DISCRETE INEQUALITIES OF WIRTINGER’S

Discrete inequalities; Difference; Eigenvalues and eigenvectors; Best constants; Orthogonal polynomials.. A historical review on thepriority in this subject was given by Mitrinovi´c and

A Volume Dedicated to Prof D S Mitrinovi´c

1 Introduction and Preliminaries

In the well-known monograph written by Hardy, Littlewood and P´olya [13, pp.184–187] the following result was mentioned as the Wirtinger’s inequality:Theorem 1.1 Let f be a periodic function with period (2π) and such that f′

1991 Mathematics Subject Classification Primary 26D15; Secondary 41A44, 33C45.

Key words and phrases. Discrete inequalities; Difference; Eigenvalues and eigenvectors; Best constants; Orthogonal polynomials.

This work was supported in part by the Serbian Scientific Foundation, grant number 04M03.

Typeset by AMS-TEX 289

Vasi´c [25, pp 141–154], including many other inequalities of the same type Theproof of W Wirtinger was first published in 1916 in the book [5] by Blaschke.However, inequality (1.1) was known before this, though with other conditions onthe function f The French and Italian mathematical literature do not mention thename of Wirtinger in connection with this inequality A historical review on thepriority in this subject was given by Mitrinovi´c and Vasi´c [24] (see also [25–26]).They have mentioned various generalisations and variations of inequality (1.1), aswell as possibility of applications of such kind of inequalities in many branches inmathematics as Calculus of Variations, Differential and Integral Equations, Spec-tral Operator Theory, Numerical Analysis, Approximation Theory, MathematicalPhysics, etc Under some condition of f , there are also many generalisations of(1.1) which give certain estimates of quotients of the form

where w is a weight function (in one or two variables) and D is a simply connectedplane domain

There are various discrete versions of Wirtinger type inequalities In this survey

we will deal only with such kind of inequalities

The paper is organised as follows In Section 2 we give a summary on the firstresults in this field given by Fan, Taussky and Todd [10] as well as some generali-sations of these discrete inequalities In Section 3 we present a general method forfinding the best possible constants An and Bn in inequalities of the form

of discrete inequalities of Wirtinger’s type for higher differences are treated inSection 4

2 Discrete Fan-Taussky-Todd Inequalities and Some

Generalisations

The basic discrete analogues of inequalities of Wirtinger were given by Fan, sky and Todd [10] Their paper has been inspiration for many investigations inthis subject We will mention now three basic results from [10]:

Taus-Theorem 2.1 If x1, x2, , xn aren real numbers and x1= 0, then

Theorem 2.2 If x0(= 0), x1, x2, , xn, xn+1(= 0) are given real numbers,then

Let A be a real symmetric matrix of the order n, and R be a diagonal matrix ofthe order n with positive diagonal elements For the generalised matrix eigenvalueproblem

(2.5) Ax = λRx, x= [ x1 xn]T,

the following results are well known (cf Agarwal [1, Ch 11]):

1◦There exist exactly n real eigenvalues λ = λν, ν = 1, , n, which need not bedistinct.

2◦

Corresponding to each eigenvalue λν there exists an eigenvector xν which can

be so chosen that n vectors x1, , xn are mutually orthogonal with respect tothe matrix R = diag (r11, , rnn), i.e.,

where a = (a1, , an), b = (b1, , bn−1) and b2 > 0 for k = 1, , n − 1, thenthe eigenvalues λν of the matrix A are real and distinct

4◦ If R = I and the eigenvalues λν of A are arranged in an increasing order, i.e.,

λ1≤ · · · ≤ λn, then for any vector x ∈ Rn, we have that

Further, for any vector x orthogonal to x1 ((x, x1) = 0), we have

If λ1 < λ2 = λ3 < λ4, then a vector x orthogonal to x1 satisfies the equality

λ2(x, x) = (Ax, x) if and only if x is a linear combination of x2and x3

5◦ If the real symmetric matrix A is positive definite, i.e., for every nonzero

x ∈ Rn, (Ax, x) > 0, then the eigenvalues λν (ν = 1, , n) are positive In

a particular case when R = I and A = H (a, b) is positive definite, then the

eigenvalues λν (ν = 1, , n) can be arranged in a strictly increasing order, 0 <

λ1< · · · < λn

Note that inequalities (2.1), (2.2) and (2.4) are based on the left inequality in (2.7)(i.e., (2.8)) The right inequality in (2.7) has not been used, so that in [10] wecannot find some opposite inequalities of (2.1), (2.2) and (2.4) As special cases ofcertain general inequalities, the opposite inequalities of (2.1), (2.2) and (2.4) werefirst proved in [19] (see also [2])

Using a method similar to one from [10], Block [6] obtained several inequalitiesrelated to (2.1), (2.2) and (2.4), as well as some generalisations of such inequalities.For example, Block has proved the following result:

Theorem 2.4 For real numbersx1, x2, , xn(= 0), xn+1= x1, the inequality

holds, with equality in (2.10) if and only if xk = A sin((2k − 1)π/(2n)), k =

1, 2, , n, where A is an arbitrary constant

Theorem 2.6 Let n = 2m and let x1, x2, , xn, xn+1 = x1 be real numberssuch that (2.3) holds Then

n − sinπ

n

(xm+ x2m)2,

with equality if and only if

xk = A cos(2kπ/n) + B sin(2kπ/n), k = 1, 2, , n,

whereA and B are arbitrary constants

Theorem 2.7 For real numbersx1, x2, , xn satisfying (2.3), the inequality

there exist the Fourier coefficients Ck and C∗

j (k = 0, 1, , m; j = 1, , m − 1)such that

+ (−1)iCm, 1 ≤ i ≤ n

For details on this method see for example [1]

New proofs of inequalities (2.1), (2.2) and (2.4) were given by Cheng [8] Hismethod is based on a connection with discrete boundary problems of the Sturm-Liouville type

(2.11) ∆ p(k − 1)∆u(k − 1)
+ q(k)u(k) + λr(k)u(k) = 0, k = 1, , n,

u(0) = λu(1), u(n + 1) = βu(n)

For some details of this method see Agarwal [1, Ch 11] Another method ofproving these inequalities was based on geometric facts in Euclidean space (cf.Shisha [32])

3 A Spectral Method and Using Orthogonal Polynomials

In this section we consider our method (see [19]) for determining the best constants

An and Bn in the inequalities

For two N -dimensional real vectors

z= [ z z ]T and w= [ w w ]T

we define the usual inner product by (z, w) =

where y ∈ RN and HN(a, b) is a three-diagonal matrix like (2.6), with N = n or

N = n − 1, depending on the conditions for the sequence x = (xk) Especially, wewill consider the following two cases:

1◦

x0= xn+1= 0 and x1, , xn are arbitrary real numbers (N = n);

2◦

x1= 0 and x2, , xn are arbitrary real numbers (N = n − 1)

For such three-diagonal matrices we can prove the following auxiliary result ([19]):Lemma 3.1 Let p = (pk) and r = (rk) be positive sequences and the matrix

b=−√pr1

1p2, , −√rn−1

pn−1pn

,

then the matrixHn(a, b) is positive definite

b=−√pr2

2p3, , −√rn−1

pn−1pn

,

then the matrixHn−1(a, b) is positive definite

We will formulate our results in terms of the monic orthogonal polynomials (πk)instead of orthonormal polynomials as we made in [19] Such an approach gives asimpler and nicer formulation than the previous one

The monic polynomials orthogonal on the real line with respect to the inner uct (f, g) = R

prod-Rf (t)g(t)dµ(t) (with a given measure dµ(t) on R) satisfy a mental three-term recurrence relation of the form

funda-(3.5) πk+1(t) = (t − αk)πk(t) − βkπk−1(t),

with π0(t) = 1 and π− 1(t) = 0 (by definition) The coefficients βk are positive.The coefficient β0, which multiplies π− 1(t) = 0 in three-term recurrence relationmay be arbitrary Sometimes, it is convenient to define it by β0=R

Rdµ(t) Thenthe norm of πk can be express in the form

(3.6) kπkk =p(π, πk) =pβ0β1· · · βk

An interesting and very important property of polynomials πk(t), k ≥ 1, is thedistribution of zeros Namely, all zeros of πn(t) are real and distinct and arelocated in the interior of the interval of orthogonality Let τν(n), ν = 1, , n,denote the zeros of πn(t) in an increasing order

which is known as the Jacobi matrix Also, the monic polynomial πn(t) can beexpressed in the following determinant form

πn(t) = det(tIn− Jn),where In is the identity matrix of the order n For some details on orthogonalpolynomials see [17] and [23]

Regarding to the conditions on the sequence x = (xk), we consider now twoimportant cases:

Case 1◦

(x0= xn+1= 0) If we take αk−1= −akand√

βk = −bk(i.e., βk= b2

k >0), k ≥ 1, then we can consider the matrix Hn(−a, −b) = −Hn(a, b), defined by(2.6), as a Jacobi matrix for certain class of orthogonal polynomials (πk) Thus,for every y ∈ Rn we have

(Hn(a, b)y, y) = (−Hn(−a, −b)y, y) = (−Jny, y)

k(t) = πk(t)/kπkk, we have (cf Milovanovi´c [18, p 178])

tπ∗

(t) = Jnπ∗(t) +pβnπ∗

n(t)en.This means that for the eigenvalue t = τν(n) of Jn, the corresponding eigenvector

is given by π∗

(τν(n)) Notice also that the same eigenvector corresponds to theeigenvalue −τν(n) of the matrix −Jn Therefore, the following theorem holds.Theorem 3.2 Let p= (pk)k∈N 0 and r= (rk)k∈N 0 be two positive sequences,

αk−1= −rk−1+ rk

pk

2 k

pkpk+1 (k ≥ 1),and let(πk) be a sequence of polynomials satisfying (3.5) Then for any sequence

of real numbersx0(= 0), x1, , xn,xn+1(= 0), inequalities

Corollary 3.3 For each sequence of the real numbersx0(= 0), x1, , xn,xn+1

(= 0), the following inequalities hold :

n

X

x2k

Equality in the left inequality (3.9) holds if and only if

xk= C sin kπ

n + 1, k = 1, , n,whereC is an arbitrary constant

Equality in the right inequality (3.9) holds if and only if

xk = C(−1)ksin kπ

n + 1, k = 1, , n,whereC is an arbitrary constant

Proof For pk= rk = 1 we obtain αk= −2 and βk = 1 for each k Consequently,the recurrence relation (3.5) becomes

Bn= −τ1(n)= 4 sin2 nπ

2(n + 1) = 4 cos

2(n + 1).Since kSkk = pπ/2 for each k, using (3.10) and (3.11) we find the extremalsequences for the left and the right inequality in (3.9) For example, for the rightinequality we have

from which follows

xk= C(−1)ksin kπ

n + 1 (k = 1, , n),where C is an arbitrary constant 

Remark 3.1 Theorem 2.2 is contained in Corollary 3.3

In a more general case we can take

pk = (a + bk)2 and rk = (a + bk)(a + b(k + 1)),

with a, b ≥ 0 When b = 0 we obtain Corollary 3.3 However, if b 6= 0, because ofhomogeneity in (3.8), it is enough to put b = 1 In that case, we obtain the samepolynomials as in Corollary 3.3

Corollary 3.4 For each sequence of the real numbersx0(= 0), x1, , xn,xn+1

(= 0), the following inequalities

Equality in the right inequality (3.12) holds if and only if

xk= C(−1)k

k + a sin

kπ

n + 1, k = 1, , n,whereC is an arbitrary constant

Remark 3.2 The corresponding inequalities for a = 0 were considered in [19].Corollary 3.5 For each sequence of the real numbersx0(= 0), x1, , xn,xn+1

polyno-Equality in the left (right ) inequality (3.12) holds if and only if

xk = CLk−1(x)/(k − 1)! (k = 1, , n),wherex = An (x = Bn) and C is an arbitrary constant

In this case we have αk = −(2k + 1) and βk = k2, so that the relation (3.5)becomes

πk+1(t) = (t + 2k + 1)πk(t) − k2πk−1(t)

Putting t = −x and πk(−x) = (−1)kLk(x), this relation reduces to one, which responds to the monic Laguerre polynomials orthogonal on (0, +∞) with respect

cor-to the measure dµ(x) = e− xdx The norm of Lk(x) is given by kLkk = k!

In a more general case we can take

(3.13) r0= 0, rk= 1

B(s + 1, k), pk=

1(k + s)B(s + 1, k) (k ≥ 1),where s > −1 and B(p, q) is the beta function (B(p, q) = Γ(p)Γ(q)/Γ(p + q), Γ

is the gamma function) Then we have αk = −(2k + s + 1) and βk = k(k + s),and the corresponding recurrence relation, after changing variable t = −x and

Corollary 3.6 Let s > −1 and let r = (rk)k∈N 0 and p = (pk)k∈N be given by(3.13) For each sequence of real numbers x0(= 0), x1, , xn, xn+1(= 0), wehave

Case 2◦ (x1= 0) Here, in fact, we consider the inequalities

for any sequence of the real numbers x1(= 0), x2, , xn.

Using Lemma 3.1 (Part 2◦

pk+1pk+2 (k ≥ 1),and also αk−1= −ak,√

k(t) = πk(t)/kπkk, we have, as in the previous case,

tπ∗(t) = Jn−1π∗(t) +pβn−1π∗n−1(t)en−1,but now

Hn−1(a, b) = −Hn−1(−a, −b) = −Jn−1−prn

n

Dn−1,where Dn−1= diag (0, , 0, 1) So, we obtain that

Hn−1(a, b)π∗(t) + tπ∗(t) =pβn−1π∗n−1(t) −prn

n

π∗n−2(t)en−1,

from which we conclude that the eigenvalues of Hn−1(a, b), in notation λν = −τν,

ν = 1, , n − 1, are the zeros of the polynomial

n−1(t) −rpn

n

π∗ n−2(t)

The corresponding eigenvectors are π∗

Theorem 3.7 Let p= (pk)k∈N and r= (rk)k∈N be two positive sequences,αk−1

and βk (k ≥ 1) be given by (3.17), and let (πk) be a sequence of polynomialssatisfying (3.5) Then for any sequence of real numbers x1(= 0), x2, , xn,inequalities (3.16) hold, with An = min

ν {−τν} Bn = max

ν {−τν}, where τν, ν =

1, , n − 1, are zeros of the polynomial Rn−1(t) given by (3.19)

Equality in the left (right ) inequality (3.16) holds if and only if

x1= 0, xk= √Cp

k · πk−2(t)

kπk−2k , k = 2, , n,wheret = −An (t = −Bn), kπkk is given by (3.6) and C is an arbitrary constant.Some corollaries of this theorem are the following results:

Corollary 3.8 For each sequence of real numbers x1(= 0), x2, , xn, thefollowing inequalities hold:

Equality in the right inequality (3.20) holds if and only if

xk= C(−1)ksin2(k − 1)π

2n − 1 , k = 1, , n,whereC is an arbitrary constant

Here we have (as in Corollary 3.3) that

πk(t) = Sk(x) = sin(k + 1)θ

sin θ , t + 2 = 2x,and

Rn−1(t) = Sn−1(x) − Sn−2(x) = cos((2n − 1)θ/2)

cos(θ/2) ,and therefore

(3.23) x1= 0, xk = C(−1)k L

s k−2(Bn)Γ(k + s − 1), k = 2, , n,

whereC is an arbitrary constant.

k−2Ls k−2(Bn)p(k − 2)!Γ(k + s − 1)

= (−1)k pΓ(s + 1)

Γ(k + s − 1)L

s k−2(Bn),

we obtain the extremal sequence (3.23) for which the equality is attained in(3.22) 

Remark 3.3 A few members of the monic generalised Laguerre polynomials Ls+1k (x)are

At the end of this section we mention some results of Losonczi [15] He consideredinequalities of the form

of the variables xk Hence there are 6 different cases in (3.24) corresponding to

i = 1, 2 or i = 3, 4 and the + or − sign Losonczi found the best constants α±

n

X

|xk|2

holds, with the best constant γ = 2 cos(π/(r + 2)), where r = [n/m] The case m = 1was previously proved by Fej´er [11] It is clear that the inequalities (3.25) are related to(3.24).

4 Inequalities for Higher Differences

In this section we give a short account on generalisations of Wirtinger’s typeinequalities to higher differences The first results for the second difference wereproved by Fan, Taussky and Todd [10]:

Theorem 4.1 If x0(= 0), x1, x2, , xn, xn+1(= 0) are given real numbers,then

Theorem 4.2 Ifx0, x1, , xn, xn+1 are given real numbers such that x0 = x1,

xn+1= xn and (2.3) holds, then

A converse inequality of (4.1) was proved by Lunter [16], Yin [36] and Chen [7](see also Agarwal [1])

Theorem 4.3 If x0(= 0), x1, x2, , xn, xn+1(= 0) are given real numbers,then

Theorem 4.4 If x0, x1, , xn, xn+1 are given real numbers such that x0= x1

Proof In this case, the n × n symmetric matrix corresponding to the quadraticform

This matrix is the square of the n × n matrix

The eigenvalues of Hnare

λν = λν(Hn) = 4 cos2(n − ν + 1)π

2n , ν = 1, , n,and therefore, the largest eigenvalue of Hn is