(1)http://jipam.vu.edu.au/ Volume 7, Issue 1, Article 14, 2006 A POTPOURRI OF SCHWARZ RELATED INEQUALITIES IN INNER PRODUCT SPACES (II) S.S

http://jipam.vu.edu.au/

Volume 7, Issue 1, Article 14, 2006

A POTPOURRI OF SCHWARZ RELATED INEQUALITIES IN INNER PRODUCT SPACES (II)

S.S. DRAGOMIR

SCHOOL OFCOMPUTERSCIENCE ANDMATHEMATICS

VICTORIAUNIVERSITY

PO BOX14428, MELBOURNECITY

VICTORIA8001, AUSTRALIA. sever.dragomir@vu.edu.au URL:http://rgmia.vu.edu.au/dragomir

Received 28 June, 2005; accepted 17 September, 2005 Communicated by A. Lupa¸s

ABSTRACT. Further inequalities related to the Schwarz inequality in real or complex inner prod- uct spaces are given.

Key words and phrases: Schwarz inequality, Inner product spaces, Reverse inequalities.

2000 Mathematics Subject Classification. 46C05, 26D15.

1. INTRODUCTION

Let(H;h·,·i)be an inner product space over the real or complex number fieldK. One of the most important inequalities in inner product spaces with numerous applications is the Schwarz inequality, that may be written in two forms:

(1.1) |hx, yi|² ≤ kxk²kyk², x, y ∈H (quadratic form) or, equivalently,

(1.2) |hx, yi| ≤ kxk kyk, x, y ∈H (simple form).

The case of equality holds in (1.1) or in (1.2) if and only if the vectors x and y are linearly dependent.

In the previous paper [6], several results related to Schwarz inequalities have been estab- lished. We recall few of them below:

ISSN (electronic): 1443-5756

c 2006 Victoria University. All rights reserved.

196-05

1. Ifx, y ∈H\ {0}andkxk ≥ kyk,then

(1.3) kxk kyk −Rehx, yi ≤











1

2r²_kxk

kyk

^r−1

kx−yk² if r≥1

1 2

_kxk

kyk

^1−r

kx−yk² if r <1.

2. If(H;h·,·i)is complex,α ∈CwithReα,Imα >0andx, y ∈H are such that (1.4)

x−Imα Reα ·y

≤r then

(1.5) kxk kyk −Rehx, yi ≤ 1

2 ·Reα Imαr². 3. Ifα∈K\ {0},then for anyx, y ∈H

(1.6) kxk kyk −Re α²

|α|² hx, yi

≤ 1

2 ·[|Reα| kx−yk+|Imα| kx+yk]²

|α|² .

4. Ifp≥1,then for anyx, y ∈H one has

(1.7) kxk kyk −Rehx, yi ≤ 1 2 ×







(kxk+kyk)^2p − kx+yk^2p1_p ,

kx−yk^2p− |kxk − kyk|^2p1_p . 5. Ifα, γ >0andβ ∈Kwith|β|² ≥αγ then forx, a∈H witha6= 0and (1.8)

x− β αa

≤ |β|²−αγ¹₂

α kak,

one has

kxk kak ≤ ReβRehx, ai+ ImβImhx, ai

√αγ (1.9)

≤ |β| |hx, ai|

√αγ and

(1.10) kxk²kak²− |hx, ai|² ≤ |β|²−αγ

αγ |hx, ai|².

The aim of this paper is to provide other results related to the Schwarz inequality. Applica- tions for reversing the generalised triangle inequality are also given.

2. QUADRATICSCHWARZRELATED INEQUALITIES

The following result holds.

Theorem 2.1. Let(H;h·,·i)be a complex inner product space andx, y ∈H, α∈[0,1].Then

(2.1)

αkty−xk²+ (1−α)kity−xk² kyk²

≥ kxk²kyk²−[(1−α) Imhx, yi+αRehx, yi]² ≥0 for anyt∈R.

Proof. Firstly, recall that for a quadratic polynomialP :R→R,P(t) =at²+ 2bt+c, a >0, we have that

(2.2) inf

t∈R

P(t) =P

−b a

= ac−b² a . Now, consider the polynomialP :R→Rgiven by

(2.3) P (t) :=αkty−xk²+ (1−α)kity−xk². Since

kty−xk² =t²kyk²−2tRehx, yi+kxk² and

kity−xk² =t²kyk²−2tImhx, yi+kxk², hence

P (t) = t²kyk²−2t[αRehx, yi+ (1−α) Imhx, yi] +kxk².

By the definition ofP (see (2.3)), we observe thatP (t)≥0for everyt∈R, therefore ¹₄∆≤0, i.e.,

[(1−α) Imhx, yi+αRehx, yi]²− kxk²kyk² ≤0, proving the second inequality in (2.1).

The first inequality follows by (2.2) and the theorem is proved.

The following particular cases are of interest.

Corollary 2.2. For anyx, y ∈H one has the inequalities:

kty−xk²kyk² ≥ kαk²kyk²−[Rehx, yi]² ≥0;

(2.4)

kity−xk²kyk² ≥ kαk²kyk²−[Imhx, yi]² ≥0;

(2.5) and (2.6) 1

2

kty−xk² +kity−xk²

kyk² ≥ kxk²kyk² −

Imhx, yi+ Rehx, yi 2

2

≥0, for anyt∈R.

The following corollary may be stated as well:

Corollary 2.3. Letx, y ∈ H andM_i, m_i ∈R, i∈ {1,2}such thatM_i ≥ m_i >0, i∈ {1,2}. If either

(2.7) RehM₁y−x, x−m₁yi ≥0 and RehM₂iy−x, x−im₂yi ≥0, or, equivalently,

x− M₁+m₁

2 y

≤ 1

2(M₁−m₁)kyk and (2.8)

x− M₂+m₂ 2 iy

≤ 1

2(M₂−m₂)kyk hold, then

(0≤)kxk²kyk²−[(1−α) Imhx, yi+αRehx, yi]² (2.9)

≤ 1 4kyk⁴

α(M₁−m₁)²+ (1−α) (M₂−m₂)² for anyα∈[0,1].

Proof. It is easy to see that, ifx, z, Z ∈H,then the following statements are equivalent:

(i) RehZ−x, x−zi ≥0, (ii)

x− ^z+Z₂

≤ ¹₂kZ −zk.

Utilising this property one may simply realize that the statements (2.7) and (2.8) are equivalent.

Now, on making use of (2.8) and (2.1), one may deduce the desired inequality (2.9).

Remark 2.4. If one assumes thatM₁ =M₂ =M, m₁ =m₂ =min either (2.7) or (2.8), then (0≤)kxk²kyk²−[(1−α) Imhx, yi+αRehx, yi]²

(2.10)

≤ 1

4kyk⁴(M −m)² for eachα∈[0,1].

Remark 2.5. Corollary 2.3 may be seen as a potential source of some reverse results for the Schwarz inequality. For instance, ifx, y ∈HandM ≥m >0are such that either

(2.11) RehM y −x, x−myi ≥0 or

x− M +m

2 y

≤ 1

2(M −m)kyk hold, then

(2.12) (0≤)kxk²kyk²−[Rehx, yi]² ≤ 1

4(M −m)²kyk⁴. Ifx, y ∈H andN ≥n >0are such that either

(2.13) RehN iy−x, x−niyi ≥0 or

x− N +n 2 iy

≤ 1

2(N −n)kyk hold, then

(2.14) (0≤)kxk²kyk²−[Imhx, yi]² ≤ 1

4(N −n)²kyk⁴. We notice that (2.12) is an improvement of the inequality

(0≤)kxk²kyk²− |hx, yi|² ≤ 1

4(M −m)²kyk⁴ that has been established in [4] under the same condition (2.11) given above.

The following result may be stated as well.

Theorem 2.6. Let(H;h·,·i)be a real or complex inner product space andx, y ∈H, α∈[0,1]. Then

(2.15)

αkty−xk²+ (1−α)ky−txk² αkyk²+ (1−α)kxk²

≥

(1−α)kxk²+αkyk² αkxk²+ (1−α)kyk²

−[Rehx, yi]² ≥0 for anyt∈R.

Proof. Consider the polynomialP :R→Rgiven by

(2.16) P (t) :=αkty−xk²+ (1−α)ky−txk². Since

kty−xk² =t²kyk²−2tRehx, yi+kxk² and

ky−txk² =t²kxk²−2tRehx, yi+kyk², hence

P (t) =

αkyk²+ (1−α)kxk²

t²−2tRehx, yi+

αkxk²+ (1−α)kyk²

for anyt∈R.

By the definition of P (see (2.16)), we observe that P(t) ≥ 0 for every t ∈ R, therefore

1

4∆≤0,i.e., the second inequality in (2.15) holds true.

The first inequality follows by (2.2) and the theorem is proved.

Remark 2.7. We observe that if either α = 0 or α = 1, then (2.15) will generate the same reverse of the Schwarz inequality as (2.4) does.

Corollary 2.8. Ifx, y ∈H,then

(2.17) kty−xk²+ky−txk²

2 · kxk²+kyk²

2 ≥ kxk²+kyk² 2

!2

−[Rehx, yi]² ≥0

for anyt∈Rand (2.18) kx±yk²

αkyk²+ (1−α)kxk²

≥

(1−α)kxk²+αkyk² αkxk²+ (1−α)kyk²

−[Rehx, yi]² ≥0 for anyα∈[0,1].

In particular, we have

(2.19) kx±yk²· kxk²+kyk² 2

!

≥ kxk² +kyk² 2

!2

−[Rehx, yi]² ≥0.

In [7, p. 210], C.S. Lin has proved the following reverse of the Schwarz inequality in real or complex inner product spaces(H;h·,·i) :

(0≤)kxk²kyk²− |hx, yi|² ≤ 1

r² kxk²kx−ryk² for anyr∈R,r6= 0andx, y ∈H.

The following slightly more general result may be stated:

Theorem 2.9. Let(H;h·,·i)be a real or complex inner product space. Then for anyx, y ∈H andα ∈K(C,R)withα 6= 0we have

(2.20) (0≤)kxk²kyk²− |hx, yi|² ≤ 1

|α|² kxk²kx−αyk². The case of equality holds in (2.20) if and only if

(2.21) Rehx, αyi=kxk².

Proof. Observe that

I(α) :=kxk²kx−αyk²− |α|²

kxk²kyk²− |hx, yi|²

=kxk²

kxk²−2 Re [ ¯αhx, yi] +|α|²kyk²

− |α|²kxk²kyk²+|α|²|hx, yi|²

=kxk⁴−2kxk²Re [ ¯αhx, yi] +|α|²|hx, yi|². Since

(2.22) Re [ ¯αhx, yi]≤ |α| |hx, yi|, hence

I(α)≥ kxk⁴−2kxk²|α| |hx, yi|+|α|²|hx, yi|² (2.23)

= kxk²− |α| |hx, yi|²

≥0.

Conversely, if (2.21) holds true, thenI(α) = 0,showing that the equality case holds in (2.20).

Now, if the equality case holds in (2.20), then we must have equality in (2.22) and in (2.23) implying that

Re [hx, αyi] =|α| |hx, yi| and |α| |hx, yi|=kxk²

which imply (2.21).

The following result may be stated.

Corollary 2.10. Let (H;h·,·i) be as above and x, y ∈ H, α ∈ K\ {0} andr > 0 such that

|α| ≥r.If

(2.24) kx−αyk ≤rkyk,

then

(2.25)

q

|α|²−r²

|α| ≤ |hx, yi|

kxk kyk(≤1). Proof. From (2.24) and (2.20) we have

kxk²kyk²− |hx, yi|² ≤ r²

|α|² kxk²kyk², that is,

|α|²−r²

|α|² kxk²kyk² ≤ |hx, yi|²,

which is clearly equivalent to (2.25).

Remark 2.11. Since forΓ, γ ∈Kthe following statements are equivalent (i) RehΓy−x, x−γyi ≥0,

(ii)

x− ^γ+Γ₂ ·y

≤ ¹₂|Γ−γ| kyk, hence by Corollary 2.10 we deduce

(2.26) 2 [Re (Γ¯γ)]¹²

|Γ +γ| ≤ |hx, yi|

kxk kyk,

providedRe (Γ¯γ)>0,an inequality that has been obtained in a different way in [3].

Corollary 2.12. Ifx, y ∈H, α∈K\ {0}andρ >0such thatkx−αyk ≤ρ,then (2.27) (0≤)kxk²kyk²− |hx, yi|² ≤ ρ²

|α|²kxk². 3. OTHER INEQUALITIES

The following result holds.

Proposition 3.1. Letx, y ∈H\ {0}andε∈(0,¹₂]. If

(3.1) (0≤) 1−ε−√

1−2ε≤ kxk

kyk ≤1−ε+√

1−2ε, then

(3.2) (0≤)kxk kyk −Rehx, yi ≤εkx−yk².

Proof. Ifx=y,then (3.2) is trivial.

Supposex6=y.Utilising the inequality (2.5) from [6], we can state that kxk kyk −Rehx, yi

kx−yk² ≤ 2kxk kyk (kxk+kyk)² for anyx, y ∈H\ {0}, x6=y.

Now, if we assume that

2kxk kyk

(kxk+kyk)² ≤ε, then, after some manipulation, we get that

εkxk²+ 2 (ε−1)kxk kyk+εkyk² ≥0, which, forε∈(0,¹₂]andy 6= 0,is clearly equivalent to (3.1).

The proof is complete.

The following result may be stated:

Proposition 3.2. Let(H;h·,·i)be a real or complex inner product space. Then for anyx, y ∈H andϕ ∈Rone has:

kxk kyk −[cos 2ϕ·Rehx, yi+ sin 2ϕ·Imhx, yi]

≤ 1

2[|cosϕ| kx−yk+|sinϕ| kx+yk]². Proof. For ϕ ∈ R, consider the complex number α = cosϕ −isinϕ. Thenα² = cos 2ϕ− isin 2ϕ,|α|= 1and by the inequality (1.6) we deduce the desired result.

From a different perspective, we may consider the following results as well:

Theorem 3.3. Let(H;h·,·i)be a real or complex inner product space,α ∈Kwith|α−1|= 1.

Then for anye∈H withkek= 1andx, y ∈H,we have

(3.3) |hx, yi −αhx, ei he, yi| ≤ kxk kyk. The equality holds in (3.3) if and only if there exists aλ∈Ksuch that

(3.4) αhx, eie=x+λy.

Proof. It is known that foru, v ∈H,we have equality in the Schwarz inequality

(3.5) |hu, vi| ≤ kuk kvk

if and only if there exists aλ ∈Ksuch thatu=λv.

If we apply (3.5) foru=αhx, eie−x, v =y,we get

(3.6) |hαhx, eie−x, yi| ≤ kαhx, eie−xk kyk with equality iff there exists aλ ∈Ksuch that

αhx, eie=x+λy.

Since

kαhx, eie−xk² =|α|²|hx, ei|²−2 Re [α]|hx, ei|²+kxk²

= |α|²−2 Re [α]

|hx, ei|²+kxk²

= |α−1|²−1

|hx, ei|² +kxk²

=kxk²

and

hαhx, eie−x, yi=αhx, ei he, yi − hx, yi

hence by (3.6) we deduce the desired inequality (3.3).

Remark 3.4. Ifα= 0in (3.3), then it reduces to the Schwarz inequality.

Remark 3.5. Ifα6= 0,then (3.3) is equivalent to (3.7)

hx, ei he, yi − 1 αhx, yi

≤ 1

|α|kxk kyk.

Utilising the continuity property of modulus for complex numbers, i.e.,|z−w| ≥ ||z| − |w||

we then obtain

|hx, ei he, yi| − 1

|α||hx, yi|

≤ 1

|α|kxk kyk, which implies that

(3.8) |hx, ei he, yi| ≤ 1

|α|[|hx, yi|+kxk kyk]. Fore= _kzk^z , z6= 0,we get from (3.8) that

(3.9) |hx, zi hz, yi| ≤ 1

|α|[|hx, yi|+kxk kyk]kzk² for anyα∈K\ {0}with|α−1|= 1andx, y, z ∈H.

Forα= 2,we get from (3.9) the Buzano inequality [1]

(3.10) |hx, zi hz, yi| ≤ 1

2[|hx, yi|+kxk kyk]kzk² for anyx, y, z∈H.

Remark 3.6. In the case of real spaces the condition|α−1|= 1is equivalent to eitherα= 0 orα = 2.Forα= 2we deduce from (3.7) that

(3.11) 1

2[hx, yi − kxk kyk]≤ hx, ei he, yi ≤ 1

2[hx, yi+kxk kyk]

for anyx, y ∈H ande∈H withkek= 1,which implies Richard’s inequality [8]:

(3.12) 1

2[hx, yi − kxk kyk]kzk² ≤ hx, zi hz, yi ≤ 1

2[hx, yi+kxk kyk]kzk², for anyx, y, z∈H.

The following result concerning a generalisation for orthornormal families of the inequality (3.3) may be stated.

Theorem 3.7. Let (H;h·,·i) be a real or complex inner product space, {e_i}_i∈F a finite or- thonormal family, i.e., he_i, e_ji = δ_ij fori, j ∈ F, where δ_ij is Kronecker’s delta andα_i ∈ K, i∈F such that|α_i−1|= 1for eachi∈F.Then

(3.13)

hx, yi −X

i∈F

α_ihx, e_ii he_i, yi

≤ kxk kyk.

The equality holds in (3.13) if and only if there exists a constantλ∈Ksuch that

(3.14) X

i∈F

α_ihx, e_iie_i =x+λy.

Proof. As above, by Schwarz’s inequality, we have

(3.15)

* X

i∈F

α_ihx, e_iie_i−x, y +

≤

X

i∈F

α_ihx, e_iie_i−x

kyk with equality if and only if there exists aλ∈Ksuch thatP

i∈F α_ihx, e_iie_i =x+λy.

Since

X

i∈F

α_ihx, e_iie_i−x

2

=kxk²−2 Re

* x,X

i∈F

α_ihx, e_iie_i +

+

X

i∈F

α_ihx, e_iie_i

2

=kxk²−2X

i∈F

α_ihx, e_ii hx, e_ii+X

i∈F

|α_i|²|hx, e_ii|²

=kxk²+X

i∈F

|hx, e_ii|² |α_i|²−2 Reα_i

=kxk²+X

i∈F

|hx, e_ii|² |α_i−1|²−1

=kxk²,

hence the inequality (3.13) is obtained.

Remark 3.8. If the space is real, then the nontrivial case one can get from (3.13) is for all α_i = 2,obtaining the inequality

(3.16) 1

2[hx, yi − kxk kyk]≤X

i∈F

hx, e_ii he_i, yi ≤ 1

2[hx, yi+kxk kyk]

that has been obtained in [5].

Corollary 3.9. With the above assumptions, we have

X

i∈F

α_ihx, e_ii he_i, yi

≤ |hx, yi|+

hx, yi −X

i∈F

α_ihx, e_ii he_i, yi (3.17)

≤ |hx, yi|+kxk kyk, x, y ∈H, where|α_i−1|= 1for eachi∈F and{e_i}_i∈F is an orthonormal family inH.

4. APPLICATIONS FOR THE TRIANGLE INEQUALITY

In 1966, Diaz and Metcalf [2] proved the following reverse of the triangle inequality:

(4.1)

n

X

i=1

x_i

≥r

n

X

i=1

kx_ik,

provided the vectorsx_i in the inner product space (H;h·,·i)over the real or complex number fieldKare nonzero and

(4.2) 0≤r≤ Rehx_i, ai

kx_ik for each i∈ {1, . . . , n}, wherea∈H,kak= 1.The equality holds in (4.2) if and only if (4.3)

n

X

i=1

x_i =r

n

X

i=1

kx_ik

! a.

The following result may be stated:

Proposition 4.1. Let e ∈ H with kek = 1, ε ∈ 0,¹₂

and xi ∈ H, i ∈ {1, . . . , n} with the property that

(4.4) (0≤) 1−ε−√

1−2ε≤ kx_ik ≤1−ε+√ 1−2ε for eachi∈ {1, . . . , n}.Then

(4.5)

n

X

i=1

kx_ik ≤

n

X

i=1

x_i

+ε

n

X

i=1

kx_i−ek².

Proof. Utilising Proposition 3.1 forx=x_i andy=e,we can state that kx_ik −Rehx_i, ei ≤εkx_i−ek²

for eachi∈ {1, . . . , n}.Summing overifrom1ton,we deduce that (4.6)

n

X

i=1

kx_ik ≤Re

* _n X

i=1

x_i, e +

+ε

n

X

i=1

kx_i−ek².

By Schwarz’s inequality in(H;h·,·i),we also have Re

* _n X

i=1

xi, e +

≤

Re

* _n X

i=1

xi, e +

≤

* _n X

i=1

xi, e +

(4.7)

≤

n

X

i=1

x_i

kek=

n

X

i=1

x_i .

Therefore, by (4.6) and (4.7) we deduce (4.5).

In the same spirit, we can prove the following result as well:

Proposition 4.2. Let (H;h·,·i) be a real or complex inner product space and e ∈ H with kek= 1.Then for anyϕ∈Rone has the inequality:

(4.8)

n

X

i=1

kx_ik ≤

n

X

i=1

x_i

+1 2

n

X

i=1

[|cosϕ| kx_i−ek+|sinϕ| kx_i+ek]².

Proof. Applying Proposition 3.2 forx=x_i andy=e,we have:

(4.9) kx_ik ≤cos 2ϕ·Rehx_i, ei+ sin 2ϕ·Imhx_i, ei + 1

2[|cosϕ| kx_i−ek+|sinϕ| kx_i+ek]² for anyi∈ {1, . . . , n}.

Summing in (4.5) overifrom1ton,we have:

(4.10)

n

X

i=1

kx_ik ≤cos 2ϕ·Re

* _n X

i=1

x_i, e +

+ sin 2ϕ·Im

* _n X

i=1

x_i, e +

+ 1 2

n

X

i=1

[|cosϕ| kx_i−ek+|sinϕ| kx_i+ek]².

Now, by the elementary inequality for the real numbersm, p, M andP, mM +pP ≤ m²+p²¹₂

M²+P²¹₂ ,

we have

cos 2ϕ·Re

* _n X

i=1

xi, e +

+ sin 2ϕ·Im

* _n X

i=1

xi, e + (4.11)

≤ cos²2ϕ+ sin²2ϕ¹₂





"

Re

* _n X

i=1

xi, e +#2

+

"

Im

* _n X

i=1

xi, e +#2



1 2

=

* _n X

i=1

xi, e +

≤

n

X

i=1

xi

kek=

n

X

i=1

xi

,

where for the last inequality we used Schwarz’s inequality in(H;h·,·i).

Finally, by (4.10) and (4.11) we deduce the desired result (4.8).

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[1] M.L. BUZANO, Generalizzazione della disiguaglianza di Cauchy-Schwarz (Italian), Rend. Sem.

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