Inequalities in Hilbert Spaces

Inequalities in Hilbert Spaces Jan Wigestrand Master of Science in Mathematics Submission date: March 8 Supervisor: Eugenia Malinnikova, MATH Norwegian University of Science and Technology Department of Mathematical Sciences

NTNU Norwegian University of Science and Technology> Faculty of Information Technology, Mathematics and Electrical Engineering> Department of Mathematical Sciences Master Thesis Inequalities in Hilbert Spaces by Jan Wigestrand Supervisor: Eugenia Malinnikova Trondheim, March 5, 8

Contents Contents Acknowledgments Introduction Notation i iii v vii Classical inequalities Hilbert Spaces Hilbert spaces Examples of Hilbert spaces 3 Riesz s representation theorem 4 Cauchy-Schwarz inequality 5 Cauchy-Schwarz inequality 5 Examples of Cauchy-Schwarz inequality 6 3 Bessel s inequality 7 3 Bessel s inequality 7 3 Examples of Bessel s inequality 8 33 Application of Bessel s inequality 9 4 Selberg s inequality 3 4 Selberg s inequality 3 4 Examples of Selberg s inequality 6 Positive semidefinite matrices and inequalities 9 Positive semidefinite matrices 9 Definition and basic properties of positive semidefinite matrices 9 Further properties of positive semidefinite matrices Inequalites 3 Generalized Bessel s inequality 3 i

ii CONTENTS Selberg s inequality 4 3 Generalized Selberg s inequality 5 3 Frames 7 3 Definition and Cauchy-Schwarz upper bound 7 3 Upper bound 7 33 Lower bound 9 34 Tight frames 3 35 Examples of tight frames 3 Conclusion 35 Appendix A 39 References 4

Acknowledgments I would like to thank Eugenia Malinnikova, for her most valuable help in writing this thesis The start date is March 6, 7 The deadline is March 6, 8 Jan Wigestrand, Trondheim, March 5, 8 iii

Introduction The main result in this thesis is a new generalization of Selberg s inequality in Hilbert spaces with a proof, see page 5 In Chapter we define Hilbert spaces and give a proof of the Cauchy-Schwarz inequality and the Bessel inequality As an example of application of the Cauchy- Schwarz inequality and the Bessel inequality, we give an estimate for the dimension of an eigenspace of an integral operator Next we give a proof of Selberg s inequality including the equality conditions following [Furuta] In Chapter we give selected facts on positive semidefinite matrices with proofs or references Then we use this theory for positive semidefinite matrices to study inequalities First we give a proof of a generalized Bessel inequality following [Akhiezer,Glazman], then we use the same technique to give a new proof of Selberg s inequality We conclude with a new generalization of Selberg s inequality with a proof In the last section of Chapter we show how the matrix approach developed in Chapter and Chapter can be used to obtain optimal frame bounds We introduce a new notation for frame bounds, see page vii v

Notation a b frame bounds ( \text{\large{$a$}} \text{\large{$b$}} ) P the set {,, 3, } of all positive integers N the set {,,, 3, } of all nonnegative integers R the set of all real numbers C the set of all complex numbers z = a + ib (a R, b R, i = ) H Hilbert space A complex conjugate transpose matrix of A, A = A T A A is positive semidefinite A > A is positive definite A is the square root of a positive semidefinite matrix A I identity matrix U unitary matrix, UU = U U = I u v u and v are orthogonal vectors λ eigenvalue diag(λ,, λ n ) diagonal matrix λ max (A) largest eigenvalue of matrix A λ min> (A) smallest positive eigenvalue of matrix A vii

Chapter Classical inequalities Hilbert Spaces We will study inequalities in Hilbert spaces and in this section we give the definitions and examples of Hilbert spaces Hilbert spaces Definition A vector space H with a map, : H H C (for real vector spaces, : H H R) is called an inner product space if the following properties are satisfied: (I) x, x = x =, (I) x, x x H, (I3) x + y, z = x, z + y, z x H y H z H, (I4) αx, x = α x, x x H α C (for real vector spaces α R), (I5) x, y = y, x x H y H (the bar denotes complex conjugation) If in addition H is complete, that is ( (I6) lim x n x m, x n x m = n,m ( x H lim x x n, x x n = n then H is called a Hilbert space ) x n H n P m P ), From now on H will denote a Hilbert space The norm in H is defined by

CHAPTER CLASSICAL INEQUALITIES (I7) x = x, x x H (I8) Every Hilbert space has an orthonormal basis, see [Folland,p76] It means that there exists a system {e α } α Å of elements in H that is linearly independent, that is e α, e β = if α β and e α = e α, e α = for each α and for each x H we have x = α Å x, e α e α (the series converges in H ) If we have a separable Hilbert space we can replace {e α } α Å by {e j } j and the sentence above can be reformulated in the following way: It means that there exists a system {e j } j of elements in H that is linearly independent, that is e j, e k = if j k and e j = e j, e j = for each j and for each x H we have x = j x, e j e j (the series converges in H ) From now on a Hilbert space will be synonymous with a separable Hilbert space unless otherwise specified When the basis of H is finite we say that H is finite dimensional otherwise we say that H has infinite dimension Examples of Hilbert spaces (a) R 3 (real vector space) is a three-dimensional Hilbert space x, y = x y + x y + x 3 y 3 x, y are vectors in R 3 x, y = x T y x, y are vectors in R 3 x T = [x x x 3 ], x =,, is an orthonormal basis for R3 (b) R n (real vector space) is an n-dimensional Hilbert space x, y = x j y j x, y are vectors in R n x x x, y = x T y x, y are vectors in R n, x T = [x x x n ], x = An orthonormal basis for R n consists of n vectors each of dimension n x x x 3 x n

HILBERT SPACES 3,,, is an orthonormal basis for R n (c) x, y A = (Ax) T y = x T A T y = x T Ay x, y are vectors in R n, A R n n, A is positive definite (x T Ax > x, for more details see Chapter ) (I3), (I4) are clearly satisfied We see that if A is positive definite, then (I) and (I) are satisfied A is positive definite implies that A is symmetric (A T = A) We use the property that A is symmetric to show that (I5) is satisfied x, y A = (Ax) T y = x T A T y = (x T A T y) T = y T Ax = (A T y) T x = y, x A T = y, x A x, y are vectors in R n, A R n n It follows that (I5) is satisfied and that x, y A is an inner product Since A is symmetric the eigenvectors from different eigenspaces are orthogonal We can find an orthonormal basis for A by first finding a basis for each eigenspace of A, then apply the Gram-Schmidt process to each of these bases (d) C n (complex vector space) is an n-dimensional Hilbert space x, y = x j y j x, y are vectors in C n x x x, y = x T y x, y are vectors in C n, x T = [x x x n ], x = An orthonormal basis for C n consists of n vectors each of dimension n ( + i) ( + i),,, is an orthonormal basis for C n ( + i) x n (e) l = { x = {ξ, ξ, } : ξ j < ξ j C j P } l is an infinite-dimensional Hilbert space x, y = ξ j η j x = {ξ, ξ, }, y = {η, η, } ξ j C η j C j P

4 CHAPTER CLASSICAL INEQUALITIES An orthonormal basis for l consists of infinitely many vectors each of infinite dimension,, is an orthonormal basis for l (f) L ( X, M, µ ) { = f : X C : f is measurable and ( f dµ ) } / < where ( X, M, µ ) is a measure space and f is a measurable function on X L (X, M, µ) is an infinite-dimensional Hilbert space x, y = x(t)y(t) dµ(t) x(t) L (µ) y(t) L (µ) (g) An orthonormal basis for L [, π] is π, π cos t, π sin t, π cos t, π sin t, 3 Riesz s representation theorem We will use the Riesz representation theorem in Chapter 33 example (b) Let H be the set of all bounded linear functionals on a Hilbert space H For all F H we define F H = F(x) sup x H, x = Theorem (Riesz s representation theorem) F H there exists a unique y H such that F(x) = x, y x H () Moreover, we have y = F H A proof for Theorem can be found in [Schechter,p3]

CAUCHY-SCHWARZ INEQUALITY 5 Cauchy-Schwarz inequality The Cauchy-Schwarz inequality is one of the most used inequalities in mathematics Probably the most used inequality in advanced mathematical analysis The inequality is often used without explicit referring to it See Chapter 33 (c) for an example where the Cauchy-Schwarz inequality is used Cauchy-Schwarz inequality Theorem (Cauchy-Schwarz inequality) In an inner product space X, x, y x y x X y X () The equality () holds if and only if x and y are linearly dependent Proof y = is trivial Let y, then for any α C we have x αy = x αy, x αy = x α y, x α x, y + α y x, y Choose α = y and we have x x, y y x, y y x, y + y 4 y = x x, y y, and x, y x y follows If x, y = x y, then we can choose α C α =, such that α x, y = x, y, and we have x y α y x = x y α y x, x y α y x = x x y, y α y x x, y α x y y, x + αα y y x, x = x x y y y x x y x y x y + y y x x = According to (I), we must have x y = α y x, so x and y are linearly dependent If y = βx, β C, then x, βx = x, βx x, βx = x, βx βx, x = β x, βx x, x = x βx, and we have x, βx = x βx

6 CHAPTER CLASSICAL INEQUALITIES Examples of Cauchy-Schwarz inequality (a) In R 3 we have x, y x y x, y are vectors in R 3 We have equality if x and y are linearly dependent This can be seen from Lagrange identity which gives us x, y = x y x y, x y is the vector product, x, y are vectors in R 3 (b) In R n we have x j y j x y x, y are vectors in R n j j x x T y x x y x, y are vectors in R n, x T = [x x x n ], x = (c) x T Ay x T Ax y T Ay x, y are vectors in R n A is positive definite, A R n n (d) In C n we have x j y j x j y j x, y are vectors in C n (e) In l we have ξ j η j ξ j η j x = {ξ, ξ, }, y = {η, η, } ξ j C η j C j P (f) In L (µ) we have x(t)y(t) dµ(t) x(t) dµ(t) y(t) dµ(t) x(t) L (µ) y(t) L (µ) (g) In L (µ), Assume that µ < + and g and f L (µ), then x n the Cauchy-Schwarz inequality implies f dµ f dµ µ(x) If µ is a probability measure, then µ(x) =

3 BESSEL S INEQUALITY 7 3 Bessel s inequality Another widely used inequality for vectors in inner product spaces is the Bessel inequality The Cauchy-Schwarz inequality follows from the Bessel inequality In this section we prove the inequality and use it to give an estimate for the dimension of an eigenspace of an integral operator 3 Bessel s inequality Theorem 3 (Bessel s inequality) Let {e j } j be an orthonormal system in a Hilbert space H Then x, e j x x H (3) j Proof Let α k = x, e k, then for any n P we have x α k e k = x k= We have α k e k, x k= α k e k k= = x α k e k, x x, = x = x = x k= α k x, e k k= x, e k + k= x, e k k= α k e k + k= α k x, e k + k= α k k= α k k= x, e k α k k= x, e k = x x α k e k x k= k= Let n in the last inequality We have a sequence of nonnegative numbers, where the sum of the numbers is bounded from above Hence (3) follows The inner products x, e j in (3) are called the Fourier coefficients of x with respect to the orthonormal system {e j } j Remark We will look at a more general system later

8 CHAPTER CLASSICAL INEQUALITIES Theorem 4 Let {e j } j be an orthonormal system in a Hilbert space H Then {e j } j is an orthonormal basis if and only if for all x H we have x = x, e j (Parseval s identity) (4) j A proof for Theorem 4 can be found in [Weidmann,p39] If we have an orthonormal system with only one element ( e = y y ), then the Bessel inequality becomes the Cauchy-Schwarz inequality 3 Examples of Bessel s inequality (a) In R 3 we have x = 3 x, e j = 3 x where x = is a vector in R3 j and e =, e =, e 3 = is an orthonormal basis for R3 x (b) In R n we have x = x, e j = x n x where x = is a vector j in R n and e =, e =,, e n = is an orthonormal basis for R n x (c) In C n we have x = x, e j = x x j where x = is a vector in C n and e = ( + i) x x x 3 ( + i), e =,, e n = is ( + i) x n x n

3 BESSEL S INEQUALITY 9 an orthonormal basis for C n (d) In l we have x = ξ j x l where x = {ξ, ξ, } ξ j C j P (e) In L [, π] we have x = a + n= a n + n= b n x L [, π] where a = x, e = a n = x, e n = b n = x, e n = π π x dt, a R and π π x cos nt dt, a n R n P and π π x sin nt dt, b n R n P e = π, e = π cos t, e = π sin t, e 3 = π cos t, e 4 = π sin t, is an orthonormal basis for L [, π] a π + n= a n π cos nt + n= b n π sin nt is the Fourier series of x (f) If e is not included in example (a),(b),(c),(e) above, then we do not have an orthonormal basis and we have inequality instead of equality ( x instead of x =) 33 Application of Bessel s inequality (a) Let S = sin nx n= n < a a S converges We will show that S can not be a Fourier series of a Riemann integrable function f (x) From the Bessel inequality we have n= n a π π are the Fourier coefficients This is impossible since n= n a when a Hence S can not be a Fourier series, see [Gelbaum,Olmsted,p7] f (x) dx where n a = (b) Let H be a closed subspace of L [, ] that is contained in C[, ], where C[, ] is defined as the space of continuous functions on [, ] We will show that H is finite dimensional, see [Folland,p78 (ex66)] H is a Hilbert space since H is a closed subspace of L [, ] Both H with L -norm and C[, ] with f [,] = sup { f (x) : x [, ] } are Banach spaces, see [Griffel,p8] Consider the inclusion H C[, ] as a linear map of Banach spaces

CHAPTER CLASSICAL INEQUALITIES This map f f is closed We have to check that { ( f, f ) H C[, ] : f H } is a closed subset of H C[, ] Suppose that ( f n, f n ) is a Cauchy sequence in H C[, ], then f n is a Cauchy sequence in C[, ] and there exists an f such that f n f (uniformly on [, ]) Then f n f in H, that is f n f when n since f n f = f n f dt f n f [,] f n is a Cauchy sequence in H implies that f n is a Cauchy sequence in C[, ] since H C[, ] By the closed graph theorem the inclusion is bounded, thus there exists a C such that f [,] C f for any f H where f [,] = sup { f (x) : x [, ] } ( ) / and f = f dµ Let x [, ] and consider a linear functional F x : H H where F x ( f ) = f (x) It is bounded since F x ( f ) = f (x) f [,] C f for all f H Hence it is a continuous linear functional on H By Riesz representation theorem there exists a unique g x H such that f (x) = f, g x = f (t)g x(t) dt for all f H Further f (x) = f, g x C f f H g x, g x C g x g x C g x g x C Let { f j } n be an orthonormal system of functions in H Then by using Riesz representation theorem and Bessel s inequality and g x C from previous result we have f j(x) = f j, g x = g x, f j g x C x [, ] ( ) n = f j(x) dx C dx = C ( ) n C Thus dim H C and H is finite dimensional

3 BESSEL S INEQUALITY Remark C[, ] contains a subspace of polynomials where, x, x, are linearly independent It is infinite dimensional and is contained in L [, ], but not in H which is a closed subspace of L [, ], that is contained in C[, ] The closure (L [, ]) of this subspace is not contained in C[, ] (c) Let K(x, y) be a continuous function on [a, b] [a, b] A continuous function f on [a, b] is called an eigenfunction for K with respect to a real eigenvalue r if f (y) = r b K(x, y) f (x) dx a We will without loss of generality use [, ] instead of [a, b] { Let E r = f C[, ] : f (y) = r } K(x, y) f (x) dx We will give an estimate for the dimension of E r, see [Lang,p8 (ex7)] Let H r = { f L [, ] : f (y) = r K(x, y) f (x) dx } Let h H r We want to show that h is continuous h(y) h(y ) r K(x, y) K(x, y ) h(x) dx ( ) r K(x, y) K(x, y ) ( dx dx) h(x) when y y since K(x, y) is uniformly continuous and h L [, ] So we have that H r E r C[, ] since all functions in H r are continuous as we shown above Clearly E r H r L [, ] Hence E r = H r If we can show that H r is a closed subspace of L [, ], then we can use results from example (b) above to give an estimate for the dimension of E r Let f n f in L [, ] and f n H r and g H r Then we have f n (y) g(y) r K(x, y) f n(x) g(x) dx ( ) r K(x, ( ) y) dx f n(x) g(x) dx when fn g g is continuous and f n g on [,] Then f = g almost everywhere and f H r Hence H r is a closed subspace of L [, ] Assume that dim E r n where n P, then there exists an orthonormal

CHAPTER CLASSICAL INEQUALITIES system { f j } n in H From ( ) and ( ) in example (b) above we have that n = f j(x) dx rk(x, y) dx n r Thus dim E r r K (x, y) dydx K (x, y) dydx

4 SELBERG S INEQUALITY 3 4 Selberg s inequality Selberg s inequality is not so well known as the Cauchy-Schwarz and the Bessel inequality It is an interesting inequality and it is also a generalization of the Cauchy-Schwarz and the Bessel inequality 4 Selberg s inequality Theorem 5 (Selberg s inequality) In a Hilbert space H, x, y j k= y j, y k x x H y j y j H (5) The equality (5) holds if and only if (C) x = α j y j, α j C, and for each pair ( j, k), j k, (C) y j, y k =, or (C) α j = α k and α j y j, α k y k See [Furuta,p8] Proof For any α j C we have x α j y j = x α j y j, x α j y j = x α j y j, x x, α j y j + α j y j, α j y j = x α j y j, x α j x, y j + α j α k y j, y k k= From ( α j α k ) we have α j α k α j + α k, and we have x α j x, y j α j x, y j + α j y j, y k + k= α k y j, y k k=

4 CHAPTER CLASSICAL INEQUALITIES We can choose α j = x, y j k= y, and we have j, y k = x x, y j x, y j k= y j, y k x, y j x, y j k= y j, y k + k= x, y j y j, y k ( k= y j, y k ) + k= x, y k y j, y k ( y k, y j ) = x x, y j k= y j, y k x, y j k= y j, y k + x, y j k= y j, y k ( k= y j, y k ) + k= x, y k y j, y k ( y j, y k ) = x = x x, y j k= y j, y k + x, y j k= y j, y k + x, y j k= y j, y k + x, y j k= y j, y k + k= x, y k y j, y k x, y j k= y j, y k, and x, y j k= y j, y k x follows We will show that x = (C) x, y j k= y j, y k = x α j y j α j α k y j, y k = α j y j, y k + α k y j, y k ( )

4 SELBERG S INEQUALITY 5 (C) If (C), then for each pair ( j, k), j k, where (C) is true, we have α k y k, α j y j = α j y k, y j ( ) And for each pair ( j, k), j k, where (C) is true, we have ( ) n k= α k y k, y j k= y = j, y k = k= α k y k, y j k= y = j, y k ( k= α k y k, y j ) k= α k y j, y k α j α j k= y j, y k α j = We use ( ), and we have = ( n k= α ky k, α j y j ) k= α jα k y j, y k k= α = ky k, α j y j x = α j y j, α k y k = k= k= k= α k y k, y j αj k= y j, y k α j k= α j α k y j, y k ( k= α ky k, α j y j ) k= α jα k y j, y k k= y k, y j α j α j α k y j, y k, and Hence (C) implies x, y j k= y j, y k = x If x, y j k= y j, y k = x, then choose α j = x, y j k= y j, y k From the proof of the inequality (5) we have that equality (5) holds when we have = x α j y j and α j α k y j, y k = α j y j, y k + α k y j, y k k= k= k= For each pair ( j, k), j k we have α j y j, y k + α k y j, y k and α j α k y j, y k α j y j, y k + α k y j, y k Hence x, y j k= y j, y k = x implies ( ) If ( ), then for each pair ( j, k), j k, assume that (C) is not true

6 CHAPTER CLASSICAL INEQUALITIES Then α j y j, α k y k, and α j α k y j, y k y j, y k = α j + α k α j α k y j, y k y j, y k = α j + α k α j α k = α j + α k α j = α k Hence ( ) implies (C) When we use (C) we need only to calculate maximum (n )n pairs since we have symmetry If we have only one element (n=), y= y, then the Selberg inequality becomes the Cauchy-Schwarz inequality If we have an orthogonal system {y j } n with several elements (n ), y j, y k = ( ) if j k, let e j = y j y j, then the Selberg inequality becomes the Bessel j inequality 4 Examples of Selberg s inequality (a) In R 3, x =, y =, y =, y 3 = We have x = + + and for each pair ( j, k) in (5) we have (C) since y y, y y 3, y y 3 By Selberg s inequality we have equality in (5) since (C) is satisfied And it follows that we have Parseval s identity

4 SELBERG S INEQUALITY 7 (b) In R 3, x =, y =, y =, y 3 = We have x = + and for each pair ( j, k) in (5) we have (C) By Selberg s inequality we have equality in (5) since (C) is satisfied (c) In R 3, x =, y =, y =, y 3 = We have x = + +, and for the following pairs (, ), (, 3) in (5) we have (C) since y y, y y 3 And for the following pair (, 3) in (5), we have (C) By Selberg s inequality we have equality in (5) since (C) is satisfied (d) In L [, ], x = + t + t, y =, y = t, y 3 = t We have x = + t + t and for each pair ( j, k) in (5) we have (C) By Selberg s inequality we have equality in (5) since (C) is satisfied (e) In R 3, if x = ay + by + cy 3, a R, b R, c R, then the Selberg inequality can be written as ( ay,y + by,y + cy 3,y ) y,y + y,y + y,y 3 + ( ay,y + by,y + cy 3,y ) y,y + y,y + y,y 3 + ( ay,y 3 + by,y 3 + cy 3,y 3 ) y 3,y + y 3,y + y 3,y 3 a y, y + ab y, y + ac y, y 3 + b y, y + bc y, y 3 + c y 3, y 3 If y =, y =, y 3 = and x = a + b + c, 3 3 then we have the following Selberg s inequality, (9a+5b+c) 6 + (5a+3b+7c) 5 + (a+7b+7c) 36 9a + ab + 4ac + 3b + 4bc + 7c If a = b = c, then we have equality ( 77a = 77a ) (f) In L [, ], x = a + bt + ct, y =, y = t, y 3 = t, a R, b R, c R

8 CHAPTER CLASSICAL INEQUALITIES We have y y, y y 3 and we have the following Selberg s inequality, (a+ 3 c) 8 3 + ( 3 b) 3 + ( 3 a+ 5 c) 6 5 a + 4 3 ac + 3 b + 5 c (a+ 3 c) 8 3 + ( 3 a+ 5 c) 6 5 a + 4 3 ac + 5 c If a = c, then we have equality ( 56 5 a = 56 5 a)

Chapter Positive semidefinite matrices and inequalities In this chapter we give a new proof of Selberg s inequality It is based on the theory of positive semidefinite matrices This approach gives a new generalization of Selberg s inequality Positive semidefinite matrices Positive semidefinite matrices are closely related to nonnegative real numbers Definition and basic properties of positive semidefinite matrices A n n matrix A is normal, if A A = AA A n n matrix A is Hermitian, if A = A Hermitian matrices are normal matrices A n n matrix A is positive semidefinite, if A is Hermitian and x Ax for all nonzero x C n We will then write A A n n matrix A is positive definite, if A is Hermitian and x Ax> for all nonzero x C n We will then write A > If A B, then we will write A B The following is a list of some properties for positive semidefinite matrices that are needed in this chapter Let A, B, C, F, I, S, U denote n n matrices and x, y denote n vectors (i) If A is Hermitian, then A = U diag(λ,, λ n )U where U is a unitary 9

CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES matrix and λ j are nonnegative real numbers on diagonal matrix Proof See [Zhang,p65] (ii) If A is Hermitian, then S AS is Hermitian Proof (S AS) = S A S = S AS (iii) A implies S AS Proof x Ax implies y S ASy (iv) A implies det(a) Proof Let Ax = λx where λ is an eigenvalue and x is an eigenvector of A corresponding to λ Then for each λ, we have x Ax = x λx λ = x Ax x x det(a) = λ λ λ n (v) A and det(a) > A Proof For any y C n there exists an x C n such that Ax = y x = A y and x Ax = ( A y ) A A y = y ( A ) y = y A y (vi) If A B, then S AS S BS Proof A B S (A B)S S AS S BS (vii) If A, then there exists a matrix B such that B = A B is denoted by A Proof See [Zhang,p6] (viii) If A B and A exists and B exists, then B A Proof If C I, then I = C CC C IC = C A B I A BA A B A I B A Further properties of positive semidefinite matrices We also need properties of products of two and three positive semidefinite matrices The product of two positive semidefinite matrices is not always positive semidefinite (ix) (A and B ) AB [ ] [ ] Proof A =, B =, AB = 3 [ ] 5 5, then A, B, AB 3 4

POSITIVE SEMIDEFINITE MATRICES (x) If A then A Proof A ( y A ) A ( A y ) = ( A y ) A ( A y ) = x Ax (xi) A and B, then AB is Hermitian AB = BA Proof (AB) = B A = BA (xii) If A and B and AB = BA, then A B = B A Proof We have A, B AB is Hermitian Let A = U CU and B = U FU where C and F are diagonal matrices and U is a unitary matrix U is the same for A and B since A and B commute, see [Zhang,p6] Then we have A B = U C UU F U = U C F U = U F C U = U F UU C U = B A (xiii) If A and B and AB is Hermitian, then AB Proof y ABy = y A A By = ( A x ) A B B y = ( A y ) B ( A y ) = x Bx (xiv) A, B, B is invertible, C and ABC is Hermitian implies ABC Proof Let F = ( B AB ) ( B CB ) F is Hermitian since F = B (ABC) B, see (ii) We have ( B AB ) and ( B CB ), see (iii) From (xiii) we have F and finally from (iii) we have ABC = B FB (xv) If A, then λ max (A)I A λ max (A) is the largest eigenvalue of matrix A Proof Let B = U AU where B is a diagonal matrix and U is a unitary matrix Then we have λ max (A)I B Hence λ max (A)I A (xvi) If A, then λ min> (A) A A λ min> (A) is the smallest positive eigenvalue of matrix A Proof Let B = U AU where B is a diagonal matrix and U is a unitary matrix Then we have B λ min> (A)B = B λ min> (B)B = diag ( λ λ min>(b)λ,, λ n λ min> (B)λ n ) Hence λ min> (A) A A (xvii) We define the Gram matrix for {y, y,, y n } in a Hilbert space H in the following way:

CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES y, y y, y y n, y y, y y, y y n, y G = y, y n y, y n y n, y n () We see that G is Hermitian u Gu = β y + β y + + β n y n, β y + β y + + β n y n β β for all nonzero u =, β j C, that is G β n If y, y,, y n are linearly independent, then G > (xviii) If we have a diagonal matrix D = d d is a nonnegative real number, then we have u D u = β β for all nonzero u =, β j C, that is D Remark 3 If G = β n d n where each d j β j d j [ ] y, y y, y, then det(g) is same the y, y y, y Cauchy-Schwarz inequality

INEQUALITES 3 Inequalites In this section we will use theory for positive semidefinite matrices to study inequalities First we give a proof of a generalized Bessel s inequality following [Akhiezer,Glazman], then we use the same technique to give a new proof of Selberg s inequality We conclude this section with a new generalization of Selberg s inequality with a proof Generalized Bessel s inequality We will need the following generalized Bessel s inequality to prove Selberg s inequality in the next subsection Theorem 6 (Generalized Bessel s inequality) Let {y j } n be a linearly independent system in a Hilbert space H and let G be the corresponding Gram matrix Then x, y x, y where v = x, y n See [Akhiezer,Glazman,p4] Proof Let x = From v G v x x H () α j y j + h where α j C, h H, h y j, for any j {,,, n} x, y k = k= k= x = α j y j + h, = α j y j, = k= α j y j, y k + α j y j + h α k y k + h, k= α j α k y k, y j + h α α h, y k, we have v = G w where w = k= α k y k + α j y j, h + h, h k= α n

4 CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES = w G w + h = w G w + h For a linearly independent system {y j } n, the matrix G exists, and we have v G v = (G w) G G w = w G w = w G w We have equality when h = If y j y k, j k and y j =, then G = I and the generalized Bessel inequality can be written as x, y j x Selberg s inequality Here we give a new proof of Selberg s inequality based on the theory of positive semidefinite matrices Theorem 7 (Selberg s inequality) In a Hilbert space H, Proof We have x, y d x, y d v =, D = x, y n x, y j k= y j, y k x x H y j y j H (3) x, y j k= y j, y k x v D v x where d n, d j = y j, y k See Appendix A We will prove that v D v x Let x = α j y j + h where α j C, h H, h y j, for any j {,,, n} From the proof of the generalized Bessel s inequality we have x = w G w + h and v = G w v D v x (G w) D G w w G w + h k=

INEQUALITES 5 w G D G w w G w + h w GD G w w G w + h Hence it is sufficient to show that G GD G First we will show that G D u Gu = β y + β y + + β n y n, β y + β y + + β n y n = = u Du = k= k= β j β k y j, y k k= β j y j, y k for all nonzero u = β j d j = k= k= ( β j + β k ) y j, y k β β, β j C β n β j y j, y k Hence G D Further G GD G G ( I D G ) G ( D (D G) ) We have G, D, D is invertible, D G, and G GD G is Hermitian, so (xiv) is satisfied Hence G GD G β j β k y j, y k Remark 4 The proof for Selberg s inequality is valid for all systems {y j } including linearly dependent systems that has singular Gram matrices 3 Generalized Selberg s inequality Here we introduce a new inequality based on the results from the proof of Selberg s inequality in Chapter Theorem 8 (Generalized Selberg s inequality) In a Hilbert space H If y,, y n H, G is the Gram matrix for y,, y n, E G, and E is invertible, then v E v x x H (4)

6 CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES x, y x, y where v = x, y n Proof Let x = α j y j + h where α j C, h H, h y j, for any j {,,, n} From the proof of the generalized Bessel s inequality we have x = w G w + h, and v = G w v E v x (G w) E G w w G w + h w G E G w w G w + h w GE G w w G w + h Hence it is sufficient to show that G GE G Further G GE G G ( I E G ) G ( E (E G) ) We have G, E, E is invertible, E G, and G GE G is Hermitian, so (xiv) is satisfied Hence G GE G Remark 5 The Inequality in Theorem 8 is called Generalized Selberg s inequality because by choosing E = d d Selberg s inequality d n where d j = y j, y k we have k=

3 FRAMES 7 3 Frames In this section we show how the matrix approach developed in Chapter and Chapter can be used to obtain optimal frame bounds We introduce a new notation for frame bounds, see page vii 3 Definition and Cauchy-Schwarz upper bound We will follow [Christensen] in this subsection Definition A countable system of elements {y j } j in a Hilbert space H is a frame for H if there exist constants < a b <, such that a x x, y j b x x H (5) j The constants a, b are called frame bounds a is a lower frame bound, and b is an upper frame bound Frame bounds are not unique The optimal lower frame bound is the supremum over all lower frame bounds, and the optimal upper frame bound is the infimum over all upper frame bounds We can have infinitely many elements {y j } in a frame We will here assume that we have finitely many elements {y j } in a frame An important task is to estimate the frame bounds We start with the upper bound The first estimate is quite obvious From the Cauchy-Schwarz inequality we have m x, y j m y j x, which gives b = m y j From the Cauchy-Schwarz inequality it follows that we always have an upper bound 3 Upper bound We will in this subsection use the generalized Selberg inequality to find the optimal upper frame bound Lemma Let {y j } m be a system of elements in a Hilbert space H and let

8 CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES G be the corresponding Gram matrix Then for all x H we have Moreover, we have m m x, y j λ max (G) x (6) x, y j b x λ max (G) b (7) Proof First we will prove (6) From (xv) in Chapter we have λ max (G) I G, and then by applying generalized Selberg inequality we have (6) To prove (7) assume that m x, y j b x Choose x = α m α α j y j such that G w = λ max (G) w where w =, then m x, y j = (G w) G w, and x = w G w We have m x, y j b x (λ max (G)) w w bλ max (G) w w λ max (G) b (6) means that λ max (G) is an upper bound (6) and (7) means that λ max (G) is an optimal upper bound Remark 6 We have not used any properties of frame for describing the optimal upper bound (6) and (7) holds for any finite system {y j } m From the Selberg inequality we have α m

3 FRAMES 9 m m x, y j max y j, y k,,m x (8) k= This implies λ max (G) max,,m m y j, y k (9) k= If G = [ ], then λ max (G) < max m k= y j, y k,,m Remark 7 b = max,,m m k= y j, y k is an upper frame bound for {y j } m that always exists It may not be optimal but it is easier to compute than λ max (G) 33 Lower bound To find the optimal lower bound we use the condition that {y j } m is a frame Proposition Let {y j } m be a sequence in a Hilbert space H Then {y j } m is a frame for span{y j} m Proof See [Christensen,p4] Corollary A system of elements {y j } m in a Hilbert space H is a frame for H if and only if span{y j } m = H Proof See [Christensen,p4] Lemma Let {y j } m be a frame for a Hilbert space H and let G be the corresponding Gram matrix Then for all x H we have λ min> (G) x m x, y j ()

3 CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES where λ min> (G) is the smallest positive eigenvalue of G Moreover, we have a x m x, y j a λ min> (G) () Proof Let x = Then we have m α j y j, since from Corollary, we have x span{y,, y m } α m α x, y j = (G w) G w, and x = w G w where w = α m λ min> (G) x m x, y j λ min> (G) w G w (G w) G w w λ min> (G) G w w G w We have λ min> (G) G G, see (xvi) in Chapter Next, assume a x m x, y j Choose x = α α G w = λ min> (G) w where w =, then α m m x, y j = (G w) G w, and x = w G w We have a x m x, y j aλ min> (G) w w (λ min> (G)) w w m α j y j such that

3 FRAMES 3 a λ min> (G) () means that λ min> (G) is a lower bound () and () means that λ min> (G) is an optimal lower bound 34 Tight frames Definition 3 A frame is a tight frame if (5) is satisfied with a = b, that is if the optimal upper frame bound and the optimal lower frame bound are equal a is then called the frame bound When we have a tight frame {y j } m in a Hilbert space H, (5) becomes m x, y j = a x x H () Proposition Assume {y j } m bound a Then is a tight frame for a Hilbert space H with frame x = a m x, y j y j x H (3) Proof See [Christensen,p5] Theorem 9 (Casazza,Fickus,Kovačević,Leon,Tremain) Given an n-dimensional Hilbert space H and a sequence of positive scalars {a j } m, there exists a tight frame {y j} m for H of lengths y m = a m for all j =,, m if and only if, max,,m a j n Proof See [Casazza,Fickus,Kovačević,Leon,Tremain,p33] m a j (4) Theorem {y j } m is a tight frame for a Hilbert space H if and only if span{y j} m = H and λ min> (G) = λ max (G) G is the corresponding Gram matrix

3 CHAPTER POSITIVE SEMIDEFINITE MATRICES AND INEQUALITIES Proof Follows from Corollary, Lemma and Lemma 35 Examples of tight frames The following are examples of tight frames for R 3 (a) y =, y =, y 3 =, G = (b) y =, y =, y 3 =, y 4 =, 3 4 4 4 4 3 4 4 4 4 G = 4 3 4 4 4 4 4 3 4 4 (c) y =, y =, y 3 =, y 4 =, y 5 =, y 6 =, G = (d) y =, y =, y 3 =, y 4 =, y 5 =, y 6 =, y 7 =,

3 FRAMES 33 G = 4 4 4 3 4 4 4 4 3 4 4 4 4 4 4 3 4 4 3 4 We see that span{y j } m = R3 and λ min> (G) = λ max (G) in all the four examples By Theorem we have tight frames in all the four examples In example (a) a =, in example (b) a =, in example (c) a = and in example (d) a =

Conclusion We have shown that by using a generalized form of nonnegative real numbers called positive semidefinite matrices we get a nontrivial generalization of the Selberg inequality 35

Appendices 37

Appendix A Here we will show the first equivalence in the proof of Theorem 7 in Chapter v T D v x [ [ x, y x, y x, y n ] d d d x, y d x, y x, y d n x, y n ] x, y x x, y n d n x, y x, y x x, y n d x, y x, y + d x, y x, y + + d n x, y n x, y n x x, y k= y, y k + x, y k= y, y k + + x, y n k= y n, y k x x, y j k= y j, y k x 39

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