Extension of Jensen’s Inequality for Operators without Operator Convexity

and Applied Analysis 3 If one of the following conditions ii ψ ◦ φ−1 is concave and ψ−1 is operator monotone, ii′ ψ ◦ φ−1 is convex and −ψ−1 is operator monotone, is satisfied, then the reverse inequality is valid in 1.7 . In this paper we study an extension of Jensen’s inequality given in Theorem A. As an application of this result, we give an extension of Theorem B for a version of the quasiarithmetic mean 1.5 with an n-tuple of positive linear mappings which is not unital. 2. Extension of Jensens Operator Inequality In Theorem A we prove that Jensen’s operator inequality holds for every continuous convex function and for every n-tuple of self-adjoint operators A1, . . . , An , for every n-tuple of positive linear mappings Φ1, . . . ,Φn in the case when the interval with bounds of the operator A ∑n i 1 Φi Ai has no intersection points with the interval with bounds of the operator Ai for each i 1, . . . , n, that is, when mA,MA ∩ mi,Mi ∅, for i 1, . . . , n, 2.1 wheremA andMA,mA ≤ MA, are the bounds of A, andmi andMi,mi ≤ Mi, are the bounds of Ai, i 1, . . . , n. It is interesting to consider the case when mA,MA ∩ mi,Mi ∅ is valid for several i ∈ {1, . . . , n}, but not for all i 1, . . . , n. We study it in the following theorem. Theorem 2.1. Let A1, . . . , An be an n-tuple of self-adjoint operators Ai ∈ B H with the bounds mi and Mi, mi ≤ Mi, i 1, . . . , n. Let Φ1, . . . ,Φn be an n-tuple of positive linear mappings Φi : B H → B K , such that ∑ni 1 Φi 1H 1K. For 1 ≤ n1 < n, one denotes m min{m1, . . . , mn1}, M max{M1, . . . ,Mn1}, and ∑n1 i 1 Φi 1H α 1K, ∑n i n1 1 Φi 1H β 1K, where α, β > 0, α β 1. If m,M ∩ mi,Mi ∅, for i n1 1, . . . , n, 2.2 and one of two equalities 1 α n1 ∑ i 1 Φi Ai 1 β n ∑ i n1 1 Φi Ai n ∑ i 1 Φi Ai 2.3 is valid, then 1 α n1 ∑ i 1 Φi ( f Ai ) ≤ n ∑ i 1 Φi ( f Ai ) ≤ 1 β n ∑ i n1 1 Φi ( f Ai ) 2.4 holds for every continuous convex function f : I → provided that the interval I contains all mi,Mi, i 1, . . . , n. If f : I → is concave, then the reverse inequality is valid in 2.4 . 4 Abstract and Applied Analysis Proof. We prove only the case when f is a convex function. Let us denote A 1 α n1 ∑ i 1 Φi Ai , B 1 β n ∑ i n1 1 Φi Ai , C n ∑ i 1 Φi Ai . 2.5 It is easy to verify that A B or B C or A C implies A B C. a Let m < M. Since f is convex on m,M and mi,Mi ⊆ m,M for i 1, . . . , n1, then f t ≤ M − t M −m m t −m M −m M , t ∈ mi,Mi for i 1, . . . , n1, 2.6 but since f is convex on all mi,Mi and m,M ∩ mi,Mi ∅ for i n1 1, . . . , n, then f t ≥ M − t M −m m t −m M −m M , t ∈ mi,Mi for i n1 1, . . . , n. 2.7 Sincemi1H ≤ Ai ≤ Mi1H , i 1, . . . , n1, it follows from 2.6 that f Ai ≤ M1H −Ai M −m f m Ai −m1H M −m f M , i 1, . . . , n1. 2.8 Applying a positive linear mapping Φi and summing, we obtain n1 ∑ i 1 Φi ( f Ai ) ≤ Mα1K − ∑n1 i 1 Φi Ai M −m f m ∑n1 i 1 Φi Ai −mα1K M −m f M , 2.9 since ∑n1 i 1 Φi 1H α1K. It follows that 1 α n1 ∑ i 1 Φi ( f Ai ) ≤ M1K −A M −m f m A −m1K M −m f M . 2.10 Similarly to 2.10 in the casemi1H ≤ Ai ≤ Mi1H , i n1 1, . . . , n, it follows from 2.7 1 β n ∑ i n1 1 Φi ( f Ai ) ≥ M1K − B M −m f m B −m1K M −m f M . 2.11 Combining 2.10 and 2.11 and taking into account that A B, we obtain 1 α n1 ∑ i 1 Φi ( f Ai ) ≤ 1 β n ∑ i n1 1 Φi ( f Ai ) . 2.12 Abstract and Applied Analysis 5 It follows that 1 α n1 ∑and Applied Analysis 5 It follows that 1 α n1 ∑ i 1 Φi ( f Ai ) n1 ∑ i 1 Φi ( f Ai ) β α n1 ∑ i 1 Φi ( f Ai ) ( by α β 1 ) ≤ n1 ∑ i 1 Φi ( f Ai ) n ∑ i n1 1 Φi ( f Ai ) ( by 2.12 )