# Some inequalities and majorization for products of τ-measurable operators – Journal of Inequalities and Applications

#### ByZahra Maleki Khouzani and Seyed Mahmoud Manjegani

Aug 26, 2022

The Young inequality is a well-known inequality, valid for (p>1) and (frac{1}{p}+frac{1}{q}=1) is usually stated as

$$alpha beta le frac{1}{p}alpha ^{p}+ frac{1}{q}beta ^{q},$$

(3)

for any positive real numbers α and β, equality holds if and only if (alpha ^{p}=beta ^{q}). Several generalisations of the Young inequality where α and β are replaced by Hilbert space operators or by singular values, norms, or traces of operators are known. For more references and further discussion on the subject of the Young inequality for matrices and operators, we refer the reader to [1, 6, 9, 20]. In particular, we remark that it was the fundamental paper by T. Ando [1], which initiated the study of the Young inequality for the singular values of (ntimes n) matrices. The present paper adds to these results by formulating new Young-type norm inequalities in (L_{p}(mathcal{M})). Farenick and Manjegani [9] proved the case of equality in the Young inequality for operators in a semi-finite von Neumann algebra.

### Proposition 2.1

([9])

Let x, y be operators in (mathcal{M}).

$$mu _{t} bigl( biglvert xy^{*} bigrvert bigr) leq mu _{t} biggl(frac{1}{p} vert x vert ^{p} + frac{1}{q} vert y vert ^{q} biggr), quad textit{for all } t>0 .$$

(4)

Moreover, if (x, y in L_{1}(mathcal{M})) and bounded, then equality holds if and only if

$$vert x vert ^{p}= vert y vert ^{q}.$$

In the above proposition, x and y are bounded operators playing important role in the proof of the equality case in (4). In [19, Theorem 3.3], the authors extended (4) to singular values of τ-measurable operators. Since τ-measurable operators are not necessarily bounded with finite trace, the same proof for the case of equality in last proposition does not work. Still, we do not know for operators in (L_{p}( mathcal{M}) ): Does (mu _{t} ( | xy^{*}| ) = mu _{t} (frac{1}{p} | x | ^{p} + frac{1}{q} | y | ^{q} )) for all (t>0) imply that (| x| ^{p}=| y| ^{q})?

Let (1le p<infty ), an operator (xin mathcal{M}) is said to be locally integrable if there exists (delta >0) such that

$$int _{0}^{delta}mu _{t}(x)^{p} ,dt< infty .$$

The set containing all these operators is denoted by (mathfrak{L}_{mathrm{loc}}^{p}(mathcal{M})). Note that, in particular, all bounded operators (ain mathcal{M}) are of this class. Moreover,

$$int _{0}^{delta}mu _{t}(x)^{p} ,dtge mu _{delta}(x)^{p-1} int _{0}^{ delta}mu _{t}(x) ,dt$$

implies that (mathfrak{L}_{mathrm{loc}}^{p}(mathcal{M})subset mathfrak{L}_{mathrm{loc}}^{1}( mathcal{M})) for each (pge 1). In [19], Larotonda proved the following theorem that is answer to the above question for τ-compact operators.

### Theorem 2.2

([19])

Let (a, bin mathcal{K}(L_{0}(mathcal{M}))^{+}) with (abin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})). If

$$mu _{t} bigl( vert ab vert bigr) = mu _{t} biggl(frac{1}{p} a^{p} + frac{1}{q} b^{q} biggr) quad textit{for all}~~~t>0,$$

then (a^{p}=b^{q}). If (p=q=2), it suffices to assume that (abin mathfrak{L}_{mathrm{loc}}^{1}(mathcal{M})).

According to part (4) of Theorem 1.1, we have the following corollary.

### Corollary 2.3

Let (x, yin mathcal{K}(L_{0}(mathcal{M}))) with (xyin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})). If

$$mu _{t} bigl( biglvert xy^{*} bigrvert bigr) = mu _{t} biggl(frac{1}{p} vert x vert ^{p} + frac{1}{q} vert y vert ^{q} biggr) quad textit{for all}~~~t>0,$$

then (| x| ^{p}=| y| ^{q}).

This powerful version of the Young inequality for singular values is important in that all results on τ-measurable operators, such as norms inequalities and majorization, can be deduced from it. So, it makes sense to find more general cases of this inequality. However, in considering the young inequality, it is not known whether the above theorem has a formulation in which it is true that

$$mu _{t}bigl( vert azb vert bigr) leq mu _{t}biggl(frac{1}{p} vert az vert ^{p} + frac{1}{q} vert bz vert ^{q} biggr),$$

when a, b are positive τ-measurable operators, and (z in L_{0}(mathcal{M}) ). However, this topic exceeds the scope of this paper.

As a result of [19, Theorem 3.3], for every (x, y in L_{1}(mathcal{M})), we have

$$biglVert xy^{*} bigrVert _{1} le bigglVert frac{1}{p} vert x vert ^{p} + frac{1}{q} vert y vert ^{q} biggrVert _{1} le frac{1}{p} bigl| | x | ^{p} bigr| _{1}+ frac{1}{q}bigl| | y | ^{q} bigr| _{1}.$$

(5)

If x and y are bounded operators or (xyin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds if and only if (| x| ^{p}=| y| ^{q}) using Proposition 2.1 or Theorem 2.2. More general, using similar argument in [20, Corollary 2.5] and part (4) in Theorem 1.1, we have Hölder and Young inequalities within the (| cdot | _{1}) in (L_{1}(mathcal{M})).

### Corollary 2.4

If (x, y in L_{1}(mathcal{M})), then for all positive real numbers p, q, and r with (frac{1}{p}+frac{1}{q}=frac{1}{r}),

$$biglVert xy^{*} bigrVert _{1}^{r} le bigl( biglVert | x | ^{p} bigrVert _{1} bigr)^{frac{r}{p}} bigl( biglVert | y | ^{q} bigrVert _{1} bigr)^{frac{r}{q}} le frac{r}{p} bigl| | x | ^{p} bigr| _{1}+ frac{r}{q}bigl| | y | ^{q} bigr| _{1}.$$

Moreover, if x and y are bounded operators or (xyin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds if and only if (| x| ^{p}=| y| ^{q}).

### Theorem 2.5

(Cases of Equality in Tracial Hölder and Young Inequalities)

Assume that a, b are positive bounded operators in (L_{p}(mathcal{M}) ) and (p>1). The following statements are equivalent.

1. 1

(| ab| _{1}=| a^{p}| _{1}^{1/p} | b^{q}| _{1}^{1/q});

2. 2

(| |ab|| _{1}=| a^{p}| _{1}^{1/p} | b^{q}| _{1}^{1/q});

3. 3

(| ab| _{1}=frac{1}{p}| a^{p}| _{1}+ frac{1}{q}| b^{q}| _{1});

4. 4

(| |ab|| _{1}=frac{1}{p}| a^{p}| _{1}+ frac{1}{q}| b^{q}| _{1});

5. 5

(b^{q}=a^{p}).

### Proof

Using similar method used in the proof of [9, Theorem 3.6]. □

In the rest of this section, first, we demonstrate an extension of refinement of the Young inequality for τ-measurable operators [16, 17]. Kittaneh and Manasrah gave a refinement of the Young inequality as follows in [17]:

For (m=1, 2, ldots ) ,

$$bigl(a^{nu }b^{1-nu}bigr)^{m} +bigl( min {nu , 1-nu }bigr)^{m} bigl(sqrt{a^{m}}- sqrt{b^{m}}bigr)^{2} leq bigl(nu a + (1-nu ) b bigr)^{m},$$

(6)

where a, b are positive real numbers, and (0le nu le 1).

We generalize (6) for positive τ-measurable operators in the following theorem.

### Theorem 2.6

Let a, b be positive operators in (L_{1}(mathcal{M}) ). Then for (m=1, 2, ldots ) ,

$$biglVert a^{nu}b^{1-nu} bigrVert _{1}^{m}+min {nu , 1-nu }^{m} bigl( Vert a Vert _{1}^{frac{m}{2}} – Vert b Vert _{1}^{frac{m}{2}} bigr) ^{2} le biglVert nu a+(1-nu )b bigrVert _{1} ^{m} .$$

(7)

### Proof

By Corollary 2.4, we have

$$biglVert a^{nu}b^{1-nu} bigrVert _{1} le Vert a Vert _{1}^{ nu}. Vert b Vert _{1}^{1-nu} .$$

Thus by (6),

begin{aligned}& bigl( biglVert a^{nu}b^{1-nu} bigrVert _{1} bigr)^{m}+bigl(min { nu , 1-nu } bigr)^{m} bigl( Vert a Vert _{1}^{frac{m}{2}}- Vert b Vert _{1}^{frac{m}{2}} bigr) ^{2}\& quad le bigl( Vert a Vert _{1}^{nu} Vert b Vert _{1}^{1- nu} bigr)^{m}+bigl(min { nu , 1-nu }bigr)^{m} bigl( Vert a Vert _{1}^{frac{m}{2}}- Vert b Vert _{1}^{frac{m}{2}} bigr)^{2}\& quad le bigl(nu Vert a Vert _{1}+(1-nu ) Vert b Vert _{1} bigr)^{m}. end{aligned}

□

### Remark 2.7

A result similar to Theorem 2.6 has been proved in [22, Theorem 3.8].

The following result is another extension of refinement of the Young inequality for τ-measurable operators. Let us first recall the inequality introduced by Xing-ka in [15].

### Lemma 2.8

([15, Lemma 1.1])

Suppose that a and b are nonnegative real numbers.

1. 1.

if (0 leq nu leq frac{1}{2} ), then

$$bigl[ (nu a)^{nu }b ^{1-nu} bigr] ^{2} + nu ^{2} ( a-b)^{2} leq nu ^{2} a^{2} + (1-nu )^{2} b^{2}.$$

(8)

2. 2.

if (frac{1}{2} leq nu leq 1 ), then

$$bigl[ a^{nu } bigl( (1-nu ) b bigr) ^{1-nu} bigr] ^{2} + (1- nu )^{2} (a-b)^{2} leq nu ^{2} a^{2} +(1-nu )^{2} b^{2}.$$

(9)

### Lemma 2.9

(Cases of Equality in (8) and (9))

Equality holds in (8) if and only if (b=nu a), and equality holds in (9) if and only if (a=(1-nu )b).

### Proof

As we see in the proof of Lemma 2.8,

begin{aligned} nu ^{2} a^{2}+(1-nu )^{2}b^{2} -nu ^{2}(a-b)^{2} & = bbigl[2nu (nu a)+(1-2 nu )b bigr] geq b(nu a)^{2nu}b^{1-2nu}=bigl[(nu a)^{nu}b^{1-nu}bigr]^{2}. end{aligned}

Thus, equality holds in (8) if and only if

$$bbigl[2nu (nu a)+(1-2nu )bbigr]= b(nu a)^{2nu}b^{1-2nu}.$$

This relationship exists if and only if (b=nu a). Using the similar method, we can prove the case of equality in (9). □

### Theorem 2.10

Let (a, b in L_{2}(mathcal{M}) ) be positive operator.

1. 1.

If (0 leq nu leq frac{1}{2} ), then

$$tau bigl( bigl[ (nu a)^{nu }b^{1-nu} bigr] ^{2} bigr) + nu ^{2} bigl( sqrt{tau bigl(a^{2}bigr)} – sqrt{tau bigl(b^{2}bigr)} bigr) ^{2} leq tau bigl( nu ^{2} a^{2} +(1-nu )^{2} b^{2} bigr) ,$$

(10)

2. 2.

If (frac{1}{2} leq nu leq 1 ), then

begin{aligned} & tau bigl( bigllbrace a^{nu } bigl[ (1-nu ) b bigr] ^{1- nu} bigrrbrace ^{2} bigr) + (1-nu )^{2} bigl( sqrt{tau bigl(a^{2}bigr)} – sqrt{tau bigl(b^{2}bigr)} bigr) ^{2} \ &quad leq tau bigl( nu ^{2} a^{2} +(1- nu )^{2} b^{2} bigr). end{aligned}

(11)

Moreover, if a and b are bounded operators or (abin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds in (10) if and only if (b=nu a), and equality holds in (11) if and only if (a=(1-nu )b).

### Proof

According to Lemma 2.8, we have

$$bigl[ bigl(nu mu _{t}(a)bigr)^{nu }mu _{t}(b)^{1-nu} bigr] ^{2} + nu ^{2} bigl( mu _{t}(a) – mu _{t}(b) bigr) ^{2} leq nu ^{2} mu _{t}(a)^{2} +(1-nu )^{2} mu _{t}(b)^{2}, quad forall t>0.$$

Therefore,

begin{aligned} tau bigl( nu ^{2} a^{2} +(1-nu )^{2} b^{2} bigr) = & nu ^{2} tau bigl(a^{2} bigr) + (1-nu )^{2} tau bigl(b^{2}bigr) \ = & int _{0}^{infty } bigl[ nu ^{2} mu _{t} bigl(a^{2}bigr) + (1-nu )^{2} mu _{t} bigl(b^{2}bigr) bigr] ,dt \ geq & int _{0}^{infty } bigl[ bigl(nu mu _{t}(a)bigr)^{nu }mu _{t}(b)^{1- nu} bigr] ^{2} ,dt \ &{}+ nu ^{2} int _{0}^{infty } bigl[ mu _{t}(a)^{2} + mu _{t}(b)^{2} -2mu _{t}(a)mu _{t}(b) bigr] ,dt \ = & int _{0}^{infty } bigl[ bigl(nu mu _{t}(a) bigr)^{2nu} mu _{t}(b)^{2(1- nu )} bigr] ,dt \ &{} + nu ^{2} biggl[ tau bigl(a^{2}bigr) + tau bigl(b^{2}bigr) – 2 int _{0}^{s} mu _{t}(a)mu _{t}(b) ,dt biggr] \ = & I. end{aligned}

Now, the combination of Theorem 1.1 part (11) and the Hölder inequality gives us

begin{aligned} I geq & int _{0}^{infty }mu _{t} bigl((nu a)^{2nu} b^{2(1-nu )}bigr) ,dt \ &{} + nu ^{2} biggl[ tau bigl(a^{2}bigr) + tau bigl(b^{2}bigr) – 2 biggl( int _{0}^{s} mu _{t} bigl(a^{2}bigr) ,dt biggr)^{frac{1}{2}} biggl( int _{0}^{s} mu _{t} bigl(b^{2}bigr) ,dt biggr)^{frac{1}{2}} biggr] \ = & tau bigl( bigl[ (nu a)^{nu }b^{(1-nu )} bigr] ^{2} bigr) + nu ^{2} bigl[ tau bigl(a^{2}bigr) + tau bigl(b^{2}bigr) – 2 tau bigl(a^{2}bigr)^{ frac{1}{2}} tau bigl(b^{2} bigr)^{frac{1}{2}} bigr] \ = & tau bigl( bigl[ (nu a)^{nu }b^{(1-nu )} bigr] ^{2} bigr) + nu ^{2} bigl[ sqrt{tau bigl(a^{2}bigr)} – sqrt{tau bigl(b^{2}bigr)} bigr] ^{2}. end{aligned}

Thus,

$$tau bigl( bigl[ (nu a)^{nu }b^{1-nu} bigr] ^{2} bigr) + nu ^{2} bigl( sqrt{tau bigl(a^{2}bigr)} – sqrt{tau bigl(b^{2}bigr)} bigr) ^{2} leq tau bigl( nu ^{2} a^{2} +(1-nu )^{2} b^{2} bigr) .$$

Now, if (frac{1}{2} leq nu leq 1 ), using the similar argument, we have the second inequality. The case of equality follows by the similar argument used in the proof of [9, Theorem 3.4]. □

### Corollary 2.11

Let a, b be positive τmeasurable operators. If (0 leq nu leq frac{1}{2} ), then

$$biglVert (nu a)^{nu }b^{1-nu} bigrVert _{2}^{2} + nu ^{2} bigl( Vert a Vert _{2} – Vert b Vert _{2} bigr) ^{2} leq nu ^{2} Vert a Vert _{2}^{2} +(1-nu )^{2} Vert b Vert _{2}^{2} ,$$

and if (frac{1}{2} leq nu leq 1 ), then

$$biglVert a^{nu } bigl[ (1-nu ) b bigr] ^{1-nu} bigrVert _{2}^{2} + (1-nu )^{2} bigl( Vert a Vert _{2} – Vert b Vert _{2} bigr) ^{2} leq nu ^{2} Vert a Vert _{2}^{2} +(1-nu )^{2} Vert b Vert _{2}^{2}.$$

The next theorems give a converse of the Young inequality and its refinement for τ-measurable operators.

### Theorem 2.12

([20, Corollary 3.7])

Let a, b be positive invertible operators in (mathcal{M}). Then for (nu >1 ),

$$biglVert nu a + (1-nu )b bigrVert _{1} leq biglVert a^{nu }b^{1- nu} bigrVert _{1}.$$

(12)

Moreover, if a and b bounded operators in (L_{1}(mathcal{M})) or (abin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds if and only if (a=b ).

### Theorem 2.13

([20, Theorem 3.20])

Let a, b be positive invertible operator in (mathcal{M}) and (nu >1 ). Then

begin{aligned} & biglVert nu a + (1-nu )b bigrVert _{1} +min {1,nu -1 }Bigl( sqrt{ Vert a Vert _{1}}-sqrt{ Vert b Vert _{1}}Bigr)^{2} \ & quad leq biglVert a^{nu }b^{1-nu} bigrVert _{1} leq biglVert nu a + (1-nu )b bigrVert _{1} +max {1, nu -1}Bigl( sqrt{ Vert a Vert _{1}}-sqrt{biglVert a Vert _{1}}Bigr)^{2}. end{aligned}

Moreover, if a and b bounded operators in (L_{1}(mathcal{M})), then equality holds if and only if (mu _{t}(a)=mu _{t}(b)) for all (t>0).

Concerning all results discussed so far, we can prove the following theorems for τ-measurable operators.

### Theorem 2.14

Let (a, b in L_{1}(mathcal{M})) be positive invertible operators. Then for (nu >1 ),

$$biglVert nu a + (1-nu )b bigrVert _{1} leq biglVert a^{nu }b^{1- nu} bigrVert _{1}.$$

(13)

Moreover, if a and b are bounded operators or (abin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds if and only if (a=b).

### Theorem 2.15

Let (a, b in L_{1}(mathcal{M})) be positive invertible operators. Then for (nu >1 ),

begin{aligned} & biglVert nu a + (1-nu )b bigrVert _{1} +min {1,nu -1 }Bigl( sqrt{ Vert a Vert _{1}}-sqrt{ Vert b Vert _{1}}Bigr)^{2}\ & quad leq biglVert a^{nu }b^{1-nu} bigrVert _{1} leq biglVert nu a + (1-nu )b bigrVert _{1} +max {1, nu -1}Bigl( sqrt{ Vert a Vert _{1}}-sqrt{ Vert a Vert _{1}}Bigr)^{2}. end{aligned}

Moreover, if a and b are bounded operators or (abin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then equality holds if and only if (mu _{t}(a)=mu _{t}(b)) for all (t>0).

Hu and Xue [14] obtained another improvement of reverses of the scalar Young type inequalities for non-negative real numbers a and b in the following form.

If (0le nu le frac{1}{2}), then

$$nu ^{2} a^{2} +(1-nu )^{2}b^{2} + r_{0}a bigl(sqrt{(1-nu )b} – sqrt{a}bigr)^{2} leq (1-nu )^{2}(a – b )^{2} +a^{2nu} bigl[ (1-nu ) b bigr]^{2},$$

(14)

where (r_{0}= min {2nu , 1-2nu } ).

If (frac{1}{2} leq nu leq 1 ), then

$$nu ^{2} a^{2} +(1-nu )^{2}b^{2} + r_{0}b (sqrt{ b} – sqrt{nu a})^{2} leq nu ^{2}(a – b )^{2} + (nu a)^{2nu} b ^{2-2nu},$$

(15)

where (r_{0}= min { 2nu -1 , 2-2nu } ).

Let a, b be positive τ-measurable operators. Then, we have the same inequalities for singular values if we replace the positive real numbers a and b in the above equation with (mu _{t}(a)) and (mu _{t}(b)), respectively. We are interested in proving some versions of those inequalities and the case of equality for trace and norm of τ-measurable operators, but it is unclear for us.

In the following, we use the method of Bhatia and Davis [4] to extend some inequalities for τ-measurable operators. Zhou, Wang, and Wu established the Schwarz inequality for τ-measurable operators [24]. Using the similar argument in the proof of [24, Theorem 1], we prove the following theorem that is an important key feature in the next results.

### Theorem 2.16

Let x, y are bounded τmeasurable operators such that xy is selfadjoint. Then for every (s>0),

$$int _{0}^{s}mu _{t}(xy) ,dtle int _{0}^{s}mu _{t}(yx) ,dt.$$

### Proof

This theorem is a special case of [23, Lemma 3.1] when put (p=1). Since xy is a self-adjoint τ-measurable operator, for every positive integer n and (t>0), part (2) of Theorem 1.1 implies that

$$mu _{t}(xy)^{2n}=mu _{t}bigl( vert xy vert bigr)^{2n}=mu _{t}bigl( vert xy vert ^{2}bigr)^{n}.$$

Let f be an increasing function on ([0, infty )) with (f(0)=0) and (tto f(e^{t})) is convex. Then,

begin{aligned} int _{0}^{s} f bigl( mu _{t} bigl((xy)^{2n}bigr) bigr) ,dt = & int _{0}^{s} f bigl(mu _{t} bigl(bigl((xy)^{*}(xy)bigr)^{n}bigr) bigr) ,dt \ le & int _{0}^{s} f bigl( mu _{t}(yx)^{2n-1} Vert x Vert cdot Vert y Vert bigr) ,dt, quad text{ [Theorem 1.1(11)]} end{aligned}

particularly if (f(t)=t^{frac{1}{2n-1}}), then

$$int _{0}^{s} mu _{t}(xy)^{frac{2n}{2n-1}} ,dtle int _{0}^{s} mu _{t}(yx) bigl( Vert x Vert cdot Vert y Vert bigr)^{frac{1}{2n-1}} ,dt= bigl( Vert x Vert cdot Vert y Vert bigr)^{frac{1}{2n-1}} int _{0}^{s} mu _{t}(yx) ,dt.$$

Taking the (lim_{nto infty} inf ) of both sides, by the Fatou lemma, we get

begin{aligned} int _{0}^{s}mu _{t}(xy) ,dt = & int _{0}^{s} lim inf mu _{t}(xy)^{ frac{2n}{2n-1}} ,dt \ le & lim inf int _{0}^{s} mu _{t}(xy)^{frac{2n}{2n-1}} ,dt \ le & lim inf int _{0}^{s}mu _{t}(yx) bigl( Vert x Vert cdot Vert y Vert bigr)^{ frac{1}{2n-1}} ,dt \ =& lim inf bigl( Vert x Vert cdot Vert y Vert bigr)^{frac{1}{2n-1}} int _{0}^{s} mu _{t}(yx) ,dt \ =& int _{0}^{s}mu _{t}(yx) ,dt, end{aligned}

which proves the result. □

### Corollary 2.17

Let x, y are bounded τmeasurable operators such that xy is selfadjoint. Then, for every (r>0),

$$biglVert vert xy vert ^{r} bigrVert _{p}le biglVert vert yx vert ^{r} bigrVert _{p}, quad textit{for every } p>0.$$

### Proof

By taking (f(t)=frac{pr}{2n-1}) in the proof of Theorem 2.16. □

### Corollary 2.18

([24])

Let x, y, z be bounded τmeasurable operators and (r >0 ). Then

$$bigl| bigl| x^{ast }z y bigr| ^{r} bigr| _{1}^{2} leq bigl| bigl| x x^{ast }z bigr| ^{r} bigr| _{1} cdot bigl| bigl| z yy^{ast }bigr| ^{r}bigr| _{1}.$$

(16)

### Corollary 2.19

([24])

Let a, b, z are bounded τmeasurable operators such that a and b are positive. Then for (0 leq nu leq 1 ),

$$biglVert a^{nu }z b^{1-nu} bigrVert _{1} leq Vert a z Vert _{1} ^{nu } Vert z b Vert _{1}^{1-nu}.$$

In the following, we are going to present some new versions of majorization and norm type Young’s inequality.

Let a, b be positive τ-measureable operators. We say that a is submajorized (weakly majorized) by b in symbol (aprec _{w} b) [13], if (int _{0}^{s}mu _{t}(a),dtle int _{0}^{s}mu _{t}(b),dt) for all (s>0). Moreover, a is said to be majorized by b and is indicated by (aprec b), if (aprec _{w} y) and (int _{0}^{infty}mu _{t}(a),dt= int _{0}^{infty}mu _{t}(b),dt).

### Proposition 2.20

([18])

Let f be a continuous increasing function on (mathbb{R}^{+} ) such that (f(0)=0) and (tlongrightarrow f(e^{t})) is convex, then for positive operators a and b in (L_{0}(mathcal{M}) ) and for (s in mathbb{R}_{0}^{+} ),

$$int _{0}^{s} f bigl(mu _{t} bigl( vert ab vert ^{r}bigr) bigr),dtle int _{0}^{s} f bigl(mu _{t} bigl( a^{r}b^{r}bigr) bigr),dt.$$

### Theorem 2.21

Let a, b, z are bounded τmeasurable operators such that a and b are positive. Then for (p>0),

$$biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} prec _{w} frac{1}{2} bigl( biglvert z^{*}a bigrvert ^{p} + biglvert bz^{*} bigrvert ^{p} bigr).$$

(17)

### Proof

Suppose that (s in mathbb{R}_{0}^{+} ), we have for all (t>0 ),

begin{aligned} int _{0}^{s} mu _{t} bigl( biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} bigr) ,dt = & int _{0}^{s} mu _{t} bigl( bigl( a^{frac{1}{2}} z b^{ frac{1}{2}}bigr)^{*} bigl(a^{frac{1}{2}} z b^{frac{1}{2}} bigr) bigr) ^{ frac{p}{2}} ,dt quad text{ [Theorem 1.1(2)]} \ = & int _{0}^{s} mu _{t} bigl( b^{frac{1}{2}} z^{*} a^{ frac{1}{2}} a^{frac{1}{2}} z b^{frac{1}{2}} bigr) ^{frac{p}{2}} ,dt \ le & int _{0}^{s} mu _{t} bigl( bz^{*}az bigr) ^{ frac{p}{2}} ,dt quad text{[Theorem 2.16]} \ =& int _{0}^{s} mu _{t} bigl( bz^{*}bigl(z^{*}abigr)^{*} bigr) ^{ frac{p}{2}} ,dt \ = & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert cdot biglvert z^{*}a bigrvert bigr)^{frac{p}{2}} ,dt quad text{ [Theorem 1.1(4)]} \ le & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}} cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr) ,dt quad text{ [Proposition 2.20]} \ leq & frac{1}{2} int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert ^{p} + biglvert z^{*}a bigrvert ^{p} bigr) ,dt, quad text{ [Proposition 2.1]} end{aligned}

which proves the result. □

### Corollary 2.22

Let a, b be positive bounded operators in (L_{p}(mathcal{M}) ), and z be a bounded operator in (L_{0}(mathcal{M}) ). Then

$$biglVert bigl(a^{frac{1}{2}} z b^{frac{1}{2}} bigr)^{p} bigrVert _{1} le frac{1}{2} bigl( biglVert biglvert z^{*}a bigrvert ^{p} + biglvert bz^{*} bigrvert ^{p} bigrVert _{1} bigr),$$

(18)

and

$$biglVert a^{frac{1}{2}} z b^{frac{1}{2}} bigrVert _{p}^{p} le frac{1}{2} bigl( biglVert | z^{*}a | bigrVert _{p}^{p} + biglVert | bz^{*}| bigrVert _{p}^{p} bigr).$$

(19)

Moreover, the equality holds if (|bz^{*}|=|z^{*}a|).

### Proof

From the argument in the proof of the above theorem, for all (s in mathbb{R}_{0}^{+} ) and (t>0 ), we have

$$int _{0}^{s} mu _{t} bigl( biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} bigr) ,dt le int _{0}^{s} mu _{t} bigl( bz^{*}az bigr) ^{ frac{p}{2}} ,dt.$$

Setting (f(t)=t^{frac{p}{2}}) in Proposition 2.20 implies

begin{aligned} int _{0}^{s} mu _{t} bigl( biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} bigr) ,dt le & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert ^{ frac{p}{2}}cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr) ,dt \ le & frac{1}{2} int _{0}^{s} mu _{t}bigl( biglvert bz^{*} bigrvert ^{p}+ biglvert z^{*}a bigrvert ^{p}bigr) ,dt, \ le & frac{1}{2} biggl( int _{0}^{s} mu _{t}bigl( biglvert bz^{*} bigrvert bigr)^{p} ,dt+ int _{0}^{s}mu _{t}bigl( biglvert z^{*}a bigrvert bigr)^{p} ,dt biggr) \ = & frac{1}{2} biggl( int _{0}^{s} mu _{t}bigl( vert zb vert bigr)^{p} ,dt+ int _{0}^{s}mu _{t}bigl( vert az vert bigr)^{p} ,dt biggr) end{aligned}

by [19, Theorem 3.3] and parts (2) and (12) of Theorem 1.1. Letting (sto infty ), then we obtain the first and second inequalities.

If the equality holds, then from the above argument, we have

$$int _{0}^{infty }mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}} cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr) ,dt = int _{0}^{infty } frac{1}{2}mu _{t} bigl( biglvert bz^{*} bigrvert ^{p}+ biglvert z^{*}a bigrvert ^{p} bigr) ,dt.$$

(20)

By the Young inequality for singular values,

$$mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}}cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr)le frac{1}{2}mu _{t} bigl( biglvert bz^{*} bigrvert ^{p}+ biglvert z^{*}a bigrvert ^{p} bigr),$$

(21)

for every (t>0). Therefore, (20) shows, when coupled with (21), that

$$mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}}cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr)= frac{1}{2}mu _{t} bigl( biglvert bz^{*} bigrvert ^{p}+ biglvert z^{*}a bigrvert ^{p} bigr),$$

for almost all (t>0). However, for (xin L_{0}(mathcal{M}) ) as the nonincreasing function (mu _{t}(x)) are right continuous, (mu _{t} ( | bz^{*}| ^{frac{p}{2}}cdot | z^{*}a | ^{frac{p}{2}} )= frac{1}{2}mu _{t} ( | bz^{*} | ^{p}+| z^{*}a | ^{p} )) for all (t>0) which implies (|bz^{*}|=|z^{*}a|) by Corollary 2.3. □

### Corollary 2.23

Let a, b be positive bounded operators in (L_{p}(mathcal{M}) ), and z be bounded operator in (L_{0}(mathcal{M}) ). Then, for (p>0),

$$biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} prec frac{1}{2} bigl( biglvert z^{*}a bigrvert ^{p} + biglvert bz^{*} bigrvert ^{p} bigr),$$

(22)

if (|z^{*}a|=|bz^{*}|).

### Theorem 2.24

Let a, b be positive bounded operators in (L_{p}(mathcal{M}) ), and z be a bounded operator in (L_{0}(mathcal{M}) ). Then, for (p>0),

$$biglVert a^{frac{1}{2}} z b^{frac{1}{2}} bigrVert _{p} le Vert az Vert _{p}^{ frac{1}{2}} Vert zb Vert _{p}^{frac{1}{2}} le frac{1}{2} bigl( Vert az Vert _{p} + Vert zb Vert _{p} bigr).$$

(23)

### Proof

Suppose that (s in mathbb{R}_{0}^{+} ), we have for all (t>0 )

begin{aligned} int _{0}^{s} mu _{t} bigl( biglvert a^{frac{1}{2}} z b^{frac{1}{2}} bigrvert ^{p} bigr) ,dt = & int _{0}^{s} mu _{t} bigl( bigl( a^{frac{1}{2}} z b^{ frac{1}{2}}bigr)^{*} bigl(a^{frac{1}{2}} z b^{frac{1}{2}} bigr) bigr) ^{ frac{p}{2}} ,dt quad text{ [Theorem 1.1(2)]} \ = & int _{0}^{s} mu _{t} bigl( b^{frac{1}{2}} z^{*} a^{ frac{1}{2}} a^{frac{1}{2}} z b^{frac{1}{2}} bigr) ^{frac{p}{2}} ,dt \ = & int _{0}^{s} mu _{t} bigl( bz^{*}az bigr) ^{frac{p}{2}} ,dt quad text{[Theorem 2.16]} \ =& int _{0}^{s} mu _{t} bigl( bz^{*}bigl(z^{*}abigr)^{*} bigr) ^{ frac{p}{2}} ,dt \ = & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert cdot biglvert z^{*}a bigrvert bigr)^{frac{p}{2}} ,dt quad text{ [Theorem 1.1(4)]} \ le & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}} cdot biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr) ,dt quad text{ [Proposition 2.20]} \ leq & int _{0}^{s} mu _{t} bigl( biglvert bz^{*} bigrvert ^{frac{p}{2}} bigr)cdot mu _{t} bigl( biglvert z^{*}a bigrvert ^{frac{p}{2}} bigr) ,dt quad text{ [Theorem 1.1(13)]} \ =& int _{0}^{s} mu _{t} bigl( vert zb vert bigr)^{frac{p}{2}} cdot mu _{t} bigl( vert az vert bigr)^{frac{p}{2}} ,dt \ le & biggl( int _{0}^{s} mu _{t} bigl( vert zb vert ^{p} bigr) ,dt biggr)^{frac{1}{2}}cdot biggl( int _{0}^{s} mu _{t} bigl( vert az vert ^{p} bigr),dt biggr)^{frac{1}{2}} . end{aligned}

Letting (sto infty ), we obtain

$$biglVert a^{frac{1}{2}} z b^{frac{1}{2}} bigrVert _{p}le Vert zb Vert _{p}^{frac{1}{2}}. Vert az Vert _{p}^{frac{1}{2}}le frac{1}{2} bigl( Vert az Vert _{p} + Vert zb Vert _{p} bigr).$$

□

### Theorem 2.25

Let a, b be positive bounded operators in (L_{p}(mathcal{M}) ), and z be a bounded operator in (L_{0}(mathcal{M}) ). Then, for (0 leq nu leq 1 ),

$$biglVert a^{nu }z b^{1-nu} bigrVert _{p} leq Vert a z Vert _{p} ^{nu }cdot Vert z b Vert _{p}^{1-nu} leq nu Vert az Vert _{p} + (1-nu ) Vert zb Vert _{p} .$$

(24)

Moreover, (Vert a^{nu }z b^{1-nu} Vert _{p} = nu | az |_{p} + (1- nu ) | zb |_{p}) if (|az|_{p}=|zb|_{p}).

### Proof

The first part is [22, Lemma 2.1], and the second part of inequality is obtained from the classical Young inequality. Since some partial proof are needed for the case of equality, we write the proof with our notations. The inequality is trivial statement for (nu =0,1 ). We will use induction, for all indices (nu =frac{k}{2^{n}} ), (k=0,1, ldots , 2^{n} ). The general case then followed by continuity. Theorem 2.24 shows the result for (nu =frac{1}{2}). Suppose that inequality (24) is valid for all dyadic rational numbers with denominator (frac {k}{2^{n-1}} ). Let (nu = frac{2k+1}{2^{n}} ) be any dyadic rational. Then (nu = eta + omega ), where (omega = frac{1}{2^{n}} ) and (eta = frac {2k}{2^{n}} ). Two such rational numbers are η and (lambda = eta + 2omega = nu + omega ). Let (s in mathbb{R}_{0}^{+} ), then for all (t>0 ), we have

begin{aligned} int _{0}^{s} mu _{t}bigl( biglvert a^{nu }z b^{1-nu} bigrvert bigr)^{p} ,dt &= int _{0}^{s} mu _{t} bigl( biglvert a^{eta + omega} z b^{1-lambda + omega} bigrvert ^{p} bigr) ,dt \ & = int _{0}^{s} mu _{t} bigl( biglvert a^{omega }bigl( a^{eta }z b^{1- lambda} bigr) b^{omega } bigrvert ^{p} bigr) ,dt \ &= int _{0}^{s} mu _{t} bigl( biglvert a^{omega }bigl( a^{eta }z b^{1- lambda} bigr) b^{omega } bigrvert ^{2} bigr)^{frac{p}{2} } ,dt \ &= int _{0}^{s} mu _{t} bigl( bigl( b^{omega }b^{1-lambda}z^{*} a^{ eta }a^{omega} bigr) bigl( a^{omega }a^{eta }z b^{1-lambda} b^{omega }bigr) bigr)^{frac{p}{2} },dt \ &= int _{0}^{s} mu _{t} bigl( bigl(a^{omega }a^{omega }a^{eta }z b^{1- lambda} bigr) bigl( b^{omega }b^{omega }b^{1-lambda}z^{*} a^{eta}bigr) bigr)^{ frac{p}{2} },dt quad text{[Theorem 2.16]} \ &= int _{0}^{s} mu _{t} bigl( bigl(a^{lambda }z b^{1-lambda}bigr) bigl( b^{1- eta}z^{*} a^{eta}bigr) bigr)^{frac{p}{2} },dt \ &= int _{0}^{s} mu _{t} bigl( bigl(a^{lambda }z b^{1-lambda}bigr) bigl(a^{ eta }z b^{1-eta} bigr)^{*} bigr)^{frac{p}{2} },dt \ &le int _{0}^{s} mu _{t} bigl( biglvert a^{lambda }z b^{1-lambda} bigrvert ^{frac{p}{2}} biglvert a^{eta }z b^{1-eta} bigrvert ^{frac{p}{2} } bigr),dt, quad text{ [Proposition 2.20]} end{aligned}

where it has been obtained using (16) and replacing (lambda = eta + 2omega ). Now, using induction hypothesis and the Hölder inequality implies

begin{aligned} int _{0}^{s} mu _{t}bigl( biglvert a^{nu }z b^{1-nu} bigrvert bigr)^{p} ,dt leq &biggl( int _{0}^{s} mu _{t} bigl( biglvert a^{lambda }z b^{1- lambda} bigrvert ^{p} bigr) ,dt biggr) ^{frac{1}{2}} \ &{}times biggl( int _{0}^{s} mu _{t} bigl( biglvert a^{eta }z b^{1- eta} bigrvert ^{p} bigr) ,dt biggr) ^{frac{1}{2}} ,dt \ leq& biggl( int _{0}^{s} mu _{t} bigl( vert a z vert ^{p} bigr) ,dt biggr) ^{frac{lambda}{2}} biggl( int _{0}^{infty} mu _{t} bigl( vert z b vert ^{p} bigr) ,dt biggr) ^{ frac{1-lambda}{2}} \ &{}times biggl( int _{0}^{s} mu _{t} bigl( vert a z vert ^{p} bigr) ,dt biggr) ^{frac{eta}{2}} biggl( int _{0}^{s} mu _{t} bigl( vert z b vert ^{p} bigr) ,dt biggr) ^{frac{1-eta}{2}} \ =& biggl( int _{0}^{s} mu _{t} bigl( vert a z vert ^{p} bigr) ,dt biggr) ^{nu } biggl( int _{0}^{s} mu _{t} bigl( vert z b vert ^{p} bigr) ,dt biggr) ^{1-nu}. end{aligned}

The general case then followed by continuity, and the proof is complete. (sto infty ) implies the first part of inequality in (24). If the equality holds, then the above argument gives us case of equality in the Hölder inequality, which implies (|az|_{p}=|zb|_{p}). □

### Corollary 2.26

Let a, b be nonnegative bounded operators in (L_{p}(mathcal{M}) ), and z be a bounded operator in (L_{p}(mathcal{M}) ). Then, for (0 leq nu leq 1 ), we have

$$biglVert a^{nu}zb^{nu} bigrVert _{p}le Vert z Vert _{p}^{1-nu} Vert azb Vert _{p}^{nu} .$$

### Proof

It suffices to prove this when a is strictly positive (invertible); the general case follows from this by continuity. Using (24), we obtain

$$biglVert a^{nu}zb^{nu} bigrVert _{p}= biglVert bigl(a^{-1}bigr)^{1-nu}azb^{1-(1-nu )} bigrVert _{p}le Vert z Vert _{p}^{1-nu} Vert azb Vert _{p}^{nu} .$$

□

Here are some other types of the Young inequality. Note that these inequalities can also be expressed for the generalized s-numbers.

### Theorem 2.27

Let x, y be operators in (L_{0}(mathcal{M}) ). Then for (p, q, r in mathbb{R}^{+} ) that (frac{1}{p}+frac{1}{q}=frac{1}{r} ),

$$frac{1}{r} biglvert xy^{*} bigrvert ^{r}prec _{w} frac{1}{p} vert x vert ^{p}+ frac{1}{q} vert y vert ^{q} .$$

(25)

Moreover, if (x, y in L_{1}(mathcal{M}) ) are bounded operators or (xyin mathfrak{L}_{mathrm{loc}}^{2}(mathcal{M})), then

$$frac{1}{r} biglvert xy^{*} bigrvert ^{r}prec frac{1}{p} vert x vert ^{p}+ frac{1}{q} vert y vert ^{q} ,$$

(26)

if and only if (| x| ^{p}=| y| ^{q}).

### Proof

For all (s in mathbb{R}_{0}^{+} ) and (t >0 ), we have

begin{aligned} int _{0}^{s} mu _{t}bigl( biglvert xy^{*} bigrvert ^{r} bigr),dt =& int _{0}^{s} mu _{t} bigl( biglvert xy^{*} bigrvert ^{2} bigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t} bigl(yx^{*}xy^{*}bigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t}bigl( vert x vert ^{2} vert y vert ^{2} bigr)^{ frac{r}{2}} ,dt quad text{[24, Theorem 1]} \ leq & int _{0}^{s} mu _{t}bigl( vert x vert ^{r} vert y vert ^{r}bigr) ,dt quad text{ [Proposition 2.20]} \ le & int _{0}^{s} mu _{t}biggl( frac{r}{p} vert x vert ^{p}+ frac{r}{q} vert y vert ^{q}biggr) ,dt , quad text{ [Proposition 2.1]} end{aligned}

which proves (25). If x, y in (L_{1}(mathcal{M}) ), then (26) holds if

$$tau bigl( biglvert xy^{*} bigrvert ^{r} bigr)=tau biggl(frac{r}{p} vert x vert ^{p}+ frac{r}{q} vert y vert ^{q}biggr) .$$

As we see in the proof of Corollary 2.22, this equality holds if and only if

$$mu _{t}bigl( vert x vert ^{r} vert y vert ^{r}bigr)=mu _{t}biggl(frac{r}{p} vert x vert ^{p}+ frac{r}{q} vert y vert ^{q}biggr),$$

which implies the result by applying Corollary 2.3. □

### Theorem 2.28

Let a, b be positive bounded operators, and z be a bounded operator in (L_{0}( mathcal{M})). Then for (p, q, r in mathbb{R}^{+} ) that (frac{1}{p}+frac{1}{q}=1 ),

$$vert azb vert ^{r} prec _{w} frac{1}{p} biglvert b^{2}z^{*} bigrvert ^{frac{pr}{2}}+ frac{1}{q} biglvert z^{*}a^{2} bigrvert ^{frac{qr}{2}}$$

(27)

Moreover, (|azb|^{r} prec frac{1}{p}| b^{2}z^{*}| ^{frac{pr}{2}}+ frac{1}{q}| z^{*}a^{2}| ^{frac{qr}{2}}) if

$$biglvert zb^{2} bigrvert ^{p}= biglvert a^{2}z bigrvert ^{q}.$$

### Proof

For all (s in mathbb{R}_{0}^{+} ), in accordance with the previous proof process, we have

begin{aligned} int _{0}^{s} mu _{t}bigl( vert azb vert ^{r} bigr),dt =& int _{0}^{s} mu _{t} bigl( (azb)^{*} (azb) bigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t} bigl(bz^{*}aazbbigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t} bigl(bbz^{*} aazbigr)^{frac{r}{2}} ,dt quad text{[Theorem 2.16]} \ =& int _{0}^{s} mu _{t}bigl( biglvert b^{2}z^{*} bigrvert cdot biglvert z^{*}a^{2} bigrvert bigr)^{frac{r}{2}} ,dt quad text{ [Theorem 1.1(4)]} \ le & int _{0}^{s} mu _{t}bigl( biglvert b^{2}z^{*} bigrvert ^{frac{r}{2}} cdot biglvert z^{*}a^{2} bigrvert ^{frac{r}{2}} bigr) ,dt quad text{ [Proposition 2.20]} \ le & int _{0}^{s} mu _{t} biggl( frac{1}{p} biglvert b^{2}z^{*} bigrvert ^{ frac{pr}{2}}+frac{1}{q} biglvert z^{*}a^{2} bigrvert ^{frac{qr}{2}} biggr) ,dt . end{aligned}

The similar argument used in the proof of Theorem 2.27 implies the majorization case. □

### Corollary 2.29

Let a, b be positive bounded operators in (L_{1}( mathcal{M}) ), and z be a bounded operator in (L_{0}( mathcal{M})). Then for (p, q, r in mathbb{R}^{+} ) that (frac{1}{p}+frac{1}{q}=1 ),

1. 1

(| (azb)^{r}| _{1} le | frac{1}{p}| z^{*}a^{2} | ^{frac{pr}{2}}+frac{1}{q}| b^{2}z^{*}| ^{frac{qr}{2}} | _{1}).

2. 2

(| (azb)^{r}| _{1}^{2} le | (a^{2}z^{*})^{r} |_{1} | (z^{*}b^{2})^{r} |_{1} leq frac{1}{p}| (a^{2}z^{*})^{r} |_{1}^{p} + frac{1}{q}| (z^{*}b^{2})^{r} |_{1}^{q}).

3. 3

(| (azb)^{r}| _{1} le | (a^{2}z^{*})^{frac{r}{2} } |_{p} | (z^{*}b^{2})^{frac{r}{2} } |_{q} leq frac{1}{p}| (a^{2}z^{*})^{ frac{r}{2} }|_{p}^{p} + frac{1}{q}| (z^{*}b^{2})^{frac{r}{2} } |_{q}^{q}).

Moreover, the equality holds if (|a^{2}z^{*}|^{p}=|z^{*}b^{2}|^{q} ).

### Theorem 2.30

Let a, b be positive bounded operators, and z be a bounded operator in (L_{0}( mathcal{M})). Then for (p, q, r in mathbb{R}^{+} ) that (frac{1}{p}+frac{1}{q}=frac{1}{r} ),

$$frac{1}{r} biglvert az^{2}b bigrvert ^{r} prec _{w}frac{1}{p} biglvert bz^{*} bigrvert ^{p}+ frac{1}{q} vert az vert ^{q}.$$

(28)

Moreover, (frac{1}{r} |az^{2}b|^{r} prec frac{1}{p}| bz^{*}| ^{p}+ frac{1}{q}| az| ^{q}) if

$$biglvert bz^{*} bigrvert ^{p}= vert az vert ^{q} .$$

### Proof

For all (s in mathbb{R}_{0}^{+} ) and (t>0 ), we have

begin{aligned} int _{0}^{s} mu _{t}bigl( biglvert az^{2}b bigrvert ^{r} bigr),dt =& int _{0}^{s} mu _{t} bigl( bigl(az^{2}bbigr)^{*} bigl(az^{2}bbigr) bigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t}bigl(b bigl(z^{2}bigr)^{*}aaz^{2}b bigr)^{frac{r}{2}} ,dt \ =& int _{0}^{s} mu _{t} bigl(zbbz^{*}z^{*} aazbigr)^{frac{r}{2}} ,dt quad text{[Theorem 2.16]} \ =& int _{0}^{s} mu _{t}bigl( biglvert bz^{*} bigrvert ^{2}cdot vert az vert ^{2}bigr)^{ frac{r}{2}} ,dt quad text{ [Theorem 1.1(4)]} \ le & int _{0}^{s} mu _{t}bigl( biglvert bz^{*} bigrvert ^{r}cdot vert az vert ^{r}bigr) ,dt quad text{ [Proposition 2.20]} \ le & int _{0}^{s} mu _{t} biggl( frac{r}{p} biglvert bz^{*} bigrvert ^{p}+ frac{r}{q} vert az vert ^{q} biggr) ,dt, end{aligned}

which implies the result. The similar argument in the proof of Theorem 2.27 implies the majorization case. □

### Corollary 2.31

Let a, b be positive bounded τmeasurable operators, and z be a bounded operator in (L_{0}( mathcal{M})). Then for (p, q, r in mathbb{R}^{+} ) that (frac{1}{p}+frac{1}{q}=frac{1}{r} ),

1. 1

(frac{1}{r}| (az^{2}b)^{r}| _{1} le | frac{1}{p}| bz^{*}| ^{p}+frac{1}{q}| az| ^{q}| _{1}).

2. 2

(| (az^{2}b)^{r}| _{1}^{2} le | (zb)^{r} |_{1} | (az)^{r} |_{1} leq frac{1}{p}| (zb)^{r} |_{1}^{p} + frac{1}{q}| (az)^{r} |_{1}^{q}).

3. 3

(| (az^{2}b)^{r}| _{r}^{r} le | zb|_{p} ^{r}| az |_{q}^{r} leq frac{r}{p}|zb|_{p}^{p} + frac{r}{q}| az |_{q}^{q}).

Moreover, equality holds if (|az|^{q}=|zb|^{p} ).