On Schwarz type inequalities

We show Schwarz type inequalities and consider their converses. A continuous function $f : [0, \infty) \rightarrow [0, \infty)$ is said to be semi-operator monotone on $(a,b)$ if $\{f( t^{\frac{1}{2}} ) \}^{2}$ is operator monotone on $(a^{2},b^{2})$. Let $T$ be a bounded linear operator on a complex Hilbert space ${\mathcal H}$ and $T = U \vert T \vert$ be the polar decomposition of $T$. Let $0 \leq A, B \in B( {\mathcal H})$ and $\Vert Tx \Vert \leq \Vert Ax\Vert, \Vert T^{*} y \Vert \leq \Vert By \Vert$ for $x, y \in {\mathcal H}$. (1) If a non-zero function $f$ is semi-operator monotone on $(0, \infty)$, then $\vert \langle Tx, y \rangle \vert \leq \Vert f(A) x \Vert \Vert g(B) y \Vert$ for $x, y \in {\mathcal H}$, where $g(t) = t/f(t)$. (2) If $f, g$ are semi-operator monotone on $(0, \infty)$, then $\vert \langle U f(\vert T \vert)g(\vert T \vert)x, y \rangle \vert \leq \Vert f(A) x \Vert \Vert g(B) y \Vert$ for $x, y \in {\mathcal H}$. Also, we show converses of these inequalities, which imply that semi-operator monotonicity is necessary.