Reverses of the first Hermite-Hadamard type inequality for the square operator modulus in Hilbert spaces
Subject Areas : Operator theory
1 - Mathematics, College of Engineering \& Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia|DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science \& Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg 2050, South Africa
Keywords: Operator convex functions, Hermite-Hadamard inequality, midpoint inequality, operator power and logarithmic functions,
Abstract :
Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right)$ be a complex Hilbert space. Denote by $\mathcal{B}\left( H\right)$ the Banach $C^{\ast }$-algebra of bounded linear operators on $H$. For $A\in \mathcal{B}\left(H\right)$ we define the modulus of $A$ by $\left\vert A\right\vert :=\left(A^{\ast }A\right) ^{1/2}$ and \ $\func{Re}A:=\frac{1}{2}\left( A^{\ast}+A\right).$ In this paper we show among other that, if $A,$ $B\in \mathcal{B}\left( H\right)$ with $0\leq m\leq \left\vert \left( 1-t\right)A+tB\right\vert ^{2}\leq M$ for all $t\in \left[ 0,1\right],$ then \begin{align*} 0& \leq \int_{0}^{1}f\left( \left\vert \left( 1-t\right) A+tB\right\vert^{2}\right) dt-f\left( \frac{\left\vert A\right\vert ^{2}+\func{Re}\left(B^{\ast }A\right) +\left\vert B\right\vert ^{2}}{3}\right) \\ & \leq 2\left[ \frac{f\left( m\right) +f\left( M\right) }{2}-f\left( \frac{m+M}{2}\right) \right] 1_{H} \end{align*} for operator convex functions $f:[0,\infty )\rightarrow \mathbb{R}$. Applications for power and logarithmic functions are also provided.
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