Some Inequalities for Power Series of Selfadjoint Operators in Hilbert Spaces via Reverses of the Schwarz Inequality

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Some Inequalities for Power Series of Selfadjoint Operators in Hilbert Spaces via Reverses of the Schwarz Inequality

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Dragomir, Sever S (2009) Some Inequalities for Power Series of Selfadjoint Operators in Hilbert Spaces via Reverses of the Schwarz Inequality. Research report collection, 12 (Supp).

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SOME INEQUALITIES FOR POWER SERIES OF SELFADJOINT OPERATORS IN HILBERT SPACES VIA REVERSES OF THE

SCHWARZ INEQUALITY

S.S. DRAGOMIR

Abstract. In this paper we obtain some operator inequalities for functions de…ned by power series with real coe¢ cients and, more speci…cally, with nonnegative coe¢ cients. In order to obtain these inequalities some recent reverses of the Schwarz inequality for vectors in inner product spaces are utilized. Nat- ural applications for some elementary functions of interest are also provided.

1. Introduction Letf(z) :=P₁

n=0anzⁿ be an analytic function de…ned by a power series with nonnegative coe¢ cients an; n 0 and convergent on the open diskD(0; R) C, R >0:As the most natural examples of such functions we have

f(z) = 1 1 z =

X1 n=0

zⁿ; z2D(0;1) andf(z) = exp (z) = X1 n=0

1

n!zⁿ; z2C: Other function as power series representations with nonnegative coe¢ cients are, for instance

cosh (z) = X1 n=0

1

(2n)!z²ⁿ; z2C; sinh (z) =

X1 n=0

1

(2n+ 1)!z²ⁿ⁺¹; z2C; 1

2ln 1 +z

1 z =

X1 n=1

1

2n 1z²ⁿ ¹; z2D(0;1) ; sin ¹(z) =

X1 n=0

n+¹₂

p (2n+ 1)n!z²ⁿ⁺¹; z2D(0;1) ;

tanh ¹(z) = X1 n=1

1

2n 1z²ⁿ ¹; z2D(0;1) ;

2F₁( ; ; ; z) = X1 n=0

(n+ ) (n+ ) ( )

n! ( ) ( ) (n+ ) zⁿ; ; ; >0; z2D(0;1):

Date: 19 January, 2008.

1991Mathematics Subject Classi…cation. Primary 47A12, 47A30; Secondary 47A63.

Key words and phrases. Bounded linear operators, Commutators, Cartesian decomposition, Accretive operators.

1

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Now, by the help of the power seriesf(z) =P₁

n=0a_nzⁿ we can naturally con- struct another power series which will have as coe¢ cients the absolute values of the coe¢ cient of the original series, namely, f_A(z) :=P₁

n=0ja_njzⁿ: It is obvious that this new power series will have the same radius of convergence as the original series.

As some natural examples that are useful for applications, we can point out that, if

f(z) = X1 n=1

( 1)ⁿ

n zⁿ= ln 1

1 +z; z2D(0;1) ; g(z) =

X1 n=0

( 1)ⁿ

(2n)!z²ⁿ= cosz; z2C; h(z) =

X1 n=0

( 1)ⁿ

(2n+ 1)!z²ⁿ⁺¹= sinz; z2C; l(z) =

X1 n=0

( 1)ⁿzⁿ= 1

1 +z; z2D(0;1);

then the corresponding functions constructed by the use of the absolute values of the coe¢ cients are obviously

fA(z) = ln 1

1 z; gA(z) = coshz;

h_A(z) = sinhz and l_A(z) = 1 1 z

and they are de…ned on the same domain as the generating functions.

Before we are able to state the new inequalities for functions of selfadjoint operators we need to recallthe continuous functional calculus for such operators.

Let A be a selfadjoint linear operator on a complex Hilbert space (H;h:; :i): The Gelfand map establishes a -isometrically isomorphism between the set C(Sp(A))of allcontinuous functionsde…ned on thespectrumofA;denotedSp(A); an the C -algebra C (A) generated by A and the identity operator 1_H on H as follows (see for instance [7, p. 3]):

For anyf; g2C(Sp(A))and any ; 2Cwe have (i) ( f + g) = (f) + (g) ;

(ii) (f g) = (f) (g)and f = (f) ; (iii) k (f)k=kfk:= sup_t₂_Sp(A)jf(t)j;

(iv) (f0) = 1Hand (f1) =A;wheref0(t) = 1andf1(t) =t;fort2Sp(A): With this notation we de…ne

f(A) := (f) for allf 2C(Sp(A))

and we call it thecontinuous functional calculus for a selfadjoint operator A:

IfAis a selfadjoint operator andf is a real valued continuous function onSp(A), then f(t) 0 for any t 2 Sp(A) implies that f(A) 0; i:e: f(A) is a positive operator onH:Moreover, if bothf andg are real valued functions onSp(A)then the following important property holds:

(P) f(t) g(t) for anyt2Sp(A) implies thatf(A) g(A) in the operator order ofB(H):

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SO M E IN EQ U A LIT IES FO R POW ER SER IES 3

For a recent monograph devoted to various inequalities for functions of selfadjoint operators, see [7] and the references therein. For other results, see [9], [10], [11] and [8].

The main aim of the present paper is to provide some operator inequalities for functions de…ned by power series with real coe¢ cients and, more speci…cally, with nonnegative coe¢ cients. In order to obtain these inequalities some recent reverses of the Schwarz inequality for vectors in inner product spaces are utilized. Natural applications for some elementary functions of interest are also considered.

2. Some Operator Inequalities

We use the following vector inequality obtained by the author in [1] (see also [5, pp. 28-29]):

Lemma 1. Let(H;h; i)be an inner product space overKandz; y2H; ; 2K ( 6= )so that either

(2.1) Reh y z; z yi 0;

or, equivalently,

(2.2) z +

2 y 1

2j j kyk holds. Then we have the inequalities

(2.3) (0 )kzk kyk jhz; yij kzk kyk Re +

j + jhz; yi kzk kyk Re +

j + jhz; yi 1 4

j j² j + j kyk²: The constant ¹₄ in the last inequality is best possible.

Perhaps a more useful particular case of the above lemma is

Corollary 1. Let (H;h; i) be an inner product space over K andz; y2H; N >

n >0 so that either

(2.4) RehN y z; z nyi 0;

or, equivalently,

(2.5) z n+N

2 y 1

2(N n)kyk holds. Then we have the inequalities

(2.6) (0 )kzk kyk jhz; yij kzk kyk jRehz; yij

kzk kyk Rehz; yi 1 4

(N n)² n+N kyk²: The constant ¹₄ in the last inequality is best possible.

The following result holds:

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Theorem 1. Let consider the power seriesf(z) =P₁

n=0a_nzⁿ with real coe¢ cients a_nthat is convergent on the open diskD(0; R)withR >0:If the selfadjoint operator T on the Hilbert space H has the spectrumSp(T) [m; M]and0< m < M with

M²

pmM < R, then for any x2H with kxk= 1 we have the inequality (2.7) maxfkf(T)xk;kf_A(T)xkg hf_A(T)x; xi

+1 8 fA

M²

pmM 2fA

pmM +fA

m² pmM : Proof. SinceSp(T) [m; M]then we have0< m^kI T^k M^kIin the operator order ofB(H)and for any natural numberk;which implies that

M^kx T^kx; T^kx m^kx 0 or, equivalently,

T^kx m^k+M^k

2 x 1

2 M^k m^k for anyx2H withkxk= 1:

Now, if we apply the inequality (2.6), then we get

(2.8) T^kx T^kx; x +1

4

(M^k m^k)² M^k+m^k for any natural numberkand for anyx2H withkxk= 1:

SinceM^k+m^k 2 p

mM ^k; then we have (2.9) (M^k m^k)²

M^k+m^k

M^2k 2m^kM^k+m^2k 2 p

mM

k

=1 2

"

M² pmM

k

2 p

mM ^k+ m² pmM

k# for any natural numberk:

Utilising (2.8) and (2.9) we get (2.10) T^kx T^kx; x +1

8

"

M² pmM

k

2 p

mM ^k+ m² pmM

k#

for any natural numberkand for anyx2H withkxk= 1:

Now, if we multiply (2.10) withjakjsum overkfrom0tonand use the triangle inequality, we deduce

(2.11) max ( _n

X

k=0

akT^kx ; Xn k=0

jakjT^kx )

Xn k=0

ja_kj T^kx

* _n X

k=0

ja_kjT^kx; x +

+1 8

" _n X

k=0

jakj M² pmM

k

2 Xn k=0

jakj p

mM ^k+ Xn k=0

jakj m² pmM

k# for any natural numbernand for anyx2H withkxk= 1:

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Now, observe that

0< m² pmM <p

mM < M < M² pmM < R;

therefore all the series in the equation (2.11) are convergent and taking the limit overn! 1in (2.11) we deduce the desired result (2.7).

The following inequality also holds [2] (see also [5, p. 21]):

Lemma 2. Let(H;h; i)be an inner product space overKandz; y2H; ; 2K with Re ( ) >0 so that either (2.1) or, equivalently, (2.2) holds true. Then we have the inequalities

(2.12) kxk kyk 1 2

Re + hx; yi Re ( )

1 2

j + j

Re ( )jhx; yij: The constant ¹₂ is best possible in both inequalities.

n >0 so that either (2.4) or, equivalently, (2.5) holds true. Then we have

(2.13) kxk kyk 1

2

N+n

pN n Rehx; yi: The constant ¹₂ is best possible.

We are able to provide now another upper bound for kf(T)xk with x 2 H, kxk= 1as follows:

n=0anzⁿ with real coe¢ cients anthat is convergent on the open diskD(0; R)withR >0:If the selfadjoint operator T on the Hilbert space H has the spectrumSp(T) [m; M]and0< m < M with M

qM

m < R, then for anyx2H withkxk= 1 we have the inequality (2.14) maxfkf(T)xk;kfA(T)xkg

1 2

*"

f_A rM

m T

! +f_A

rm

M T

# x; x

+ : Proof. SinceSp(T) [m; M]then

M^kx T^kx; T^kx m^kx 0 for anyx2H withkxk= 1:

Now, if we apply the inequality (2.13), then we can state that (2.15) T^kx 1

2

M^k+m^k

pM^km^k T^kx; x

= 1 2

2 4

rM m

!k

+ rm

M

k3

5 T^kx; x

for any natural numberkand for anyx2H withkxk= 1:

Now, if we multiply (2.10) withja_kjsum overkfrom0tonand use the triangle inequality, we deduce

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(2.16) max ( _n

X

k=0

akT^kx ; Xn k=0

jakjT^kx )

Xn k=0

jakj T^kx

* _n X

k=0

jakjT^kx; x +

1 2

* _n X

k=0

jakj rM

m

!k

T^kx+ Xn k=0

jakj rm

M

k

T^kx; x +

for any natural numbernand for anyx2H withkxk= 1:

Since

M rm

M < M < M rM

m < R

therefore all the series in the equation (2.16) are convergent and taking the limit overn! 1in (2.16) we deduce the desired result (2.14).

In order to obtain another reverse inequality for the operatorf(T)we need the following result [3] (see also [5, p. 4]):

Lemma 3. Let(H;h; i)be an inner product space overKandz; y2H; ; 2K so that either (2.1) or, equivalently, (2.2) holds true. Then we have the inequality (2.17) (0 )kxk²kyk² jhx; yij² 1

4 j j²kyk⁴: The constant ¹₄ is best possible.

The case for positive bounds that will be used in the following is incorporated in Corollary 3. Let (H;h; i) be an inner product space over K andz; y2H; N >

n >0 so that either (2.4) or, equivalently, (2.5) holds true. Then we have (2.18) (0 )kxk²kyk² jhx; yij² 1

4 (N n)²kyk⁴: The constant ¹₄ is best possible.

The following result also holds:

n=0a_nzⁿ with real coe¢ cients a_nthat is convergent on the open diskD(0; R)withR >0:If the selfadjoint operator T on the Hilbert space H has the spectrumSp(T) [m; M]and0< m < M with M < R, then for anyx2H withkxk= 1 we have the inequality

(2.19) maxfkf(T)xk;kfA(T)xkg hfA(T)x; xi+1

2[fA(M) fA(m)]: Proof. SinceSp(T) [m; M]then

M^kx T^kx; T^kx m^kx 0 for anyx2H withkxk= 1:

Observe that from (2.18) we get kxk kyk

r

jhx; yij²+1

4 (N n)²kyk⁴ jhx; yij+1

2 (N n)kyk²

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provided that either (2.4) or, equivalently, (2.5) holds true.

Now if we apply this inequality forT^k we can write that T^kx T^kx; x +1

2 M^k m^k for any natural numberkand for anyx2H withkxk= 1:

Finally, if we apply a similar argument to the one from the above theorems we derive the desired result (2.19). The details are omitted.

Finally, in order to obtain the last reverse inequality, we need to use the following Klamkin-McLenaghan inequality for vectors in inner product spaces that has been obtained in [6]:

Lemma 4. Let(H;h; i)be an inner product space overKandz; y2H; ; 2K with Re ( ) >0 so that either (2.1) or, equivalently, (2.2) holds true. Then we have the inequality

(2.20) (0 )kxk²kyk² jhx; yij² h

j + j 2p

Re ( )i

jhx; yij kyk²: As a particular case of interest we have

n >0 so that either (2.4) or, equivalently, (2.5) holds true. Then we have (2.21) (0 )kxk²kyk² jhx; yij² p

N p

n ²jhx; yij kyk²:

For a generalisation of (2.12), other similar results and the equlity case analysis, see [4].

Theorem 4. Let consider the power seriesf(z) =P1

n=0anzⁿ with real coe¢ cients anthat is convergent on the open diskD(0; R)withR >0:If the selfadjoint operator T on the Hilbert space H has the spectrumSp(T) [m; M]and0< m < M with M < R, then for anyx2H withkxk= 1 we have the inequality

(2.22) maxfkf(T)xk;kf_A(T)xkg hf_A(T)x; xi+f_A(M) f_A p mM : Proof. Notice that from the inequality (2.21) we have

kxk kyk r

jhx; yij²+ p

N p

n ²jhx; yij kyk² jhx; yij+ p

N p

n p

jhx; yij kyk provided that either (2.4) or, equivalently, (2.5) holds true.

Now if we apply this inequality forT^k we can write that T^kx T^kx; x + p

M^k p m^k

q

jhT^kx; xij T^kx; x + p

M^k p m^k p

M^k for any natural numberkand for anyx2H withkxk= 1:

Finally, if we apply a similar argument to the one from the above theorems we derive the desired result (2.19). The details are omitted.

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Problem 1. It is an open problem whether or not the above inequalities (2.7), (2.14), (2.19) and (2.22) are sharp, and if they are not, which bound is better for di¤ erent examples of functions.

3. Applications 1. Consider the functionf(z) =₁¹_z =P₁

n=0zⁿ; z2D(0;1):IfT is a bounded linear operator on the Hilbert space (X;h; i)with the spectrumSp(T) [m; M] and0< M³< m < M <1 then by the inequality (2.7) we have

(3.1) (0 ) (I T) ¹x D

(I T) ¹x; xE 1

8

" p p mM

mM M²

2

1 p

mM +

pmM pmM m²

#

while by (2.13) we have (3.2) (I T) ¹x 1

2

*2 4 I

rM m T

! 1

x+ I

rm

M T

1

x 3 5; x

+

for anyx2H withkxk= 1:

If0< m < M <1;then by (2.19) and (2.22) we have the simpler inequalities (3.3) (0 ) (I T) ¹x D

(I T) ¹x; xE 1 2

M m

(1 m) (1 M) and

(3.4) (0 ) (I T) ¹x D

(I T) ¹x; xE M p mM (1 M) 1 p

mM respectively, for anyx2H withkxk= 1:

2. Consider the function f(z) = ln ₁¹_z = P₁

n=1 zⁿ

n; z 2 D(0;1): If T is a bounded linear operator on the Hilbert space (X;h; i) with the spectrum Sp(T) [m; M] and0< M³< m < M <1then by the inequality (2.7) we have (3.5) (0 ) ln (I T) ¹x D

ln (I T) ¹x; xE 1

8 ln 2 64

mM 1 p

mM

2

pmM M² p

mM m² 3 75

while by (2.13) we have (3.6) ln (I T) ¹x 1

2

* ln

2 4 I

rM m T

! 1

I rm

M T

13 5x; x

+

If0< m < M <1;then by (2.19) and (2.22) we have the simpler inequalities (3.7) (0 ) ln (I T) ¹x D

ln (I T) ¹x; xE 1

2 ln 1 m

1 M

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and

(3.8) (0 ) ln (I T) ¹x D

ln (I T) ¹x; xE

ln 1 p mM

1 M

!

respectively, for anyx2H withkxk= 1:

3. Consider the exponential function f(z) = exp (z) = P₁

n=0 zⁿ

n!; z 2 C. If T is a bounded linear operator on the Hilbert space (X;h; i)with the spectrum Sp(T) [m; M] and0< m < M <1then by the inequality (2.7) we have (3.9) (0 )kexp (T)xk hexp (T)x; xi

1

8 exp M²

pmM 2 exp p

mM + exp m² pmM while by (2.13) we have

(3.10) kexp (T)xk 1 2

*"

exp rM

m T

! + exp

rm M T

# x; x

+

Also, by (2.19) and (2.22) we have the simpler inequalities (3.11) (0 )kexp (T)xk hexp (T)x; xi 1

2 [exp (M) exp (m)]

and

(3.12) (0 )kexp (T)xk hexp (T)x; xi exp (M) exp p mM for anyx2H withkxk= 1:

4. Consider the functionf(z) = cosz =P₁

n=0 ( 1)ⁿ

(2n)!z²ⁿ; z2C:Obviously, we have fA(z) = coshz=P₁

n=0 1

(2n)!z²ⁿ; z2C: IfT is a bounded linear operator on the Hilbert space(X;h; i)with the spectrumSp(T) [m; M]and 0< m < M <

1then by the inequality (2.7) we have (3.13) kcos (T)xk hcosh (T)x; xi

+1

8 cosh M²

pmM 2 cosh p

mM + cosh m² pmM while by (2.13) we have

(3.14) kcos (T)xk 1 2

*"

cosh rM

m T

! + cosh

rm M T

# x; x

+

Moreover, by (2.19) and (2.22) we have the simpler inequalities (3.15) kcos (T)xk hcosh (T)x; xi+1

2 [cosh (M) cosh (m)]

and

(3.16) kcos (T)xk hcosh (T)x; xi+ cosh (M) cosh p mM respectively, for anyx2H withkxk= 1:

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If we apply the same inequalities only for the function with positive coe¢ cients coshthen we also have

(3.17) 0 kcosh (T)xk hcosh (T)x; xi 1

8 cosh M²

pmM 2 cosh p

mM + cosh m² pmM and

(3.18) kcosh (T)xk 1 2

*"

cosh rM

m T

! + cosh

rm

M T

# x; x

+

and

(3.19) 0 kcosh (T)xk hcosh (T)x; xi 1

2 [cosh (M) cosh (m)]

and

(3.20) 0 kcosh (T)xk hcosh (T)x; xi cosh (M) cosh p mM respectively, for anyx2H withkxk= 1:

Remark 1. The reader may obtain other operator inequalities by choosing appro- priate examples of functions de…ned by series with positive coe¢ cients. However, the details are omitted.

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[2] S.S. Dragomir, Reverses of Schwarz, triangle and Bessel inequalities in inner product spaces, J. Inequal. Pure & Appl. Math. 5(2004), No. 3, Art. 76, pp. 1-18.

[http://jipam.vu.edu.au/article.php?sid=432].

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1-59454-202-3

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Research Group in Mathematical Inequalities & Applications, School of Engineering

& Science, Victoria University, PO Box 14428, Melbourne VIC 8001, Australia.

E-mail address: [email protected] URL:http://rgmia.vu.edu.au/dragomir/