GODIXNIK NA SOFI SKI UNIVERSITET SV. KLIMENT OHRIDSKI FAKULTET PO MATEMATIKA I INFORMATIKA Tom 04 ANNUAL OF SOFIA UNIVERSITY ST. KLIMENT OHRIDSKI FACULTY OF MATHEMATICS AND INFORMATICS Volume 04 ON A DIFFERENTIAL INEQUALITY ROSSEN NIKOLOV We show that the existing methods for estimating the distance of a two-dimensional normed space with modulus of smoothness of power type to the Euclidean space generated by the John sphere of the former, are not exact. To this end, we construct a class of simple counterexamples in the plane. Keywords: Banach spaces geometry, moduli of convexity and smoothness, Banach Mazur distance. 000 Math. Subject Classification: 46B03, 46B0. and. INTRODUCTION The moduli of convexity and smoothness of a Banach space X: { } δ X ε := inf x+y : x = y =, x y = ε, 0 ε, { } x+τy + x τy ρ X τ := sup : x = y =, τ 0, respectively, are fundamental concepts in Banach space theory. The duality between them is given by Lindenstrauss formula, see e.g. [6, p. 6] { τε } ρ X τ = sup δ Xε : 0 ε. Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 39
According to the Nordlander Theorem [7], a Hilbert space H is in a sense the most convex and the most smooth among Banach spaces, that is, for any Banach space X δ X ε δ H ε = ε /4 = ε /8+Oε 4 and ρ X τ ρ H τ = +τ = τ /+Oτ 4. The Taylor expansion is written down not only for sake of greater clarity, but also because the asymptotic behaviors at 0 play an important role. For technical reasons, further we concentrate on the asymptotic behavior at 0 of the modulus of smoothness. The results concerning the modulus of convexity can be derived through the Lindenstrauss formula. Let a 0 and let X a be the class of all Banach spaces X such that ρ X τ = +a τ +oτ. Several authors independently showed that X 0 contains only Hilbert spaces [5, 9, 8]. So, it stands to reason that for small a > 0 the spaces in X a might be close to a Hilbert space in some sense. This is indeed so and the sense is made precise below. Recall that the Banach-Mazur distance between two isomorphic Banach spaces Y and Z is given by dy,z = inf{ T. T : T : Y Z arbitrary isomorphism}. Now, for a Banach space X one defines d X := inf{dy,l : Y X, dimy = }, where l denotes the two-dimensional Hilbert space, or, in other words, the Euclidean plane. Obviously, d X measures how far from an ellipse the two-dimensional sections of the sphere of X are. The famous Jordan-von Neumann Theorem [3] reads d X = if and only if X is a Hilbert space. The measure d X is very useful for estimating the type and cotype of X. The standard way of estimating d X is through the use of the John Sphere, [4]. In the pioneering work [4] John showed that d X for any X. Elaborating on this idea, let Y be a two-dimensional space. It is clear that there is an ellipse, say E, of maximal volume contained in the unit ball of Y. Define jy := max x E x. 40 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
Then, of course, Let for a 0 Then d X sup{jy : Y X, dimy = }. ga := sup{jy : Y X a, dimy = }. d X ga, a 0, X X a. From the above considerations we know that g0 =. Rakov [8] estimated jy for Y X a, his estimate as a 0 reads ga +k a with some constant k which is not important here. An asymptotically sharp estimate was given in [, ]: ga + a ++ a. From the method of [, ] it is not clear if is exact. In this work we show that it is not. We will explain briefly the method of [, ]. Let Y be a two-dimensional space in X a for some a > 0. We may assume that Y is realized in such a way in R that the unit circle x +x = of R is the John sphere of X. Denote the standard basis of R by e =,0 and e = 0,. Define rσ := e cosσ +e sinσ. Then in this polar annotation the unit sphere S Y of Y is { } S Y = cosσ,sinσ : σ [ π,π]. rσ If x,y S Y, x = r θe cosθ + e sinθ, y = r ϕe cosϕ + e sinϕ, then Lemma 3. from [] states that if r θ exists, then x+τy + x τy lim τ 0 τ Since Y X a we have that for almost all θ [0,π] sup ϕ [0,π] = sin θ ϕ r rθrθ+r θ. ϕ sin θ ϕ r rθrθ+r θ +a. 3 ϕ From John Theorem it follows that on each arc of the unit circle of length π/ there is a contact point, that is such that r =. Therefore, without loss of generality, there is α 0,π/] such that r0 = rα =, 4 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 4
and jy = max θ [0,α] rθ. 5 So, we may consider the system 3, 4 and try to estimate jy. However, with ϕ in 3 the problem is non-local and probably very difficult. In order to handle it [, ] substitute ϕ = θ+π/ and use rϕ to derive from 3 rθrθ+r θ +a for almost all θ [0,π]. 6 Then the system 6, 4 is used to estimate jy through 5. We will demonstrate that for a close to zero the outlined approach can never produce the exact value of supjy for Y X a, denoted above as ga. For sake of clarity, for a 0 denote by G a the class of all π-periodic functions r = rθ, 0 θ π, such that i 0 < rθ, r0 = rπ/ = ; ii r θ is absolutely continuous and 0 rθrθ + r θ + a almost everywhere; iii the region B r inside the curve { } S r = cosθ,sinθ : σ [ π,π] rθ is convex. It is easy to check that iii follows from r+r 0, which is contained in ii, but we do not need this fact. Finally we introduce the class F a of all π-periodic functions, which satisfy i, ii, iii and additionally iv sup ϕ,θ sin ϕ θ r rθ rθ+r θ = +a. ϕ It is clear that F a G a. Theorem.. There exist an interval I and a class X a R,. a Xa, a I of Banach spaces, such that for all a I there are b = ba > a which satisfy: i r b G a, r b F b, r b / F a ; ii max σ r a σ = d X a < max σ r b σ = d X b, 4 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
where r a σ = cosσe +sinσe a, r b σ = cosσe +sinσe b.. CONSTRUCTION OF A CLASS OF TWO-DIMENSIONAL SPACES Pick λ [0,] and set µ = λ, ν = µ λ = λ 4λ+. For θ,θ = ± we denote with D θ,θ the Euclidean disk of radius λ centered at O θ,θ = θ µ,θ µ, i.e. D θ,θ = { x = x,x R : x θ µ +x θ µ λ }. Also let C θ,θ = { x = x,x R : x θ µ +x θ µ = λ }. and D = D, D, D, D,, B λ = convd We can say that B λ is a rotund square see Figure. Clearly B is the unit Euclidean disk. Figure : A rotund square Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 43
It is easy to see that B 0 = Q where Q is the unit square, i.e. Q = { x = x,x R : x, x }. We set Y λ = R,. λ, where. λ is the Minkowski functional of B λ, i.e. { x λ = inf t > 0 : x } t B λ. Since B λ is symmetric with respect to the coordinate system, the line x = x and the origin O0,0, we have that x,x, θ x,θ x B λ, θ,θ = ± provided x,x B λ. So x e +x e λ = θ x e +θ x e λ, where e,e is the unit vector basis in R, e =,0, e = 0,. This implies monotonicity of the basis, i.e. x e +x e λ y e +y e λ whenever x y, x y. Let us mention that d at the origin. Thus d Y λ = d Y λ,l Y λ,l is the radius of the circumcircle of B λ centered = R = µ+λ = + λ. In order to find the asymptotic behavior of ρ Yλ τ at O, we need an explicit formula for the norm of Y λ. Fix λ 0,]. Further we omit the index λ, i.e. we write Y instead of Y λ,. instead of. λ, B and S stand for the unit ball and the unit sphere of Y λ. Let x = ρcosσ,ρsinσ be the representation of x R in polar coordinates. We set σ = argx. Having in mind the symmetry of B, it is enough to find anexplicit formula for x,x = x,x, when 0 x x, i.e. π 4 argx π. Denote by zµ, the unique common point of the circle C, = { x,x R : x µ +x µ = λ } and the straight line l = { x,x R : x = }. Set γ = argz. Obviously π 4 < γ π and x = x for all points x with argx [γ, π ]. If x [π γ,γ], then the vector x/ x belongs to the circle C,. Setting fx,x = x = x,x, we obtain x f µ x + f µ = λ. Calculating the roots of the above equation we get : ν µx +x λ x +x µ x x, λ fx,x = µ x +x xx x +x, λ =. 44 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
So : x if 0 argx π γ x = fx,x if π γ argx γ x if γ argx π. We mention here that the function f is defined not only on the sector { x R : π γ argx γ}. Actually f is defined on the set E = { x R : λ x +x µ x x } { x R : π } γ argx γ. It is easy to see that E { x R : 0 argx } π for λ /, while { E = x R : k x } k, x where k < k are the roots of µ λ k µ k + µ λ = 0 for 0 < λ < /. Fact.. For all x E we have: i fx x for x E { x : 0 argx π ii If xx,x S { x : π γ argx γ}, then f x,x = λ x gx,x ; f x,x = λ x x gx,x ; f x,x = λ x gx,x, where gx,x = µx +x ν 3. Proof. We prove only i. Since fx,x is homogeneous, i.e. fkx,kx = kfx,x for all k > 0, it suffices to check i only for x S. If x { u R π : γ argu γ}, then fx = = x. If x E { u R : γ argu } π, then x =. From x f, x f C, it follows f <, i.e. fx = fx,x > = x. } Set x,y,τ = x+τy + x τy x. Evidently, ρ Y τ = sup{ x,y,τ : x,y S}. Due to the symmetry of S: ρ Y τ = sup { x,y,τ : x,y S, argx [ π 4, π } ]. Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 45
Fact.. Let x,y S, argx [γ, π ], τ µ. Then x,y,τ z,y,τ. Proof. Since y =, we get y. So τy µ. Since argx [γ, π ], we have 0 x µ and x =, x ±τy µ±τy. The monotonicity of the basis implies: x±τy = x ±τy e +±τy e µ±τy e +±τy e = z ±τy. Corollary.3. If τ µ, then ρ Y τ = sup{ x,y,τ : x A, y S}, where A is the arc { x S, π 4 argx γ}. Proposition.4. The following estimate holds: { ρ Y τ lim τ 0 τ λ sup x } x y y gx,x : x A, y S. Proof. Pick a convex compact set F E such that its interior contains the arc A. Choose τ F 0, µ in such a way that x±τy F whenever x A, y S, τ τ F. Set fx,y,τ = fx+τy+fx τy fx. From Fact.i we have fx±τy x±τy, x A, y S, τ τ F. Since fx = x for x A we get x,y,τ fx,y,τ whenever x A, y S, τ τ F. Using that, we get for τ 0,τ F ] ρ Y τ sup{ fx,y,τ : x A, y S}. Take x A, y S. Applying Taylor s formula to ϕτ = fx+τy fx and ψτ = fx τy fx, we can find θ = θ x,y,τ, θ = θ x,y,τ 0, in such a way that fx,y,τ τ = { f 4 x+θ τyy +f x+θ τyy y +f x+θ τyy } + { f 4 x+θ τyy +f x+θ τyy y +f x+θ τyy}. 46 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
Having in mind that the second derivatives are uniformly continuous on F, we get : ρ Y τ lim τ 0 τ { f xy +f xy y +f xy : x A, y S } To finish the proof, it is enough to use Fact. ii. Lemma.5. Let x = x,x A. Then sup x x y y = µx +x +λ x +x. y,y S Proof. The determinant represents the oriented area of the parallelogram, defined by the vectors x and y. Therefore, for a fixed x, the left-hand side achieves its greatest value when the distance from the point y,y to the support of the vector x is maximal. This is satisfied for some ȳ C, S or ȳ C, S. Without loss of generality we assume that ȳ C, : s µ + t + µ = λ. The tangent to C, at the point ȳ,ȳ is parallel to the support of x, i.e. the normal to C, is orthogonal to x. Therefore the scalar product < v,x >= 0, where v = ȳ µ,ȳ +µ. We get for ȳ,ȳ the system: x ȳ µ+x ȳ +µ = 0 ȳ µ +ȳ +µ = λ with solution: ȳ = µ+ λx x +x x +x ȳ = µ λx. Hence, x x ȳ ȳ = x ȳ x ȳ = µx +x λ x +x. where Proposition.6. Let 0 < λ <. Then hλ = λ ρ Y τ lim τ 0 τ hλ, λ 3λ++λ λ λ+. Proof. From Proposition.4 and Lemma.5 it follows ρ Y τ lim τ 0 τ λ µx +x +λ x sup +x : x,x A gx,x, Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 47
where gx,x = µx +x ν 3 is defined in Fact.. For brevity, we denote µx +x +λ x +x Mx,x =. gx,x On the arc A we have : i.e. x +x µx +x +µ = λ, x +x = µx +x µ λ = µx +x ν. Set x +x = t, then µx +x = t +ν, and after substituting we get Mx,x = We need to examine the function t +ν +λt t +ν ν 3 = mt = t +λt+ν t ν 3. t +λt+ν t ν 3. By the cosine formula we get +µ t µ+λ = + λ. The left-hand side expression represents the distance from the origin O0, 0 to zµ,, while the right-hand side expression is the distance to the middle of arc A. From t +µ = λ λ+ > λ 4λ+ = ν it is clear that mt is defined in this interval. Calculating m and simplifying, we obtain: m t = t +λt+ν t 3 +4λt +5νt+λν t ν 4. We now show that m < 0 for 0 < λ <, i.e. m is decreasing in the [ interval I = +µ, + λ]. Obviously I [, ]. The quadratic polynomial ut = t +λt+ν is increasing in [ λ, ], whence ut > u = +λ+λ 4λ+ = λ λ+3 > 0. Obviously ν = λ 4λ+ > 0. It follows that the coefficients of vt = t 3 +4λt + 5νt+λν are positive, which implies vt > 0 when t I. Finally, in order to find the greatest value of M we use : +µ +ν = + λ + λ λ = λ 6λ+4, +µ ν = + λ λ +λ = λ, 48 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
i.e. +µ Mµ, = m +λ +µ +ν +µ = +µ ν 3 = λ 3 λ 3λ++λ λ λ+. The above and the remark at the beginning complete the proof. Theorem.7. For 0 < λ < we have ρ Y τ lim τ 0 τ = hλ. Proof. According to Proposition.6, the function m is continuous and decreasing in the interval +µ, + λ] [. Let ǫ 0,m +µ m + λ. There exists such that Whence z ǫ x,x A, x = x ǫ,x = x ǫ, m x +x = m +µ ǫ. Mz ǫ = Mx,x = m +µ ǫ. Choose τ ǫ > 0, such that {p,q : max p x, q x τ ǫ } {u R : π γ argu γ }. If τ < τ ǫ, then z ǫ,y,τ = fz ǫ,y,τ for all y S. By Lemma.5, similarly as in Proposition.4 we get : lim τ 0 ρ Y τ τ = λ Mx,x = λ z ǫ,y,τ lim sup τ 0 y S τ m +µ which combined with Proposition.6 concludes the proof. fz ǫ,y,τ = lim sup τ 0 y S τ ǫ = hλ λ ǫ, Remark.8. Let us point out that for arbitrary small τ, { } x+τy + x τy ρ Y τ = sup, x A, y S Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 49
is not attained at the point zµ,. Indeed, for τ 0,τ ǫ there holds either π γ argz +τy γ, γ argz τy π or π γ argz τy γ, γ argz +τy π. Similarly as in Proposition.4, we have z,y,τ = τ 4 [ f z +θτyy +f z +θτyy y +f z +θτyy ], where θ = θy,τ 0,. Thus { } z +τy + z τy lim sup τ 0 τ, y S = 4 hλ. This is because r σ does not exists at σ = γ rσ is defined in the Introduction. We start by establishing Fact 3.. The function is decreasing in 0,]. 3. PROOF OF THE MAIN THEOREM hλ = λ Proof. It is sufficient to check that is decreasing. The derivative is negative if λ 3λ++λ λ λ+ hλ = λ 3λ++λ λ λ+ h λ = λ 3 λ λ++λ 3λ+ λ λ+ λ 3 λ λ++λ 3λ+ < 0, λ 0,. The latter is equivalent to the inequality λ λ+ [ λ λ++λ 3 ] +λλ < 0, λ 0,, which is true because both summands are negative for λ 0,. 50 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
Lemma 3.. Let 0 < λ <,. λ correspond to the space Y λ and r λ θ = cosθe +sinθe λ be the function which describes the sphere of rotund square. Then sλ = supr λ θr λ θ+r λθ θ λ λ+ +ay λ < +ay λ, where we have set ρ Yλ τ +ay λ = lim ρ 0 τ = hλ. Above, we assumed that θ γ = argz. Also, θ does not correspond to any other common point of the circle and the straight line, because for such points r θ does not exist. Proof. Ifx= r λ θ cosθ,sinθbelongstoasegmentofs λ, thenr λ θ+r λ θ=0. Let x A see Corollary.3 and x zµ,. From sin θ ϕ sup θ,ϕ rλ ϕ r λ θr λ θ+r λθ = +ay λ, by substituting ϕ = θ π we get r λ ϕr λθr λ θ+r λθ +ay λ. Hence, r λ θr λ θ+r λθ rλϕ+ay λ < λ λ+ +ay λ. Above we have used the inequality r λ ϕ < z, where z = +µ = λ λ+. Proof of Theorem. At the beginning we note that d Y λ = + λ is a decreasing function of λ. From Lemma 3., sλ = supr λ θr λ θ+r λθ θ λ λ+ hλ = λλ λ+ h λ. We denote the right-hand side with kλ. As λλ λ+ kλ decreases in this interval too, due to Fact 3.. is decreasing in 0,, Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 5
Thus for all λ 0, we have sλ kλ < hλ 7 and lim λ 0 +kλ = lim +hλ =. λ 0 Let a I = h,. There exists a unique λ = λa <, such that a = hλ = hλa, i.e. λa is the inverse function of a = hλ, considered in the interval 0,. We define X a = Y λa, which means X a = R,. a, where. a =. λa. Respectively let r a σ = cosσe +sinσe a = cosσe +sinσe λa = r λ σ. By definition it is clear that X a X a for all a I. Let a I is fixed and λ = λa is as above. From 7 it follows that there exists a unique λ : 0 < λ < λ, for which a = hλ = kλ. Let b = hλ, i.e. λ = λb. Obviously b > a. For r λ σ we have: r λ σ r λ σ+r λ σ sλ kλ hλ = +a = +ay λ. But this is equivalent to r b σ = r λ σ G a. Also it is clear that r b σ F b whence r b σ / F a. From the note in the beginning max σ r a σ = d X a < d X b = max σ r b σ. In the wording of the theorem we write r a and. a, instead of r a and. a. 4. REFERENCES [] Ivanov, M., Troyanski, S.: Uniformly smooth renorming of Banach spaces with modulus of convexity of power type. J. Funct. Anal., 37, 006, 373 390. [] Ivanov, M., Parales, A.J., Troyanski, S.: On the geometry of Banach spaces with modulus of convexity of power type. Studia Mathematica, 97, no., 00, 8 9. [3] Jordan, P., von Neumann, J.: On inner products in linear, metric spaces. Ann. Math., 36, 935, 79 73. [4] John, F.: Extremum problems with inequalities as subsidary conditions. In: Studies and Essays Presented to R. Courant, Interscience, 948, 87 04. [5] Kirchev, K., Troyanski, S.: On some characterisations of spaces with scalar product. C. R. Acad. Bulgare Sci., 8, 975, 445 447. [6] Lindenstrauss, J., Tzafriri, L.: Classical Banach Spaces. II: Function Spaces, Springer, 979. 5 Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53.
[7] Nordlander, G.: The modulus of convexity in normed linear spaces. Ark. Mat., 4, 960, 5 7. [8] Rakov, S.: Uniformly smooth renormings of uniformly convex Banach spaces. J. Soviet Math., 3, 985, 73 7. [9] Senechalle, D.: Euclidean and non-euclidean norms in a plane. Illinois J. Math., 5, 97, 8 89. Received on March 0, 07 Rosen Nikolov Faculty of Mathematics and Informatics St. Kl. Ohridski University of Sofia 5, J. Bourchier blvd., BG-64 Sofia BULGARIA E-mail: rumpo959@abv.bg Ann. Sofia Univ., Fac. Math and Inf., 04, 07, 39 53. 53