Inequalities of Convex Functions and Self-Adjoint Operators Zlatko Pavi

Throughout the paper we will use a real interval I with the nonempty interior and real segments [a, b] and (a, b)with a < b. We briefly summarize a development path of the operator formof Jensen’s inequality. LetH andK beHilbert spaces, let B(H) and B(K) be associated C-algebras of bounded linear operators, and let 1 H and 1 K be their identity operators. Combining the results from [1, 2], it follows that every operator convex function f : I → R satisfies the Schwarz inequality


Introduction
Throughout the paper we will use a real interval I with the nonempty interior and real segments [, ] and (, ) with  < .
We briefly summarize a development path of the operator form of Jensen's inequality.Let H and K be Hilbert spaces, let B(H) and B(K) be associated C * -algebras of bounded linear operators, and let 1  and 1  be their identity operators.

Journal of Operators
If the function  is convex, then we have the inequality and the reverse inequality Let , ,  ∈ R be coefficients such that  +  −  = 1.Let , ,  ∈ R be points where  < .We consider the affine combination  +  − .Inserting the affine combination  = + assuming that + = 1, we get the binomial form because  line {,} () ≥ ().
Lemma 1 is trivially true if  = .It is also valid for  ∈ [−1, 1] because then the observed affine combinations with  ≤ 0 become convex, and associated inequalities follow from Jensen's inequality.The similar combinations including  ∈ [−1, 1] were observed in [5, Corollary 11 and Theorem 12] additionally using a monotone function .If  =  =  = 1, then the inequality in ( 10) is reduced to simple Mercer's variant of Jensen's inequality obtained in [6].
Then the affine combination and every convex function  : I → R, where I = conv{, , } satisfies the inequality Proof.The condition  =  +  ∉ (, ) entails  ≤ 0 or  ≥ 1, and the coefficients of the binomial form of (8) satisfy So, the combination +− does not belong to (, ).
Applying the inequality in (7), we get the series of inequalities as in (11) but with the reverse inequality signs.
It is not necessary to require  ∈ [1, ∞) in Lemma 2, because it follows from the other coefficient conditions.

Operator
Variants.We write  ≤  for self-adjoint operators ,  ∈ B(H) if the inner product inequality ⟨, ⟩ ≤ ⟨, ⟩ holds for every vector  ∈ H.A selfadjoint operator  is positive (nonnegative) if it is greater than or equal to null operator ( ≥ 0).If Sp() ⊆ I and ,  : I → R are continuous functions such that () ≤ () for every  ∈ Sp(), then the operator inequality () ≤ () is valid.The bounds of a self-adjoint operator  are defined with and its spectrum Sp() is contained in [  ,   ] wherein we have the operator inequality More details on the theory of bounded operators and their inequalities can be found in [7].The operator versions of Lemmas 1 and 2 follow.Then and every convex continuous function  : [, ] → R satisfies the inequality Proof.The spectral inclusion in (16) follows from the inclusion in (9).Using the affinity of the function  line {,} and the operator inequalities  line {,} (⋅) ≥ (⋅), we can replace the discrete inequalities in (11) with the operator inequalities.

Main Results
We want to extend and generalize the inequalities in ( 17) and ( 19) including positive operators and positive linear mappings.The main results are Theorems 8 and 9.
Then every affine function () =  + V, where  and V are real constants, satisfies the operator equality Proof.Applying the affinity of the function  and the assumption achieving the equality in (20).
Lemma 6.Let Φ  : B(H) → B(K) be positive linear mappings and let   ∈ B(H) be positive linear operators so that Then the spectrum of the operator sum Proof.Applying the positive operators   and the positive mappings Φ  to the assumed spectral inequalities we get Summing the above inequalities and using the assumption ∑  =1 Φ  (  ) = 1  , we have Then the spectrum of the operator is contained in [, ], and every convex continuous function  : [, ] → R satisfies the inequality If the function  is concave, then the reverse inequality is valid in (28).
Then the spectrum of the operator satisfies the relation Sp() ∩ (, ) = 0, and every convex continuous function  : I → R, where I contains all spectra and [, ], satisfies the inequality If the function  is concave, then the reverse inequality is valid in (31).
Proof.The relation Sp() ∩ (, ) = 0 is the consequence of the relation in (18).Assuming and using the convexity of  and the affinity of  line {,} , as well as the inequalities  line {,} (  ) ≤ (  ), we get the series of inequalities as in (29) but with the reverse inequality signs.

Application to Quasi-Arithmetic Means
In applications of convexity to quasi-arithmetic means, we use strictly monotone continuous functions ,  : I → R such that the function  ∘  −1 is convex, in which case we say that  is -convex.A similar notation is used for concavity.This terminology is taken from [9,Definition 1.19].
A continuous function  : I → R is said to be operator increasing on I if  ≤  implies () ≤ () for every pair of self-adjoint operators ,  ∈ B(H) with spectra in I.A function  is said to be operator decreasing if the function − is operator increasing.
Take an operator affine combination as in Theorem 8.If  : [, ] → R is a strictly monotone continuous function, we define the -quasi-arithmetic mean of the combination  as the operator The spectrum of the operator   () is contained in [, ] because the spectrum of the operator is contained in ([, ]).The quasi-arithmetic means defined in (33) are invariant with respect to the affinity; that is, the equality holds for all pairs of real numbers  ̸ = 0 and V. Indeed, if () = () + V, then and therefore, it follows that The order of the pair of quasi-arithmetic means   and   depends on convexity of the function  ∘  −1 and monotonicity of the function .Theorem 8 can be applied to operator means as follows.
Corollary 10.Let  be an affine combination as in (32) satisfying the assumptions of Theorem 8. Let ,  : [, ] → R be strictly monotone continuous functions.
If  is either -convex with operator increasing  −1 or -concave with operator decreasing  −1 , then one has the inequality   () ≤   () . ( If  is either -convex with operator decreasing  −1 or concave with operator increasing  −1 , then one has the reverse inequality in (38).
Proof.Let us prove the case in which  is -convex with operator increasing  −1 .Put [, ] = ([, ]).Applying the inequality in (28) of Theorem 8 to the affine combination   of (34) with Sp(  ) ⊆ [, ] and the convex function Assigning the increasing function  −1 to the above inequality, we attain which finishes the proof.
Using Corollary 10 we get the following version of the harmonic-geometric-arithmetic mean inequality for operators.
Quasi-arithmetic operator means without applying operator convexity were also investigated in [4,10].