“…If f is a convex function on an interval J containing m, M, thenf (λA + (1 − λ)D) ≤ λf (A) + (1 − λ)f (D) − δ f X ≤ λf (A) + (1 − λ)f (D) (3.3)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 getf A + D 2 ≤ f (A) + f (D) 2 − δ f X ≤ f (A) + f (D) 2for 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] that if f : [0, ∞) → [0, ∞) is a convex function with f (0) ≤ 0, then f (A) + f (B) ≤ f (A + B)(3.4) 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.…”