L’Enseignement Mathématique (2) 64 (2018), 327–351 DOI 10.4171/LEM/64-3/4-6
An elementary and uni ed proof of Grothendieck’s inequality
Shmuel Friedland, Lek-Heng Lim and Jinjie Zhang
Abstract. We present an elementary, self-contained proof of Grothendieck’s inequality that uni es the real and complex cases and yields both the Krivine and Haagerup bounds, the current best-known explicit bounds for the real and complex Grothendieck constants respectively. is article is intended to be pedagogical, combining and streamlining known ideas of Lindenstrauss–Pełczyński, Krivine, and Haagerup into a proof that need only univariate calculus, basic complex variables, and a modicum of linear algebra as prerequisites.
Mathematics Subject Classi cation (2010). Primary: 47A07; Secondary: 46B28, 46B85
Keywords. Grothendieck inequality, Grothendieck constant, Krivine bound, Haagerup bound.
1. Introduction
We will let F R or C throughout this article. In 1953, Grothendieck proved D a powerful result that he called “the fundamental theorem in the metric theory of tensor products” [Gro]; he showed that there exists a nite constant K > 0 m n such that for every l; m; n N and every matrix M .Mij / F , 2 D 2 ˇ m n ˇ ˇ m n ˇ ˇX X ˇ ˇX X ˇ (1) max ˇ Mij xi ; yj ˇ K max ˇ Mij "i ıj ˇ xi yj 1ˇ h iˇ Ä "i ıj 1ˇ ˇ k kDk kD i 1 j 1 j jDj jD i 1 j 1 D D D D where ; is the standard inner product in F l , the maximum on the left is h i l taken over all xi ; yj F of unit 2-norm, and the maximum on the right is 2 taken over all "i ; ıj F of unit absolute value (i.e., "i 1, ıj 1 over i 2i D ˙ D ˙ R; "i e i , ıj e j over C ). e inequality (1) has since been christened D D Grothendieck’s inequality and the smallest possible constant K Grothendieck’s 328 S. Friedland, L.-H. Lim and J. Zhang constant. e value of Grothendieck’s constant depends on the choice of F and F we will denote it by KG . Over the last 65 years, there have been many attempts to improve and simplify the proof of Grothendieck’s inequality, and also to obtain better bounds for the F Grothendieck constant KG , whose exact value remains unknown. e following are some major milestones: (i) e central result of Grothendieck’s original paper [Gro] is that his eponymous inequality holds with =2 KR sinh.=2/ 2:301 and Ä G Ä 1:273 4= KC . Grothendieck relied on the sign function for the real Ä G case and obtained the complex case from the real case via a complexi cation argument. (ii) e power of Grothendieck’s inequality was not generally recognized until the work of Lindenstrauss and Pełczyński [LP] 15 years later, which connected the inequality to absolutely p -summing operators. ey elucidated and improved Grothendieck’s proof in the real case by computing expectations of sign functions and using Taylor expansions, although they did not get better R bounds for KG . (iii) Rietz [Rie] obtained a slightly smaller bound KR 2:261 in 1974 by G Ä averaging over Rn with normalized Gaussian measure and using a variational argument to determine an optimal scalar map corresponding to the sign function. R C (iv) Our current best known upper bounds for KG and KG are due to Krivine [Kri], who in 1979 used Banach space theory and ideas in [LP] to get
R KG 1:78221 Ä 2 log.1 p2/ I C and Haagerup [Haa], who in 1987 extended Krivine’s ideas to C to get
C 8 KG 1:40491; Ä .x0 1/ C where x0 Œ0; 1 is the unique solution to: 2 Z =2 cos2 t x dt .x 1/: 0 p1 x2 sin2 t D 8 C R C (v) Our current best known lower bounds for KG and KG are due to Davie [Dav1, Dav2], who in 1984 used spherical integrals to get
R 1 .x/ KG sup 1:67696; x .0;1/ max .x/; f .x/ 2 An elementary and uni ed proof of Grothendieck’s inequality 329
where r Ä r Z 2 x2=2 2 x2 8 1 t2=2 .x/ xe ; f .x/ e .x/ 1 e dt WD WD C x I and C 1 .x/ KG sup 1:33807; x>0 g.x/ where Ä Z 1 x2 1 t2 .x/ 1 e x e dt ; WD 2 C x Ä Z 2 Ä 1 x2 1 t2 2 x2 g.x/ .1 e / e dt .x/ 1 .1 e / : WD x C x C x (vi) Progress on improving the aforementioned bounds halted for many years. R Believing that Krivine’s bound is the exact value of KG , some were spurred R to nd matrices that yield it as the lower bound of KG [Kon1]. e belief was dispelled in 2011 in a landmark paper [Bra], which demonstrated the existence of a positive constant " such that KR < = 2 log.1 p2/ " but G C the authors did not provide an explicit better bound. To date, Krivine’s and R Haagerup’s bounds remain the best known explicit upper bounds for KG C and KG respectively. (vii) ere have also been many alternate proofs of Grothendieck’s inequality employing a variety of techniques, among them factorization of Hilbert spaces [Mau, Jam, Pis1], absolutely summing operators [DJT,LP, Pis2], geometry of Banach spaces [AK, LT], metric theory of tensor product [DFS], basic probability theory [Ble], bilinear forms on C -algebra [Kai]. In this article, we will present a proof of Grothendieck’s inequality that uni es both the (a) real and (b) complex cases; and yields both the (c) Krivine and (d) Haagerup bounds [Kri, Haa]. It is also elementary in that it requires little more than standard college mathematics. Our proof will rely on Lemma 2.1, which is a variation of known ideas in [LP, Haa, Jam]. In particular, the idea of using the sign function to establish (1) in the real case was due to Grothendieck himself [Gro] and later also appeared in [LP, Kri]; whereas the use of the sign function in the complex case rst appeared in [Haa]. To be clear, all the key ideas in our proof were originally due to Lindenstrauss–Pełczyński, Krivine, Haagerup, and König [LP, Kri, Haa, Kon2], our only contribution is pedagogical – combining, simplifying, and streamlining their ideas into what we feel is a more palatable proof. To understand the proof, readers need only know univariate calculus, basic complex variables, and a small amount of linear algebra. We will use some basic Hilbert space theory and tensor product constructions in Section4 but both notions will be explained in a self-contained and elementary way. 330 S. Friedland, L.-H. Lim and J. Zhang
2. Gaussian integral of sign function
roughout this article, our inner product over C will be sesquilinear in the second argument, i.e.,
n x; y yx for all x; y C : h i WD 2 For z F R or C , the sign function is 2 D ´ z= z if z 0; (2) sgn.z/ j j ¤ D 0 if z 0 D I and for z F n , the Gaussian function is 2 ´ n=2 2 F .2/ exp z 2=2 if F R; Gn .z/ k k D D n exp. z 2/ if F C: k k2 D Lemma 2.1 below is based on [Jam, Haa]; the complex version in particular is a slight variation of [Haa, Lemma 3.2]. It plays an important role in our proof because the right side of (3) depends only on the inner product u; v h i and not (explicitly) on the dimension n. In addition, the functions on the right are homeomorphisms and admit Taylor expansions, making it possible to expand them in powers u; v d , which will come in useful when we prove eorem 4.1. h i
n Lemma 2.1. Let u; v F with u 2 v 2 1. en 2 k k D k k D Z F (3) sgn u; z sgn z; v Gn .z/ dz Fn h i h i 8 2 ˆ arcsin u; v if F R; < h i D Z =2 cos2 t D ˆ u; v dt if F C: ˆ 2 2 1=2 :h i 0 .1 u; v sin t/ D jh ij Proof. Case I: F R. Let arccos u; v , so that Œ0; and D h i D 2 arcsin u; v =2 . Choose ˛; ˇ such that 0 < ˇ ˛ < and de ne h i D ® ¯ E.˛; ˇ/ .r cos ; r sin ; x3; : : : ; xn/ r 0; ˛ ˇ : D W Ä Ä R e Gaussian measure of a measurable set A is the integral of Gn .x/ over A. Upon integrating with respect to x3; : : : ; xn , the following term remains:
Z Z ˇ Z 1 1 .x2 x2/ 1 1 1 r2 e 2 1 C 2 dx1 dx2 d re 2 dr .ˇ ˛/=2: 2 E.˛;ˇ/ D 2 ˛ 0 D An elementary and uni ed proof of Grothendieck’s inequality 331
Hence the Gaussian measure of E.˛; ˇ/ is .ˇ ˛/=2 . Since there is an isometry n T of R such that T u e1 and T v .cos ; sin ; 0; : : : ; 0/, the left side of D D (3) may be expressed as Z R sgn T u; x sgn x; T v Gn .x/ dx: Rn h i h i e set of x where T u; x > 0 and T v; x > 0 is E. =2; =2/, which h i h i has Gaussian measure . /=2 ; ditto for T u; x < 0 and T v; x < 0. e h i h i set of x where T u; x < 0 and T v; x > 0 is E.=2; =2/, which has h i h i C Gaussian measure =2 ; ditto for T u; x > 0 and T v; x < 0. e set of x h i h i where T u; x 0 has zero Gaussian measure. Hence the value of this integral h i D is . /=2 . /=2 =2 =2 2 arcsin u; v = . C D h i 2n Case II: F C . We de ne vectors ˛; ˇ R with ˛2i 1 Re.ui /, D 2 D ˛2i Im.ui /, ˇ2i 1 Re.vi /, ˇ2i Im.vi /, i 1; : : : ; n. en ˛ and ˇ D 2nD D Dn are unit vectors in R . For any z .z1; : : : ; zn/ C , we write D 2 2n x Re.z1/; Im.z1/; : : : ; Re.zn/; Im.zn/ R : D 2 en, n n X X Re u; z Re.ui zi / Re.ui / Re.zi / Im.ui / Im.zi / h i D D C i 1 i 1 D D ˛; x x; ˛ ; D h i D h i and likewise Re. z; v / x; ˇ . By a change-of-variables and Case I, we have h i D h i Z Z C R sgn Re u; z sgn Re z; v Gn .z/ dz sgn x; ˛ sgn x; ˇ G2n.x/ dx Cn h i h i D R2n h i h i 2 2 (4) arcsin ˛; ˇ arcsin Re u; v : D h i D h i It is easy to verify that for any z C , 2 Z 2 1 i i (5) sgn.z/ sgn Re.e z/ e d: D 4 0 By (4), (5), and Fubini’s theorem, Z C sgn u; z sgn z; v Gn .z/ dz Cn h i h i Z 2 Z 2 Z 1 i Á sgn Re e u; z D 16 0 0 Cn h i i' Á i. '/ C sgn Re z; e v e C G .z/ dz d d' h i n Z 2 Z 2 1 i i' Á i. '/ arcsin Re e u; e v e C d d' D 8 0 0 h i 332 S. Friedland, L.-H. Lim and J. Zhang
Case II(a): u; v R. e integral above becomes h i 2 Z 2 ÄZ 2 1 i. '/ arcsin cos. '/ u; v e C d d' D 8 0 0 C h i 1 Z 2 ÄZ 2 ' C arcsin u; v cos teit dt d' D 8 0 ' h i 1 Z 2 ÄZ 2 arcsin u; v cos teit dt d' D 8 0 0 h i 1 Z 2 (6) arcsin u; v cos teit dt: D 4 0 h i
Since arcsin u; v cos t is an even function with period 2 , h i Z 2 arcsin u; v cos t sin t dt 0; 0 h i D the last integral in (6) becomes
1 Z 2 arcsin u; v cos t cos t dt; 4 0 h i and as arcsin u; v cos t cos t is an even function with period , it becomes h i Z =2 Z =2 arcsin u; v cos t cos t dt arcsin u; v sin t sin t dt; 0 h i D 0 h i which, upon integrating by parts, becomes
Z =2 cos2 t (7) u; v dt: 2 1=2 h i 0 1 u; v 2 sin t jh ij
Case II(b): u; v R. is reduces to Case II(a) by setting c C of unit h i … 2 modulus so that c u; v u; v and cu; v R, then by (7), h i D jh ij h i 2 Z Z C C sgn u; z sgn z; v Gn .z/ dz c sgn cu; z sgn z; v Gn .z/ dz Cn h i h i D Cn h i h i Z =2 cos2 t c cu; v dt 2 1=2 D h i 0 1 cu; v 2 sin t jh ij Z =2 cos2 t u; v dt: 2 1=2 D h i 0 1 u; v 2 sin t jh ij An elementary and uni ed proof of Grothendieck’s inequality 333
We will make a simple but useful observation1 about the quantities in (1) that we will need for the proof of Corollary 2.3 later.
Lemma 2.2. Let F R or C and d; m; n N . For any M F m n , we have D 2 2 ˇ m n ˇ ˇ m n ˇ ˇX X ˇ ˇX X ˇ (8) max ˇ Mij "i ıj ˇ max ˇ Mij "i ıj ˇ "i 1; ıj 1ˇ ˇ D "i ıj 1ˇ ˇ j jÄ j jÄ i 1 j 1 j jDj jD i 1 j 1 D D D D d and for any x1; : : : ; xm; y1; : : : ; yn F , 2 ˇ m n ˇ ˇ m n ˇ ˇX X ˇ ˇX X ˇ (9) max ˇ Mij xi ; yj ˇ max ˇ Mij xi ; yj ˇ: xi 1; yj 1ˇ h iˇ D xi yj 1ˇ h iˇ k kÄ k kÄ i 1 j 1 k kDk kD i 1 j 1 D D D D Proof. We will start with (8). Suppose there exists M F m n such that the left- 2 hand side of (8) exceeds the right-hand side. Let the maximum of the left-hand side be attained by "1;:::;"m and ı1; : : : ; ın . By our assumption, at least one " or ı must be less than 1 in absolute value and so let " < 1 without loss i j j 1j of generality. Fix "i " , i 2; : : : ; m and ıj ı , j 1; : : : ; n, but let "1 D i D D j D vary with "1 1 and consider the maximum of the left hand-side over "1 . j j Ä Since max a"1 b "1 1 is always attained on the boundary "1 1 for ¹j C j W j j Ä º j j D any a; b F , this contradicts our assumption. e proof for (9) is similar with 2 norm in place of absolute value. In the corollary below, the inequality on the left is the “original Grothendieck inequality”, i.e., as rst stated by Grothendieck2 in [Gro], and the inequality on the right is due to Haagerup [Haa].
Corollary 2.3. Let F R or C and d; m; n N . For any M F m n with D 2 2 ˇ m n ˇ ˇX X ˇ (10) max ˇ Mij "i ıj ˇ 1; "i ıj 1ˇ ˇ Ä j jDj jD i 1 j 1 D D d any x1; : : : ; xm; y1; : : : ; yn F of unit 2-norm, we have 2 ˇ m n ˇ ˇX X ˇ ˇ Mij arcsin xi ; yj ˇ if F R; ˇ h iˇ Ä 2 D i 1 j 1 D D ˇ m n ˇ ˇX X ˇ ˇ Mij H xi ; yj ˇ 1 if F C; ˇ h i ˇ Ä D i 1 j 1 D D where H denotes the function on the right side of (3) for F C . D
1 is of course follows from other well-known results but we would like to keep our exposition self-contained. 2 e better known modern version (1) is in fact due to Lindenstrauss and Pełczyński in [LP]. 334 S. Friedland, L.-H. Lim and J. Zhang
Proof. e condition (10) implies that ˇ m n ˇ ˇX X R ˇ R ˇ Mij sgn xi ; x sgn yj ; x G .z/ˇ G .z/; ˇ h i h i d ˇ Ä d i 1 j 1 D D ˇ m n ˇ ˇX X C ˇ C ˇ Mij sgn z; xi sgn z; y G .z/ˇ G .z/; ˇ h i h j i d ˇ Ä d i 1 j 1 D D for any x Rd , z Cd respectively. Integrating over Rd or Cd respectively 2 2 and applying Lemma 2.1 give the required results. Note that we have implicitly relied on (8) in Lemma 2.2 as the sgn function is not always of absolute value one and may be zero. Corollary 2.3 already looks a lot like the Grothendieck inequality (1) but the nonlinear functions arcsin and H are in the way. To obtain the Grothendieck inequality, we linearize them: First by using Taylor series to replace these functions by polynomials; and then using a ‘tensor trick’ to express the polynomials as linear functions on a larger space. is is the gist of the proof in Section4.
3. Haagerup function
We will need to make a few observations regarding the functions on the right side of (3) for the proof of Grothendieck’s inequality. Let the complex Haagerup function of a complex variable z be Z =2 cos2 t H.z/ z dt; z 1; 2 1=2 WD 0 1 z 2 sin t j j Ä j j and the real Haagerup function h as the restriction of H to Œ 1; 1 R. Observe  that h Œ 1; 1 Œ 1; 1 and is a strictly increasing continuous bijection. Since W ! Œ 1; 1 is compact, h is a homeomorphism of Œ 1; 1 onto itself. By the Taylor expansion
2 2 1=2 X1 .2k 1/ŠŠ 2k 2k .1 x sin t/ x sin t; x 1; 0 t < =2; D .2k/ŠŠ j j Ä Ä k 0 D and Z =2 .2k 1/ŠŠ cos2 t sin2k t dt ; D 4.k 1/ .2k/ŠŠ 0 C thus we get Ä 2 X1 .2k 1/ŠŠ 2k 1 (11) h.x/ x C ; x Œ 1; 1: D 4.k 1/ .2k/ŠŠ 2 k 0 D C An elementary and uni ed proof of Grothendieck’s inequality 335
Since h is analytic at x 0 and h .0/ 0, its inverse function h 1 D 0 ¤ W Œ 1; 1 Œ 1; 1 can be expanded in a power series in some neighborhood of 0 !
1 X1 2k 1 (12) h .x/ b2k 1x C : D C k 0 D One may in principle determine the coe cients using the Lagrange inversion formula: 2k 2k 1 1 Ä d  t à C b2k 1 lim : C D .2k 1/Š t 0 dt 2k h.t/ C ! For example, 4 1 4 Á3 1 4 Á7 b1 ; b3 ; b5 0; b7 : D D 8 D D 1024
But determining b2k 1 explicitly becomes di cult as k gets larger. A key step C in Haagerup’s proof [Haa] requires the nonpositivity of the coe cients beyond the rst:
(13) b2k 1 0; for all k 1: C Ä is step is in our view the most technical part of [Haa]. We have no insights on how it may be avoided but we simpli ed Haagerup’s proof of (13) in Section5 to keep to our promise of an elementary proof – using only calculus and basic complex variables. 1 It follows from (13) that eh .z/ b1z h .z/ has nonnnegative Taylor WD coe cients. Pringsheim’s theorem implies that if the radius of convergence of 1 the Taylor series of eh .z/ is r , then eh .z/, and thus h .z/, has a singular point at z r . As h0.t/ > 0 on .0; 1/ and h.1/ 1, we must have r 1. It D 1 PN D2k 1 also follows from (13) that h .t/ k 0 b2k 1t C for any t .0; 1/ and PN 2k 1 Ä 1D C 2 N N . So k 1 b2k 1 t C b1t h .t/ for any t .0; 1/ and N N . So PN2 D j C j Ä P 2 2 k 1 b2k 1 b1 1 for any N N and we have k1 0 b2k 1 2b1 1. As 1D j C j Ä P 2 1 D j C j Ä h .1/ 1 we deduce that k1 0 b2k 1 h .1/ 1, and therefore D D C D D X1 (14) b2k 1 2b1 1: j C j D k 0 D We now turn our attention back to the complex Haagerup function. Observe that H.z/ h. z / for all z D z C z 1 and arg H.z/ arg.z/ j j D j j 2 WD ¹ 2 W j j Ä º D for 0 z D . So H D D is a homeomorphism of D onto itself. Let ¤ 2 W ! H 1 D D be its inverse function. Since H.z/ sgn.z/h. z /, we get W ! D j j
1 1 X1 2k 1 (15) H .z/ sgn.z/h . z / sgn.z/ b2k 1 z C : D j j D C j j k 0 D 336 S. Friedland, L.-H. Lim and J. Zhang
P 2k 1 Dini’s theorem shows that the function '.x/ k1 0 b2k 1 x C is a strictly WD D j C j P increasing and continuous on Œ0; 1, with '.0/ 0 and '.1/ k1 0 b2k 1 D D D j C j b1 4= > 1; note that '.1/ is nite by (14). us there exists a unique D c0 .0; 1/ such that '.c0/ 1. So 2 D
X1 2k 1 8 1 1 '.c0/ b2k 1 c0 C c0 h .c0/; D D j C j D k 0 D where the last equality follows from b1 4= and (13). erefore we 1 D 1 obtain h .c0/ 8c0= 1, and if we let x0 h .c0/ .0; 1/, then D WD 2 h.x0/ .x0 1/=8 0. From the Taylor expansion of h.x/, the function C D x h.x/ .x 1/=8 is increasing and continuous on Œ0; 1. Hence x0 is the 7! C unique solution in Œ0; 1 to (16) h.x/ .x 1/ 0 8 C D and c0 .x0 1/=8. D C As Corollary 2.3 indicates, the Haagerup function H plays the analogue of arcsin in the complex case. Unlike arcsin, H is a completely obscure function,3 and any of its properties that we require will have to be established from scratch. e goal of this section is essentially to establish (11)–(16), which we will need later.
4. A uni ed proof of Grothendieck’s inequality
In this section we will need the notions of (i) tensor product and (ii) Hilbert n L n k space, but just enough to make sense of T .F / 1 .F / where F R or D k 0 ˝ D C . In keeping to our promise of an elementary proof,D we will brie y introduce these notions in a simple manner. For our purpose, it su ces to regard the tensor product of k copies of F n , denoted
n k n n .F /˝ F F ; D „ ˝ƒ‚ ˝ … k copies as the F -vector space of k -dimensional hypermatrices, n k ® ¯ .F /˝ Œai1 ik ai1 ik F; i1; : : : ; ik 1; : : : ; n ; WD W 2 2 ¹ º where scalar multiplication and vector addition of hypermatrices are de ned coordinatewise. Also, we let .F n/ 0 F . For k vectors x; y; : : : ; z F n , their ˝ WD 2 tensor product is the k -dimensional hypermatrix given by
3 We are unaware of any other occurrence of H outside its use in Haagerup’s proof of his bound in [Haa]. An elementary and uni ed proof of Grothendieck’s inequality 337
x y z Œx y z n .F n/ k: i1 i2 ik i1;i2;:::;ik 1 ˝ ˝ ˝ ˝ WD D 2 We write k x˝ x x : WD „ ˝ƒ‚ ˝ … k copies If ; is an inner product on F n , then de ning h i (17) x y z; x0 y0 z0 x; x0 y; y0 z; z0 h ˝ ˝ ˝ ˝ ˝ ˝ i WD h ih i h i and extending bilinearly (if F R) or sesquilinearly (if F C ) to all of .F n/ k D D ˝ yields an inner product on the k -dimensional hypermatrices. In particular we have k k k x˝ ; y˝ x; y : h i D h i n If e1; : : : ; en is the standard orthonormal basis of F , then ¹ º ® n k ¯ (18) ei ei .F /˝ i1; : : : ; ik 1; : : : ; n 1 ˝ ˝ k 2 W 2 ¹ º n k is an orthonormal basis of .F /˝ . For more information about hypermatrices see [Lim] and for a more formal de nition of tensor products see [FA]. If an F -vector space H is equipped with an inner product ; such that h i every Cauchy sequence in H converges with respect to the induced norm v v; v 1=2 , we call H a Hilbert space. Hilbert spaces need not be nite- k k D jh ij dimensional; we call H separable if there is a countable set of orthonormal vectors ej H j J , i.e., J is a countable index set, such that every v H ¹ 2 W 2 º 2 satis es 2 X ˇ ˇ2 (19) v ˇ v; ej ˇ : k k D h i j J 2 n k Let ; k be the inner product on .F / as de ned in (17), k be its induced h i ˝ k k norm, and Bk be the orthonormal basis in (18). Let n N . e tensor algebra 2 of F n is the F -vector space4 (20) 1 ° ± n M n k n k X1 2 T .F / .F /˝ .v0; v1; v2;:::/ vk .F /˝ ; vk < WD D W 2 k 0k kk 1 k 0 D D equipped with the inner product
X1 (21) u; v uk; vk k: h i WD h i k 0 D S It is a separable Hilbert space since k1 0 Bk is a countable set of orthonormal vectors satisfying (19). We write forD the norm induced by (21). k k 4 e direct sum in (20) is a Hilbert space direct sum, i.e., it is the closure of the vector space direct sum. 338 S. Friedland, L.-H. Lim and J. Zhang
eorem 4.1 (Grothendieck inequality with Krivine and Haagerup bounds). Let m n F R or C and l; m; n N . For any M F , any x1; : : : ; xm; y1; : : : ; yn D 2 2 2 F l of unit 2-norm, we have ˇ m n ˇ ˇ m n ˇ ˇX X ˇ F ˇX X ˇ (22) max ˇ Mij xi ; yj ˇ K max ˇ Mij "i ıj ˇ; xi yj 1ˇ h iˇ Ä "i ıj 1ˇ ˇ k kDk kD i 1 j 1 j jDj jD i 1 j 1 D D D D where 8 KR and KC WD 2 log.1 p2/ WD .x0 1/ C C are Krivine’s and Haagerup’s bounds respectively. Recall that x0 is as de ned in (16).
Proof. As we described at the end of Section2, we will ‘linearize’ the nonlinear functions arcsin and H in Corollary 2.3 by using Taylor series to replace these functions by polynomials, followed by a ‘tensor trick’ [Jam, Kri] to express polynomials as linear functions on an in nite-dimensional space. Case I: F R. Let c arcsinh.1/ log.1 p2/. Taylor expansion gives D WD D C 2k 1 X1 k c C 2k 1 (23) sin c xi ; yj . 1/ xi ; yj C h i D .2k 1/Šh i k 0 D C 2k 1 X1 k c C D .2k 1/ .2k 1/E . 1/ xi˝ C ; yj˝ C : D .2k 1/Š k k 0 D C For any l N , let T .Rl / be as in (20), and S;T Rl T .Rl / be nonlinear 2 W ! maps de ned by s 2k 1 k c C .2k 1/ 1 S.x/ Sk.x/ k 0;S2k.x/ 0; S2k 1.x/ . 1/ x˝ C ; WD WD C WD .2k 1/Š D C s 2k 1 c C .2k 1/ 1 T .x/ Tk.x/ k 0;T2k.x/ 0; T2k 1.x/ x˝ C ; WD WD C WD .2k 1/Š D C for any x Rl . To justify that S and T are indeed maps into T .Rl /, we need 2 to demonstrate that S.x/ ; T .x/ < but this follows from k k k k 1 2k 1 2 X1 2 X1 c C 2.2k 1/ X1 2 2 S.x/ Sk.x/ k x C Tk.x/ k T .x/ k k D k k D .2k 1/Šk k D k k D k k k 0 k 0 k 0 D D C D and 2k 1 X1 c C 2.2k 1/ 2 x C sinh c x < .2k 1/Šk k D k k 1 k 0 D C An elementary and uni ed proof of Grothendieck’s inequality 339 for all x Rl . Note that 2 2k 1 X1 k c C 2k 1 S.x/; T .y/ . 1/ x; y C sin c x; y ; h i D .2k 1/Šh i D h i k 0 D C which is the essence of the ‘tensor trick’. Hence (23) becomes: sin c xi ; yj S.xi /; T .yj / or c xi ; yj arcsin S.xi /; T .yj / : h i D h i h i D h i l Moreover, since xi and yj are unit vectors in R , we get 2 2 2 2 S.xi / sinh.c xi / 1 and T .yj / sinh.c yj / 1: k k D k k D k k D k k D l As the m n vectors S.x1/; : : : ; S.xm/; T .y1/; : : : ; T .yn/ in T .R / span a C subspace S T .Rl / of dimension d m n; and since any two nite- Â Ä C dimensional inner product spaces are isometric, S is isometric to Rd with the standard inner product. So we may apply Corollary 2.3 to obtain ˇ m n ˇ ˇ m n ˇ ˇX X ˇ 1ˇX X ˇ ˇ Mij xi ; yj ˇ ˇ Mij arcsin S.xi /; T .yj / ˇ ; ˇ h iˇ D c ˇ h iˇ Ä 2c i 1 j 1 i 1 j 1 D D D D which is Krivine’s bound since =2c = 2 log.1 p2/ KR . D C D
Case II: F C . Let c0 .0; 1/ be the unique constant de ned in (16) D 2 such that '.c0/ 1. By the Taylor expansion in (15) and noting that D sgn.z/ z 2k 1 zkzk 1 , j j C D C 1 X1 ˇ ˇ2k 1 (24) H c0 xi ; yj sgn c0 xi ; yj b2k 1ˇc0 xi ; yj ˇ C h i D h i C h i k 0 D X1 2k 1 k k 1 b2k 1c0 C xi ; yj xi ; yj C D C h i h i k 0 D X1 2k 1 k k 1 b2k 1c0 C xi ; yj xi ; yj C D C h i h i k 0 D X1 2k 1D k .k 1/ k .k 1/E b2k 1c0 C xi˝ xi˝ C ; yj˝ yj˝ C : D C ˝ ˝ k k 0 D l l For any l N , let Dl x C x 1 be the unit ball, let T .C / be as 2 D ¹ 2 l W k k Ä º in (20), and let S;T Dl T .C / be nonlinear maps de ned by W ! S.x/ Sk.x/ 1 ;S2k.x/ 0; D k 0 WD q D 2k 1 .k/ .k 1/ S2k 1.x/ sgn.b2k 1/ b2k 1 c0 C x˝ x˝ C ; C WD C j C j N ˝ T .x/ Tk.x/ 1 ;T2k.x/ 0; D k 0 WD q D 2k 1 .k/ .k 1/ T2k 1.x/ b2k 1 c0 C x˝ x˝ C ; C WD j C j N ˝ 340 S. Friedland, L.-H. Lim and J. Zhang
l for any x Dl . en S and T are maps into T .C / since 2
2 X1 2 X1 2k 1 2.2k 1/ X1 2 S.x/ Sk.x/ k b2k 1 c0 C x C Tk.x/ k k k D k k D j C j k k D k k k 0 k 0 k 0 D D D T .x/ 2 D k k and, as b1 > 0 and b2k 1 0 for all k 1 by (13), C Ä
X1 2k 1 2.2k 1/ 2 1 2 b2k 1 c0 C x C 2b1c0 x H .c0 x / < : j C j k k D k k k k 1 k 0 D As in Case I, the ‘tensor trick’ allows us to rewrite (24) as
1 ˝ ˛ H c0 xi ; yj S.xi /; T .yj / or c0 xi ; yj H S.xi /; T .yj / : h i D h i D h i l Moreover, since xi and yj are unit vectors in C , we get
2 X1 2k 1 S.xi / b2k 1 c0 C '.c0/ 1; k k D j C j D D k 0 D and similarly T .yj / 1. So we may apply Corollary 2.3 to get k k D ˇ m n ˇ ˇ m n ˇ ˇX X ˇ 1 ˇX X ˇ 1 ˇ Mij xi ; yj ˇ ˇ Mij H S.xi /; T .yj / ˇ ; ˇ h iˇ D c0 ˇ h i ˇ Ä c0 i 1 j 1 i 1 j 1 D D D D C which is Haagerup’s bound since 1=c0 8=.x0 1/ K . D C D
5. Nonpositivity of b2k 1 C To make the proof in this article entirely self-contained, we present Haagerup’s proof of the nonpositivity of b2k 1 that we used earlier in (13). While the main C ideas are all due to Haagerup, our small contribution here is that we avoided the use of any known results of elliptic integrals in order to stay faithful to our claim of an elementary proof, i.e., one that uses only calculus and basic complex variables. To be clear, while the functions Z =2 Z =2 2 2 1=2 2 2 1=2 (25) K.x/ .1 x sin t/ dt; E.x/ .1 x sin t/ dt WD 0 WD 0 do make a brief appearance in the proof of Lemma 5.1, the reader does not need to know that they are the complete elliptic integrals of the rst and second kinds respectively. Haagerup had relied liberally on properties of K and E that An elementary and uni ed proof of Grothendieck’s inequality 341 require substantial eort to establish [Haa]. We will only use trivialities that follow immediately from de nitions. Our point of departure from Haagerup’s proof is the following lemma about two functions h1 and h2 , which we will see in Lemma 5.2 arise respectively from the real and imaginary parts of the analytic extension of the real Haagerup function h Œ 1; 1 Œ 1; 1 to the upper half plane. W !
Lemma 5.1. Let h1; h2 Œ1; / R be de ned by W 1 ! Z =2 p 2 2 h1.x/ 1 x sin t dt; WD 0 Z =2 2 2 sin t h2.x/ .1 x / dt; p 2 WD 0 1 .1 x 2/ sin t which are clearly strictly increasing functions on Œ1; / with 1
h1.1/ 1; lim h1.x/ =2; h2.1/ 0; lim h2.x/ : D x D D x D 1 !1 !1 en (26) !1.x/ x h1.x/h0 .x/ h0 .x/h2.x/ for x 1; WD 2 1 D 2 (27) !2.x/ x h1.x/h0 .x/ h2.x/h0 .x/ 2h1.p2/h2.p2/ > WD 1 C 2 4 for 1 x p2: Ä Ä
Proof. We start by observing some properties of h10 and h20 . As
1 Z =2 sin2 t 1 Z =2 sin2 t h10 .x/ 3 dt 2 dt; D x 0 p1 x 2 sin2 t D x 0 px2 sin2 t R =2 1 h0 is strictly decreasing on .1; /. As cos t dt , limx 1 h0 .x/ 1 1 0 D 1 C 1 D p!2 2 . Clearly limx h10 .x/ 0. Furthermore, when x > 1, since x sin t 1 !1 D px2 1, we have (28) 0 < h10 .x/ for x > 1: Ä 4x2px2 1 It is straightforward to see that the functions E and K in (25) have derivatives given by
1 1 2 (29) E0.y/ E.y/ K.y/ ;K0.y/ E.y/ .1 y /K.y/ : D y D y.1 y2/ 2 Clearly, h2.x/ K.y/ E.y/, where y y.x/ p1 x . So by chain rule, D D D 342 S. Friedland, L.-H. Lim and J. Zhang
Z =2 d 1 2 2 1=2 h20 .x/ y0.x/ .K E/ y.x/ 1 .1 x / sin t dt: D dy D x 0
Hence h20 is strictly decreasing on Œ1; /, h20 .1/ =2, and limx h20 .x/ 0. 1 D !1 D To show (26), observe that
h1.x/ E.1=x/; xh0 .x/ K.1=x/ E.1=x/; D 1 D h2.x/ K.y/ E.y/; xh0 .x/ E.y/; D 2 D where again y p1 x 2 . Hence D
!1.x/ E.1=x/E.y/ ŒK.1=x/ E.1=x/ŒK.y/ E.y/ D E.1=x/K.y/ K.1=x/E.y/ K.1=x/K.y/: D C
Computing !10 , we see from (29) that !10 0. So !1 is a constant function. By 2 Á (28), limx 1 h10 .x/.1 x / 0, and so limx 1 !1.x/ =2. us !1.x/ =2 & D & D D for all x > 1 and we may set !1.1/ =2. D We now show (27) following Haagerup’s arguments. Note that !2.x/ E.1=x/ K.1=x/ E.1=x/ E.y/ K.y/ E.y/ : D C Let g.x/ E.px/ K.px/ E.px/. A straightforward calculation using (29) WD shows that 1ÄE.px/ K.px/ E.px/2 g00.x/ 0; x Œ0; 1: D 2 1 x x 2 So g is convex on Œ0; 1. Hence g.1 x/ is also convex on Œ0; 1. Let f .x/ g.x/ g.1 x/. en f is convex on Œ0; 1 and f .1=2/ 0. WD C 0 D erefore f .x/ f .1=2/ 2g.1=2/. is yields the rst inequality in (27): !2.x/ 2h1.p2/h2.p2/ for x Œ1; p2. 2 e Taylor expansions of h1 and h2 may be obtained as that in (11), Ä 2 X1 .2k/Š 1 2k (30) h1.x/ x ; D 2 22k.kŠ/2 1 2k k 0 D Ä 2 X1 .2k/Š 2k 2 k (31) h2.x/ .1 x / : D 2 22k.kŠ/2 2k 1 k 0 D 5 Approximate numerical values of h1 and h2 at x p2 and 4 are calculated D to be:
5 For example, using www.wolframalpha.com, which is freely available. Such numerical calculations cannot be completely avoided – Haagerup’s proof implicitly contains them as he used tabulated values of elliptic integrals. An elementary and uni ed proof of Grothendieck’s inequality 343
(32) h1.p2/ 1:3506438; h2.p2/ 0:5034307; h1.4/ 1:5459572; h2.4/ 1:7289033:
e second inequality in (27) then follows from 2h1.p2/h2.p2/ 2 1:35064 0:50343 > =4.
In the next two lemmas and their proofs, Arg will denote principal argument.
Lemma 5.2. Let h Œ 1; 1 Œ 1; 1 be the real Haagerup function as de ned in W ! Section 3. en h can be extended to a function h H C that is continuous C W ! on the closed upper half-plane H z C Im.z/ 0 and analytic on the D ¹ 2 W º upper half-plane H z C Im.z/ > 0 . In addition, h has the following D ¹ 2 W º C properties: (i) Im h .z/ Im h . z / for all z H z C z 1 and h .z/ 0 C C j j 2 \ ¹ 2 W j j º C ¤ for all z H 0 . 2 n¹ º (ii) For x Œ1; /, 2 1 Re h .x/ h1.x/; Im h .x/ h2.x/; C D C D
where h1; h2 are as de ned in Lemma 5.1. (iii) For all k N and all real ˛ > 1, 2 Z ˛ 2 .2k 1/Á (33) b2k 1 Im h .x/ C dx rk.˛/ C D .2k 1/ C C C 1 where .2k 1/ ˛ Á C (34) rk.˛/ Im h .˛/ : j j Ä 2k 1 C C Proof. Integrating by parts, we obtain Z =2 Z =2 h.x/ cos t d arcsin.x sin t/ sin t arcsin.x sin t/ dt; x Œ 1; 1: D 0 D 0 2 e analytic function sin z is a bijection of Œ =2; =2 Œ0; / onto H and 1 it maps the line segment t ia =2 t =2 onto the half ellipsoid ¹ C W Ä Ä º z H z 1 z 1 2 cosh a . Let arcsin be the inverse of this mapping. ¹ 2 W j j C j C j D º C en arcsin is continuous in H and analytic in H. In addition, we have: C ´ arcsin x if x Œ 1; 1; arcsin x 2 C D sgn x i arccosh x if x . ; 1/ .1; /; 2 C j j 2 1 [ 1 1 Á Im.arcsin z/ arccosh z 1 z 1 ; z H: C D 2 j j C j C j 2 344 S. Friedland, L.-H. Lim and J. Zhang
If we de ne Z =2 h .z/ sin t arcsin .z sin t/ dt; z H; C WD 0 C 2 then h is a continuous extension of h to H and is analytic in H. C (i) Since arccosh is increasing on Œ1; /, we have 1 ´ 1 Á arccosh z if z 1; Im.arcsin z/ arccosh z 1 z 1 j j j j C D 2 j j C j C j 0 if z < 1: j j erefore for z H z C z 1 , 2 \ ¹ 2 W j j º Z =2 Im h .z/ sin t Im arcsin .z sin t/ dt C D 0 C Z =2 sin t arccosh z sin t dt Im h . z /: arcsin.1= z / j j D C j j j j As Im.arcsin z/ > 0 on H, we have Im h .z/ > 0 on H. For x Œ 1; 1, C C 2 h .x/ h.x/ is zero only at x 0. For x . ; 1/ .1; /, C D D 2 1 [ 1 Z =2 Im h .x/ sin t arccosh x sin t dt > 0: C D arcsin.1= x / j j j j Hence h has no zero in H 0 . C n¹ º (ii) Let x .1; /. Integrating by parts followed by a change-of-variables 2 1 sin u x sin t in the next-to-last equality gives us: D Z arcsin.1=x/ Z =2 Re h .x/ sin t arcsin.x sin t/ dt sin t dt C D 0 C 2 arcsin.1=x/ Z arcsin.1=x/ 2 Z =2 cos t p 2 2 x dt 1 x sin u du D 0 p1 x2 sin2 t D 0 h1.x/: D A change-of-variables sin v .1 x 2/ 1=2 cos t in the next-to-last equality D gives us:
Z =2 Z =2 cos2 t Im h .x/ sin t arccosh.x sin t/ dt x dt 2 C D arcsin.1=x/ D arcsin.1=x/ px2 sin t 1 Z =2 2 2 sin v .1 x / p dv h2.x/: D 0 1 .1 x 2/ sin2 v D An elementary and uni ed proof of Grothendieck’s inequality 345
(iii) e power series (11) shows that h de nes an analytic function h.z/ in the open unit disk that is identically equal to h .z/ on z C z < 1 H. C ¹ 2 W j j º \ Since h.0/ 0 and h .0/ 0, we can nd some ı0 .0; 1 such that h.z/ D 0 ¤ 2 has an analytic inverse function (12) in z C z < ı0 . For 0 < ı < ı0 , ¹ 2 W j j º let Cı be a counterclockwise orientated circle with radius ı . It follows that h.Cı / is a simple closed curve with winding number 1. Integrating by C parts with a change-of-variables, we have 1 Z h 1.z/ 1 Z z b dz h .z/ dz: 2k 1 2k 2 2k 2 0 C D 2i h.Cı/ z C D 2i Cı h.z/ C
Note that b2k 1 R and C 2 Z zh .z/ Z 1 Z d Ä z .2k 1/ 0 dz dz dz 0: 2k 2 2k 1 2k 1 C Cı h.z/ C C Cı h.z/ C D Cı dz h.z/ C D en we get Z 1 .2k 1/ b2k 1 h.z/ C dz C D 2.2k 1/ Cı C Z 1 .2k 1/ Im h.z/ C dz D 2.2k 1/ Cı C Z 2 .2k 1/ Im h.z/ C dz D .2k 1/ C C ı0 where C is the quarter circle ıei 0 =2 . Since h.z/ ı0 ¹ W Ä Ä º identically equals h .z/ on Cı0 and h .z/ has no zeros in the set C C z C ı z ˛; 0 Arg z =2 by (i), Cauchy’s integral formula ¹ 2 W Ä j j Ä Ä Ä º yields ÄZ ˛ Z 2 .2k 1/ .2k 1/ b2k 1 Im h .z/ C dz h .z/ C dz C D .2k 1/ C C C C ı C˛0 Z iı .2k 1/ h .z/ C dz : C i˛ C Moreover, since h .z/ is real on Œı; 1 and its real part vanishes on the C imaginary axis, we are left with Z ˛ 2 .2k 1/ b2k 1 Im.h .z/ C / dz C D .2k 1/ C 1 ÄZ C 2 .2k 1/ Im h .z/ C dz : C .2k 1/ C C C˛0 By (i), h .z/ Im h .z/ Im h . z /. us C C C j j ˇZ ˇ .2k 1/ ˇ .2k 1/ ˇ ˛ Á C ˇ h .z/ C dzˇ Im h .˛/ : C C ˇ C˛0 ˇ Ä 2 346 S. Friedland, L.-H. Lim and J. Zhang
e integral expression of b2k 1 in (33) will be an important ingredient in C the proof that b2k 1 0 for k 1. We establish some further approximations C Ä for this integral in the next and nal lemma.
Lemma 5.3. Let ˛ 4 throughout.6 Let .x/ Arg h .x/ for x Œ1; /. D WD C 2 1 en Œ1; / Œ0; 2 is strictly increasing on for x 1, .1/ 0, and W 1 ! D limx .x/ =2. In addition, we have the following: !1 D (i) Let p .2k 1/.˛/= . Let WD b C c Z Â.x/ r=.2k 1/ 2 D C .2k 1/ˇ ˇ Ir h .x/ C ˇ sin .2k 1/.x/ ˇ dx WD .2k 1/ Â.x/ .r 1/=.2k 1/ j C j C C D C for r 1; 2; : : : ; p , and D Z ˛ 2 .2k 1/ˇ ˇ J h .x/ C ˇ sin .2k 1/.x/ ˇ dx: WD .2k 1/ Â.x/ p=.2k 1/ j C j C C D C en Z ˛ 2 .2k 1/ p p 1 Im h .x/ C dx I1 I2 ::: . 1/ Ip . 1/ C J: .2k 1/ C D C C C C 1
(ii) Let k 4. en p 2 and I1 > I2 > > Ip > J . Â.p2/=2 .2k 1/ 2 (iii) Let k 4 and c h .p2/ e . en I1 > 0:57c C =.2k 1/ D j C j C and I2 < 0:85I1 .
Proof. Since .x/ arctan h2.x/=h1.x/ , by (26), we get D
d.x/ h1.x/h .x/ h .x/h2.x/ (35) 20 10 > 0; x > 1: dx D h .x/ 2 j C j So .x/ is strictly increasing on for x 1. It is clear that .1/ 0. By Lemma 5.1, D limx h1.x/ =2 and limx h2.x/ , so limx .x/ =2. !1 D !1 D C1 !1 D (i) is follows from dividing the interval of the integral Œ1; ˛ into p 1 C subsets: Z ˛ 2 .2k 1/ Im.h .x/ C / dx .2k 1/ 1 C C Z ˛ 2 .2k 1/ h .x/ C sin .2k 1/.x/ dx D .2k 1/ 1 j C j C C p p 1 I1 I2 ::: . 1/ Ip . 1/ C J: D C C C 6 To avoid confusion, we write ‘˛ ’ for the upper limit of our integrals instead of ‘4’ as the same number will also appear in an unrelated context ‘k 4.’ An elementary and uni ed proof of Grothendieck’s inequality 347
(ii) We write x x./, Œ0; =2/, for the inverse function of .x/. D 2 D By (35), we have
Z r=.2k 1/ 4 C ˇ ˇ 2k 1ˇ ˇ C Ir 2 x./ˇh x./ ˇ ˇ sin .2k 1/ ˇ d; D .2k 1/ .r 1/=.2k 1/ C C C C Z Â.˛/ 4 ˇ ˇ 2k 1ˇ ˇ C J 2 x./ˇh x./ ˇ ˇ sin .2k 1/ ˇ d: D .2k 1/ p=.2k 1/ C C C C
By Lemma 5.1, h1.x/ and h2.x/ are strictly increasing function of x 2 2 2 2 Œ1; /, therefore, so is h .x/ h1.x/ h2.x/ . With this, we deduce 1 2k 1 j C j D C that x h .x/ C is strictly decreasing on Œ1; ˛ for k 4 as j C j d 2k 1 x h .x/ C dx j C j ˇ ˇ 2k 1 . 2k 1/x ˇ ˇ 2k 1 d ˇ ˇ2 ˇh .x/ˇ C C ˇh .x/ˇ ˇh .x/ˇ D C C 2 C dx C ˇ ˇ 2k 1ˇ ˇ2 Á ˇh .x/ˇ ˇh .x/ˇ .2k 1/x h1.x/h10 .x/ h2.x/h20 .x/ D C C C ˇ ˇ 2k 1ˇ ˇ2 Á ˇh .x/ˇ ˇh .x/ˇ .2k 1/ Ä C C 4 ˇ ˇ 2k 1ˇ ˇ2 7 Á ˇh .x/ˇ ˇh .˛/ˇ 0:1187 < 0; Ä C C 4 where we have used the fact that h .x/ 2 is increasing on Œ1; ˛ in the j C j next-to-last inequality and the numerical value is calculated from those of h1.4/ and h2.4/ in (32). Since sin..2k 1// is periodic with period j C j =.2k 1/, we obtain I1 > I2 > : : : > Ip . In addition, C Z Â.˛/ 4 ˇ ˇ 2k 1ˇ ˇ C J 2 x./ˇh x./ ˇ ˇ sin .2k 1/ ˇ d D .2k 1/ p=.2k 1/ C C C C Z Â.˛/ =.2k 1/ 4 C ˇ ˇ 2k 1ˇ ˇ C 2 x./ˇh x./ ˇ ˇ sin .2k 1/ ˇ d Ä .2k 1/ .p 1/=.2k 1/ C C C C < Ip: Finally, we have .˛/ arctan h1.˛/=h2.˛/ 0:8412 > =4 arctan.1/, D D and so p .2k 1/.˛/= 9.˛/= 2 for k 4. D b C c b c D (iii) Since x./ 1 for Œ0; =2/, we have 2 Z =.2k 1/ 4 C ˇ 2k 1ˇ ˇ I1 x./ h x / ˇ C ˇ sin .2k 1/ ˇ d: 2.2k 1/ j C C C 0 Recall that .x/ and x x./ are inverse functions of one another. D D For Œ0; .p2/, 2 348 S. Friedland, L.-H. Lim and J. Zhang
d ˇ ˇ 1 d ˇ ˇ2 d Á 1 log ˇh x./ ˇ log ˇh .x/ˇ d C D 2 dx C dx h1.x/h .x/ h2.x/h .x/ !2.x/ 1 10 C 20 > ; D h1.x/h .x/ h .x/h2.x/ D !1.x/ 2 20 10 for x Œ1; p2, where we have used (26), (27), and the fact that .x/ 2 is strictly increasing for x 1. Hence log h x./ log h .p2/ j C j Ä j C j ..p2/ /=2 which is equivalent to ˇ ˇ Â=2 ˇh x./ ˇ ce ; 0; .p2/ C Ä 2 Â.p2/=2 where c h .p2/ e 1:2059 and .p2/ > =9, using values of D j C j h1.p2/ and h2.p2/ in (32). It follows that for k 4, we have Z =.2k 1/ 4 C Â=2 2k 1 I1 .ce / C sin .2k 1/ d 2.2k 1/ C C 0 2k 1 Z 4c C .k 1=2/Â=.2k 1/ e C sin d D 2.2k 1/2 C 0 2k 1 Z 4c C Â=2 e sin d 2.2k 1/2 C 0 2c Á2 1 e =2 c .2k 1/ 0:57c .2k 1/ C C > C : D 1 1=4 .2k 1/2 .2k 1/2 C C C d Since d log h x./ 1=2, we get j C j ˇ ˇ 2k 1 Á C .k 1=2/=.2k 1/ˇ ˇ 2k 1 ˇh x =.2k 1/ ˇ e C ˇh x./ ˇ C ˇ C C C ˇ Ä C Moreover, since .5=p3/ > 2=9, we know that x./ 5=p3 on Œ0; 2=9. Ä Hence for k 4, it follows from the above results that Z 2=.2k 1/ 4 5 C ˇ ˇ 2k 1ˇ ˇ C Ir 2 x./ˇh x./ ˇ ˇ sin .2k 1/ ˇ d D .2k 1/ p3 =.2k 1/ C C C C 4 5 Z =.2k 1/ ˇ Áˇ 2k 1 C x./ˇh x =.2k 1/ ˇ C 2 ˇ ˇ D .2k 1/ p3 0 C C C C sin .2k 1/ d C Z =.2k 1/ 4 5 .k 1=2/=.2k 1/ C ˇ ˇ 2k 1 C 2 e C x./ˇh x./ ˇ Ä .2k 1/ p3 0 C C sin .2k 1/ d C 5 7=18 e I1 < 0:85I1: Ä p3 An elementary and uni ed proof of Grothendieck’s inequality 349
2k 1 e fact that x h .x/ C is strictly decreasing on Œ1; 4 for k 4, j C j established in the proof of (ii) above, is a crucial observation for establishing the nonpositivity of b2k 1 for k 4. Observe that since h .x/ is strictly increasing C 7 j C j for x > 1, it is enough to show that x h .x/ is strictly decreasing on Œ1; 4, j C j 2k 1 which is what we did. Note that for a xed k 1, x h .x/ C is increasing j C j for large enough x , as h .x/ behaves like C log x for x 1. j C j 1 eorem 5.4. Let the Taylor expansion of h .x/ be as in (12). en b2k 1 0 C Ä for k 1.
Proof. Let k 4 and let I1;I2;:::;Ip;J be as de ned in Lemma 5.3. By (33) with ˛ 4 and Lemma 5.3(i) and (ii), we have D p 1 p b2k 1 I1 I2 ::: . 1/ Ip . 1/ J r2k 1 5p2 C D C C C C > I1 I2 r2k 1 5p2 : C By (34) and Lemma 5.3(iii) with c 1:2059 (established in its proof), we get 0:0855 .2k 1/ 4 .2k 1/ I1 I2 > .1:206/ C ; r2k 1.4/ .1:728/ C : .2k 1/2 j C j Ä 2k 1 C C Since b2k 1 > I1 I2 r2k 1.4/, we get b2k 1 < 0 for k 9. Direct C C C computation using the Lagrange inversion formula gives us b3; b5; : : : ; b17 0, Ä proving nonpositivity for k 8. Ä
Acknowledgment. We thank the anonymous referee for his careful reading and many helpful suggestions that greatly improved our exposition. We would also like to thank Małgorzata Stawiska for suggesting that we send this article to L’Enseignement Mathématique. is work is generously supported by DARPA D15AP00109 and NSF IIS 1546413. LHL acknowledges additional support from a DARPA Director’s Fellowship and the Eckhardt Faculty Fund.
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(Reçu le 19 avril 2018)
Shmuel Friedland, Department of Mathematics, Statistics and Computer Science, University of Illinois, Chicago, 851 South Morgan Street, Chicago, IL 60607-7045, USA e-mail: [email protected]
Lek-Heng Lim, Computational and Applied Mathematics Initiative, Department of Statistics, University of Chicago, 5747 South Ellis Avenue, Chicago, IL 60637-1514, USA e-mail: [email protected]
Jinjie Zhang, Department of Statistics, University of Chicago, 5747 South Ellis Avenue, Chicago, IL 60637-1514, USA Current address: Department of Mathematics, University of California, San Diego, 9500 Gilman Drive, La Jolla, CA 92093-0112, USA e-mail: [email protected]
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