STEINHAUS PROPERTY IN BANACH LATTICES

Transcription

1 DEPARTMENT OF MATHEMATICS TECHNICAL REPORT STEINHAUS PROPERTY IN BANACH LATTICES DAMIAN KUBIAK AND DAVID TIDWELL SPRING 2015 No TENNESSEE TECHNOLOGICAL UNIVERSITY Cookevlle, TN 38505

2 STEINHAUS PROPERTY IN BANACH LATTICES DAMIAN KUBIAK AND DAVID TIDWELL Abstract. We prove that a strctly monotone order contnuous Banach lattce X over a non-atomc measure space has the Stenhaus property, that s for every quas-fnte set A X there s a dense set Y X such that for every y Y and for every postve nteger n there exsts an open ball centered at y whch contans exactly n ponts of A. We present several examples of spaces satsfyng and falng the Stenhaus property. We also delver detaled proofs of some mportant and known results connected wth the Stenhaus property. 1. Introducton Hugo Stenhaus n 1957 n a polsh journal for hgh school teachers Matematyka rased a queston [3] of exstence of a crcle n the plane surroundng exactly n ponts of the nteger lattce Z Z, n N. The answer to that queston s affrmatve and can be found n [4] or [1]. In fact, t s possble to show that there exsts a dense set Y R 2 such that for every y Y and for every nteger n there s a crcle surroundng exactly n ponts of Z Z. That problem have been generalzed by P. Zwoleńsk n 2011 [5] n the followng way. An nfnte set A n a metrc space X s sad to be quas fnte f every open ball n X contans fntely many elements of A. We say that a Banach space (X, X ) has the Stenhaus property f for every quas-fnte set A X there exsts a dense set Y X such that for every y Y and every n N there exsts an open ball B centered at y whch contans exactly n elements of A. Zwoleńsk showed that every Hlbert space has the Stenhaus property. In a preprnt from 2013, T. Kochanek found an equvalent condton for the Stenhaus property n Banach spaces. He also showed that the L p spaces over a non-atomc measure space have the Stenhaus property f 1 p < [2]. We present a generalzaton of that result n secton 4. In secton 3 gve detaled proofs of some Kochanek s results. 2. Prelmnares Let X be a Banach space over R. By S(x, r) and B(x, r) we denote the sphere and the ball centered at x X and radus r > 0, respectvely. We also wrte S(X) = S(0, 1) and B(X) = B(0, 1) to denote the unt sphere and the unt ball of X. Let (Ω, Σ, µ) be a σ-fnte complete measure space. By L 0 = L 0 (Ω) we denote the set of all (equvalence classes wth respect to the equalty µ-a.e. of) measurable extended-real valued functons on Ω. Let (E, E ) be a Banach lattce on (Ω, Σ, µ), that s E L 0 and f x y µ-a.e. on Ω, x L 0, y E then x E and x E y E. For any functon x L 0, the support of x s defned by supp(x) = {t Ω : x(t) 0}. An element x E s called order contnuous f for every 0 x n x such that x n 0 µ-a.e. t holds x n E 0. We say that a Banach lattce (E, E ) s strctly monotone f for all x, y E, y E < x E whenever 0 y x and x y. Date: May 11, Mathematcs Subject Classfcaton. 46B20, 46B04, 51M04. Key words and phrases. Stenhaus property, lattce ponts problem, Banach lattce. Some results from ths report are part of the second named author s Master Thess defended at the Tennessee Technologcal Unversty on Aprl 14,

3 2 DAMIAN KUBIAK AND DAVID TIDWELL 3. Stenhaus property n Banach spaces In ths secton we provde elaborated proofs of some results obtaned n [2]. Theorem 3.1 ([2]Th. 2). A Banach space (X, ) has the Stenhaus property f and only f for every x, y S(X) and for every δ > 0 there exsts z X such that z < δ and max{ x + z, y + z } > 1 > mn{ x + z, y + z }. Proof. Suppose that X has the Stenhaus property. Let x, y S(X), x y and δ (0, 1). Let A = {y, x, 2x, 3x,...}. Clearly A s quas-fnte. By the Stenhaus property there s u B((δ/4)x, δ/4) such that for some r > 0, B(u, r) contans exactly one element of A. Snce x u x (δ/4)x + (δ/4)x u < 1, wthout loss of generalty we can assume that x B(u, r), y / B(u, r) and r < 1. We have that u u (δ/4)x + (δ/4)x < δ/2. Hence by the trangle nequalty (1) 1 δ/2 < x u < r y u < 1 + δ/2. It follows that r = 1 ɛ for some ɛ (0, δ/2). Let ρ be any number satsfyng (2) 0 < ρ < r x u. From (1) we have that ρ < δ/2 ɛ. Defne w = y u, v = (ɛ + ρ)w/ w and z = (u + v). Clearly v = ɛ + ρ and by (1) w r. It follows that z = u + v ɛ + ρ + δ/2 < δ and Moreover by (2) y + z = y (u + v) = w(1 + (ɛ + ρ)/ w ) = w + ɛ + ρ r + ɛ + ρ > 1. x + z = x (u + v) x u + v < r ρ + ɛ + ρ = r + ɛ = 1. Hence max{ x + z, y + z } > 1 > mn{ x + z, y + z }. Now assume that for every x, y S(X) and for every δ > 0 there exsts z X such that z < δ and max{ x + z, y + z } > 1 > mn{ x + z, y + z }. Let A X be a quas-fnte set. Defne (3) G n = {x X : A B(x, r) = n for some r > 0}, n N {0}. Fx n N. Frst we show G n s open. Let x G n, that s there s r > 0 such that A B(x, r) = n. Let 0 < ɛ < mn{r x a : a A and x a < r}. It follows that A B(x, r) B(x, r ɛ). Now for y B(x, ɛ/2) and a A B(x, r), Moreover, for c A \ B(x, r), y a x a + y x < r ɛ + ɛ/2 = r ɛ/2. y c y x x c r ɛ/2. Hence A B(y, r ɛ/2) = n, that s y G n. Hence G n s open. Now we show that each G n s dence n X. Assume towards contradcton that there s n N such that G n s not dense n X. Hence there s an open ball B(x 0, r 0 ), x 0 X, r 0 > 0 such that (4) G n B(x 0, r 0 ) =. By defnton of G n, ether A B(x 0, r 0 ) > n or A B(x 0, r 0 ) < n. Snce A s quas-fnte by takng r 0 large enough, we assume, wthout loss of generalty, that A B(x 0, r 0 ) > n. Defne I 0 = A B(x 0, r 0 ), J 0 = A S(x 0, r 0 ) and L 0 = A \ (I 0 J 0 ). Clearly I 0 + J 0 > n. Let δ 0 = r 0. Suppose that for a postve nteger, for all j {1, 2,..., 1}, x j, δ j, I j, J j and L j are defned and satsfy I j + J j > n, (5) x j B(x 0, r 0 ), x j b < x j c for all b J j, c L j, and f I j then x j a < x j b for all a I j, b J j. Clearly, all the assumptons are satsfed f = 1.

4 STEINHAUS PROPERTY IN BANACH LATTICES 3 Let D = {m < n : A B(x 1, r) = m for some r > 0}, m = max D and r = max{r > 0 : A B(x 1, r) = m }. Note that r exsts because A s quas-fnte. Defne I = A B(x 1, r ), J = A S(x 1, r ) and L = A \ (I J ). Clearly I = m and I + J n. Let k = J. Snce A s quas-fnte we can defne a postve number (6) λ = mn{ x 1 c r : c L }/2. If I + J = n then A B(x 1, r + λ ) = n. Hence x 1 G n what by (5) contradcts (4) and the proof denseness of G n s fnshed. Snce J = 1 mples that I + J = n, by the above, we may assume that J > 1 and I + J > n. Assumng that mn = we defne (7) γ = mn{r x 1 a : a I }/2 and choose a postve number δ such that (8) δ < mn{δ 1, λ, γ }/(r 2 ). By J > 1 we choose two dstnct elements b (1), b (2) J. Snce (x 1 b)/r S(X) for every b J by the assumpton we get that there s z X wth z < δ such that (x 1 b (1) )/r +z < 1 and (x 1 b (2) )/r + z > 1. Defnng x = x 1 + r z we get that Moreover x b (1) < r 1 and x b (2) > r 1. x x 0 x 1 x 0 + r z r 1 z 1 + r 2 z r z < r 0 Hence x B(x 0, r 0 ). We have that for all b J and for every c L, 2 j < r 0. x b x 1 b + r z < x 1 b + λ x 1 b + ( x 1 c r )/2 = (r + x 1 c )/2 and for all c L, x c = x 1 c ( r z ) > x 1 c λ x 1 c ( x 1 c r )/2 = (r + x 1 c )/2. Furthermore, f I then for all a I,, x a x 1 a +r z < x 1 a +γ x 1 a +(r x 1 a )/2 = (r + x 1 a )/2 and for all a I and for every b J, x b = x 1 b ( r z ) > x 1 b γ x 1 b (r x 1 a )/2 = (r + x 1 a )/2. Now we see that all the assumptons whch held true for 1, hold true for. By (3.1) we get that J < J 1 f 2. Therefore we can repeat the process untl a contradcton wth non denseness of G n s found. Snce all the sets G n, n N are open and dense, by the Bare Category Theorem the set Y = n=1 G n s dense n X. It follows that for all y Y and n N there exsts an open ball B(y, r), r > 0 such that A B(y, r) = n. Hence X satsfes the Stenhaus property. By usng the above characterzaton we can prove the followng result. Theorem 3.2 ([2] Cor. 3). Every strctly convex Banach space satsfes the Stenhaus property. Proof. Suppose that X s strctly convex. Let x, y S(X), x y and δ > 0. Defne z = (δ/4)x (δ/4)y. Clearly z < δ. Recall that strct convexty of X gves αx + (1 α)y < 1 for all α (0, 1). Hence by takng α = 1 (δ/4), y + z < 1. By the trangle nequalty x + z 1. If x + z = 1, snce x x + z, strct convexty of X gves x + (1/2)z < 1. However x + (1/2)z = (1 + δ/8)x (δ/8)y 1. j=1

5 4 DAMIAN KUBIAK AND DAVID TIDWELL That s a contradcton. Hence x + z > 1. By Theorem 3.1 we get that X has the Stenhaus property. 4. Stenhaus property n some Banach lattces In ths secton we prove that a large class of Banach lattces over a non-atomc measure space satsfes the Stenhaus property. Lemma 4.1. A strctly monotone Banach lattce (E, E ) has the Stenhaus property f and only f for every x, y S(E), x, y 0, x y and for every δ > 0 there s z E wth z < δ such that max{ x + z E, y + z E } > 1 > mn{ x + z E, y + z E }. Proof. Let x, y S(E), x y and δ > 0. Let D = {t Ω : x(t) y(t)}. If sgn(x) sgn(y) then, wthout loss of generalty, there s a measurable set C D wth postve measure such that x(t) < 0, y(t) 0 and y(t) x(t) η for all t C and some η > 0. Let a measurable set B C wth µ(b) > 0 and β > 0 be such that β x(t) for t B. Let 0 < γ mn{β, η} be so small that z = (γ/2)χ B has z < δ. Note that by the choce of β, 0 z x, hence z E. Clearly x+z x, x+z x and y + z y, y + z y. Snce E s strctly monotone we get that x + z E < 1 and y + z E > 1. If sgn(x) = sgn(y) then x y. By the assumpton there exsts z E wth z < δ such that max{ x + z E, y + z E } > 1 > mn{ x + z E, y + z E }. By takng z = z sgn(x) the clam follows. Theorem 4.2. Let (E, E ) be a strctly monotone order contnuous Banach lattce on a non-atomc measure space. Then (E, E ) has the Stenhaus property. Proof. We apply Lemma 4.1. Let x, y S(E), x, y 0, x y and δ > 0. Wthout loss of generalty, there s a measurable set D wth postve measure and η > 0 such that y(t) x(t) η for t D. Snce E s order contnuous and the measure space s non-atomc there s a measurable set C D wth µ(c) > 0 such that 0 < (x + y)χ C E < δ. Let z = (x + y)χ C. We get that x + z x + ηχ C and y + z y ηχ C. Now by applyng the strct monotoncty of E the clam follows. Strct monotoncty of a Banach lattce over a non-atomc measure space s not suffcent for the Stenhaus property. An example whch shows that s the space (L 1 L )(Ω, Σ, µ) over a fnte measure space such that µ(ω) > 1 and equpped wth the norm f = f 1 + f. To see that t s enough to take parwse dsjont measurable sets A, B, C Ω wth postve measure such that (µ(a) + 1)(1 + η) < µ(b) + µ(c) for some small η > 0 and µ(b) = µ(c), and defne and f = µ(a) + 1 χ A µ(b) + µ(c) χ B C g = µ(a) + 1 χ A η 2 µ(b) + µ(c) χ B η 2 µ(b) + µ(c) χ C and take δ (0, m), where { ( m = mn 2 2 µ(a) ) 1 + η, 1 } 1 η. 2 µ(b) + µ(c) 2 µ(b) + µ(c) Observe that h < δ mples h δ µ-a.e. on Ω. Hence f + h 0, g + h 0, f + h = (f + h)χ A = (g + h)χ A = g + h and f + h 1 = g + h 1. An example of an order contnuous non strctly monotone Banach lattce over a non-atomc measure space s the space (L 1 + L )(Ω, Σ, µ) such that µ(ω) < equpped wth the norm f = nf{max{ g 1, h } : f = g + h, g L 1, h L }.

6 STEINHAUS PROPERTY IN BANACH LATTICES 5 In vew of Theorem 4.2 ths space may be consdered n order to determne whether the assumpton of order contnuty tself s also not suffcent for the Stenhaus property. Unfortunately, we were unable to establsh whether (L 1 + L ) fals the Stenhaus property or not. An example of a Banach lattce whch fals the Stenhaus property s L (Ω, Σ, µ) over a non-atomc measure space. Indeed, t s enough to consder f = χ A + (1/2)χ Ω\A and g = χ A + (1/4)χ Ω\A, where A s a measurable set wth postve measure such that µ(ω \ A) > 0. An example of a non strctly convex Banach space whch satsfes the Stenhaus property s L 1 (Ω, Σ, µ) over a non-atomc measure space. That s the case snce L 1 s a strctly monotone Banach lattce. Note that the sequence space l 1 fals the Stenhaus property. To see that t s enough to consder x = (1/2, 1/2, 0, 0,...) and y = (2/5, 3/5, 0, 0,...). References [1] R. Honsberger, Mathematcal gems from elementary combnatorcs, number theory, and geometry, The Mathematcal Assocaton of Amerca, Buffalo, N. Y., The Dolcan Mathematcal Expostons, No. 1. MR (54 #7150) [2] T. Kochanek, Stenhaus lattce-pont problem for banach spaces, ArXv Perprnt, arxv: [math.fa] (2013), 1 5. [3] W. Serpńsk, Sur quelques problèmes concernant les ponts aux coordonnées entères, Ensegnement Math. (2) 4 (1958), MR (20 #4518) [4] H. Stenhaus, One hundred problems n elementary mathematcs, Basc Books, Inc., Publshers, New York, Wth a foreword by Martn Gardner. MR (28 #1110) [5] P. Zwoleńsk, Some generalzaton of Stenhaus lattce ponts problem, Colloq. Math. 123 (2011), no. 1, MR (2012d:46062) Mathematcs Department, Tennessee Technologcal Unversty, 110 Unversty Drve, Box 5054, Cookevlle, TN E-mal address: dkubak@tntech.edu Mathematcs Department, Tennessee Technologcal Unversty, 110 Unversty Drve, Box 5054, Cookevlle, TN E-mal address: dtdwell42@students.tntech.edu