Non-symmetric polarization - PDF Free Download

Non-symmetric polarization Sunke Schlüters València, 18. October 2017 Institut für Mathematik Carl von Ossietzky Universität Oldenburg Joint work with Andreas Defant 1

Introduction: Polynomials and Polarization

Homogeneous polynomials Let throughout the talk P : C n C denote a m homogeneous polynomial. 2

Homogeneous polynomials Let throughout the talk P : C n C denote a m homogeneous polynomial. Every such polynomial can be written as P (x) = c α x α α N n 0 α =m 2

Homogeneous polynomials Let throughout the talk P : C n C denote a m homogeneous polynomial. Every such polynomial can be written as P (x) = c α x α1 1 xαn n α N n 0 α =m 2

Homogeneous polynomials Let throughout the talk P : C n C denote a m homogeneous polynomial. Every such polynomial can be written as P (x) = c α x α1 1 xαn n = c (j1,...,j m)x j1 x jm. α N n 0 α =m 1 j 1... j m n An m form which is somewhat naturally associated to P is given by L P (x (1),..., x (m) ) := c (j1,...,j m)x (1) x (m) j m, 1 j 1... j m n and the symmetrization of L P, SL P (x (1),..., x (m) ) := 1 L P (x (σ(1)),..., x (σ(m)) ), m! σ Σ m where Σ m denotes the set of all permutations of {1,..., m}, is symmetric and likewise defines P. j 1 2

Norm inequalities With a norm on C n, the space of all m homogeneous polynomials P and the space of all m linear forms L become Banach spaces, when equipped with the supremum norms and L := P := sup P (x) x 1 sup L ( x (1),..., x (m)). x (k) 1 3

The classical polarization formula Theorem (Polarization formula) For each m homogeneous polynomial P : C n C and every choice of x (1),..., x (m) C n, ( SL P x (1),..., x (m)) = 1 ( m 2 m ε 1 ε m P ε k x (k)). m! ε k =±1 k=1 As an easy consequence, The question P SL P mm m! P e m P. Can we compare the norms of a m homogeneous polynomial P with norms of non-symmetric m forms, which define P? 4

Non-symmetric m forms defining P We introduced the m form L P, which is somewhat naturally associated with P. However, L P is in general not symmetric. 5

Non-symmetric m forms defining P We introduced the m form L P, which is somewhat naturally associated with P. However, L P is in general not symmetric. The norm inequality, which arose from the polarization formula, is therefore not applicable! Bad news: We can t expect to have the same or a similar norm inequality for every m form defining P. Consider for example L : (C 2 ) 2 C, (x, y) x 1 y 2 x 2 y 1 and P (x) := L(x, x). Then L 0, but P = 0, i.e. L > 0 and P = 0. 5

Our results

Main theorem We consider 1 unconditional norms on C n, that is x k y k for all k implies x y. 6

Main theorem We consider 1 unconditional norms on C n, that is x k y k for all k implies x y. Examples are the l n p norms p. 6

Main theorem We consider 1 unconditional norms on C n, that is x k y k for all k implies x y. Examples are the l n p norms p. Theorem There exists a universal constant c 1 1 such that for every m homogeneous polynomial P : C n C and every 1 unconditional norm on C n LP c m2 1 (log n) m 1 P. Moreover, if = p for 1 p < 2, then there even is a constant c 2 = c 2 (p) 1 for which L P c m2 2 P. 6

Partial symmetrization Keeping the polarization formula in mind, we have to show that L P c SL P with a suitable constant c. We will use an iterative approach. 7

Partial symmetrization Keeping the polarization formula in mind, we have to show that L P c SL P with a suitable constant c. We will use an iterative approach. For 1 k n define the partial symmetrization S k L P (x (1),..., x (m) ) := 1 L P (x (σ(1)),..., x (σ(k)), x (k+1),..., x (σ(m)) ). k! σ Σ k 7

Partial symmetrization (cont.) Theorem There exists a universal constant c 1 1 such that for every m homogeneous polynomial P : C n C, every 1 unconditional norm on C n and 2 k m S k 1 L P (c 1 log n) k S k L P. Moreover, if = p for 1 p < 2, then there even is a constant c 2 = c 2 (p) 1 for which S k 1 L P c k 2 S k L P. 8

Ideas of the proof

Coefficients of m forms and Schur multiplication An m form L : (C n ) m C is uniquely determined by its coefficients c i = c i (L) := L(e i1,..., e im ), i I(n, m) := {1,..., n} m. With L i : (C n ) m C, (x (1),..., x (m) ) x (1) L = i I(n,m) i 1 c i L i. x (m) i m we have We call A C I(n,m) an (m dimensional) matrix and by c i (A) we denote its i th entry. For A, B C I(n,m) we define the Schur product A B (entry wise) by c i (A B) := c i (A) c i (B). 9

Coefficients of m forms and Schur multiplication An m form L : (C n ) m C is uniquely determined by its coefficients c i = c i (L) := L(e i1,..., e im ), i I(n, m) := {1,..., n} m. With L i : (C n ) m C, (x (1),..., x (m) ) x (1) L = i I(n,m) i 1 c i L i. x (m) i m For an m form L : (C n ) m C and A C I(n,m) let we have A L := i ( ci (A) c i (L) ) L i. For a norm on C n and A C I(n,m) we define the Schur norm µ m (A) as the best constant c, such that for any m form L : (C n ) m C. A L c L 9

Plan for the proof 1. Write S k 1 L P as the Schur product of a (suitable) matrix and S k L P, i.e. S k 1 L P = A k S k L P. 10

Plan for the proof 1. Write S k 1 L P as the Schur product of a (suitable) matrix and S k L P, i.e. S k 1 L P = A k S k L P. 2. Estimate the Schur norm of A k 10

Plan for the proof 1. Write S k 1 L P as the Schur product of a (suitable) matrix and S k L P, i.e. S k 1 L P = A k S k L P. 2. Estimate the Schur norm of A k 2.1 Decompose A k into a sum and product of more handily pieces. 2.2 Generalize results of Kwapień and Pe lczyński (1970) and Bennett (1976) (for the case m = 2) to any m. 2.3 Use the compatibility of addition and Schur multiplication with the Schur norm to estimate µ m (A k ). 10

Comparing coefficients To write S k 1 L P as the Schur product of a matrix and S k L P we have to compare coefficients. 11

Comparing coefficients To write S k 1 L P as the Schur product of a matrix and S k L P we have to compare coefficients. We have the following result: Proposition Let P : C n C be a m homogeneous polynomial, i I(n, m) and k {2,..., n}. Then c i (S k 1 L P ) = provided max{i 1,..., i k 1 } i k ; and otherwise. k {1 u k i u = i k } c i(s k L P ) c i (S k 1 L P ) = 0 c i (S k L P ) 11

Decomposing A k To estimate µ m (Ak ) we decompose it into more handily pieces. 12

Decomposing A k To estimate µ m (Ak ) we decompose it into more handily pieces. Lemma For 1 k m we have with A k,u := A k = Q {1,...,k} Q =u ( k 1 u=1 T u,k) ( k u=1 ) k u Ak,u ( D q,k) ( ) q Q q Q c(1 Dq,k ), where Q c denotes the complement of Q in {1,..., k}. The matrices 1, D u,v, T u,v C I(n,m) (u, v {1,..., m}, u v) are defined by c i (1) := 1, c i (D u,v ) := { 1, iu = i v 0, i u i v, c i (T u,v ) := { 1, iu i v 0, i u > i v. 12

Classical Schur multipliers In the case m = 2 Kwapień and Pe lczyński (1970) and Bennett (1976) obtained for these matrices: µ 2 (D 1,2 ) 1, µ 2 (T 1,2 ) log 2 (2n), and, moreover, for 1 p < 2 there is a constant c 3 = c 3 (p) such that µ 2 p (T 1,2 ) c 3. This can be generalized to any norm on C n and any m. We have: µ m (Du,v ) 1, µ m (T u,v ) log 2 (2n), µ m p (T u,v ) c 3. 13

Decomposing A k (cont.) Using that µ m (A B) µm (A) µm (B) and that µ m is a norm, we are able to use our decomposition and the norm estimates on the previous slide to estimate the Schur norm of A k. We obtain µ m (Ak ) k3 k( µ m (T u,k ) ) k 1 (c1 log n) k, respectively for 1 p < 2 µ m p (A k ) k3 k( µ m p (T u,k ) ) k 1 c k 2. 14

Summary We established upper bounds µ m (Ak ) c k with nice constants c. 15

Summary We established upper bounds µ m (Ak ) c k with nice constants c. Hence S k 1 L P c k S k L P. 15

Summary We established upper bounds µ m (Ak ) c k with nice constants c. Hence S k 1 L P c k S k L P. Iteratively applying this result yields L P = S 1 L P 15

Summary We established upper bounds µ m (Ak ) c k with nice constants c. Hence S k 1 L P c k S k L P. Iteratively applying this result yields L P = S 1 L P c 2 S 2 L P c 2 c 3 S 3 L P... c 2+3+...+m S m L P 15

Summary We established upper bounds µ m (Ak ) c k with nice constants c. Hence S k 1 L P c k S k L P. Iteratively applying this result yields L P = S 1 L P c 2 S 2 L P c 2 c 3 S 3 L P... c 2+3+...+m S m L P c m2 SL P c m2 e m P. 15

Simple improvement We established the identity thus S k 1 L P = A k S k L P, L P = ( m k=2 A k) SL P. 16

Simple improvement We established the identity thus S k 1 L P = A k S k L P, L P = ( m k=2 A k) SL P. In the general case, we get an log n factor out of every occurrence of T u,k in A k. In the product m A k we have the factor k=2 ( ) k 1 T u,k. u=1 m k=2 This gives, using the naive approach, a factor of (log n) m2. However, we can improve the result by checking that ( ) m k 1 T u,k = m T k 1,k, k=2 u=1 k=2 16

References G. Bennett. Unconditional convergence and almost everywhere convergence. (1976). A. Defant and S. Schlüters. Non-symmetric polarization. (2017). S. Kwapień and A. Pe lczyński. The main triangle projection in matrix spaces and its applications. (1970). F. A. Sukochev and A. A. Tomskova. (E, F ) Schur multipliers and applications. (2013). 17