## Jensen’s operator inequality.(English)Zbl 1051.47014

Given a continuous function $$f: I \to \mathbb R$$ where $$I \subset \mathbb R$$, by methods of the spectral theory we may consider the corresponding function $$f: \mathbb B(H)_{sa}^I \to \mathbb B(H)_{sa}$$ acting from the set of all (linear bounded) self-adjoint operators on an infinite-dimensional Hilbert space $$H$$ with spectra in $$I$$ to the set of all selfadjoint operators. An $$f$$ is said to be operator convex if $$f (\lambda x + (1 - \lambda) y) \leq \lambda f(x) + (1 - \lambda) f(y)$$ for each $$\lambda \in [0,1]$$ and each $$x,y \in \mathbb B(H)_{sa}^I$$. A convex function need not be operator convex. In [Math. Ann. 246, 249–250 (1980; Zbl 0407.47012)], F. Hansen noted the following remarkable result: if $$f: I \to \mathbb R$$ is operator convex, then $$f(\sum_{k=1}^n a_k^* x_k a_k) \leq \sum_{k=1}^n a_k^* f(x_k) a_k$$ for each $$n \in \mathbb N$$, $$x_k \in \mathbb B(H)_{sa}^I$$ and $$a_k \in \mathbb B(H)$$, $$k = 1, \dots, n$$ such that $$\sum_{k=1}^n a_k^* a_k = \text{Id}$$. Since then, a number of attempts of the authors were done to prove it for partial cases. The paper under review is devoted to a complete proof of the above inequality. One of the main results (Theorem 2.1) asserts, in particular, that $$f$$ is operator convex if and only if the above Jensen’s operator inequality holds. Among other related results, an analogous inequality for traces of square matrices is established to be equivalent to the operator convexity of a function.

### MSC:

 47A63 Linear operator inequalities 26A51 Convexity of real functions in one variable, generalizations 47A56 Functions whose values are linear operators (operator- and matrix-valued functions, etc., including analytic and meromorphic ones) 46L10 General theory of von Neumann algebras 47C15 Linear operators in $$C^*$$- or von Neumann algebras

Zbl 0407.47012
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