Let $J$ be an interval such that $J\neq (-\infty, \infty)$. $P(J)$ denotes the set of all operator monotone functions on $J$. We set $P_+(J)=\left\{f\in P(J)\;|\, f(t)\geq 0, \, t\in J \right\}$. If $f\in P_+(a, b)$ and $-\infty < a$, then $f$ has the natural extension to $[a, b)$, which belongs to $P_{+}[a, b).$ We therefore identify $P_+(a, b)$ with $P_+[a, b)$.

It is well-known that if $f(t)\in P_{+}(0, \infty)$, then $\frac{t}{f(t)}$ (if $f\neq0$) and ${f(t^\alpha)}^\frac{1}{\alpha}$ are both in $P_{+}(0, \infty)$, and that if $f(t), \phi(t), \varphi(t)$ are all in $P_{+}(0, \infty)$, then so are

$ \phi(f(t))\varphi(\frac{t}{f(t)})~~~~ (\mbox{if}~ f\neq0), f(t^\alpha)\phi(t^{1-\alpha}) $

and $f(t)^\alpha\phi(t)^{1-\alpha}$ for $0 < \alpha < 1$ (see [1-5]). Throughout this work, we assume that a function is continuous and increasing means "strictly increasing". Further more, for convenience, let $B(\mathcal{H})$ denote the $C^{\ast}$-algebra of all bounded linear operators acting on a Hilbert space $\mathcal{H}$. A capital letter $A$ means an element belongs to $B(\mathcal{H})$, $\Phi$ means a positive linear map from $B(\mathcal{H})$ to $B(\mathcal{H})$ and we assume $\Phi(I)=I$ always stand (see [7, 8]). In this paper, we also assume that $J=[a, b)$ or $J=(a, b)$ with $-\infty\leq a < b\leq+\infty$.

Definition 1.1[9, 10]  Let $P_+^{-1}(J), LP_+(J)$ denote the following sets, respectively,

$ P_+^{-1}(J)=\left\{h\;|\;{h\; \mbox{is increasing on} \;J }, \; h((a, b))=(0, \infty), \;h^{-1}\in P(h(J)) \right\}, $

where $h^{-1}$ stands for the inverse function of $h$.

$ LP_+(J)=\left\{h\;|\;{h\; \mbox{is defined on} \;J }, \; h(t)>0\; \mbox{on} \;(a, b), \;\log h\in P(a, b) \right\}. $

Definition 1.2  Let $h(t)$ and $g(t)$ be functions defined on $J$, and $g(t)$ is increasing, then $h$ is said to be majorized by $g$, in symbol $h\preceq g$ if the composite $h\circ g^{-1}$ is operator monotone on $g(J)$, which is equivalent to

$ \sigma(A), \sigma(B)\subset J, \qquad g(A)\le g(B)\Longrightarrow h(A)\le h(B). $

Lemma 1.1(Product lemma)(see [9, 10])  Let $h, g$ be non-negative functions defined on $J$. Suppose the product $hg$ is increasing, $(hg)(a+0)=0$ and $(hg)(b-0)=\infty.$Then

$ g\preceq hg \;\mbox{on}\; J\iff h\preceq hg \;\mbox{on}\;J. $

Moreover, for every $\psi_1, \psi_2$ in $P_+[0, \infty), $

$ g\preceq hg \;\mbox{on}\;J\Longrightarrow \psi_1(h)\psi_2(g)\preceq hg\;\mbox{on}\;J. $

Theorem 1.1  (Product theorem) (see [9, 10])

$ {P_+^{-1}(J)\cdot P_+^{-1}(J)}\subset {P_+^{-1}(J)}, \qquad{LP_+^{-1}(J)\cdot P_+^{-1}(J)}\subset {P_+^{-1}(J)}. $

Further, let $g_i(t)\in LP_+(J)$ for $1\le i\le m$ and $h_j(t)\in P_+^{-1}(J)$ for $1\le j\le n$. Then for every $\psi_i, \phi_j\in P_+[0, \infty)$, we have

$ \prod\limits_{i=1}^{m}{{{\psi }_{i}}}({{g}_{i}})\prod\limits_{j=1}^{n}{{{\phi }_{j}}}({{h}_{j}})\preceq \prod\limits_{i=1}^{m}{{{g}_{i}}}\prod\limits_{j=1}^{n}{{{h}_{j}}}\in P_{+}^{-1}(J). $

Before to prove our main results, we give the following lemmas.

Lemma 2.1(L-H inequality)(see [2, 12])  If $0\le\alpha\le1, A\ge B\geq0$, then $A^\alpha\ge B^\alpha.$

Lemma 2.2(Furuta inequality)(see [6, 9])   Let $A\ge B\ge 0$, then

(1)$ (B^\frac{r}{2}A^pB^\frac{r}{2})^\alpha\ge(B^\frac{r}{2}B^pB^\frac{r}{2})^\alpha;$

(2)$(A^\frac{r}{2}A^pA^\frac{r}{2})^\alpha\ge(A^\frac{r}{2}B^pA^\frac{r}{2})^\alpha, $

where $r\ge0$, $p\ge1$ with $0 < \alpha\le\frac{1+r}{p+r}$.

Lemma 2.3(Hansen inequality)(see [13])  Let $X$ and $A$ be bounded linear operators on $\mathcal{H}$, and such that $X\ge0, \lVert A\rVert\le1$. If $f$ is an operator monotone function on $[0, \infty)$, then

$ A^*f(X)A\le f(A^*XA). $

Theorem 2.1  Put $J\neq(-\infty, \infty)$, $\eta\in P_+(J)\cap P_+^{-1}(J)$, $f_i\in P_+(J)$, $i=1, 2, \cdots, n$, $g(t)\in P_+^{-1}(J)\cup \left\{ f_1(t) \right\}$, and $k_n(t)=f_1(t)f_2(t)\cdots f_n(t)$. If $h(t)$ is defined on $J$ such that $f_1(t)h(t)\in P_+^{-1}(J)$, then

(ⅰ) the function $\phi_n$ on $(0, \infty)$ defined by

$ \begin{aligned} \phi_n(k_n(t)h(t)g(t))=k_n(t)\eta(t) \quad (t\in J) \end{aligned} $

(2.1)

belongs to $P_+(0, \infty)$;

(ⅱ) if $A\le C\le B$, then

$ \begin{aligned} \phi_n(k_n(C)^\frac{1}{2}h(A)g(A)k_n(C)^\frac{1}{2})&\le&\phi_n(k_n(C)^\frac{1}{2}h(C)g(C)k_n(C)^\frac{1}{2})\\ &\le&\phi_n(k_n(C)^\frac{1}{2}h(B)g(B)k_n(C)^\frac{1}{2}). \end{aligned} $

(2.2)

Proof  (ⅰ) Since $f_1(t)\preceq t\preceq f_1(t)h(t)$, by product lemma $h(t)\preceq f_1(t)h(t)$, therefore $h(t)$ is nondecreasing. When $g\in P_+^{-1}(J)$, since $\eta\in P_+(J)\cap P_+^{-1}(J)$, we have $\eta(t)\preceq t\preceq g(t)$. Now putting $\psi_0(s)=s$, $\psi_1(g)=\eta$, $\psi_2(f_1h)=f_1$, obviously, we have $\psi_0, \psi_1, \psi_2\in P_+(0, \infty)$. By taking $s$ in $\psi_0(s)$ as $f_2\cdots f_n$, and from product theorem, we obtain

$ k_n\eta=\psi_0(f_2\cdots f_n)\psi_1(g)\psi_2(f_1h)\preceq k_n(t)h(t)g(t). $

Therefore we have $\phi_n$ belongs to $P_+(0, \infty)$ for $\phi_n$ given in (ⅰ).

When $g(t)=f_1(t)$, by taking $\psi_0(s)=s$, $\psi_1(g(t)h(t))=\eta(t)$, we have $\psi_0, \psi_1\in P_+(0, \infty)$, and then $\phi_n\in P_+(0, \infty)$ by product theorem.

(ⅱ) First we prove that

$ C\le B\Longrightarrow \phi_n(k_n(C)^\frac{1}{2}h(C)g(C)k_n(C)^\frac{1}{2})\le\phi_n(k_n(C)^\frac{1}{2}\\ h(B)g(B)k_n(C)^\frac{1}{2}). $

Since $\phi_n, k_n, h, g$ are all nonnegative, nondecreasing functions and $J$ is a right open interval, by considering $C+\epsilon, B+\epsilon$, we may assume that $k_n(C)^\frac{1}{2}, h(C), h(B), g(C), g(B)$ are positive semi-definite and invertible. Through (ⅰ),

$ \phi_1(f_1(t)h(t)g(t))=f_1(t)\eta(t). $

(2.3)

Since $0\le f_1(C)\le f_1(B)\Longrightarrow f_1(C)^\frac{1}{2}f_1(B)^{-1}f_1(C)^\frac{1}{2}\le1$, by Lemma 2.3, we have

$ \begin{array}{rlr} \phi_1(k_1(C)^\frac{1}{2}h(B)g(B)k_1(C)^\frac{1}{2})&=\phi_1(f_1(C)^\frac{1}{2}f_1(B)^{{-}\frac{1}{2}}\\ &f_1(B)h(B)g(B)f_1(B)^{{-}\frac{1}{2}}f_1(C)^\frac{1}{2})\\ &\ge f_1(C)^\frac{1}{2}f_1(B)^{{-}\frac{1}{2}}\phi_1(f_1(B)\\ &h(B)g(B))f_1(B)^{{-}\frac{1}{2}}f_1(C)^\frac{1}{2}\\ &=f_1(C)^\frac{1}{2}f_1(B)^{{-}\frac{1}{2}}f_1(B)\eta(B)\\ &f_1(B)^{{-}\frac{1}{2}}f_1(C)^\frac{1}{2}\\ &=f_1(C)^\frac{1}{2}\eta(B)f_1(C)^\frac{1}{2}\\&\ge f_1(C)\eta(C)=\phi_1(k_1(C)^\frac{1}{2}\\ &h(C)g(C)k_1(C)^\frac{1}{2}). \end{array} $

This implies the right part of (2.2) holds for $n=1$. Next we assume the right part of (2.2) holds for $n-1$. Since $k_{n-1}(t)\eta(t)\in P_+^{-1}(J)$ and $f_n\in P_+(J)$, so $f_n \preceq t\preceq k_{n-1}(t)\eta(t)$, and this means that there exists $\Psi_n\in P_+(0, \infty)$ such that $f_n(t)=\Psi_n(k_{n-1}(t)\eta(t))$. Put $s=k_{n-1}(t)\eta(t)$, we can obtain $\phi_n(\phi_{n-1}^{-1}(s)\Psi_n(s))=s\Psi_n(s)$. Since the following inequality holds

$ \phi_{n-1}(k_{n-1}(C)^\frac{1}{2}h(C)g(C)k_{n-1}(C)^\frac{1}{2})\le\phi_{n-1}(k_{n-1}(C)^\frac{1}{2}h(B)g(B)k_{n-1}(C)^\frac{1}{2}). $

Denote the left side of the upper inequalities as $H$, the right one as $K$, we have

$ \Psi_n(H)^\frac{1}{2}\Psi_n(K)^{-1}\Psi_n(H)^{\frac{1}{2}}\le I. $

By $H=\phi_{n-1}(k_{n-1}(C)h(C)g(C))=k_{n-1}(C)\eta(C)$, we obtain

$ \Psi_n(H)=f_n(C), \quad \phi_{n-1}^{-1}(K)=k_{n-1}(C)^\frac{1}{2}h(B)g(B)k_{n-1}(C)^\frac{1}{2}. $

(2.4)

By Lemma 2.3 again, we obtain

$ \begin{array}{lll} &&\phi_n(\Psi_n(H)^\frac{1}{2}\phi_{n-1}^{-1}(K)\Psi_n(H)^\frac{1}{2})\\ &=&\phi_n(\Psi_n(H)^\frac{1}{2}\Psi_n(K)^{{-}\frac{1}{2}}\Psi_n(K) \phi_{n-1}^{-1}(K)\Psi_n(K)^{{-}\frac{1}{2}}\Psi_n(H)^\frac{1}{2}) \\ &\ge&\Psi_n(H)^\frac{1}{2}\Psi_n(K)^{{-}\frac{1}{2}}\phi_n(\Psi_n(K)\phi_{n-1}^{-1}(K))\Psi_n(K)^{{-}\frac{1}{2}}\Psi_n(H)^\frac{1}{2}\\ &=&\Psi_n(H)^\frac{1}{2}\Psi_n(K)^{{-}\frac{1}{2}}K\Psi_n(K)\Psi_n(K)^{{-}\frac{1}{2}}\Psi_n(H)^\frac{1}{2}\\ &=&\Psi_n(H)^\frac{1}{2}K\Psi_n(H)^\frac{1}{2}\ge H\Psi_n(H). \end{array} $

From the above inequalities and (2.4), we get

$ \begin{array}{rlr} &\phi_n(f_n(C)^\frac{1}{2}k_{n-1}(C)^\frac{1}{2}h(B)g(B)k_{n-1}(C)^\frac{1}{2}f_n(C)^\frac{1}{2})\\ \ge& f_n(C)k_{n-1}(C)\eta(C)=\phi_n(k_n(C)h(C)g(C)). \end{array} $

Therefore the right part of (2.2) holds for $n$, one can proof the left part of (2.2) similarly.

Remark  In Theorem 2.1, let $n=2$, $f_1(t)=g(t)=1$, $f_2(t)=t^r(r\ge0)$, $h(t)=t^p~(p\ge1)$, and $\eta(t)=t$, then we have $\phi_2(t^{p+r})=t^{1+r}$. So Furuta inequality can be obtained by (2.2) and L-H inequality.

Lemma 2.4(see [10, 11])  Put $J\neq(-\infty, \infty)$, then $g\in LP_+(J)$ if and only if there exists a sequence $\left\{ g_n \right\}$ of a finite product of functions in $P_+(J)$ which converges pointwise to $g$ on $J$, further more, $\left\{ g_n \right\}$ converges uniformly to $g$ on every bounded closed subinterval of $J$.

Theorem 2.2  Put $J\neq(-\infty, \infty)$, $f(t)>0$ for $t\in J$ and $\eta(t), h(t), k(t), g(t)$ are nonnegative functions on $J$ such that $\eta\in P_+(J)\cap P_+^{-1}(J), f\in P_+(J), fh\in P_+^{-1}(J), \frac{k}{f}\in LP_+(J), g\in P_+^{-1}(J)\cup \left\{ f \right\}$, then

(ⅰ) the function $\phi$ on $(0, \infty)$ defined by

$ \begin{aligned} \phi(k(t)h(t)g(t))=k(t)\eta(t) \quad (t\in J) \end{aligned} $

(2.5)

belongs to $P_+(0, \infty)$;

(ⅱ) If $A\le C\le B$, then for $\varphi\in P(0, \infty)$ such that $\varphi\preceq\phi$ on $(0, \infty), $

$ \begin{align} & \varphi (k{{(C)}^{\frac{1}{2}}}h(A)g(A)k{{(C)}^{\frac{1}{2}}})\le \varphi (k{{(C)}^{\frac{1}{2}}}h(C)g(C)k{{(C)}^{\frac{1}{2}}}) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \varphi (k{{(C)}^{\frac{1}{2}}}h(B)g(B)k{{(C)}^{\frac{1}{2}}}). \\ \end{align} $

(2.6)

Proof  (ⅰ) First consider $g\in P_+^{-1}(J)$. Put $l=\frac{k}{f}$, then $k=lf$ and

$ k(t)h(t)g(t)=l(t)f(t)h(t)g(t), ~~ k(t)\eta(t)=l(t)f(t)\eta(t). $

Let $\psi_0(s)=s$, $\psi_1(f(t)h(t))=f(t)$, $\psi_2(g(t))=\eta(t)$, then $\psi_0, \psi_1, \psi_2\in P_+(0, \infty)$. By taking $s=l(t)$ and applying product theorem, we get

$ \psi_0(l(t))\psi_1(f(t)h(t))\psi_2(g(t))\preceq l(t)f(t)h(t)g(t), $

which equals to $k(t)\eta(t)\preceq k(t)h(t)g(t)$. So we have $\phi\in P_+(0, \infty)$ for $\phi$ such that

$ \phi(k(t)h(t)g(t))=k(t)\eta(t)~~ (t\in I). $

If $g=f$, taking $\psi_0(s)=s$, $\psi_1(h(t)g(t))=\eta(t)$, obviously, we have $\psi_0, \psi_1\in P_+(0, \infty)$, and then $\psi_0(k)\psi_1(hg)\preceq khg$. Hence we also have $\phi\in P_+(0, \infty)$ from product theorem.

(ⅱ) From Lemma 2.4, we obtain there exists a sequence $\left\{ l_n \right\}$, where $l_n(t)$ is a finite product of functions in $P_+(J)$, such that $l_n(t)$ converges ponitwise to $l(t)$. Put $k_n(t)=f(t)l_n(t)$ then we easily get $k_n(t)$ converges to $k(t)=f(t)l(t)$. Define $\phi_n(k_n(t)h(t)g(t))=k_n(t)\eta(t)$ $(t\in J)$, $\phi_n\in P_+(0, \infty)$. By Theorem 2.1, we have

$ \begin{array}{lll} \phi_n(k_n(C)^\frac{1}{2}h(A)g(A)k_n(C)^\frac{1}{2})&\le&\phi_n(k_n(C)^\frac{1}{2}h(C)g(C)k_n(C)^\frac{1}{2})\\ &\le&\phi_n(k_n(C)^\frac{1}{2}h(B)g(B)k_n(C)^\frac{1}{2}). \end{array} $

Since $k_n(t)h(t)g(t)\in P_+^{-1}(J)$ therefore $k_n(t)h(t)g(t)$ is increasing on $J$ and converges uniformly to $k(t)h(t)g(t)$ on every compact interval of $J$. Since $k(t)h(t)g(t)(\in P_+^{-1}(J))$ is increasing, the inverse of $k_n(t)h(t)g(t)$ converges uniformly to one of $k(t)h(t)g(t)$ on every compact interval of $J$. It is also clear that $k_n(t)\eta(t)$ converges uniformly to $k(t)\eta(t)$ on every compact interval of $J$. Therefore $\phi_n$ converges uniformly to $\phi$ on every compact interval of $J$, since $k_n(C)$ converges to $k(C)$ in the operator norms, (2.6) holds for $\phi$ hence for any $\varphi$ given by $\varphi\preceq\phi$.

Lemma 2.5(Choi inequality)(see [6, 7])  Let $\Phi$ be a positive unital linear map, then

(C1) when $A>0$ and $-1\le p\le0$, then $\Phi(A)^p\le\Phi(A^p)$;

(C2) when $A\ge0$ and $0\le p\le1$, then $\Phi(A)^p\ge\Phi(A^p)$;

(C3) when $A\ge0$ and $1\le p\le2$, then $\Phi(A)^p\le\Phi(A^p)$.

Corollary 2.1  Put $J\neq(-\infty, \infty)$, $f(t)>0 $ for $t\in J$ and $\eta(t), h(t), k(t), g(t)$ are nonnegative functions on $J$ such that $\eta\in P_+(J)\cap P_+^{-1}(J), f\in P_+(J), fh\in P_+^{-1}(J), \frac{k}{f}\in LP_+(J), g\in P_+^{-1}(J)\cup \left\{ f \right\}$, the function $\phi$ on $(0, \infty)$ defined as (2.5), $\Phi$ is a positive unital linear map. If

$ A_0^{p_0}\le A_1^{p_0}, A_1^{p_1}\le A_2^{p_1}, ~~ 0\le p_0\le p_1\le p_2, $

then for $\varphi\in P(0, \infty)$ such that $\varphi\preceq\phi$,

$ \begin{aligned} &\varphi(k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2}h(\Phi(A_0^{p_0}))g(\Phi(A_0^{p_0}))k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2})\\ \le&\varphi(k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2}h(\Phi(A_1^{p_1})^\frac{p_0}{p_1})g(\Phi(A_1^{p_1})^\frac{p_0}{p_1})k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2})\\ \le&\varphi(k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2}h(\Phi(A_2^{p_2})^\frac{p_0}{p_2})g(\Phi(A_2^{p_2})^\frac{p_0}{p_2})k(\Phi(A_1^{p_1})^\frac{p_0}{p_1})^\frac{1}{2}).\\ \end{aligned} $

(2.7)

Proof  By Choi inequality and L-H inequality, we obtain

$ \Phi(A_2^{p_2})^\frac{p_0}{p_2}=\Phi(A_2^{p_2})^{\frac{p_1}{p_2}\cdot\frac{p_0}{p_1}}\ge\Phi(A_2^{p_1})^\frac{p_0}{p_1}\ge \Phi(A_1^{p_1})^\frac{p_0}{p_1}\ge \Phi(A_1^{p_0})\ge \Phi(A_0^{p_0}), $

which contains $\Phi(A_2^{p_2})^\frac{p_0}{p_2}\ge \Phi(A_1^{p_1})^\frac{p_0}{p_1}\ge \Phi(A_0^{p_0})$, from Theorem 2.2, we thus get (2.7).

Corollary 2.2  Put

$ \eta(t)\in P_+(J)\cap P_+^{-1}(J), ~~ \frac{h(t)}{t^p}, ~~ \frac{k(t)}{t^r}\in LP_+(0, \infty), ~~ p, r\ge0 $

and $p+r\ge1$, $g(t)\in P_+^{-1}(0, \infty)$. The function $\phi$ on $(0, \infty)$ is defined by

$ \phi(k(t)h(t)g(t))=k(t)\eta(t)~~ (t\in J), \varphi\in P(0, \infty) $

such that $\varphi\preceq\phi$. Then (2.5) and (2.6) in Theorem 2.2 hold.

Proof  Put $c=\min\left\{ 1, p\right\}$, then $f(t)=t^{1-c}\in P_+(0, \infty)$. Thus we get

$ f(t)h(t)=\frac{h(t)}{t^p}t^{1+p-c}\in P_+^{-1}(0, \infty), ~~ \frac{k(t)}{f(t)}=\frac{k(t)}{t^r}t^{r+c-1}\in LP_+(0, \infty), $

which means the conditions of Theorem 2.2 is satisfied. Therefore (2.5) and (2.6) in Theorem 2.2 hold.

Corollary 2.3  Put $\eta(t)\in P_+(J)\cap P_+^{-1}(J)$, $\frac{h(t)}{t^p}, \frac{k(t)}{t^r}\in LP_+(0, \infty)$, $p, r\ge0$ and $p+r\ge1$, $s\ge1$, we obtain

$ \begin{align} & \log (k{{(C)}^{\frac{1}{2}}}h(A){{A}^{s}}k{{(C)}^{\frac{1}{2}}})\le \log (k{{(C)}^{\frac{1}{2}}}h(C){{C}^{s}}k{{(C)}^{\frac{1}{2}}}) \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \le \log (k{{(C)}^{\frac{1}{2}}}h(B){{B}^{s}}k{{(C)}^{\frac{1}{2}}}). \\ \end{align} $

(2.8)

Proof  Put $g(t)=t^s (s\ge1)$, $\eta(t)=t$ in Corollary 2.2. Then we only need to show $\log s\preceq \phi(s), s\in(0, \infty)$. The definition of $\phi$ is given in (2.5). The upper majorization relationship is equivalent to

$ \log(k(t)h(t)t^s)\preceq\phi(k(t)h(t)t^s)=k(t)t. $

It is obviously that $\log k(t), \log h(t), \log t^s$ are operator monotone on $(0, \infty)$ and $k(t)t=\frac{k(t)}{t^r}t^{r+1}\in P_+^{-1}(0, \infty)$, then

$ \log(k(t)h(t)t^s)=\log k(t)+\log h(t)+\log t^s\preceq t\preceq k(t)t, \quad t\in(0, \infty). $

Therefore (2.8) holds.