Operator inequalities reverse to the Jensen inequality (Q5951092)

The paper obtains reverse operator inequalities of Jensen's one as follows: Suppose that \(H\) is a Hilbert space, \(A_{i}=A_{i}^{*}\in B(H)\), \(1\leq i\leq n\), and \(aI\leq A_{i}\leq bI\) for \(i\in\{1,\cdots, n\}\). Further, suppose that \(R_{i}\in B(H)\) are arbitrary operators satisfying the condition \(\sum_{i=1}^{n} R_{i}^{*}R_{i}=I\). If \(f\) is a continuous and convex function on \(t\in [a,b]\), then the following inequalities hold: \[ \sum_{i=1}^{n}R^{*}_{i}f(A_{i})R_{i} -f\left(\sum_{i=1}^{n} R_{i}^{*}A_{i}R_{i}\right) \leq c_{1}\cdot I, \] where \(c_{1}=k(t_{1}-a)+f(a)-f(t_{1})\), and \(t_{1}\in [a,b]\) is a solution of the equation \(f'(t)=k\); \[ \sum_{i=1}^{n}R^{*}_{i}f(A_{i})R_{i} \leq c_{2} \cdot f\left(\sum_{i=1}^{n} R_{i}^{*}A_{i}R_{i}\right), \] where \(c_{2}=\frac{k(t_{2}-a)+f(a)} {f(t_{2})}\), and \(t_{2}\in [a,b]\) is a solution of the equation \(kf(t)=\{k(t-a)+f(a)\}f'(t)\); \[ f^{-1}\left(\sum_{i=1}^{n}R_{i}^{*} f(A_{i})R_{i}\right)- \sum_{i=1}^{n}R^{*}_{i}A_{i}R_{i} \leq c_{3}\cdot I, \] where \(c_{3}=f^{-1} (k(t_{3}-a)+f(a))-t_{3}\), and \(t_{3}\in [a,b]\) is a solution of the equation \(f'(f^{-1}(k(t-a)+f(a)))=k\); \[ f^{-1}\left(\sum_{i=1}^{n}R_{i}^{*} f(A_{i})R_{i}\right) \leq c_{4} \cdot \left(\sum_{i=1}^{n} R^{*}_{i}A_{i}R_{i}\right), \] where \(c_{4}=\frac{f^{-1} (k(t_{4}-a)+f(a)}{t_{4}}\), and \(t_{4}\in [a,b]\) is a solution of the equation \(kt=f'(f^{-1}(y))\cdot f^{-1}(y)\), in which \(y=k(t-a)+f(a)\).