x 2... be a monotone ncreasng sequence of postve numbers wth 1 = x. We have to fn a sequence of reals y 1, y 2,... such that for any polynomal p(x) = j=0 a jx j wth real coeffcents a j, there exsts a postve nteger wth p(x ) y 1. Wrte p(x) n the form p(x) = x, a, where x = (1, x,..., x ), a = (a 0, a 1,..., a ) R +1, an. stans for the scalar prouct. Usng ths notaton, we have x = (1, x,..., x ) an the nequalty p(x ) y 1 can be rewrtten as y 1 x, a y + 1. For a fxe, the locus of ponts a R +1 satsfyng ths ouble nequalty s a slab S R +1 of wth w = 2 = 2 x (, wth normal vector x. See Fg. 1. The sequence x 1, x 2,... j=0 x2j )1/2 s P -controllng f an only f the sequence of slabs S 1, S 2,... permts a translatve coverng of R +1. If x 3 for nfntely many (an, hence, for all) postve ntegers, then for the wths of the corresponng slabs S we have w > 1. Thus, these slabs permt a translatve coverng 3 of R +1, because each of them can be translate to cover any ball of ameter 1. 3 Therefore, we can assume that x > 3 for all m. In fact, we can assume wthout loss of generalty that x > 3 for all 1, otherwse we smply scar the frst m 1 members of the sequence, an prove the theorem for the resultng sequence x m x m We are gong to explot the fact that the normal vectors x = (1, x,..., x ) of the slabs S le on the moment curve (1, x, x 2,..., x ). Frst, we nee an auxlary lemma. Lemma 1. Let be a postve nteger, let 3 x 1 x 2... be a fnte or nfnte sequence of reals, an let x = (1, x, x 2,..., x ) for every. Then there exst +1 lnearly nepenent vectors u 1,..., u +1 R +1 such that for every ( = 1, 2,...) an j (j = 1, 2,..., + 1), we have () x +1, u 1 x, u 1 x +1, u j x, u j, () x, u j 1 3 x u j. Proof. Take the stanar bass e 1,..., e +1 n R +1,.e., let e enote the all-zero vector wth a sngle 1 at the -th poston. Set u j := e +1 j +e +1 for j = 1,..., an u +1 := e +1. Conton () trvally hols for j = 1 an very easy to check for j = +1. For j = 2,...,, t reuces to x x x +1 whch s equvalent to x 1 + x x j x j + x (x x +1)(x j + x ) (x j +1 + x +1)(x 1 + x ). 4,

5 The last nequalty can be rewrtten as x j +1 x j j 1 (x +1 x )( or, vng both ses by x j j 1 +1 x j j 2 x k +1x j 1 k + (x +1 x ), as j 2 x k +1x j 1 k + x k +1x j 2 k x k +1x j 2 k x j 1 +1 xj 1 0. x j 1 +1 xj 1 ) 0, Usng the fact x +1 x, an bounng from above each sum by ts largest term multple by the number of terms, we obtan that the left-han se of the last nequalty s at most jx j (j 1)xj 2 +1 xj 1 +1 xj 1 < x j 1 +1 (2j 1 xj 1 ). As x 3, the rght-han se of ths nequalty s always negatve an () hols. It remans to verfy conton (). Takng nto account that x 3, we have x, u +1 = x 1 2 x = 1 2 x u +1. On the other han, for j = 1,...,, we obtan x, u j = x j Ths completes the proof of Lemma 1. + x 1 2 x 1 3 x u j. In orer to establsh Theorem 1, t s enough to prove that there s a constant c = c(+1) such that any system of slabs S ( = 1,..., n) n R +1 whose normal vectors are (1, x,..., x ) for some 3 x 1 x 2... x n an whose total wth s at least c, permts a translatve coverng of a ball of unt ameter. Ths s an mmeate corollary of Lemma 1 an the followng asserton. Lemma 2. For every postve nteger, for any system of + 1 lnearly nepenent vectors u 1,..., u +1 n R +1, an for any γ > 0, there s a constant c wth the followng property. Gven any system of slabs S ( = 1,..., n) n R +1, whose normal vectors x satsfy the contons () x +1, u 1 x, u 1 x +1, u j x, u j, () x, u j γ x u j for every an j, an whose total wth n =1 w s at least c, the slabs S permt a translatve coverng of a ( + 1)-mensonal ball of unt ameter. 5

6 Fgure 2: We place the slabs one by one. Proof. Instea of coverng a ball of unt ameter, t wll be more convenent to cover the smplex Δ wth one vertex n the orgn 0 an the others at the ponts (vectors) u j (j = 1,..., + 1). By properly scalng these vectors, f necessary, we can assume that Δ contans a ball of unt ameter. We place the slabs one by one. See Fg. 2. We place S 1, a translate of S 1, so that one of ts bounary hyperplanes passes through 0 an the other one cuts a smplex Δ 1 out of the cone Γ of all lnear combnatons of the vectors u 1,..., u +1 wth postve coeffcents. Accorng to our assumptons, we have x 1, u j > 0 for every j. Therefore, S 1 oes not separate Γ nto two cones: S 1 Γ s nee a smplex Δ 1. Suppose that we have alreay place S 1,..., S, the translates of S 1,..., S, so that ther unon covers a smplex Δ wth one vertex at the orgn, an the others along the + 1 half-lnes that span the cone Γ. We also assume that the facet of Δ opposte to the orgn s a bounary hyperplane of S. Let p (j) enote the vertex of Δ that belongs to the open half-lne parallel to u j emanatng from 0 (j = 1,..., + 1). Next, we place a translate S +1 of S +1 so that one of ts bounary hyperplanes, enote by π, passes through p (1), an the other one, π, cuts the half-lne parallel to u 1 at a pont p +1 (1) wth p +1 (1) > p (1). That s, p +1 (1) s further away from the orgn than p (1) s. Let p +1 (2),..., p +1 ( + 1) enote the ntersecton ponts of π wth the half-lnes parallel to u 2,..., u +1, respectvely, an let Δ +1 be the smplex nuce by the vertces 0, p +1 (1),..., p +1 ( + 1). We have to verfy that Δ +1 s entrely covere by the slabs S 1,..., S +1. By the nucton hypothess, Δ was covere by the slabs S 1,..., S. Thus, t s suffcent to check that the 6

7 hyperplane π ntersects every ege 0p (j) of Δ, for j = 1,..., + 1. Let α j u j be the ntersecton pont of π wth the half-lne parallel to u j, an let p (j) = β j u j. We have to prove that α j β j. By efnton, we have x +1, p (1) α j u j = 0 an x, p (1) β j u j = 0. From here, we get α j = x +1, p (1) x, p (1) β j x +1, u j x, u j = x +1, p (1) x+1, u j x, p (1) x, u j = x +1, u 1 x+1, u j x, u 1 x, u j. In vew of assumpton () of the lemma, the rght-han se of the above chan of equatons s at most 1, as requre. Observe that urng the whole proceure the uncovere part of the cone Γ always remans convex an, hence, connecte. In the nth step, n =1S Δ n. By the constructon, p (1) les at least w farther away from the orgn along the half-lne parallel to u 1 than p 1 (1) oes. Thus, we have n p n (1) w c. Usng the fact that x n, p n (j) p n (1) = 0 for every j 2, an takng nto account assumpton (), we obtan p n (j) x n, p n (j) x n =1 = x n, p n (1) x n γ p n (1) γc. Thus, f c s suffcently large, we have p n (j) u j. Ths means that Δ n contans the smplex Δ efne n the frst paragraph of ths proof. Hence, t also contans a ball of unt ameter, as requre. 3 Proof of Theorem 2 In ths secton, we exten the technque use n the proof of Theorem 1 to establsh Theorem 2. As n the proof Theorem 1, we can wrte any functon l = a kf k L(f 0,..., f ) as l(x) = x, a, where x = (f 0 (x), f 1 (x),..., f (x)) an a = (a 0, a 1,..., a ) R +1. As before, we only have to prove the f part of the theorem, whch s equvalent to the fact that the slabs S R +1 wth normal vector x = (f 0 (x ),..., f (x )) an wth w = 2 x = 2 ( f k 2(x )) 2 1/2 f (x ), for = 1, 2,..., permt a translatve coverng of R +1. Agan, t s enough to conser the case when lm x =, otherwse each slab S contans a ball of ameter at least 2 f (lm x ) > 0. 7

8 We follow the scheme of the proof of Theorem 1. Accorng to Lemma 2, t s enough to show that there exst + 1 lnearly nepenent vectors u 1,..., u +1 that satsfy contons () an () wth x = (f 0 (x ),..., f (x )) an wth a sutable constant γ > 0. We can assume wthout loss of generalty that x 1, an hence all x s, are so large that they satsfy x 1 x 0 an the nequaltes f j (x) f (x) f j(x 1 ) f (x 1 ) 1, (1) for every x x 1 an j = 0,..., 1. To see ths, observe that f j (x)/f (x) = F j (x)/f ε(x) s monotone ecreasng n x, because F j s monotone ecreasng, whle f s monotone ncreasng. Let e 1,..., e +1 be the stanar bass n R +1. For 1 j + 1, set +1 u j := e k 1 2 e +2 j. k=1 In other wors, all coornates of u j are 1, wth the excepton of the (+2 j)-th coornate, whch s 1 2. By efnton, we have x, u j 1f 2 (x ) an u j < + 1. It follows from (1) that for j, so that f j (x ) f (x ) 1 Hence, for every an j, ( 1/2 x fk 2 (x )) 2f (x ). x, u j 1 2 f (x ) x 1 2 2( + 1) x u j. Therefore, conton () n Lemma 2 s satsfe wth γ = 1 2 2(+1). It remans to verfy conton (). For the rest of the argument, fx j (1 j + 1). We have to show that for every ( = 1, 2,...), the nequalty x +1, u 1 x, u 1 x +1, u j x, u j hols. For j = 1, the statement s trval. Therefore, we may suppose that j > 1. Next, we want to get r of f (x) n the left han se, keepng both the numerator an enomnator postve. The above nequalty equvalent to the followng: x +1, u x +1, u j x, u x, u j x +1, u j x, u j. 8

9 Usng the notaton φ(x) = 1 1 f k (x) f 1 +1 j(x), ψ(x) = f k (x) 1 2 f +1 j(x), the above nequalty may be rewrtten as φ(x +1 ) φ(x ) f (x +1 ) + ψ(x +1 ). (2) f (x ) + ψ(x ) Before checkng that (2) s true, let us summarze the propertes of the functons φ an ψ we nee: 1. φ(x +1 )/φ(x ) f 1 ε (x +1 )/f 1 ε (x ) for the constant ε > 0 from Theorem 2, 2. ψ(x +1 ) cf 1 ε (x +1 ) for a constant c > 0, an 3. ψ(x +1 ) ψ(x ). By the monotoncty of F k, we have f k (x +1 )/f k (x ) f 1 ε (x +1 )/f 1 ε (x ), for k = 0,..., 1. Now property 1 follows from the fact that, f a 0,..., a 1, b 0,..., b 1, t are postve numbers satsfyng a 0 /b 0 t,..., a 1 /b 1 t, then (a a 1 )/(b b 1 ) t. Usng that lm x F k (x) = 0 for k = 0,..., 1, we get property 2. Property 3 s a rect consequence of our assumpton that each f k (k = 0, 1,...) s monotone ncreasng for x x 0. We have to verfy (2). In vew of property 1, t s suffcent to show f 1 ε (x +1 ) f 1 ε (x ) f (x +1 ) + ψ(x +1 ), f (x ) + ψ(x ) whch s equvalent to ( ( f ψ(x )f 1 ε (x +1 ) ψ(x +1 )f 1 ε (x ) f (x )f 1 ε (x +1 ) ) ) ε (x +1 ) 1, f (x ) or, n a slghtly fferent form, ψ(x )f 1 ε (x +1 ) ψ(x +1 )f 1 ε (x ) f (x )f 1 ε ( ( (x +1 ) ε f (x +1 ) f 1 ε f 1 ε (x ) (x ) ) ) ε 1 ε 1. Replacng the left-han se by a larger quantty (takng property 3 nto account) an the rght-han se by a smaller one (applyng the nequalty (1 + x) α 1 + αx, val for all α, x 0), we obtan the stronger nequalty ψ(x +1 )(f 1 ε ( (x +1 ) f 1 ε (x )) f (x )f 1 ε (x +1 ) 9 ε 1 ε f 1 ε (x +1 ) f 1 ε f 1 ε (x ) (x ) ). (3)

10 Thus, t s suffcent to prove (3). Rearrangng the terms, we obtan ψ(x +1 ) ε 1 ε f ε (x )f 1 ε (x +1 ). By property 2, we have ψ(x +1 ) cf 1 ε (x +1 ), so that t s enough to check that cf 1 ε (x +1 ) ε 1 ε f (x ε )f 1 ε (x +1 ), that s, c ε f ε 1 ε (x ). As f (x) s an ncreasng functon for x x 0, the last nequalty s satsfe f we choose x 1 (an, hence, all other x ) suffcently large. Ths completes the proof of (3), (2), an so the proof of Theorem 2. 4 Conclung remarks 1. As was mentone n the Introucton, Conjecture 1 s known to be true n the plane. Moreover, n [11] a stronger statement was prove: there exsts a constant c such that every collecton of strps wth total wth at least c permts a translatve coverng of a sk of ameter 1. In vew of ths, one can make the followng even boler conjecture. Conjecture 2. For any postve nteger, there exsts a constant c = c() such that every collecton of slabs n R of total wth at least c permts a translatve coverng of a unt ameter -mensonal ball. Suppose Conjecture 1 s true for a postve nteger. Answerng a queston n [11], Imre Z. Ruzsa [10] prove that then, for the same value of, Conjecture 2 also hols. Thus, the two conjectures are equvalent. 2. Gven a class F of functons R R, we say that a sequence of postve numbers x 1 x 2... s strongly F-controllng f there exst reals y 1, y 2,... wth the property that, for every ε > 0 an every f F, one can fn an wth f(x ) y ε. It s easy to see that the conton n Theorem 1 s suffcent to guarantee that the sequence x 1, x 2,... s strongly P -controllng. Theorem 2 can also be strengthene analogously. 3. The am of ths paper was to fn necessary an suffcent contons for a sequence of numbers to be L-controllng, where L = L(f 1,..., f ) s the class of functons that can be obtane as lnear combnatons of f 1,..., f. We reuce ths problem to a queston about coverng R wth translates of certan slabs. However, the two problems are not necessarly equvalent. For example, we have notce that the slabs obtane at ths reucton ha some specal propertes: apart from ther wths, ther normal vectors were also prescrbe. Ths enable us to cover R wth ther translates, even f we o not know whether such a coverng exsts for every system of slabs wth the same wths. Nevertheless, n a more complcate sense, the two problems are equvalent. 10

11 Theorem 3. Gven a postve nteger, an a sequence of postve numbers x 1, x 2,..., efne a famly F = F(, x 1, x 2,...) of -tuples of functons f 1,..., f : R R as F = {(f 1,..., f ) : fj 2 (x ) = x 2 j=1 for all }. Then a sequence of slabs wth wths x 1, x 2,... permts a translatve coverng of R f an only f x 1, x 2,... s L(f 1,..., f )-controllng for every -tuple (f 1,..., f ) F, where L(f 1,..., f ) = {a 1 f a f : a 1,..., a R}. References [1] Th. Bang, On coverng by parallel-strps, Mat. Tsskr. B (1950), [2] Th. Bang, A soluton of the plank problem, Proc. Amer. Math. Soc. 2 (1951), [3] P. Brass, W. Moser, an J. Pach, Research Problems n Dscrete Geometry, Sprnger, Heelberg, [4] P. Erős an J. Pach, On a problem of L. Fejes Tóth, Dscrete Math. 30 (1980), no. 2, [5] L. Fejes Tóth, Remarks on the ual of Tarsk s plank problem n Hungaran), Matematka Lapok 25 (1974), [6] H. Groemer, On coverngs of convex sets by translates of slabs, Proc. Amer. Math. Soc. 82 (1981), no. 2, [7] H. Groemer, Coverng an packng propertes of boune sequences of convex sets, Mathematka 29 (1982), [8] H. Groemer, Some remarks on translatve coverngs of convex omans by strps, Cana. Math. Bull. 27 (1984), no. 2, [9] A. Kupavsk an J. Pach, Translatve coverng of the space wth slabs, manuscrpt. [10] I. Z. Ruzsa, personal communcaton. [11] E. Maka Jr. an J. Pach, Controllng functon classes an coverng Euclean space, Stu. Scent. Math. Hungarca 18 (1983), [12] A. Tarsk, Uwag o stopnu równoważnośc welok atów (n Polsh), Parametr 2 (1932),