Some new refinements of the Schwarz inequality in inner product spaces are given

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

Volume 5, Issue 3, Article 60, 2004

REFINEMENTS OF THE SCHWARZ AND HEISENBERG INEQUALITIES IN HILBERT SPACES

S.S. DRAGOMIR

SCHOOL OFCOMPUTERSCIENCE ANDMATHEMATICS

VICTORIAUNIVERSITY OFTECHNOLOGY

PO BOX14428 MELBOURNEVIC 8001

AUSTRALIA.

sever.dragomir@vu.edu.au

URL:http://rgmia.vu.edu.au/SSDragomirWeb.html

Received 24 August, 2004; accepted 17 September, 2004 Communicated by S. Saitoh

ABSTRACT. Some new refinements of the Schwarz inequality in inner product spaces are given.

Applications for discrete and integral inequalities including the Heisenberg inequality for vector- valued functions in Hilbert spaces are provided.

Key words and phrases: Schwarz inequality, Triangle inequality, Heisenberg Inequality.

2000 Mathematics Subject Classification. Primary 46C05; Secondary 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

(1.1) |hx, yi|² ≤ kxk²kyk², x, y ∈H

with equality iffxandyare linearly dependent.

In 1966, S. Kurepa [1] established the following refinement of the Schwarz inequality in inner product spaces that generalises de Bruijn’s result for sequences of real and complex numbers [2].

Theorem 1.1. LetH be a real Hilbert space andH_C the complexification ofH.Then for any pair of vectorsa∈H, z ∈H_C

(1.2) |hz, ai|² ≤ 1

2kak² kzk²+|hz,zi|¯

≤ kak²kzk².

In 1985, S.S. Dragomir [3, Theorem 2] obtained a different refinement of (1.1), namely:

ISSN (electronic): 1443-5756

156-04

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Theorem 1.2. Let (H;h·,·i)be a real or complex inner product space and x, y, e ∈ H with kek= 1.Then we have the inequality

(1.3) kxk kyk ≥ |hx, yi − hx, ei he, yi|+|hx, ei he, yi| ≥ |hx, yi|.

In the same paper [3, Theorem 3], a further generalisation for orthonormal families has been given (see also [4, Theorem 3]).

Theorem 1.3. Let {ei}_i∈H be an orthonormal family in the Hilbert space H. Then for any x, y ∈H

kxk kyk ≥

hx, yi −X

i∈I

hx, e_ii he_i, yi

+X

i∈I

|hx, e_ii he_i, yi|

(1.4)

≥

hx, yi −X

i∈I

hx, e_ii he_i, yi

+

X

i∈I

hx, e_ii he_i, yi

≥ |hx, yi|.

The inequality (1.3) has also been obtained in [4] as a particular case of the following result.

Theorem 1.4. Letx, y, a, b∈Hbe such that

kak² ≤2 Rehx, ai, kbk² ≤2 Rehy, bi. Then we have:

(1.5) kxk kyk ≥ 2 Rehx, ai − kak²¹₂

2 Rehy, bi − kbk²¹₂

+|hx, yi − hx, bi − ha, yi+ha, bi|. Another refinement of the Schwarz inequality for orthornormal vectors in inner product spaces has been obtained by S.S. Dragomir and J. Sándor in [5, Theorem 5].

Theorem 1.5. Let{e_i}i∈{1,...,n} be orthornormal vectors in the inner product space(H;h·,·i).

Then

(1.6) kxk kyk − |hx, yi| ≥

n

X

i=1

|hx, e_ii|²

n

X

i=1

|hy, e_ii|²

!¹₂

−

n

X

i=1

hx, e_ii he_i, yi

≥0

and

(1.7) kxk kyk −Rehx, yi ≥

n

X

i=1

|hx, e_ii|²

n

X

i=1

|hy, e_ii|²

!¹₂

−

n

X

i=1

Re [hx, e_ii he_i, yi]≥0.

For some properties of superadditivity, monotonicity, strong superadditivity and strong monotonicity of Schwarz’s inequality, see [6]. Here we note only the following refinements of the Schwarz inequality in its different variants for linear operators [6]:

a) Let H be a Hilbert space and A, B : H → H two selfadjoint linear operators with A≥B ≥0,then we have the inequalities

(1.8) hAx, xi¹² hAy, yi¹² − |hAx, yi| ≥ hBx, xi¹² hBy, yi¹² − |hBx, yi| ≥0 and

(1.9) hAx, xi hAy, yi − |hAx, yi|² ≥ hBx, xi hBy, yi − |hBx, yi|² ≥0 for anyx, y ∈H.

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b) LetA:H →Hbe a bounded linear operator onHand letkAk= sup{kAxk,kxk= 1}

the norm of A.Then one has the inequalities

(1.10) kAk²(kxk kyk − |hx, yi|)≥ kAxk kAyk − |hAx, Ayi| ≥0 and

(1.11) kAk⁴ kxk²kyk²− |hx, yi|²

≥ kAxk²kAyk²− |hAx, Ayi|² ≥0.

c) Let B : H → H be a linear operator with the property that there exists a constant m >0such thatkBxk ≥mkxkfor anyx∈H.Then we have the inequalities

(1.12) kBxk kByk − |hBx, Byi| ≥m²(kxk kyk − |hx, yi|)≥0 and

(1.13) kBxk²kByk²− |hBx, Byi|² ≥m⁴ kxk²kyk²− |hx, yi|²

≥0.

For other results related to Schwarz’s inequality in inner product spaces, see Chapter XX of [8] and the references therein.

Motivated by the results outlined above, it is the aim of this paper to explore other avenues in obtaining new refinements of the celebrated Schwarz inequality. Applications for vector- valued sequences and integrals in Hilbert spaces are mentioned. Refinements of the Heisenberg inequality for vector-valued functions in Hilbert spaces are also given.

2. SOME NEW REFINEMENTS

The following result holds.

Theorem 2.1. Let(H;h·,·i)be an inner product space over the real or complex number field Kandr₁, r₂ >0.Ifx, y ∈Hsatisfy the property

(2.1) kx−yk ≥r₂ ≥r₁ ≥ |kxk − kyk|,

then we have the following refinement of Schwarz’s inequality

(2.2) kxk kyk −Rehx, yi ≥ 1

2 r₂²−r²₁

(≥0).

The constant ¹₂ is best possible in the sense that it cannot be replaced by any larger quantity.

Proof. From the first inequality in (2.1) we have

(2.3) kxk²+kyk² ≥r²₂+ 2 Rehx, yi. Subtracting in (2.3) the quantity2kxk kyk,we get

(2.4) (kxk − kyk)² ≥r₂²−2 (kxk kyk −Rehx, yi). Since, by the second inequality in (2.1) we have

(2.5) r²₁ ≥(kxk − kyk)²,

hence from (2.4) and (2.5) we deduce the desired inequality (2.2).

To prove the sharpness of the constant ¹₂ in (2.2), let us assume that there is a constantC > 0 such that

(2.6) kxk kyk −Rehx, yi ≥C r₂²−r²₁ , provided thatxandysatisfy (2.1).

Lete∈H withkek= 1and forr₂ > r₁ >0,define

(2.7) x= r₂ +r₁

2 ·e and y= r₁−r₂ 2 ·e.

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Then

kx−yk=r₂and |kxk − kyk|=r₁, showing that the condition (2.1) is fulfilled with equality.

If we replacexandyas defined in (2.7) into the inequality (2.6), then we get r²₂−r₁²

2 ≥C r₂²−r²₁ ,

which implies thatC ≤ ¹₂,and the theorem is completely proved.

The following corollary holds.

Corollary 2.2. With the assumptions of Theorem 2.1, we have the inequality:

(2.8) kxk+kyk −

√2

2 kx+yk ≥

√2 2

q

r²₂−r²₁.

Proof. We have, by (2.2), that

(kxk+kyk)²− kx+yk² = 2 (kxk kyk −Rehx, yi)≥r²₂−r₁² ≥0 which gives

(2.9) (kxk+kyk)² ≥ kx+yk²+

q

r²₂−r²₁ 2

. By making use of the elementary inequality

2 α²+β²

≥(α+β)², α, β ≥0;

we get

(2.10) kx+yk²+

q

r²₂−r₁² 2

≥ 1 2

kx+yk+ q

r²₂−r₁² 2

.

Utilising (2.9) and (2.10), we deduce the desired inequality (2.8).

If (H;h·,·i) is a Hilbert space and {e_i}_i∈I is an orthornormal family in H, i.e., we recall thathe_i, e_ji = δ_ij for anyi, j ∈ I,whereδ_ij is Kronecker’s delta, then we have the following inequality which is well known in the literature as Bessel’s inequality

(2.11) X

i∈I

|hx, e_ii|² ≤ kxk² for each x∈H.

Here, the meaning of the sum is X

i∈I

|hx, e_ii|² = sup

F⊂I

( X

i∈F

|hx, e_ii|², F is a finite part ofI )

.

The following result providing a refinement of the Bessel inequality (2.11) holds.

Theorem 2.3. Let (H;h·,·i)be a Hilbert space and {e_i}_i∈I an orthornormal family in H.If x∈H, x6= 0,andr₂, r₁ >0are such that:

(2.12)

x−X

i∈I

hx, e_iie_i

≥r₂ ≥r₁ ≥ kxk − X

i∈I

|hx, e_ii|²

!¹₂

(≥0), then we have the inequality

(2.13) kxk − X

i∈I

|hx, e_ii|²

!¹₂

≥ 1

2· r²₂−r₁² P

i∈I|hx, e_ii|²¹₂

(≥0).

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The constant ¹₂ is best possible.

Proof. Consider y := P

i∈Ihx, e_iie_i.Obviously, since H is a Hilbert space,y ∈ H.We also note that

kyk=

X

i∈I

hx, e_iie_i

= v u u t

X

i∈I

hx, e_iie_i

2

= s

X

i∈I

|hx, e_ii|², and thus (2.12) is in fact (2.1) of Theorem 2.1.

Since

kxk kyk −Rehx, yi=kxk X

i∈I

|hx, e_ii|²

!¹₂

−Re

* x,X

i∈I

hx, e_iie_i +

= X

i∈I

|hx, e_ii|²

!¹₂ 

kxk − X

i∈I

|hx, e_ii|²

!¹₂

,

hence, by (2.2), we deduce the desired result (2.13).

We will prove the sharpness of the constant for the case of one element, i.e., I = {1}, e₁ =e∈H,kek= 1.For this, assume that there exists a constantD >0such that

(2.14) kxk − |hx, ei| ≥D· r₂²−r²₁

|hx, ei|

providedx∈H\ {0}satisfies the condition

(2.15) kx− hx, eiek ≥r₂ ≥r₁ ≥ kxk − |hx, ei|.

Assume thatx =λe+µf withe, f ∈ H,kek = kfk = 1 ande ⊥ f.We wish to see if there exists positive numbersλ, µsuch that

(2.16) kx− hx, eiek=r₂ > r₁ =kxk − |hx, ei|. Since (forλ, µ >0)

kx− hx, eiek=µ and

kxk − |hx, ei|=p

λ²+µ²−λ hence, by (2.16), we getµ=r₂ and

q

λ²+r₂²−λ=r₁ giving

λ²+r²₂ =λ²+ 2λr₁+r²₁ from where we get

λ = r²₂−r₁² 2r1

>0.

With these values forλandµ,we have

kxk − |hx, ei|=r₁, |hx, ei|= r₂²−r²₁ 2r1

and thus, from (2.14), we deduce

r₁ ≥D· r²₂−r₁²

r²₂−r²₁ 2r1

,

givingD≤ ¹₂.This proves the theorem.

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The following corollary is obvious.

Corollary 2.4. Letx, y ∈Hwithhx, yi 6= 0andr₂ ≥r₁ >0such that

kykx− hx, yi kyk ·y

≥r₂kyk ≥r₁kyk (2.17)

≥ kxk kyk − |hx, yi|(≥0). Then we have the following refinement of the Schwarz’s inequality:

(2.18) kxk kyk − |hx, yi| ≥ 1

2 r²₂−r₁² kyk²

|hx, yi|(≥0). The constant ¹₂ is best possible.

The following lemma holds.

Lemma 2.5. Let(H;h·,·i)be an inner product space andR ≥1.Forx, y ∈H,the subsequent statements are equivalent:

(i) The following refinement of the triangle inequality holds:

(2.19) kxk+kyk ≥Rkx+yk;

(ii) The following refinement of the Schwarz inequality holds:

2 R²−1

kx+yk².

Proof. Taking the square in (2.19), we have (2.21) 2kxk kyk ≥ R²−1

kxk²+ 2R²Rehx, yi+ R²−1 kyk². Subtracting from both sides of (2.21) the quantity2 Rehx, yi,we obtain

2 (kxk kyk −Rehx, yi)≥ R²−1 kxk²+ 2 Rehx, yi+kyk²

= R²−1

kx+yk²,

which is clearly equivalent to (2.20).

By the use of the above lemma, we may now state the following theorem concerning another refinement of the Schwarz inequality.

Theorem 2.6. Let(H;h·,·i)be an inner product space over the real or complex number field andR ≥1, r≥0.Ifx, y ∈Hare such that

(2.22) 1

R (kxk+kyk)≥ kx+yk ≥r, then we have the following refinement of the Schwarz inequality

2 R²−1 r².

The constant ¹₂ is best possible in the sense that it cannot be replaced by a larger quantity.

Proof. The inequality (2.23) follows easily from Lemma 2.5. We need only prove that ¹₂ is the best possible constant in (2.23).

Assume that there exists aC > 0such that

(2.24) kxk kyk −Rehx, yi ≥C R² −1

r² providedx, y, Randrsatisfy (2.22).

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Considerr = 1, R >1and choosex= ^1−R₂ e, y = ^1+R₂ ewithe∈H,kek= 1.Then x+y =e, kxk+kyk

R = 1

and thus (2.22) holds with equality on both sides.

From (2.24), for the above choices, we have ¹₂(R²−1) ≥ C(R²−1), which shows that

C ≤ ¹₂.

Finally, the following result also holds.

Theorem 2.7. Let(H;h·,·i)be an inner product space over the real or complex number field Kandr∈(0,1].Forx, y ∈H,the following statements are equivalent:

(i) We have the inequality

(2.25) |kxk − kyk| ≤rkx−yk;

(ii) We have the following refinement of the Schwarz inequality

2 1−r²

kx−yk².

The constant ¹₂ in (2.26) is best possible.

Proof. Taking the square in (2.25), we have

kxk²−2kxk kyk+kyk² ≤r² kxk²−2 Rehx, yi+kyk² which is clearly equivalent to

1−r² kxk²−2 Rehx, yi+kyk²

≤2 (kxk kyk −Rehx, yi) or with (2.26).

Now, assume that (2.26) holds with a constantE >0,i.e., (2.27) kxk kyk −Rehx, yi ≥E 1−r²

kx−yk², provided (2.25) holds.

Definex= ^r+1₂ e, y= ^r−1₂ ewithe∈H,kek= 1.Then

|kxk − kyk|=r, kx−yk= 1 showing that (2.25) holds with equality.

If we replace x and y in (2.27), then we get E(1−r²) ≤ ¹₂ (1−r²), implying that E ≤

1

2.

3. DISCRETEINEQUALITIES

Assume that(K; (·,·))is a Hilbert space over the real or complex number field. Assume also thatp_i ≥0,i∈HwithP∞

i=1p_i = 1and define

`²_p(K) :=

(

x:= (x_i)_i∈

N

x_i ∈K, i∈N and

∞

X

i=1

p_ikx_ik² <∞ )

.

It is well known that`²_p(K)endowed with the inner producth·,·i_pdefined by hx,yi_p :=

∞

X

i=1

p_i(x_i, y_i)

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and generating the norm

kxk_p :=

∞

X

i=1

p_ikx_ik²

!¹₂

is a Hilbert space overK.

We may state the following discrete inequality improving the Cauchy-Bunyakovsky-Schwarz classical result.

Proposition 3.1. Let (K; (·,·)) be a Hilbert space and p_i ≥ 0 (i∈N) with P∞

i=1p_i = 1.

Assume thatx,y∈`²_p(K)andr₁, r₂ >0satisfy the condition (3.1) kxi−yik ≥r2 ≥r1 ≥ |kxik − kyik|

for each i ∈ N. Then we have the following refinement of the Cauchy-Bunyakovsky-Schwarz inequality

(3.2)

∞

X

i=1

p_ikx_ik²

∞

X

i=1

p_iky_ik²

!¹₂

−

∞

X

i=1

p_iRe (x_i, y_i)≥ 1

2 r²₂ −r₁²

≥0.

The constant ¹₂ is best possible.

Proof. From the condition (3.1) we simply deduce

∞

X

i=1

p_ikx_i −y_ik² ≥r²₂ ≥r₁² ≥

∞

X

i=1

p_i(kx_ik − ky_ik)² (3.3)

≥





∞

X

i=1

p_ikx_ik²

!¹₂

−

∞

X

i=1

p_iky_ik²

!¹₂



2

.

In terms of the normk·k_p,the inequality (3.3) may be written as

(3.4) kx−yk_p ≥r2 ≥r1 ≥

kxk_p− kyk_p . Utilising Theorem 2.1 for the Hilbert space

`²_p(K),h·,·i_p

,we deduce the desired inequality (3.2).

Forn = 1 (p1 = 1),the inequality (3.2) reduces to (2.2) for which we have shown that ¹₂ is

the best possible constant.

By the use of Corollary 2.2, we may state the following result as well.

Corollary 3.2. With the assumptions of Proposition 3.1, we have the inequality

(3.5)

∞

X

i=1

p_ikx_ik²

!¹₂ +

∞

X

i=1

p_iky_ik²

!¹₂

−

√2 2

∞

X

i=1

p_ikx_i+y_ik²

!¹₂

≥

√2 2

q

r²₂−r₁².

The following proposition also holds.

i=1p_i = 1.

Assume thatx,y∈`²_p(K)andR≥1, r ≥0satisfy the condition

(3.6) 1

R(kx_ik+ky_ik)≥ kx_i+y_ik ≥r

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for eachi∈N. Then we have the following refinement of the Schwarz inequality

(3.7)

∞

X

i=1

p_ikx_ik²

∞

X

i=1

p_iky_ik²

!¹₂

−

∞

X

i=1

2 R²−1 r².

The constant ¹₂ is best possible in the sense that it cannot be replaced by a larger quantity.

Proof. By (3.6) we deduce

(3.8) 1

R

" _∞ X

i=1

p_i(kx_ik+ky_ik)²

#¹₂

≥

∞

X

i=1

p_ikx_i+y_ik²

!¹₂

≥r.

By the classical Minkowsky inequality for nonnegative numbers, we have

(3.9)

∞

X

i=1

pikxik²

!¹₂ +

∞

X

i=1

pikyik²

!¹₂

≥

" _∞ X

i=1

pi(kxik+kyik)²

#¹₂ ,

and thus, by utilising (3.8) and (3.9), we may state in terms ofk·k_pthe following inequality

(3.10) 1

R

kxk_p+kyk_p

≥ kx+yk_p ≥r.

Employing Theorem 2.6 for the Hilbert space`²_p(K)and the inequality (3.10), we deduce the desired result (3.7).

Since, forp= 1, n= 1,(3.7) is reduced to (2.23) for which we have shown that ¹₂ is the best constant, we conclude that ¹₂ is the best constant in (3.7) as well.

Finally, we may state and prove the following result incorporated in

i=1p_i = 1.

Assume thatx,y∈`²_p(K)andr∈(0,1]such that

(3.11) |kx_ik − ky_ik| ≤rkx_i−y_ik for eachi∈N, holds true. Then we have the following refinement of the Schwarz inequality

(3.12)

∞

X

i=1

p_ikx_ik²

∞

X

i=1

p_iky_ik²

!¹₂

−

∞

X

i=1

2 1−r²

∞

X

i=1

p_ikx_i−y_ik².

The constant ¹₂ is best possible in (3.12).

Proof. From (3.11) we have

" _∞ X

i=1

p_i(kx_ik − ky_ik)²

#¹₂

≤r

" _∞ X

i=1

p_ikx_i−y_ik²

#¹₂ . Utilising the following elementary result

∞

X

i=1

pikxik²

!¹₂

−

∞

X

i=1

pikyik²

!¹₂

≤

∞

X

i=1

pi(kxik − kyik)²

!¹₂ , we may state that

kxk_p− kyk_p

≤rkx−yk_p.

Now, by making use of Theorem 2.7, we deduce the desired inequality (3.12) and the fact that

1

2 is the best possible constant. We omit the details.

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4. INTEGRALINEQUALITIES

Assume that (K; (·,·)) is a Hilbert space over the real or complex number field K. If ρ : [a, b] ⊂ R → [0,∞) is a Lebesgue integrable function with Rb

a ρ(t)dt = 1, then we may consider the spaceL²_ρ([a, b] ;K)of all functions f : [a, b] → K, that are Bochner measurable and Rb

a ρ(t)kf(t)k²dt < ∞. It is known thatL²_ρ([a, b] ;K) endowed with the inner product h·,·i_ρdefined by

hf, gi_ρ:=

Z b a

ρ(t) (f(t), g(t))dt

and generating the norm

kfk_ρ:=

Z b a

ρ(t)kf(t)k²dt ¹2

is a Hilbert space overK.

Now we may state and prove the first refinement of the Cauchy-Bunyakovsky-Schwarz integral inequality.

Proposition 4.1. Assume thatf, g∈L²_ρ([a, b] ;K)andr₂, r₁ >0satisfy the condition (4.1) kf(t)−g(t)k ≥r₂ ≥r₁ ≥ |kf(t)k − kg(t)k|

for a.e. t∈[a, b].Then we have the inequality

(4.2)

Z b a

ρ(t)kf(t)k²dt Z b

a

ρ(t)kg(t)k²dt ¹2

− Z b

a

ρ(t) Re (f(t), g(t))dt ≥ 1

2 r₂²−r₁²

(≥0).

Proof. Integrating (4.1), we get

(4.3)

Z b a

ρ(t) (kf(t)−g(t)k)²dt ¹2

≥r₂ ≥r₁ ≥ Z b

a

ρ(t) (kf(t)k − kg(t)k)²dt

1 2

. Utilising the obvious fact

(4.4)

Z b a

ρ(t) (kf(t)k − kg(t)k)²dt ¹2

≥

Z b a

ρ(t)kf(t)k²dt ¹2

− Z b

a

ρ(t)kg(t)k²dt ¹2

,

we can state the following inequality in terms of thek·k_ρnorm:

(4.5) kf −gk_ρ≥r₂ ≥r₁ ≥

kfk_ρ− kgk_ρ .

Employing Theorem 2.1 for the Hilbert spaceL²_ρ([a, b] ;K),we deduce the desired inequality (4.2).

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To prove the sharpness of ¹₂ in (4.2), we choose a = 0, b = 1, f(t) = 1, t ∈ [0,1]and f(t) =x, g(t) =y, t∈[a, b], x, y ∈K.Then (4.2) becomes

kxk kyk −Rehx, yi ≥ 1

2 r²₂−r²₁ provided

kx−yk ≥r₂ ≥r₁ ≥ |kxk − kyk|,

which, by Theorem 2.1 has the quantity ¹₂ as the best possible constant.

The following corollary holds.

Corollary 4.2. With the assumptions of Proposition 4.1, we have the inequality

(4.6)

Z b a

ρ(t)kf(t)k²dt

1 2

+ Z b

a

ρ(t)kg(t)k²dt

1 2

−

√2 2

Z b a

ρ(t)kf(t) +g(t)k²dt ¹₂

≥

√2 2

q

r²₂−r₁². The following two refinements of the Cauchy-Bunyakovsky-Schwarz (CBS) integral inequality also hold.

Proposition 4.3. Iff, g∈L²_ρ([a, b] ;K)andR ≥1, r ≥0satisfy the condition

(4.7) 1

R(kf(t)k+kg(t)k)≥ kf(t) +g(t)k ≥r for a.e. t∈[a, b],then we have the inequality

(4.8)

Z b a

ρ(t)kf(t)k²dt Z b

a

ρ(t)kg(t)k²dt

1 2

− Z b

a

ρ(t) Re (f(t), g(t))dt≥ 1

2 R²−1 r².

The proof follows by Theorem 2.6 and we omit the details.

Proposition 4.4. Iff, g∈L²_ρ([a, b] ;K)andζ ∈(0,1]satisfy the condition (4.9) |kf(t)k − kg(t)k| ≤ζkf(t)−g(t)k

for a.e. t∈[a, b],then we have the inequality

(4.10)

Z b a

ρ(t)kf(t)k²dt Z b

a

ρ(t)kg(t)k²dt ¹2

− Z b

a

ρ(t) Re (f(t), g(t))dt ≥ 1

2 1−ζ² Z b

a

ρ(t)kf(t)−g(t)k²dt.

The proof follows by Theorem 2.7 and we omit the details.

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5. REFINEMENTS OFHEISENBERGINEQUALITY

It is well known that if(H;h·,·i)is a real or complex Hilbert space andf : [a, b] ⊂ R→H is an absolutely continuous vector-valued function, thenf is differentiable almost everywhere on[a, b],the derivativef⁰ : [a, b]→His Bochner integrable on[a, b]and

(5.1) f(t) =

Z t a

f⁰(s)ds for any t∈[a, b].

The following theorem provides a version of the Heisenberg inequalities in the general setting of Hilbert spaces.

Theorem 5.1. Letϕ : [a, b] → H be an absolutely continuous function with the property that bkϕ(b)k² =akϕ(a)k².Then we have the inequality:

(5.2)

Z b a

kϕ(t)k²dt ²

≤4 Z b

a

t²kϕ(t)k²dt· Z b

a

kϕ⁰(t)k²dt.

The constant4is best possible in the sense that it cannot be replaced by any smaller constant.

Proof. Integrating by parts, we have successively Z b

a

kϕ(t)k²dt=tkϕ(t)k²

b

a

− Z b

a

t d

dt kϕ(t)k² dt (5.3)

=bkϕ(b)k²−akϕ(a)k²− Z b

a

td

dthϕ(t), ϕ(t)idt

=− Z b

a

t[hϕ⁰(t), ϕ(t)i+hϕ(t), ϕ⁰(t)i]dt

=−2 Z b

a

tRehϕ⁰(t), ϕ(t)idt

= 2 Z b

a

Rehϕ⁰(t),(−t)ϕ(t)idt.

If we apply the Cauchy-Bunyakovsky-Schwarz integral inequality Z b

a

Rehg(t), h(t)idt≤ Z b

a

kg(t)k²dt Z b

a

kh(t)k²dt ¹2

forg(t) =ϕ⁰(t), h(t) = −tϕ(t), t∈[a, b],then we deduce the desired inequality (4.5).

The fact that4is the best constant in (4.5) follows from the fact that in the (CBS) inequality, the case of equality holds iffg(t) = λh(t)for a.e. t∈[a, b]andλa given scalar inK. We omit

the details.

For details on the classical Heisenberg inequality, see, for instance, [7].

Utilising Proposition 4.1, we can state the following refinement of the Heisenberg inequality obtained above in (5.2):

Proposition 5.2. Assume thatϕ : [a, b]→His as in the hypothesis of Theorem 5.1. In addition, if there existr₂, r₁ >0so that

kϕ⁰(t) +tϕ(t)k ≥r₂ ≥r₁ ≥ |kϕ⁰(t)k − |t| kϕ(t)k|

for a.e. t∈[a, b],then we have the inequality Z b

a

kϕ⁰(t)k²dt ¹2

− 1 2

Z b a

kϕ(t)k²dt≥ 1

2(b−a) r²₂−r₁²

(≥0).

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The proof follows by Proposition 4.1 on choosingf(t) =ϕ⁰(t), g(t) =−tϕ(t)andρ(t) =

1

b−a, t∈[a, b].

On utilising Proposition 4.3 for the same choices of f, gand ρ,we may state the following results as well:

Proposition 5.3. Assume thatϕ : [a, b]→His as in the hypothesis of Theorem 5.1. In addition, if there existR ≥1andr >0so that

1

R (kϕ⁰(t)k+|t| kϕ(t)k)≥ kϕ⁰(t)−tϕ(t)k ≥r for a.e. t∈[a, b],then we have the inequality

Z b a

a

kϕ⁰(t)k²dt ¹₂

−1 2

Z b a

kϕ(t)k²dt

≥ 1

2(b−a) R²−1

r²(≥0). Finally, we can state

Proposition 5.4. Letϕ: [a, b]→Hbe as in the hypothesis of Theorem 5.1. In addition, if there existsζ ∈(0,1]so that

|kϕ⁰(t)k − |t| kϕ(t)k| ≤ζkϕ⁰(t) +tϕ(t)k for a.e. t∈[a, b],then we have the inequality

Z b a

a

kϕ⁰(t)k²dt

1 2

−1 2

Z b a

kϕ(t)k²dt

≥ 1

2 1−ζ² Z b

a

kϕ⁰(t) +tϕ(t)k²dt(≥0).

This follows by Proposition 4.4 and we omit the details.

REFERENCES

[1] S. KUREPA, On the Buniakowsky-Cauchy-Schwarz inequality, Glasnik Mat. Ser III, 1(21) (1966), 147–158.

[2] N.G. DE BRUIJN, Problem 12, Wisk. Opgaven, 21 (1960), 12–13.

[3] S.S. DRAGOMIR, Some refinements of Schwarz inequality, Suppozionul de Matematic˘a ¸si Aplica¸tii, Polytechnical Institute Timi¸soara, Romania, 1-2 November 1985, 13–16.

[4] S.S. DRAGOMIRANDJ. SÁNDOR, Some inequalities in prehilbertian spaces, Studia Univ., Babe¸s- Bolyai, Mathematica, 32(1) (1987), 71–78 MR 89h: 46034.

[5] S.S. DRAGOMIRAND J. SÁNDOR, On Bessels’ and Gram’s inequalities in prehilbertian spaces, Periodica Math. Hungarica, 29(3) (1994), 197–205.

[6] S.S. DRAGOMIRANDB. MOND, On the superadditivity and monotonicity of Schwarz’s inequality in inner product spaces, Contributions, Macedonian Acad. of Sci. and Arts, 15(2) (1994), 5–22.

[7] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge University Press, Cambridge, United Kingdom, 1952.

[8] D.S. MITRINOV ´C, J.E. PE ˇCARI ´C ANDA.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, Dordrecht/Boston/London, 1993.