Let \(x\in \overline{\mathcal{M}} \) and \(d_{\mu(x)}(t)\) be the classical distribution function of \(s\rightarrow\mu_{s}(x)\). By Proposition 1.2 of [10], we deduce
$$\lambda_{t}(x)=d_{\mu(x)}(t)=m \bigl( \bigl\{ s\in(0, \infty): \mu_{s}(x)>t \bigr\} \bigr),\quad t>0, $$
where m is the Lebesgue measure on \((0, \infty)\). Since \(s\rightarrow\mu_{s}(x)\) is non-increasing and continuous from the right (see, Lemma 2.5 of [7]), we have
$$\lambda_{t}(x)=\inf \bigl\{ s>0: \mu_{s}(x)\leq t \bigr\} , \quad t>0. $$
Moreover,
$$ \mu_{\lambda_{s}(x)}(x) \leq s,\quad s>0. $$
(3.1)
The following lemma, which includes a basic property of generalized singular number, plays a central role in our investigation.
Lemma 3.1
Let
\(x_{i}\in \overline{\mathcal{M}}\), \(i=1, 2, \ldots, n\). Then
$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ \max \bigl\{ \mu_{s_{1}}(x_{1}), \mu_{s_{2}}(x_{2}), \ldots, \mu_{s_{n}}(x_{n}) \bigr\} : s_{i}\geq0, \sum_{i=1}^{n}s_{i} \leq t \Biggr\} . \end{aligned}$$
Moreover,
$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ \max \bigl\{ \mu_{s_{1}}(x_{1}), \mu_{s_{2}}(x_{2}), \ldots, \mu_{s_{n}}(x_{n}) \bigr\} : s_{i}\geq0, \sum_{i=1}^{n}s_{i}=t \Biggr\} . \end{aligned}$$
Proof
Let \(s_{i}\geq0\) with \(\sum_{i=1}^{n}s_{i}\leq t\). By (2.4), we get
$$\tau \Biggl(\bigoplus_{i=1}^{n}e_{(\mu_{s_{i}}(x_{i}), \infty)} \bigl(|x_{i}| \bigr) \Biggr)\leq\sum_{i=1}^{n}s_{i} \leq t. $$
Therefore, according to the definition of generalized singular number, we obtain
$$\begin{aligned} \mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)\leq\Biggl\| \bigoplus_{i=1}^{n}x_{i}e_{[0,\mu _{s_{i}}(x_{i})]}\bigl(|x_{i}|\bigr) \Biggr\| \leq\max_{i=1, 2, \ldots, n} \bigl\{ \mu_{s_{i}}(x_{i}) \bigr\} . \end{aligned}$$
For the reverse inclusion, from (2.3), we get
$$\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ s\geq0: \tau \Biggl(e_{(s, \infty)} \Biggl( \Biggl|\bigoplus_{i=1}^{n} x_{i} \Biggr| \Biggr) \Biggr)\leq t \Biggr\} . $$
Since
$$e_{(s, \infty)} \Biggl(\Biggl|\bigoplus_{i=1}^{n} x_{i}\Biggr| \Biggr)=e_{(s, \infty)} \Biggl(\bigoplus _{i=1}^{n} |x_{i}| \Biggr) =\bigoplus _{i=1}^{n} e_{(s, \infty)} \bigl(|x_{i}| \bigr), $$
we have
$$\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr)=\inf \Biggl\{ s\geq0: \sum_{i=1}^{n} \tau \bigl(e_{(s, \infty )} \bigl(|x_{i}| \bigr) \bigr)\leq t \Biggr\} . $$
Let \(s_{i}=\tau(e_{(s, \infty)}(|x_{i}|))\). It follows from inequality (3.1) that \(\mu_{s_{i}}(x_{i})\leq s\). Hence
$$\max_{i=1, 2, \ldots, n} \bigl\{ \mu_{s_{i}}(x_{i}) \bigr\} \leq\mu_{t} \Biggl(\bigoplus_{i=1}^{n} x_{i} \Biggr). $$
□
Remark 3.2
-
(1)
Let \(x\in \overline{\mathcal{M}}\). If \(x_{i}=x\), \(i=1, 2, \ldots, n\), it follows from Lemma 3.1 that \(\mu_{t}(\bigoplus_{i=1}^{n} x_{i})=\mu_{\frac{t}{n}}(x)\), \(t>0\).
-
(2)
Let \(x\in \overline{\mathcal{M}}\). If \(x_{1}=x\) and \(x_{i}=0\), \(i= 2, 3, \ldots, n\), it follows from Lemma 3.1 that \(\mu_{t}(\bigoplus_{i=1}^{n} x_{i})=\mu_{t}(x)\), \(t>0\).
-
(3)
Let \(x_{1}, x_{2}, y_{1}, y_{2}\in\overline{\mathcal{M}}\) such that \(\mu_{t}(x_{i})\leq\mu_{t}(y_{i})\), \(t>0\), \(i=1, 2\). From Lemma 3.1, we deduce \(\mu_{t}(x_{1}\oplus x_{2})\leq\mu_{t}(y_{1}\oplus y_{2})\), \(t>0\).
As an application of Lemma 3.1 we now obtain the desired generalized singular number inequality (1.3) for τ-measurable operators.
Lemma 3.3
Let
\(x, y, z\in \overline{\mathcal{M}} \). If
\(\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )\geq0\), then
$$2\mu_{t}(z)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$
Proof
Let \(N=\bigl ( {\scriptsize\begin{matrix}{} 0&z\cr z^{*}&0 \end{matrix}} \bigr )\), \(M=\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )\), and \(U=\bigl ( {\scriptsize\begin{matrix}{} 1&0\cr 0&-1 \end{matrix}} \bigr )\). Then
$$\begin{aligned} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} &= \begin{pmatrix} x&-z\\ -z^{*}&y \end{pmatrix} \\ &= \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} -2N. \end{aligned}$$
Hence \(N=\frac{1}{2}(M-UMU^{*})\). Let \(N=N^{+}-N^{-}\) be the Jordan decomposition of N. It follows from Lemma 6 of [11] that \(\mu_{t}(N^{+})\leq\mu_{t}(\frac{1}{2}M)\), \(t>0\), and
$$\mu_{t} \bigl(N^{-} \bigr)\leq\mu_{t} \biggl( \frac{1}{2} UMU^{*} \biggr)\leq \|U\|\bigl\| U^{*}\bigr\| \mu_{t} \biggl( \frac{1}{2}M \biggr) \leq\mu_{t} \biggl(\frac{1}{2}M \biggr),\quad t>0. $$
By Theorem 6 of [12], we have
$$\mu_{t}(N)=\mu_{t} \bigl(N^{+}-N^{-} \bigr)\leq \mu_{t} \bigl(N^{+}\oplus N^{-} \bigr), \quad t>0. $$
Therefore, from Lemma 3.1 we obtain
$$\mu_{2t}(N)\leq\mu_{2t} \bigl(N^{+}\oplus N^{-} \bigr)\leq \mu_{2t} \biggl(\frac{1}{2}M \oplus\frac{1}{2}M \biggr)= \mu_{t} \biggl(\frac{1}{2}M \biggr),\quad t>0, $$
i.e.,
$$2\mu_{2t} \left ( \begin{pmatrix} 0&z\\ z^{*}&0 \end{pmatrix} \right )=2\mu_{2t}(N)\leq \mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$
It is clear that \(\bigl ( {\scriptsize\begin{matrix}{} 0&1\cr 1&0 \end{matrix}} \bigr )\bigl ( {\scriptsize\begin{matrix}{} 0&z\cr z^{*}&0 \end{matrix}} \bigr )=\bigl ( {\scriptsize\begin{matrix}{} z&0\cr 0&z^{*} \end{matrix}} \bigr )^{*}\) and \(\|\bigl ( {\scriptsize\begin{matrix}{} 0&1\cr 1&0 \end{matrix}} \bigr )\|=1\). Then Lemma 2.5 of [7] and Lemma 3.1 imply that
$$2\mu_{t}(z) =2\mu_{2t} \left ( \begin{pmatrix} z&0\\ 0&z^{*} \end{pmatrix} \right )\leq2 \mu_{2t}(N)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\ z^{*}&y \end{pmatrix} \right ),\quad t>0. $$
□
Combing Lemma 3.3 with the following theorem we see that inequalities (1.1), (1.2), and (1.3) hold for τ-measurable operators.
Theorem 3.4
The following statements are equivalent:
-
(1)
Let
\(0\leq x, y \in \overline{\mathcal{M}}\). Then
\(\mu_{t}(x-y)\leq\mu_{t}(x\oplus y)\), \(t>0\).
-
(2)
For any
\(x, y\in\overline{\mathcal{M}}\), \(2\mu_{t}(xy^{*})\leq\mu_{t}(x^{*}x+y^{*}y)\), \(t>0\).
-
(3)
Let
\(x, y, z\in \overline{\mathcal{M}} \). If
\(\bigl ( {\scriptsize\begin{matrix}{} x&z\cr z^{*}&y \end{matrix}} \bigr )\geq0\), then
$$2\mu_{t}(z)\leq\mu_{t} \left ( \begin{pmatrix} x&z\\z^{*}&y \end{pmatrix} \right ),\quad t>0. $$
Proof
(1) ⇒ (2): For any \(x, y\in\overline{\mathcal{M}}\), we write \(X=\bigl ( {\scriptsize\begin{matrix}{} x&0\cr y&0 \end{matrix}} \bigr )\), \(Y=\bigl ( {\scriptsize\begin{matrix}{} x&0\cr -y&0 \end{matrix}} \bigr )\). Then \(X^{*}X=\bigl ( {\scriptsize\begin{matrix}{} x^{*}x+y^{*}y&0\cr 0&0 \end{matrix}} \bigr )\) and \(Y^{*}Y=\bigl ( {\scriptsize\begin{matrix}{} x^{*}x+y^{*}y&0\cr 0&0 \end{matrix}} \bigr )\). It follows from Lemma 3.1 and (1) that
$$\begin{aligned} 2\mu_{t} \left ( \begin{pmatrix} yx^{*}&0\\0&xy^{*} \end{pmatrix} \right ) &=2\mu_{t} \left ( \begin{pmatrix} 0&xy^{*}\\yx^{*}&0 \end{pmatrix} \right )=\mu_{t} \bigl(XX^{*}-YY^{*} \bigr) \\ &\leq\mu_{t} \bigl(XX^{*}\oplus YY^{*} \bigr) \\ &=\inf \bigl\{ \max \bigl(\mu_{a} \bigl(XX^{*} \bigr), \mu_{b} \bigl(YY^{*} \bigr) \bigr): a, b\geq0, a+b=t \bigr\} \\ &=\inf \bigl\{ \max \bigl(\mu_{a} \bigl(X^{*}X \bigr), \mu_{b} \bigl(Y^{*}Y \bigr) \bigr): a, b\geq0, a+b=t \bigr\} \\ &=\inf_{ a, b\geq0, a+b=t} \bigl\{ \max \bigl(\mu_{a} \bigl(x^{*}x+y^{*}y \bigr), \mu_{b} \bigl(x^{*}x+y^{*}y \bigr) \bigr) \bigr\} \\ &=\mu_{t} \left ( \begin{pmatrix} x^{*}x+y^{*}y&0\\ 0&x^{*}x+y^{*}y \end{pmatrix} \right ),\quad t>0. \end{aligned}$$
Lemma 3.1 ensures that \(2\mu_{t}(xy^{*})\leq\mu_{t}(x^{*}x+y^{*}y)\), \(t>0\).
(2) ⇒ (1): Let \(0\leq x, y \in \overline{\mathcal{M}}\) and let
$$S= \begin{pmatrix} x^{\frac{1}{2}}&-y^{\frac{1}{2}}\\0&0 \end{pmatrix},\qquad T= \begin{pmatrix} x^{\frac{1}{2}}& y^{\frac{1}{2}}\\0&0 \end{pmatrix}. $$
From (2) we have \(2\mu_{t}(ST^{*})\leq\mu_{t}(S^{*}S+T^{*}T)\), \(t>0\). Then the result follows from Lemma 3.1.
From Lemma 3.3 we have (1) ⇒ (3).
(3) ⇒ (1): For any \(0\leq x, y\in\overline{\mathcal{M}}\), we have the following unitary similarity transform:
$$\frac{1}{\sqrt{2}} \begin{pmatrix} 1&1\\-1&1 \end{pmatrix} \begin{pmatrix} \frac{x+y}{2}&\frac{x-y}{2}\\ \frac{x-y}{2} &\frac{x+y}{2} \end{pmatrix} \frac{1}{\sqrt{2}} \begin{pmatrix} 1&-1\\1&1 \end{pmatrix} = \begin{pmatrix} x&0\\ 0 &y \end{pmatrix} \geq0. $$
According to (3), we obtain
$$\mu_{t}(x-y)\leq\mu_{t} \left ( \begin{pmatrix} \frac{x+y}{2}&\frac{x-y}{2}\\ \frac{x-y}{2} &\frac{x+y}{2} \end{pmatrix} \right )\leq \mu_{t} \begin{pmatrix} x&0\\0 &y \end{pmatrix},\quad t>0. $$
□
Lemma 3.5
Let
\(0\leq x, y\in \overline{\mathcal{M}}\)
and
\(0\leq r\leq1\). Then
$$2\mu_{t} \bigl(x^{1+r}+y^{1+r} \bigr)\geq \mu_{t} \bigl((x+y)^{\frac {1}{2}} \bigl(x^{r}+y^{r} \bigr) (x+y)^{\frac{1}{2}} \bigr),\quad t>0. $$
Proof
Let \(0\leq x, y\in \overline{\mathcal{M}}\) and \(0\leq r\leq1\). Since \(1\leq1+r\leq2\), the function \(t\rightarrow t^{1+r}\) is operator convex. Hence
$$\frac{x^{1+r}+y^{1+r}}{2}\geq \biggl(\frac{x+y}{2} \biggr)^{1+r} = \frac{1}{2}(x+y)^{\frac{1}{2}} \biggl(\frac{x+y}{2} \biggr)^{r}(x+y)^{\frac{1}{2}}. $$
Note that \(t\rightarrow t^{r}\) (\(0\leq r\leq1\)) is operator concave, we obtain \(\frac{x^{r}+y^{r}}{2}\leq(\frac{x+y}{2})^{r}\). Therefore,
$$x^{1+r}+y^{1+r}\geq\frac{1}{2}(x+y)^{\frac{1}{2}} \bigl(x^{r}+y^{r} \bigr) (x+y)^{\frac{1}{2}}. $$
This completes the proof. □
Based on Lemma 3.5 we now obtain the desired generalized singular number inequality (1.4) for τ-measurable operators.
Theorem 3.6
Let
\(0\leq r\leq1\)
and
\(0\leq x, y\in L^{1}(\mathcal{M})\). Then
$$ \mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad t>0. $$
(3.2)
Proof
Let \(0\leq v\leq1\). If we replace x, y by \(x^{\frac{1}{1+v}}\), \(y^{\frac{1}{1+v}}\), respectively, in Lemma 3.5, we deduce
$$2\mu_{t}(x+y)\geq\mu_{t} \bigl( \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr)^{\frac{1}{2}} \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr)^{\frac{1}{2}} \bigr). $$
It follows from Lemma 2 of [13] and the fact \(x, y\in L^{1}(\mathcal{M})\) that
$$ 2\mu_{t}(x+y)\geq\mu_{t} \bigl( \bigl(x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}} \bigr) \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) \bigr). $$
(3.3)
Note that
$$\begin{aligned} &\mu_{t} \left ( \begin{pmatrix} (x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}) (x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}})&0\\ 0&0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}&y^{\frac{1}{2+2v}}\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{2+2v}}&0\\ y^{\frac{1}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}&0\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}&0\\y^{\frac{1}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}&0\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{2+2v}}&y^{\frac{1}{2+2v}}\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) x^{\frac{1}{2+2v}}&x^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) y^{\frac{1}{2+2v}}\\y^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) x^{\frac{1}{2+2v}} &y^{\frac{1}{2+2v}}(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}}) y^{\frac{1}{2+2v}} \end{pmatrix} \right ). \end{aligned}$$
Combining Lemma 3.1, Lemma 3.3, and inequality (3.3) we deduce
$$\begin{aligned} \mu_{t}(x+y)&\geq\mu_{t} \bigl(x^{\frac{1}{2+2v}} \bigl(x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}} \bigr) y^{\frac{1}{2+2v}} \bigr) \\ &=\mu_{t} \bigl(x^{\frac{2v+1}{2+2v}}y^{\frac{1}{2+2v}}+x^{\frac {1}{2+2v}}y^{\frac{2v+1}{2+2v}} \bigr),\quad 0\leq v\leq1. \end{aligned}$$
Therefore,
$$\mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad \frac{1}{2}\leq r\leq \frac{3}{4}. $$
On the one hand, we have
$$\begin{aligned} &\mu_{t} \left ( \begin{pmatrix} (x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}) (x^{\frac{v}{1+v}}+y^{\frac{v}{1+v}})&0\\ 0 &0 \end{pmatrix} \right ) \\ &\quad=\mu_{t} \left ( \begin{pmatrix} x^{\frac{v}{2+2v}}&0\\y^{\frac{v}{2+2v}} &0 \end{pmatrix} \begin{pmatrix} x^{\frac{1}{1+v}}+y^{\frac{1}{1+v}}&0\\ 0 &0 \end{pmatrix} \begin{pmatrix} x^{\frac{v}{2+2v}}&y^{\frac{v}{2+2v}}\\ 0 &0 \end{pmatrix} \right ). \end{aligned}$$
Repeating the arguments above we get
$$\mu_{t} \bigl(x^{r}y^{1-r}+x^{1-r}y^{r} \bigr)\leq\mu_{t}(x+y),\quad \frac{3}{4}\leq r\leq1. $$
By the symmetry property of inequality (3.2) with respect to \(r=\frac{1}{2}\), we see that inequality (3.2) holds for all \(0\leq r\leq1\). □
Let \(0\leq x, y\in \overline{\mathcal{M}}\). Then Lemma 3.1 and Theorem 3.4 imply that
$$\mu_{t} \bigl((x-y)\oplus0 \bigr)\leq\mu_{t}(x\oplus y), \quad t>0. $$
If \(x, y\in \overline{\mathcal{M}}\) with \(\mu_{t}(x)\leq\mu_{t}(y)\), \(t>0\), Lemma 3.1 gives us that
$$\mu_{t}(x)=\mu_{t}(x\oplus0)\leq\mu_{t}(y\oplus y),\quad t>0. $$
Some examples of such inequalities related to ones discussed above are presented below.
Lemma 3.7
Let
\(x, y\in\overline{\mathcal{M}}^{sa}: =\{z\in\overline{\mathcal{M}}; z=z^{*}\}\)
such that
\(\pm y\leq x\). If
\(x\geq0\), then
$$\mu_{t}(y)\leq\mu_{t}(x\oplus x) $$
and
$$\int_{0}^{t}\mu_{t}(y)\,ds\leq \int_{0}^{t}\mu_{s}(x)\,ds,\quad t>0. $$
Proof
Since \(\pm y\leq x\), we have \(-x\leq y\leq x\). Then Theorem 1 of [14] indicates that \(2|y|\leq x+uxu^{*}\) for some unitary \(u\in\overline{\mathcal{M}}^{sa}\). From Theorem 4.4 and Lemma 2.5 of [7], we deduce
$$2\mu_{t}(y)\leq\mu_{t} \bigl(x+uxu^{*} \bigr)\leq \mu_{\frac{t}{2}} \bigl(uxu^{*} \bigr)+\mu_{\frac {t}{2}}(x)\leq2 \mu_{\frac{t}{2}}(x) =2\mu_{t}(x\oplus x),\quad t>0, $$
and
$$2 \int_{0}^{t}\mu_{s}(y)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x+uxu^{*} \bigr)\,ds \leq2 \int_{0}^{t}\mu_{s}(x)\,ds,\quad t>0. $$
□
We conclude this section with a series of inequalities which are related to the Heinz mean inequality for a generalized singular number of τ-measurable operators.
Proposition 3.8
Let
\(x, y\in\overline{\mathcal{M}}\). Then
$$ \mu_{t} \bigl(x^{*}y+y^{*}x \bigr)\leq\mu_{t} \bigl( \bigl(x^{*}x+y^{*}y \bigr)\oplus \bigl(x^{*}x+y^{*}y \bigr) \bigr),\quad t>0, $$
(3.4)
and
$$ \mu_{t} \bigl(yx^{*}+xy^{*} \bigr)\leq\mu_{t} \bigl( \bigl(x^{*}x+y^{*}y \bigr)\oplus \bigl(x^{*}x+y^{*}y \bigr) \bigr),\quad t>0. $$
(3.5)
Proof
Since \((x\pm y)^{*}(x\pm y)\geq0\), we have \(\pm(x^{*}y+y^{*}x)\leq x^{*}x+y^{*}y\). Thus inequality (3.4) follows from Lemma 3.7. Inequality (3.5) follows from Theorem 6 of [12] and Theorem 3.4(2). □
Corollary 3.9
Let
\(x, y\in\overline{\mathcal{M}}\)
and
\(0< r\leq\infty\). Then
$$\int_{0}^{t}\mu_{s} \bigl(x^{*}y+y^{*}x \bigr)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x^{*}x+y^{*}y \bigr)\,ds,\quad t>0, $$
(3.6)
and
$$ \int_{0}^{t}\mu_{s} \bigl(yx^{*}+xy^{*} \bigr)\,ds\leq \int_{0}^{t}\mu_{s} \bigl(x^{*}x+y^{*}y \bigr)\,ds,\quad t>0. $$
(3.7)
Proof
It follows from Lemma 3.7 and the proof of Proposition 3.8. □
Proposition 3.10
Let
\(x, y\in\overline{\mathcal{M}}\). Then
$$ \mu_{t}(x+y)\leq\mu_{t} \bigl( \bigl(|x|+|y| \bigr) \oplus \bigl( \bigl|x^{*} \bigr|+ \bigl|y^{*} \bigr| \bigr) \bigr), \quad t>0. $$
(3.8)
Proof
Let \(x\in\overline{\mathcal{M}}\). Note that \(\bigl ( {\scriptsize\begin{matrix}{} |x|&\pm x^{*}\cr \pm x &|x^{*}| \end{matrix}} \bigr )\geq0\). Then
$$\begin{aligned} \begin{pmatrix} |x|+|y|& \pm(x+y)^{*}\\ \pm(x+y) &|x^{*}|+|y^{*}| \end{pmatrix}\geq0. \end{aligned}$$
Thus
$$\begin{aligned} \pm \begin{pmatrix} 0& (x+y)^{*}\\x+y &0 \end{pmatrix}\leq \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix}. \end{aligned}$$
By Lemma 3.7, we obtain
$$\begin{aligned} \mu_{t} \bigl((x+y)\oplus(x+y)^{*} \bigr)={}&\mu_{t} \left ( \begin{pmatrix} 0& (x+y)^{*}\\x+y &0 \end{pmatrix} \right ) \\ \leq{}&\mu_{t} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right . \\ &{}\left .\oplus \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ) \\ ={}&\mu_{\frac{t}{2}} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ),\quad t>0. \end{aligned}$$
According to Lemma 2.5 of [7] and Lemma 3.1, we get
$$\mu_{t} \bigl((x+y)\oplus(x+y)^{*} \bigr)=\mu_{\frac{t}{2}}(x+y),\quad t>0. $$
This implies that
$$\mu_{t}(x+y)\leq\mu_{t} \left ( \begin{pmatrix} |x|+|y|&0\\0 &|x^{*}|+|y^{*}| \end{pmatrix} \right ),\quad t>0. $$
□