Operator Jensen inequality for superquadratic functions

Linear Algebra and its Applications xxx (2013) xxx–xxx

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Operator Jensen inequality for superquadratic functions Mohsen Kian ∗ Department of Mathematics, Faculty of Basic Sciences, University of Bojnord, Bojnord, Iran

ARTICLE INFO

ABSTRACT

Article history: Received 27 September 2012 Accepted 7 December 2012 Available online xxxx

We present a Jensen operator inequality for superquadratic functions. In particular we extend the inequality f (Ax, x)  f (A)x, x for superquadratic functions and give some applications for our result. © 2013 Elsevier Inc. All rights reserved.

Submitted by Volker Mehrmann AMS classification: 47A63 26D15 Keywords: Superquadratic function Operator Jensen inequality Convex function Positive operator Matrix inequality

1. Introduction In what follows, assume that B(H ) is the C ∗ -algebra of all bounded linear operators on a Hilbert space H and I is the identity operator. If dim H = n, we identify B(H ) with the algebra Mn (C) of all n × n matrices with complex entries. Let f : J → R be a continuous real function. We denote by σ (J ) the set of all self-adjoint operators on H whose spectra are contained in J. For A ∈ σ (J ), we mean by f (A) the continuous functional calculus at A. A function f : [0, ∞) → R is said to be superquadratic provided that for all s  0 there exists a constant Cs ∈ R such that f (t )

 f (s) + Cs (t − s) + f (|t − s|)

(1)

for all t  0. This notion was introduced in [2]. Some basic properties of this class of functions were given in the same paper see also [1]. In particular, it was shown that: ∗ Tel.: +98 9151520429. E-mail addresses: [email protected], [email protected]. 0024-3795/$ - see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.laa.2012.12.011

Please cite this article in press as: M. Kian, Operator Jensen inequality for superquadratic functions, Linear Algebra Appl. (2013), http://dx.doi.org/10.1016/j.laa.2012.12.011

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M. Kian / Linear Algebra and its Applications xxx (2013) xxx–xxx

Lemma A [2, Lemma 2.2]. If f is a superquadratic function with Cs as in (1), then (i) f (0)  0; (ii) If f (0) = f  (0) = 0 and f is differentiable at s, then Cs (iii) If f  0, then f is convex and f (0) = f  (0) = 0.

= f  (s);

A powerful tool in the theory  n of inequalities, n the classical Jensen inequality, asserts thatifnf : J → R is a convex function, then f λ a  i i i=1 i=1 λi f (ai ) for all ai ∈ J and λi ∈ [0, 1] with i=1 λi = 1. There are several inequalities which are special kinds of this inequality. So, many mathematicians paid their attention to get generalizations and reformulation of this inequality. The reader is referred to [8] for more information about convex functions and the classical Jensen inequality. Mond and Peˇcaric´ [7] gave an operator extension of the Jensen inequality as follows. Theorem B [7]. If f f (Ax, x)

: J → R is a convex function, then

 f (A)x, x

(2)

for any A ∈ σ (J ) and any unit vector x ∈ H . This remarkable result have been applied to obtain various inequalities for functions of self-adjoint operators on Hilbert spaces. For a recent monograph concerning such inequalities we refer the reader to [3]. Many other versions of the Jensen operator inequality can be found in [4–6]. The Hölder–McCarthy inequality is a special case of (2). Theorem C (Hölder–McCarthy inequality [3]). Let A be a positive operator on H . If x vector, then (i) Ax, xr  Ar x, x for all r > 1; (ii) Ax, xr  Ar x, x for all 0 < r < 1; (iii) If A is invertible, then Ax, xr  Ar x, x

for all r

∈ H is a unit

< 0.

The main aim of this paper is to present an operator Jensen inequality similar to (2) for superquadratic functions. Our result would be an improvement of (2) in the case of nonnegative superquadratic functions. Some applications are given as well. 2. Main results A convex function lies above each of its tangent lines. In other words, if f function, then for each point (s, f (s)), there exists a real number Cs such that Cs (t

: J → R is a convex

− s) + f (s)  f (t ) for all t ∈ J

and if f is differentiable at s, then Cs = f  (s). But in the case where f : [0, ∞) → R is a superquadratic function, as it was mentioned in [2], f lies above its tangent lines plus a translation of f itself, i.e., for each point (s, f (s)), there exists a real number Cs such that f (|t

− s|) + Cs (t − s) + f (s)  f (t ) for all t  0.

Now let f

: [0, ∞) → R be a superquadratic function. It follows from the condition (1) that

f (λa + (1 − λ)b) for all a, b

 λf (a) + (1 − λ)f (b) − λf ((1 − λ)|b − a|) − (1 − λ)f (λ|b − a|)

(3)

 0 and all λ ∈ [0, 1].

Please cite this article in press as: M. Kian, Operator Jensen inequality for superquadratic functions, Linear Algebra Appl. (2013), http://dx.doi.org/10.1016/j.laa.2012.12.011

M. Kian / Linear Algebra and its Applications xxx (2013) xxx–xxx

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To interpret the last inequality as a matrix inequality, we set ⎞

⎛ A

=⎝

a 0

⎠

⎛ and x

0 b

Then we have Ax, x

= ⎝√

⎞

√ λ

1−λ

⎠.

= λa + (1 − λ)b and ⎛

|A − Ax, x| = ⎝

⎞

(1 − λ)|a − b|

0

0

λ|a − b|

⎠.

It follows

f (|A − Ax, x|)x, x = λf ((1 − λ)|b − a|) + (1 − λ)f (λ|b − a|). Therefore, as a matrix Jensen inequality for a superquadratic function f f (Ax, x)

: [0, ∞) → R, we have

 f (A)x, x − f (|A − Ax, x|)x, x

(4)

for any positive matrix A and any unit vector x. We give a proof for (4) in the next theorem. Theorem 2.1 (Jensen operator inequality for superquadratic functions). If f continuous superquadratic function, then f (Ax, x)

 f (A)x, x − f (|A − Ax, x|)x, x

for any positive operator A and any unit vector x Proof. Let A f (A) So for all x

: [0, ∞) → R is a

∈H.

 0. For each s  0 it follows from (1) that

 f (s)I + Cs A − Cs sI + f (|A − sI |). ∈ H we have

f (A)x, x  f (s)x, x + Cs Ax, x − Cs sx, x + f (|A − sI |)x, x. Let x

(5)

∈ H be a unit vector. Applying (5) with s = Ax, x we get f (Ax, x)

 f (A)x, x − f (|A − Ax, x|)x, x.



Remark 2.2. We note that if the function f in Theorem 2.1 is nonnegative, then (4) gives an improvement of (2) for the convex function f , i.e., f (Ax, x)

 f (A)x, x − f (|A − Ax, x|)x, x  f (A)x, x.

On the other hand, if the superquadratic function f is negative and −f is convex, then (4) give a converse of (2) for the concave function f , i.e.,

f (A)x, x  f (Ax, x)  f (A)x, x − f (|A − Ax, x|)x, x. As a multiple version of (4), we deduce the following result for a sequence of operators. Please cite this article in press as: M. Kian, Operator Jensen inequality for superquadratic functions, Linear Algebra Appl. (2013), http://dx.doi.org/10.1016/j.laa.2012.12.011

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M. Kian / Linear Algebra and its Applications xxx (2013) xxx–xxx

Corollary 2.3. Let A1 , . . . , An be positive operators in B(H ) and let x1 , . . . , xn n 2 i=1 xi = 1. If f : [0, ∞) → R is a superquadratic function, then ⎛ f⎝

⎞

n i=1

Ai xi , xi ⎠ 

⎛  f ⎝Ai f (Ai )xi , xi  −  i=1 i=1

n

n

∈ H be such that

⎞   − Aj xj , xj ⎠ xi , xi .  j=1 n

Proof. It follows from Theorem 2.1 when A is the positive operator A = A1 · · · ⊕ H ) and x is the unit vector x = (x1 , . . . , xn ) ∈ H ⊕ · · · ⊕ H . 

⊕ · · · ⊕ An ∈ B(H ⊕

In the next result we consider a more general case. The reader may compare it to [6, Lemma 2.3]. Proposition 2.4. Let x then f (Ax, x) 

∈ H be a vector with x  1. If f : [0, ∞) → R is a superquadratic function,

1



 

f (A)x, x − f A − 

2 − x 2

1

x

 



Ax, x x, x − x 2 f 2



1 − x 2

x 2



Ax, x

(6)

for any positive operator A.

Proof. We proceed by a similar argument as in [6, Lemma 2.3]. Let A Put y = xx so that y = 1. We have

 0 and x ∈ H with x  1.

f (Ax, x) = f ( x 2 Ay, y + (1 − x 2 )0)

 x 2 f (Ay, y) + (1 − x 2 )f (0) − x 2 f ((1 − x 2 )Ay, y) − (1 − x 2 )f ( x 2 Ay, y) (by (3)) 

1 − x 2 2 2  x f (Ay, y)− x f Ax, x −(1 − x 2 )f (Ax, x) (by f (0)  0)

x 2  x 2 (f (A)y, y − f (|A − Ay, y|)y, y) 

1 − x 2 2 − x f Ax, x − (1 − x 2 )f (Ax, x)

x 2

(by Theorem 2.1)



   1   Ax, x x, x = f (A)x, x − f A −  

x 2 

1 − x 2 − x 2 f Ax, x − (1 − x 2 )f (Ax, x),

x 2 whence we get the desired inequality.  Remark 2.5. As we see if x = 1, then (6) turns out to be (4). Also we remark that if the superquadratic function f is nonnegative, then Proposition 2.4 gives an improvement of [6, Lemma 2.3]. In fact, we have the series of inequalities







  1 − x 2 1   f (A)x, x − f A − Ax, x x, x − x 2 f Ax, x 2 2 2 − x

x

x 2 

    1 − x 2 1    f (A)x, x − f A − Ax, x x, x − x 2 f Ax, x 2

x

x 2  f (A)x, x.

f (Ax, x) 

1



Please cite this article in press as: M. Kian, Operator Jensen inequality for superquadratic functions, Linear Algebra Appl. (2013), http://dx.doi.org/10.1016/j.laa.2012.12.011

M. Kian / Linear Algebra and its Applications xxx (2013) xxx–xxx

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3. Some applications There are several examples of superquadratic functions. But for our applications we consider among them the power functions. As it was noticed in [2], if f is nonnegative, then condition (1) is stronger than the concept of convexity. So, using Theorem 2.1 we can improve some inequalities concerning convex functions. Applying (4) to the superquadratic function f (t ) = t r (r  2), we obtain an improvement of the Hölder–McCarthy’s inequality. Corollary 3.1. Let A be a positive operator on H . If x

∈ H is a unit vector, then

Ax, xr  Ar x, x − |A − Ax, x|r x, x, for all r  2. In addition, utilizing (4) to the superquadratic function f (t ) converse of the Hölder–McCarthy’s inequality. Corollary 3.2. Let A be a positive operator on H . If x

= −t r (1 < r  2), we obtain a

∈ H is a unit vector, then

(Ax, xr )Ar x, x  Ax, xr + |A − Ax, x|r x, x, for all 1 < r  2. Let us give an example. Example 3.3. Consider the nonnegative superquadratic function f (t ) ⎛ ⎜

A

then A

1 0

−1

√

⎞

⎛

⎟

⎜√ √ ⎟ ⎟ 2/ 3 ⎟ , ⎠ 0

⎜ ⎟ ⎜ 0 3 1 ⎟ and x = ⎜ =⎜ ⎝ ⎝ ⎠ −1 1 2

1/

3

= t 3 on [0, ∞). If

⎞

∈ M3 (C) is positive and x is a unit vector in C3 . By a simple calculation we have

f (Ax, x)

= 12.65, f (A)x, x = 19.34, f (|A − Ax, x|)x, x = 1.47,

whence f (Ax, x)

 f (A)x, x − f (|A − Ax, x|)x, x  f (A)x, x,

i.e., Corollary 3.1 is really an improvement of Theorem C for √ r  2. Considering the superquadratic function g (t ) = −t t which is negative, gives an example for Corollary 3.2. With the same values for A and x we have g (Ax, x)

= −3.5564, g (A)x, x = −3.8069, g (|A − Ax, x|)x, x = −0.812,

whence g (Ax, x) + g (|A − Ax, x|)x, x

 g (A)x, x  g (Ax, x).

Corollary 3.4. Let A1 , . . . , An be positive operators and w1 , . . . , wn be positive scalars such that wi = 1. If f : [0, ∞) → R is a superquadratic function, then

n

i=1

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M. Kian / Linear Algebra and its Applications xxx (2013) xxx–xxx

⎞ ⎞ ⎛

⎛  n n n n    f⎝ wi Ai x, x⎠  wi f (Ai )x, x − wi f ⎝Ai − wj Aj x, x⎠ x, x ,   i =1 i =1 i =1 j =1

(7)

for all unit vectors x ∈ H .

Proof. Let x ∈ H be a unit vector. Putting xi Corollary 2.3 we obtain inequality (7). 

=

√

wi x

∈ H so that

n

i=1

xi 2 = 1 and applying

As another consequence we can obtain refinements of some norm inequalities. Corollary 3.5. If f

: [0, ∞) → R is a nonnegative superquadratic function, then

⎞ ⎛  n     ⎠ ⎝ f  wi Ai   i=1 



  ⎧

⎛  n  n ⎨      wi f (Ai ) − inf wi f ⎝Ai  ⎩  i=1  x =1 i=1

⎞ ⎫  ⎬   ⎠ − wj Aj x, x x, x ⎭  j=1 n

for all positive operators A1 , . . . , An and all positive numbers w1 , . . . , wn with

n

i=1

wi

= 1.

Proof. Taking supremum from both sides of (7) and taking into account that f is non-decreasing, we obtain the desired inequality.  Example 3.6. Applying Corollary 3.5 to f (t ) r    n     w A i i   i=1



= t r (r  2) we obtain

  ⎧

  n  n ⎨    r  wi Ai  wi Ai   − xinf

=1 ⎩  i=1  i=1

r ⎫  ⎬  − wj Aj x, x x, x ⎭  j=1 n

for all positive operators A1 , . . . , An and all positive numbers w1 , . . . , wn with

n

i=1

wi

= 1.

Acknowledgement The author would like to express his hearty thanks to the referee for his/her valuable comments. The author would like to thank the Tusi Math. Research Group (TMRG). References [1] S. Abramovich, S. Ivelic, ´ J. Peˇcaric, ´ Improvement of Jensen-Steffensen’s inequality for superquadratic functions, Banach J. Math. Anal. 4 (2010) 159–169. [2] S. Abramovich, G. Jameson, G. Sinnamon, Refining Jensen’s inequality, Bull. Math. Soc. Sci. Math. Roumanie 47 (2004) 3–14. [3] S.S. Dragomir, Inequalities for functions of selfadjoint operators on Hilbert spaces, arXiv:1203.1667v1[math.FA]. [4] T. Furuta, H. Mici ´ c, ´ J. Peˇcaric, ´ Y. Seo, Mond–Peˇcaric´ Method in Operator Inequalities, Zagreb Element, 2005. [5] M. Kian, M.S. Moslehian, Operator inequalities related to Q-class functions, Math. Slovaca, in press. [6] J.S. Matharua, M.S. Moslehian, J.S. Aujla, Eigenvalue extensions of Bohr’s inequality, Linear Algebra Appl. 435 (2011) 270–276. [7] B. Mond, J. Peˇcaric, ´ Convex inequalities in Hilbert space, Houston J. Math. 19 (1993) 405–420. [8] J. Peˇcaric, ´ F. Proschan, Y.L. Tong, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press Inc., Boston, MA, 1992.

Please cite this article in press as: M. Kian, Operator Jensen inequality for superquadratic functions, Linear Algebra Appl. (2013), http://dx.doi.org/10.1016/j.laa.2012.12.011