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On the range of the parameters for the grand Furuta inequality to be valid

Abstract

In order to investigate the precise range D of the parameters p, s, t, and r for which the grand Furuta inequality is valid. We use the method of reductio ad absurdum. We find the following results: The area 1<p, 0<s<1, 0<t<1+r, and ts<r is contained in the complement of the range D. The condition 1s seems essential for the grand Furuta inequality.

MSC:47A63, 15A45.

1 Introduction

A bounded linear operator T on a Hilbert space is said to be positive semidefinite (denoted by 0T) if 0(Th,h) for all vectors h. We write 0<T if T is positive semidefinite and invertible.

Furuta obtained an epoch-making extension of the Löwner-Heinz inequality [1, 2].

Theorem 1.1 [3]

Let 0p, 1q, and 0r with p+r(1+r)q. If 0BA holds, then

( A r 2 B p A r 2 ) 1 q A p + r q .

It is well known that Theorem 1.1 is equivalent to the next theorem, which is often called the essential case of the Furuta inequality.

Theorem 1.2 Let 1p and 0r. If 0BA holds, then

( A r 2 B p A r 2 ) 1 + r p + r A 1 + r .

The following result by Tanahashi is a full description of the best possibility of the range

p+r(1+r)qand1q

as far as all parameters are positive. We would like to emphasize that the theorem can be divided into two cases.

Theorem 1.3 [4]

Let p, q, r be positive real numbers. If (1+r)q<p+r or 0<q<1, then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality

( A r 2 B p A r 2 ) 1 q A p + r q .

The next proposition is corresponding to the case (1+r)q<p+r of the previous theorem by putting q= p + r ( 1 + r ) α .

Proposition 1.4 Let 0<p, 0r. If 1<α, then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality

( A r 2 B p A r 2 ) 1 + r p + r α A ( 1 + r ) α .

On the other hand, the following ‘α-free’ proposition corresponds to the case 0<q<1 of Theorem 1.3 by putting q= p + r 1 + r .

Proposition 1.5 Let 0<p<1 and 0<r. Then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality

( A r 2 B p A r 2 ) 1 + r p + r A 1 + r .

Since the condition q= p + r 1 + r is the essential case for the Furuta inequality, our interest in Proposition 1.5 is not at all less than Proposition 1.4.

Furuta gave a unifying extension of both Theorem 1.1 and the Ando-Hiai inequality [5], which is often called the grand Furuta inequality.

Theorem 1.6 [6]

Let 1p, 1s, 0t1, and tr. If 0BA with 0<A, then the following inequality holds:

{ A r 2 ( A t 2 B p A t 2 ) s A r 2 } 1 t + r ( p t ) s + r A 1 t + r .
(1)

Again, Tanahashi showed that the outside powers in this theorem are the best possible.

Theorem 1.7 [7]

Let 1p, 1s, 0t1, and tr. If 1<α, then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality

{ A r 2 ( A t 2 B p A t 2 ) s A r 2 } 1 t + r ( p t ) s + r α A ( 1 t + r ) α .

Remark 1.8 In [7], Theorem 1.7 is originally stated as follows:

Let p, r, s, t be real numbers satisfying 1<s, 0<t<1, tr, 1p. If

1 t + r ( p t ) s + r <α,

then there exist invertible matrices A, B with 0BA which do not satisfy

{ A r 2 ( A t 2 B p A t 2 ) s A r 2 } α A { ( p t ) s + r } α .

These are just a matter of rephrasing, although their α differs each other. Theorem 1.7 can be naturally considered as an extension of Proposition 1.4. Indeed, if we put s=1, t=0 in Theorem 1.7, then we obtain Proposition 1.4 restricted to 1p. On the other hand, being different from Theorem 1.3, even if all parameters are positive, Theorem 1.7 does not show that the range

1p,1s,0t1,tr

cannot be expanded anymore for the grand Furuta inequality to be valid. Thus the clarification of the best possibility of the grand Furuta inequality is less satisfactory than that of the Furuta inequality. So our problem is to determine the range:

{ ( p , s , t , r ) R + 4 ; the inequality (1) holds whenever  0 < B A } .
(2)

Although it would be a nice theorem if one could precisely determine the range (2) all at once, it seems difficult to the author. Therefore, we should treat several main cases of the problem.

It is quite natural to expect an ‘α-free’ version which can be regarded as corresponding to Proposition 1.5. The following result obtained by Koizumi and the author is such an attempt.

Theorem 1.9 [8]

Let 0<p, 0<s, 0<t1, and tr. Suppose that

t<pand 1 t + r ( p t ) s + r sp<1.

Then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality (1).

Remark 1.10 The quantity 1 t + r ( p t ) s + r sp in the above assumption has an essential meaning. It also appears in a certain functional inequality (cf. [9]).

  1. (a)

    If 1p, 1s, 0t1, and tr, then 1 t + r ( p t ) s + r 1 1 t + r ( p t ) s + r sp.

  2. (b)

    If 0<p, 0<s<1, sp<1, and 0<tr, then 1 t + r ( p t ) s + r sp<1.

Remark 1.11 The case (ii) of [[8], Theorem 2.1] by Koizumi and the author treats the case 0<p=t<1, 0<s, t<r. However, we have A 1 < A 2 by the notations in [8] and the proof for (i) is not applicable to (ii). It seems still open.

The main purpose of this article is to show that the area 1<p, 0<s<1, 0<t<1+r, and ts<r is contained in the complement of the range (2).

2 Preliminaries

In this section, we will outline Tanahashi’s argument in [4] and [7] without proofs.

Let A, B be 2×2 matrices with 0<BA and B= ( 1 0 0 b ) , and let U be a unitary which diagonalizes A as U AU= ( d 1 0 0 d 2 ) . Assume A and B satisfy the grand Furuta inequality (1). Put α=1t+r and ψ=(pt)s+r. Then

{ U A r 2 U ( U A t 2 U U B p U U A t 2 U ) s U A r 2 U } α ψ U A α U,

hence we have

{ ( d 1 r 2 0 0 d 2 r 2 ) [ ( d 1 t 2 0 0 d 2 t 2 ) U ( 1 0 0 b p ) U ( d 1 t 2 0 0 d 2 t 2 ) ] s ( d 1 r 2 0 0 d 2 r 2 ) } α ψ ( d 1 α 0 0 d 2 α ) .
(3)

Denote

( d 1 t 2 0 0 d 2 t 2 ) U ( 1 0 0 b p ) U ( d 1 t 2 0 0 d 2 t 2 ) =k ( A 1 A 3 A 3 A 2 ) ,

where k is a positive scalar to be specified later.

Lemma 2.1 Suppose that A 1 < A 2 and A 3 <0. Let

V= 1 A 1 + A 2 + 2 ε 1 ( ε 1 A 1 + A 2 + ε 1 A 1 + A 2 + ε 1 ε 1 ) ,

where

2 ε 1 = A 1 A 2 + ( A 1 A 2 ) 2 + 4 A 3 2 .

Then A 3 = ( A 1 + A 2 + ε 1 ) ε 1 , V is unitary and

V ( A 1 A 3 A 3 A 2 ) V= ( A 2 + ε 1 0 0 A 1 ε 1 ) .

The formula (3) implies

{ ( d 1 r 2 0 0 d 2 r 2 ) k s V ( ( A 2 + ε 1 ) s 0 0 ( A 1 ε 1 ) s ) V ( d 1 r 2 0 0 d 2 r 2 ) } α ψ ( d 1 α 0 0 d 2 α ) .

Write the left-hand matrix as

k s α ψ ( A 1 + A 2 + 2 ε 1 ) α ψ ( B 1 B 3 B 3 B 2 ) α ψ ,

where

B 1 = d 1 r { ε 1 ( A 2 + ε 1 ) s + ( A 1 + A 2 + ε 1 ) ( A 1 ε 1 ) s } , B 2 = d 2 r { ( A 1 + A 2 + ε 1 ) ( A 2 + ε 1 ) s + ε 1 ( A 1 ε 1 ) s } , B 3 = d 1 r 2 d 2 r 2 ( A 1 + A 2 + ε 1 ) ε 1 { ( A 2 + ε 1 ) s ( A 1 ε 1 ) s } .

Lemma 2.2 Keep the situation as above. Assume that B 2 < B 1 . Then the following inequality holds:

ε 2 { γ d 1 α ( B 2 ε 2 ) α ψ } { ( B 1 + ε 2 ) α ψ γ d 2 α } ( B 1 B 2 + ε 2 ) { γ d 1 α ( B 1 + ε 2 ) α ψ } { γ d 2 α ( B 2 ε 2 ) α ψ } ,
(4)

where

ε 2 = 1 2 ( B 1 + B 2 + ( B 1 B 2 ) 2 + 4 B 3 2 ) , γ = { k s ( A 1 + A 2 + 2 ε 1 ) } α ψ .

3 Results

We begin with an elementary inequality for real numbers.

Lemma 3.1 If 1<p, then

2 p 1 p+ 2 p 1<0.

As Theorem 1.7 is regarded as an extension of Proposition 1.4, our main theorem can be considered as an extension of Proposition 1.5, which shows one of the largest pieces of the complement of the range (2). The advantage is that the assumptions on the parameters other than 0<s<1 are very mild. Note that, if we change 0<s<1 to 1s, then the inequality holds for 0<t1.

Theorem 3.2 Let 1<p, 0<s<1, 0<t<1+r, and ts<r. Then there exist 2×2 matrices A and B with 0<BA that do not satisfy the inequality

{ A r 2 ( A t 2 B p A t 2 ) s A r 2 } 1 t + r ( p t ) s + r A 1 t + r .

Yamazaki’s simplified proof of Theorem 1.7 in [10] is not applicable in our context. The method of our proof is the same as Tanahashi’s argument, whose outline is explained in the previous section. We would like to emphasize there are several branching points such that the conditions about parameters in the assumption are to be reflected to powers or coefficients in calculations. Moreover, we have to estimate all terms up to the order of x 2 in the sequel. On the other hand, in the existing literature, such as [7, 8] and [11], it is sufficient to estimate only main and second terms.

Proof As in the preliminaries, we set α=1t+r and ψ=(pt)s+r. Note that 0<α and 0<ψ. We will consider matrices

A= ( x 2 + 1 x x 3 ) andB= ( 1 0 0 2 ) .

Then we have 0<BA. The eigenvalues of A are x 2 + 4 ± x 4 + 4 2 . Let

c= x 2 2 + x 4 + 4 2 x andU= 1 c 2 + 1 ( c 1 1 c ) .

Then U is unitary and U AU= ( d 1 0 0 d 2 ) , where

d 1 = x 2 + 4 + x 4 + 4 2 and d 2 = x 2 + 4 x 4 + 4 2 .

Assume A and B satisfy the grand Furuta inequality (1). Then we have

A 1 = d 1 t ( c 2 + 2 p ) , A 2 = d 2 t ( 1 + 2 p c 2 ) , A 3 = d 1 t 2 d 2 t 2 ( 2 p 1 ) c , k = 1 c 2 + 1

by the notation of the previous section. It is easy to see that d 1 x 2 , d 2 2, cx, A 1 x 2 2 t and A 2 2 p t x 2 as x. Therefore, for sufficiently large x, we have A 1 < A 2 . Further, we shall see later B 1 x 2 r + 2 + 2 ( 1 t ) s , B 2 x 2 + 2 s as x. Since ts<r, we have B 2 < B 1 for sufficiently large x.

Now we estimate each term of the inequality (4) with respect to x. The estimation of the factor γ d 2 α ( B 2 ε 2 ) α ψ in the right-hand side is a little delicate so that we need some calculation. Terms in other factors can be roughly estimated. As is a usual notation, f(x)=o( x β ) means that

f ( x ) x β 0(x+).

One can establish the following formulas:

x 4 + 4 = x 2 + 2 x 2 + o ( x 2 ) , d 1 α = x 2 α ( 1 + 2 α x 2 + o ( x 2 ) ) , d 2 α = 2 α ( 1 α 2 x 2 + o ( x 2 ) ) , c = x ( 1 1 x 2 + o ( x 2 ) ) , ( c 2 + 1 ) s = x 2 s ( 1 s x 2 + o ( x 2 ) ) .

Next

A 1 = x 2 2 t ( 1 + 2 p 2 2 t x 2 + o ( x 2 ) ) , A 2 = 2 p t x 2 ( 1 + 2 p + 1 + 2 p 1 t + 1 2 p x 2 + o ( x 2 ) ) , A 3 2 = 2 t ( 2 p 1 ) 2 x 2 2 t ( 1 3 t + 4 2 x 2 + o ( x 2 ) ) ,

hence

A 2 A 1 = 2 p t x 2 ( 1 2 p + t x 2 t + 2 p + 1 + 2 p 1 t + 1 2 p x 2 + o ( x 2 ) ) , ( A 2 A 1 ) ( 4 A 3 2 ( A 1 A 2 ) 2 ) 2 1 x 2 + 4 t = x 2 t o ( x 2 )

and

ε 1 = 1 2 ( A 2 A 1 ) ( 1 + 1 + 4 A 3 2 ( A 1 A 2 ) 2 ) = 1 2 ( A 2 A 1 ) { ( 1 2 1 ) 4 A 3 2 ( A 1 A 2 ) 2 + ( 1 2 2 ) ( 4 A 3 2 ( A 1 A 2 ) 2 ) 2 + } = A 3 2 A 2 A 1 + x 2 t o ( x 2 ) = 2 t ( 2 p 1 ) 2 x 2 2 t ( 1 + o ( 1 ) ) ( 2 p t x 2 ( 1 + o ( 1 ) ) ) 1 = 2 p ( 2 p 1 ) 2 x 2 t ( 1 + o ( 1 ) ) .

Since ε 1 is small, the estimations of A 1 + A 2 +2 ε 1 and A 1 + A 2 + ε 1 (resp. A 2 + ε 1 ) in these details are the same as A 2 A 1 (resp. A 2 ), and the main term of A 1 ε 1 is the same as that of A 1 . Hence

ε 1 ( A 1 ε 1 ) s = 2 p ( 2 p 1 ) 2 x 2 t + 2 ( 1 t ) s ( 1 + o ( 1 ) ) = x 2 ( s + 1 ) o ( x 2 )

and

( A 1 + A 2 + ε 1 ) ( A 2 + ε 1 ) s + ε 1 ( A 1 ε 1 ) s = 2 ( p t ) ( s + 1 ) x 2 ( s + 1 ) ( 1 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p x 2 + o ( x 2 ) ) .

Further, since 0<s<1, 0<t<1+r, and ts<r,

B 1 = 2 p t x 2 + 2 r + 2 ( 1 t ) s ( 1 + o ( 1 ) ) , B 2 = 2 r + ( p t ) ( s + 1 ) x 2 ( s + 1 ) B 2 = ( 1 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 + o ( x 2 ) ) , B 3 2 = 2 r t + 2 ( p t ) s ( 2 p 1 ) 2 x 2 + 2 r 2 t + 4 s ( 1 + o ( 1 ) ) .

The main term of B 1 B 2 is the same as B 1 , hence

B 3 2 ( B 1 B 2 ) 2 1 x 2 + 2 r + 2 t 4 t s , ( B 1 B 2 ) ( 4 B 3 2 ( B 1 B 2 ) 2 ) 2 1 x 2 + 2 r + 4 t 2 s 6 t s = x 2 t + 2 s + 2 t s o ( x 2 ) .

Therefore,

ε 2 = 1 2 ( B 1 B 2 ) ( 1 + 1 + 4 B 3 2 ( B 1 B 2 ) 2 ) = 1 2 ( B 1 B 2 ) { ( 1 2 1 ) 4 B 3 2 ( B 1 B 2 ) 2 + ( 1 2 2 ) ( 4 B 3 2 ( B 1 B 2 ) 2 ) 2 + } = B 3 2 B 1 B 2 + x 2 t + 2 s + 2 t s o ( x 2 ) = 2 r t + 2 ( p t ) s ( 2 p 1 ) 2 x 2 + 2 r 2 t + 4 s ( 1 + o ( 1 ) ) ( 2 p t x 2 + 2 r + 2 ( 1 t ) s ( 1 + o ( 1 ) ) ) 1 = 2 r p + 2 ( p t ) s ( 2 p 1 ) 2 x 2 t + 2 s + 2 t s ( 1 + o ( 1 ) ) ,

so the estimation of B 2 ε 2 is the same as B 2 in its detail, where the assumption 0<s<1 is crucial. Hence

( B 2 ε 2 ) α ψ = { 2 r + ( p t ) ( s + 1 ) x 2 ( s + 1 ) ( 1 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 + o ( x 2 ) ) } α ψ = 2 { r + ( p t ) ( s + 1 ) } α ψ x 2 ( s + 1 ) α ψ { 1 + α ψ ( 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 ) + ( α ψ 2 ) ( 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 ) 2 + ( α ψ 3 ) ( 2 p + t x 2 t + ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 ) 3 + + o ( x 2 ) } = 2 { r + ( p t ) ( s + 1 ) } α ψ x 2 ( s + 1 ) α ψ { 1 α ψ 2 p + t x 2 t + ( α ψ 2 ) 2 2 ( p + t ) x 4 t ( α ψ 3 ) 2 3 ( p + t ) x 6 t + + ( 1 ) k ( α ψ k ) 2 k ( p + t ) x 2 k t + α ψ ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 + o ( x 2 ) } ,

where k is the nonnegative integer determined by kt1<(k+1)t.

The main term of B 1 + ε 2 is obviously the same as that of B 1 and

( B 1 + ε 2 ) α ψ = ( 2 p t x 2 + 2 r + 2 ( 1 t ) s ) α ψ ( 1 + o ( 1 ) ) .

Furthermore,

γ = { x 2 s ( 1 s x 2 + o ( x 2 ) ) 2 p t x 2 ( 1 2 p + t x 2 t + 2 p + 1 + 2 p 1 t + 1 2 p x 2 + o ( x 2 ) ) } α ψ = 2 ( p t ) α ψ x 2 ( s + 1 ) α ψ { 1 α ψ 2 p + t x 2 t + ( α ψ 2 ) 2 2 ( p + t ) x 4 t ( α ψ 3 ) 2 3 ( p + t ) x 6 t + + ( 1 ) k ( α ψ k ) 2 k ( p + t ) x 2 k t + α ψ 2 p + 1 + 2 p 1 t + 1 2 p s 2 p x 2 + o ( x 2 ) } .

For the following four factors in the formula (4), it is sufficient to estimate their main terms only,

γ d 1 α ( B 2 ε 2 ) α ψ = 2 ( p t ) α ψ x 2 ( s + 1 ) α ψ + 2 α ( 1 + o ( 1 ) ) , ( B 1 + ε 2 ) α ψ γ d 2 α = ( 2 p t x 2 + 2 r + 2 ( 1 t ) s ) α ψ ( 1 + o ( 1 ) ) , B 1 B 2 + ε 2 = 2 p t x 2 + 2 r + 2 ( 1 t ) s ( 1 + o ( 1 ) ) , γ d 1 α ( B 1 + ε 2 ) α ψ = 2 ( p t ) α ψ x 2 ( s + 1 ) α ψ + 2 α ( 1 + o ( 1 ) ) .

Now we have the estimation of the most delicate factor in the formula (4), whose main terms are canceled by subtraction. We have

γ d 2 α ( B 2 ε 2 ) α ψ = 2 ( p t ) α ψ + α x 2 ( s + 1 ) α ψ { 1 α ψ 2 p + t x 2 t + ( α ψ 2 ) 2 2 ( p + t ) x 4 t ( α ψ 3 ) 2 3 ( p + t ) x 6 t + + ( 1 ) k ( α ψ k ) 2 k ( p + t ) x 2 k t + α ψ 2 p + 1 + 2 p 1 t + 1 2 p s 2 p 1 ψ 2 p x 2 + o ( x 2 ) } 2 { r + ( p t ) ( s + 1 ) } α ψ x 2 ( s + 1 ) α ψ { 1 α ψ 2 p + t x 2 t + ( α ψ 2 ) 2 2 ( p + t ) x 4 t ( α ψ 3 ) 2 3 ( p + t ) x 6 t + + ( 1 ) k ( α ψ k ) 2 k ( p + t ) x 2 k t + α ψ ( s + 1 ) ( 2 p + 1 + 2 p 1 t + 1 ) 2 p 1 r 2 p x 2 + o ( x 2 ) } = 2 ( p t ) α ψ + α x 2 ( s + 1 ) α ψ α ψ { s ( 2 p 1 p + 2 p 1 ) 2 p x 2 + o ( x 2 ) } .

Applying these estimations to the inequality (4), we can obtain

2 r p + 2 ( p t ) s ( 2 p 1 ) 2 x 2 t + 2 s + 2 t s 2 ( p t ) α ψ x 2 ( s + 1 ) α ψ + 2 α ( 2 p t x 2 + 2 r + 2 ( 1 t ) s ) α ψ ( 1 + o ( 1 ) ) 2 p t x 2 + 2 r + 2 ( 1 t ) s 2 ( p t ) α ψ x 2 ( s + 1 ) α ψ + 2 α 2 ( p t ) α ψ + α x 2 ( s + 1 ) α ψ α ψ { s ( 2 p 1 p + 2 p 1 ) 2 p x 2 + o ( x 2 ) } ,

and hence

2 r p + 2 ( p t ) s ( 2 p 1 ) 2 ( 1 + o ( 1 ) ) 2 p t 2 α x 2 r + 2 ( 1 t ) s x 2 ( s + 1 ) α ψ x 2 t 2 s 2 t s x { 2 + 2 r + 2 ( 1 t ) s } α ψ α s ( 2 p 1 p + 2 p 1 ) 2 p ψ ( 1 + o ( 1 ) ) .
(5)

The power of x in the right-hand may be positive, 0 or negative. However, by Lemma 3.1, the assumption 1<p implies that 2 p 1 p+ 2 p 1<0. By letting x in (5), we have

0< 2 r p + 2 ( p t ) s ( 2 p 1 ) 2 <

or

0< 2 r p + 2 ( p t ) s ( 2 p 1 ) 2 0.

This is a contradiction and completes the proof of Theorem 3.2. □

Corollary 3.3 Let 1<p, 0<s<1, 0<t1, and tr. Then there exist 2×2 matrices A, B with 0<BA that do not satisfy the inequality

{ A r 2 ( A t 2 B p A t 2 ) s A r 2 } 1 t + r ( p t ) s + r A 1 t + r .

Proof It is obvious that 0<t<1+r and ts<r. □

References

  1. Löwner K: Über monotone Matrixfunktionen. Math. Z. 1934, 38: 177-216. 10.1007/BF01170633

    MathSciNet  Article  MATH  Google Scholar 

  2. Heinz E: Beiträge zur Störungstheorie der Spektralzerlegung. Math. Ann. 1951, 123: 415-438. 10.1007/BF02054965

    MathSciNet  Article  MATH  Google Scholar 

  3. Furuta T:AB0 assures ( B r A p B r ) 1 / q B ( p + 2 r ) / q for r0, p0, q1 with (1+2r)qp+2r. Proc. Am. Math. Soc. 1987, 101: 85-88.

    MathSciNet  MATH  Google Scholar 

  4. Tanahashi K: Best possibility of the Furuta inequality. Proc. Am. Math. Soc. 1996, 124: 141-146. 10.1090/S0002-9939-96-03055-9

    MathSciNet  Article  MATH  Google Scholar 

  5. Ando T, Hiai H: Log majorization and complementary Golden-Thompson type inequalities. Linear Algebra Appl. 1994, 197/198: 113-131.

    MathSciNet  Article  MATH  Google Scholar 

  6. Furuta T: Extension of the Furuta inequality and Ando-Hiai log-majorization. Linear Algebra Appl. 1995, 219: 139-155.

    MathSciNet  Article  MATH  Google Scholar 

  7. Tanahashi K: The best possibility of the grand Furuta inequality. Proc. Am. Math. Soc. 2000, 128: 511-519. 10.1090/S0002-9939-99-05261-2

    MathSciNet  Article  MATH  Google Scholar 

  8. Koizumi T, Watanabe K: Another consequence of Tanahashi’s argument on best possibility of the grand Furuta inequality. Cent. Eur. J. Math. 2013,11(2):368-375. 10.2478/s11533-012-0061-3

    MathSciNet  MATH  Google Scholar 

  9. Watanabe K: A certain functional inequality derived from an operator inequality. J. Math. Inequal. 2014 http://files.ele-math.com/articles/jmi-08-03.pdf, 8(1):69-81. http://files.ele-math.com/articles/jmi-08-03.pdf

    MathSciNet  Article  MATH  Google Scholar 

  10. Yamazaki T: Simplified proof of Tanahashi’s result on the best possibility of generalized Furuta inequality. Math. Inequal. Appl. 1999, 2: 473-477.

    MathSciNet  MATH  Google Scholar 

  11. Watanabe, K: On the range of the parameters for the grand Furuta inequality to be valid II. J. Math. Inequal. (2014, in press). http://files.ele-math.com/preprints/jmi-1330-pre.pdf

    Google Scholar 

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Acknowledgements

The author was supported in part by Grants-in-Aid for Scientific Research, Japan Society for the Promotion of Science.

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Watanabe, K. On the range of the parameters for the grand Furuta inequality to be valid. J Inequal Appl 2014, 297 (2014). https://doi.org/10.1186/1029-242X-2014-297

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Keywords

  • Löwner-Heinz inequality
  • Furuta inequality
  • matrix inequality