Operator inequalities of Jensen type

We present some generalized Jensen type operator inequalities involving sequences of self-adjoint operators. Among other things, we prove that if $f:[0,\infty) \to \mathbb{R}$ is a continuous convex function with $f(0)\leq 0$, then {equation*} \sum_{i=1}^{n} f(C_i) \leq f(\sum_{i=1}^{n}C_i)-\delta_f\sum_{i=1}^{n}\widetilde{C}_i\leq f(\sum_{i=1}^{n}C_i) {equation*} for all operators $C_i$ such that $0 \leq C_i\leq M \leq \sum_{i=1}^{n} C_i $ \ $(i=1,...,n)$ for some scalar $M\geq0$, where $ \widetilde{C_i} = 1/2 - |\frac{C_i}{M}- 1/2 |$ and $\delta_f = f(0)+f(M) - 2 f(\frac{M}{2})$.


Introduction and Preliminaries
Let B(H ) be the C * -algebra of all bounded linear operators on a complex Hilbert space H and I denote the identity operator. If dim H = n, then we identify B(H ) with the C * -algebra M n (C) of all n × n matrices with complex entries. Let us endow the real space B h (H ) of all self-adjoint operators in B(H ) with the usual operator order ≤ defined by the cone of positive operators of B(H ).
If T ∈ B h (H ), then m = inf{ T x, x : x = 1} and M = sup{ T x, x : x = 1} are called the bounds of T . We denote by σ(J) the set of all selfadjoint operators on H with spectra contained in J. All real-valued functions are assumed to be continuous in this paper. A real valued function f defined on an interval J is said to be operator convex if f (λA+(1−λ)B) ≤ λf (A)+(1−λ)f (B) for all A, B ∈ σ(J) and all λ ∈ [0, 1]. If the function f is operator convex, then the so-called Jensen operator inequality f (Φ(A)) ≤ Φ(f (A)) holds for any unital positive linear map Φ on B(H ) and any A ∈ σ(J). The reader is referred to [3,4,8] for more information about operator convex functions and other versions of the Jensen operator inequality. It should be remarked that if f is a real convex function, but not operator convex, then the Jensen operator inequality may not hold. To see this, consider the convex (but not operator convex) function f (t) = t 4 defined on [0, ∞) and the positive mapping Φ : M 3 (C) → M 2 (C) defined by Φ((a ij ) 1≤i,j≤3 ) = (a ij ) 1≤i,j≤2 for any A = (a ij ) 1≤i,j≤3 ∈ M 3 (C). If holds for every convex function f : J → R provided that the interval J contains all m i , M i ; see also [7]. Another variant of the Jesnen operator inequality is the so-called Jensen-Mercer operator inequality [5] asserting that if f is a real convex function on an interval [m, M], then Recently, in [9] an extension of the Jensen-Mercer operator inequality was presented as follows: Theorem B.[9, Corollary 2.3] Let f be a convex function on an interval J. Let (1. 3) The authors of [9] used inequality (1.3) to obtain some operator inequalities. In particular, they gave a generalization of the Petrović operator inequality as follows: Some other operator extensions of (1.4) can be found in [1,2,11]. In this paper, as a continuation of [9], we extend inequality (1.3), refine (1.3) and improve some of our results in [9]. Some applications such as further refinements of the Petrović operator inequality and the Jensen-Mercer operator inequality are presented as well.

Results
To presenting our results, we introduce the abbreviation: We need the following lemma may be found in [7,Lemma 2]. We give a proof for the sake of completeness.
If f is concave on [m, M], then inequality (2.1) is reversed.
In the next theorem we present a generalization of [9, Theorem 2.1].
holds for every convex function f : J → R, where If f is concave, then the reverse inequalities are valid in (2.6).
Proof. We prove only the case when f is convex. Let [m, M] ⊆ J. It follows from the convexity of f on J that and similarly Applying the positive linear mappings Φ i and Φ i , respectively, to both sides of (2.8) and (2.9) and summing we get On the other hand, taking into account that m ≤ 1 Adding two inequalities (2.12) and (2.13) and putting (2.10) and (2.11)) which is the first inequality in (2.6).
(2) f 1 γ in which Before giving an example, we present some special cases of Theorem 2.2 which are useful in our applications. The next corollary provides a refinement of [ (2.14) where In particular, If f is concave on J, then inequalities (2.14) and (2.15) are reversed.
Another special case of Theorem 2.2 leads to a refinement of [9, Corollary 2.3]. ( Hence there is no relationship between the right hand sides of inequalities in Corollary 2.5.

Corollary 2.5. Let f be a convex function on an interval J. Let
Corollary 2.7. Let f be a convex function on an interval J.
for two real numbers m < M and If f is concave, then inequalities (2.17) and (2.18) are reversed.
Proof. We prove only inequality (2.17) in the convex case. It follows from Using the same reasoning as in the proof of Theorem 2.2 we get which give the first inequality in (2.17). It is easy to see that δ f,n X n ≥ 0, whence the second inequality derived.

Applications
Using the results in Section 2, we provide some applications which are refinements of some well-known operator inequalities. As the first, we give a refinement of the operator Jensen-Mercer inequality.
The next result provides a refinement of the Petrović inequality for operators.
Corollary 3.2. If f : [0, ∞) → R is a convex function and B 1 , · · · , B n are positive operators such that n i=1 B i = MI for some scalar M > 0, then Summing above inequalities over i we get As another consequence of Theorem 2.2, we present a refinement of the Jensen operator inequality for real convex functions. The authors of [9] introduce a subset Ω of B h (H ) × B h (H ) defined by We have the following result.
If f is concave, then inequalities in (3.1) are reversed.
Proof. Putting B i = C i = A i +D i 2 and using inequality (1) of Corollary 2.5, we conclude the desired result.
Note that utilizing Corollary 2.5, we even be able to obtain a converse of the Jensen operator inequality. For this end, under the assumptions in the Corollary 3.3 we have Note that the function f need not to be operator convex. Let us give an example to illustrate these inequalities. for each A = (a ij ) 1≤i,j≤3 ∈ M 3 (C). Consider the convex function f (t) = e t on [0, ∞). If It should be mentioned that in the case when f is operator convex, under the assumptions in Corollary 3.3 we have even more: for all (A, D) ∈ Ω and all λ ∈ [0, 1], where X = If f is concave, then inequality (3.3) is reversed.
Proof. Put n = 1 and let Φ be the identity map in Corollary 3.3 to get for any (A, D) ∈ Ω, which implies (3.3) by the continuity of f .
Regarding to obtain an operator version of (3.4), it is shown in [9] for all strictly positive operators A, B for which A ≤ M ≤ A + B and B ≤ M ≤ A + B for some scalar M. We give a refined extension of this result as follows.
Theorem 3.6. If f : [0, ∞) → R is a convex function with f (0) ≤ 0 then for all positive operators C i such that C i ≤ M ≤ n i=1 C i (i = 1, . . . , n) for some scalar M ≥ 0. If f is concave, then the reverse inequality is valid in (3.5).
In particular, if f is convex, then for all positive operators A, B such that A ≤ MI ≤ A + B and B ≤ MI ≤ A + B for some scalar Proof. Without loss of generality let M > 0. Lemma 2.1 implies that since f (0) ≤ 0. Summing the above inequalities over i we get Combining two inequalities (3.6) and (3.7), we reach to the desired inequality (3.5). is satisfied, then If f is concave and any one of the following conditions is satisfied, then inequality (3.8) is reversed.