Open Access

An application of matrix inequalities to certain functional inequalities involving fractional powers

Journal of Inequalities and Applications20122012:221

https://doi.org/10.1186/1029-242X-2012-221

Received: 9 May 2012

Accepted: 20 September 2012

Published: 3 October 2012

Abstract

We will show certain functional inequalities involving fractional powers, making use of the Furuta inequality and Tanahashi’s argument.

MSC:26D07, 26A09, 39B62, 47A63.

Keywords

inequalities fractional powers matrix inequalities Furuta inequality

1 Introduction

Let x be an arbitrary positive real number. One can easily see the inequality
( x 3 2 1 ) ( x 2 1 ) 6 5 ( x 5 2 1 ) ( x 1 ) ,
for instance, is reduced to a simple polynomial inequality by putting t = x 1 2 . However, at least to the author, it seems not easy to give an elementary proof of the inequality
x 2 2 + 3 4 ( x 2 1 ) ( x 2 + 3 2 1 ) 1 2 ( x 2 + 3 1 ) ( x 1 ) ,

which has a very similar form to the preceding one although their corresponding numerical parts are different.

The purpose of this article is to show the following theorem.

Theorem 1.1 Let 0 p , 1 q and 0 r with p + r ( 1 + r ) q . If 0 < x , then
x 1 + r p + r q 2 ( x p 1 ) ( x p + r q 1 ) p q ( x p + r 1 ) ( x 1 ) .
(1)

An elementary approach to proving the inequality (1) might be to consider the power series expansion.

Put t = x 1 , c = 1 + r p + r q 2 and
f ( t ) = p q ( ( 1 + t ) p + r 1 ) t ( 1 + t ) c ( ( 1 + t ) p 1 ) ( ( 1 + t ) p + r q 1 ) .
Then we can expand f ( t ) around t = 0 as
f ( t ) = p q { p + r + ( p + r 2 ) t + ( p + r 3 ) t 2 + ( p + r 4 ) t 3 + ( p + r 5 ) t 4 + } t 2 { 1 + c t + ( c 2 ) t 2 + ( c 3 ) t 3 + ( c 4 ) t 4 + } { p + ( p 2 ) t + ( p 3 ) t 2 + ( p 4 ) t 3 + ( p 5 ) t 4 + } { p + r q + ( p + r q 2 ) t + ( p + r q 3 ) t 2 + ( p + r q 4 ) t 3 + ( p + r q 5 ) t 4 + } t 2 = a 4 t 4 + a 5 t 5 + a 6 t 6 + .
Thus, the constant term and the coefficients of t, t 2 and t 3 are 0. Further, one can obtain
a 4 = p ( p + r ) 24 q ( r 2 + 2 p r + 1 ( p + r q ) 2 ) , a 5 = p ( p + r ) ( p + r 3 ) 48 q ( r 2 + 2 p r + 1 ( p + r q ) 2 )
and
a 6 = p ( p + r ) 5760 q { 3 ( p + r q ) 4 + 10 ( p + r q ) 2 { 3 ( p + r ) ( p + r 8 ) + p 2 + 41 } 33 ( p + r ) 4 + 240 ( p + r ) 3 + 30 ( p + r ) 2 ( p 2 15 ) 240 ( p + r ) ( p 2 1 ) + ( 3 p 2 + 413 ) ( p 2 1 ) } .

Thus, if the assumption for the parameters p, q and r in Theorem 1.1 is satisfied, then we have 0 < a 4 . However, the signature of a 5 and a 6 depends on parameters, and one cannot see any signs of a simple rule among the coefficients of higher order terms. Although f ( t ) is non-negative on a sufficiently small neighborhood of t = 0 , it seems difficult to show that f ( t ) is non-negative entirely on 1 < t < by such an argument as above.

Let us recall some fundamental concepts on related matrix inequalities. A capital letter means a matrix whose entries are complex numbers. A square matrix T is said to be positive semidefinite (denoted by 0 T ) if 0 ( T x , x ) for all vectors x. We write 0 < T if T is positive semidefinite and invertible. For two selfadjoint matrices T 1 and T 2 of the same size, a matrix inequality T 1 T 2 is defined by 0 T 2 T 1 .

The celebrated Löwner-Heinz theorem includes:

Theorem 1.2 [1, 2]

Let 0 p 1 . If 0 B A , then B p A p .

For 1 < p , 0 B A does not always ensure B p A p . Furuta obtained an epoch-making extension of the Löwner-Heinz inequality by using the Löwner-Heinz inequality itself.

Theorem 1.3 [3]

Let 0 p , 1 q and 0 r with p + r ( 1 + r ) q . If 0 B A , then
( A r 2 B p A r 2 ) 1 q A p + r q .
(2)
The following result by Tanahashi is a full description of the best possibility of the range
p + r ( 1 + r ) q and 1 q

as far as all parameters are positive.

Theorem 1.4 [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 < B A that do not satisfy the inequality
( A r 2 B p A r 2 ) 1 q A p + r q .

One notices the coincidence between the assumption on parameters in Theorem 1.1 and Theorem 1.3. As a matter of fact, the inequality (1) is a particular conclusion of the Furuta inequality. We should point out that Tanahashi’s argument in [4] is almost sufficient to deduce the former from the latter. In the next section, we will prove Theorem 1.1 using Theorem 1.3 and Tanahashi’s argument.

2 Proof of Theorem 1.1

As we mentioned above, our proof of Theorem 1.1 has a major part which is parallel to [4]. Our matrix A is a little different from that in [4], we use a variable y instead of ε and δ. It simplifies the argument to an extent, though the improvement is not essential.

Throughout this paper, we assume that 1 < a < b and 0 < y . We will consider matrices
A = ( a ( a 1 ) y ( a 1 ) y b + y )
and
B = ( 1 0 0 b ) .

Then we have 0 < B A . The eigenvalues of A are a + b + y ± d 2 , where d = a 2 + b 2 + y 2 2 a b + 2 ( a + b 2 ) y .

Lemma 2.1 0 < d < ( a + b + y ) 2 and a b y d 0 .

Proof Obviously,
d = ( a b ) 2 + y ( y + 2 ( a + b 2 ) ) > 0 , d = ( a + b + y ) 2 4 ( a b + y ) < ( a + b + y ) 2 .

If a b y d = 0 , then we would have a = 1 or y = 0 , which is contrary to the assumption. □

Let
c = 2 ( a 1 ) y a b y d
and
U = 1 c 2 + 1 ( c 1 1 c ) .
Then U is unitary and
U A U = 1 2 ( d 1 0 0 d 2 ) ,
where
d 1 = a + b + y + d , d 2 = a + b + y d .
By the assumption and Theorem 1.3, A and B satisfy the inequality (2). Then
( U A r 2 U U B p U U A r 2 U ) 1 q U A p + r q U ,
hence we have
{ ( d 1 r 2 0 0 d 2 r 2 ) U ( 1 0 0 b p ) U ( d 1 r 2 0 0 d 2 r 2 ) } 1 q 2 p q ( d 1 p + r q 0 0 d 2 p + r q ) .
(3)
Denote
( d 1 r 2 0 0 d 2 r 2 ) U ( 1 0 0 b p ) U ( d 1 r 2 0 0 d 2 r 2 ) = 1 c 2 + 1 ( A 1 A 3 A 3 A 2 ) ,
where
A 1 = d 1 r ( c 2 + b p ) , A 2 = d 2 r ( 1 + c 2 b p ) , A 3 = d 1 r 2 d 2 r 2 c ( 1 b p ) = ( ( a + b + y ) 2 d ) r 2 c ( 1 b p ) = ( 4 a b + 4 y ) r 2 c ( 1 b p ) .

Lemma 2.2 Let p, q, r be positive real numbers. Then A 2 < A 1 and A 3 < 0 .

Proof Since d 2 < d 1 and 0 < r , we have d 2 r < d 1 r . Moreover,
( c 2 + b p ) ( 1 + c 2 b p ) = ( c 2 1 ) ( 1 b p ) , 1 b p < 0
and
c 2 1 = 2 ( a b ) 2 + 2 y 2 + 4 ( b a ) y + 2 ( b a + y ) d ( a b y d ) 2 < 0 ,

hence we have 1 + c 2 b p < c 2 + b p . Thus A 2 < A 1 .

It is obvious that 1 b p < 0 and 0 < c , and hence A 3 < 0 . □

Let
V = 1 A 1 A 2 + 2 ε 1 ( A 1 A 2 + ε 1 ε 1 ε 1 A 1 A 2 + ε 1 ) ,
where
2 ε 1 = A 1 + A 2 + ( A 1 A 2 ) 2 + 4 A 3 2 .
Then it is easy to see that 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 1 + ε 1 0 0 A 2 ε 1 ) .

The following lemma is one of the most important points in Tanahashi’s argument. Although the substance is presented in the whole proof of [4], Theorem], we should restate and prove it in our context for the readers’ convenience.

Lemma 2.3
ε 1 { γ d 1 p + r q ( A 2 ε 1 ) 1 q } { ( A 1 + ε 1 ) 1 q γ d 2 p + r q } ( A 1 A 2 + ε 1 ) { γ d 1 p + r q ( A 1 + ε 1 ) 1 q } { γ d 2 p + r q ( A 2 ε 1 ) 1 q } ,
(4)

where γ = ( c 2 + 1 2 p ) 1 q .

Proof The formula (3) implies
( c 2 + 1 ) 1 q V ( ( A 1 + ε 1 ) 1 q 0 0 ( A 2 ε 1 ) 1 q ) V 2 p q ( d 1 p + r q 0 0 d 2 p + r q ) .
(5)
Write the left-hand matrix as
( c 2 + 1 ) 1 q ( A 1 A 2 + 2 ε 1 ) 1 ( B 1 B 3 B 3 B 2 ) ,
where
B 1 = ( A 1 A 2 + ε 1 ) ( A 1 + ε 1 ) 1 q + ε 1 ( A 2 ε 1 ) 1 q , B 2 = ε 1 ( A 1 + ε 1 ) 1 q + ( A 1 A 2 + ε 1 ) ( A 2 ε 1 ) 1 q , B 3 = A 1 A 2 + ε 1 ε 1 { ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q } .
Then, by the formula (5), we have
0 ( γ ( A 1 A 2 + 2 ε 1 ) d 1 p + r q B 1 B 3 B 3 γ ( A 1 A 2 + 2 ε 1 ) d 2 p + r q B 2 ) .
So, its determinant is also non-negative. We expand it to obtain
0 γ 2 ( A 1 A 2 + 2 ε 1 ) 2 d 1 p + r q d 2 p + r q γ ( A 1 A 2 + 2 ε 1 ) d 1 p + r q B 2 γ ( A 1 A 2 + 2 ε 1 ) d 2 p + r q B 1 + B 1 B 2 B 3 2 .
(6)
Now,
B 1 B 2 B 3 2 = { ( A 1 A 2 + ε 1 ) ( A 1 + ε 1 ) 1 q + ε 1 ( A 2 ε 1 ) 1 q } { ε 1 ( A 1 + ε 1 ) 1 q + ( A 1 A 2 + ε 1 ) ( A 2 ε 1 ) 1 q } ( A 1 A 2 + ε 1 ) ε 1 { ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q } 2 = ( A 1 A 2 + 2 ε 1 ) 2 ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q .
Hence, the formula (6) implies
0 ( A 1 A 2 + 2 ε 1 ) { γ 2 ( A 1 A 2 + 2 ε 1 ) d 1 p + r q d 2 p + r q γ d 1 p + r q B 2 γ d 2 p + r q B 1 } + ( A 1 A 2 + 2 ε 1 ) 2 ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q .
Cancel the common positive factor A 1 A 2 + 2 ε 1 and substitute the definitions for B 1 and B 2 . Then a simple calculation shows that
ε 1 { γ 2 d 1 p + r q d 2 p + r q γ d 1 p + r q ( A 1 + ε 1 ) 1 q γ d 2 p + r q ( A 2 ε 1 ) 1 q + ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q } ( A 1 A 2 + ε 1 ) { γ 2 d 1 p + r q d 2 p + r q γ d 1 p + r q ( A 2 ε 1 ) 1 q γ d 2 p + r q ( A 1 + ε 1 ) 1 q + ( A 1 + ε 1 ) 1 q ( A 2 ε 1 ) 1 q } .
By factorizing, we have
ε 1 { γ d 1 p + r q ( A 2 ε 1 ) 1 q } { γ d 2 p + r q ( A 1 + ε 1 ) 1 q } ( A 1 A 2 + ε 1 ) { γ d 1 p + r q ( A 1 + ε 1 ) 1 q } { γ d 2 p + r q ( A 2 ε 1 ) 1 q } .

This completes the proof of Lemma 2.3. □

Now, we estimate each term of the inequality (4) with respect to y + 0 . A key point in making use of the inequality (4) is that both estimations of the factor ε 1 on the left-hand side and the factor γ d 1 p + r q ( A 1 + ε 1 ) 1 q on the right-hand side contain a common subfactor y. After the cancellation of this y, we will derive the desired functional inequality by letting y + 0 , a 1 + 0 and applying l’Hopital’s rule. Terms in other factors can be roughly estimated.

In the following, o means o ( y ) , that is
o y 0 ( y + 0 ) ,

and o ( 1 ) denotes a term such that o ( 1 ) 0 ( y + 0 ).

One can establish the following formulae:
d = ( b a ) { 1 + a + b 2 ( b a ) 2 y + o ( y ) } , d 1 p + r q = ( 2 b ) p + r q { 1 + p + r q b 1 b ( b a ) y + o ( y ) } , d 2 p + r q = ( 2 a ) p + r q { 1 + p + r q a + 1 a ( b a ) y + o ( y ) } , c = 2 ( a 1 ) y a b y ( b a + a + b 2 b a y + o ( y ) ) = y a 1 b a { 1 b 1 ( b a ) 2 y + o ( y ) } , c 2 + 1 = 1 + a 1 ( b a ) 2 y + o ( y ) , ( c 2 + 1 ) 1 q d 1 p + r q = { 1 + a 1 q ( b a ) 2 y + o ( y ) } ( 2 b ) p + r q { 1 + p + r q b 1 b ( b a ) y + o ( y ) } = ( 2 b ) p + r q { 1 + 1 q b ( b a ) 2 ( ( a 1 ) b + ( p + r ) ( b 1 ) ( b a ) ) y + o ( y ) } , ( c 2 + 1 ) 1 q d 2 p + r q = ( 2 a ) p + r q ( 1 + o ( 1 ) ) , A 1 = ( 2 b ) r { 1 + r ( b 1 ) b ( b a ) y + o ( y ) } { b p + a 1 ( b a ) 2 y + o ( y ) } = 2 r b p + r { 1 + 1 b ( b a ) 2 ( r ( b 1 ) ( b a ) + b 1 p ( a 1 ) ) y + o ( y ) } , A 2 = ( 2 a ) r ( 1 + o ( 1 ) ) , A 3 2 = ( 4 a b + 4 y ) r y a 1 ( b a ) 2 ( 1 + o ( 1 ) ) ( 1 b p ) 2 = y 4 r a r b r a 1 ( b a ) 2 ( 1 b p ) 2 ( 1 + o ( 1 ) ) , ε 1 = 1 2 ( A 1 A 2 ) ( 1 + 1 + 4 A 3 2 ( A 1 A 2 ) 2 ) = A 3 2 A 1 A 2 + o = y 4 r a r b r ( a 1 ) ( b a ) 2 ( 1 b p ) 2 ( 1 + o ( 1 ) ) 2 r b p + r ( 1 + o ( 1 ) ) ( 2 a ) r ( 1 + o ( 1 ) ) + o = y 2 r a r b r ( a 1 ) ( 1 b p ) 2 ( b a ) 2 ( b p + r a r ) ( 1 + o ( 1 ) ) , ( A 1 + ε 1 ) 1 q = ( 2 r b p + r { 1 + 1 b ( b a ) 2 ( r ( b 1 ) ( b a ) + b 1 p ( a 1 ) ) y + o ( y ) } + y 2 r a r b r ( a 1 ) ( 1 b p ) 2 ( b a ) 2 ( b p + r a r ) ( 1 + o ( 1 ) ) ) 1 q = 2 r q b p + r q { 1 + 1 q b ( b a ) 2 ( r ( b 1 ) ( b a ) + b 1 p ( a 1 ) + a r b 1 p ( a 1 ) ( 1 b p ) 2 b p + r a r ) y + o ( y ) } , ( A 2 ε 1 ) 1 q = 2 r q a r q ( 1 + o ( 1 ) ) , A 1 A 2 + ε 1 = 2 r ( b p + r a r ) ( 1 + o ( 1 ) ) , γ d 1 p + r q ( A 2 ε 1 ) 1 q = 2 r q ( b p + r q a r q ) ( 1 + o ( 1 ) ) , γ d 2 p + r q ( A 1 + ε 1 ) 1 q = 2 r q ( a p + r q b p + r q ) ( 1 + o ( 1 ) ) , γ d 2 p + r q ( A 2 ε 1 ) 1 q = 2 r q ( a p + r q a r q ) ( 1 + o ( 1 ) ) .
Now, we have the estimation of the most delicate factor in the formula (4), whose constant term is canceled by subtraction.
γ d 1 p + r q ( A 1 + ε 1 ) 1 q = 2 p q ( 2 b ) p + r q { 1 + 1 q b ( b a ) 2 ( ( a 1 ) b + ( p + r ) ( b 1 ) ( b a ) ) y + o ( y ) } 2 r q b p + r q { 1 + 1 q b ( b a ) 2 ( r ( b 1 ) ( b a ) + b 1 p ( a 1 ) + a r b 1 p ( a 1 ) ( 1 b p ) 2 b p + r a r ) y + o ( y ) } = 2 r q b p + r q 1 q ( b a ) 2 { ( a 1 ) b + p ( b 1 ) ( b a ) b 1 p ( a 1 ) a r b 1 p ( a 1 ) ( 1 b p ) 2 b p + r a r } y ( 1 + o ( 1 ) )
Substitute these estimations for the inequality (4), cancel the positive factor y, and let y + 0 , then we have
2 r a r b r ( a 1 ) ( 1 b p ) 2 ( b a ) 2 ( b p + r a r ) 2 r q ( b p + r q a r q ) 2 r q ( b p + r q a p + r q ) 2 r ( b p + r a r ) 2 r q b p + r q 1 q ( b a ) 2 { ( a 1 ) b + p ( b 1 ) ( b a ) b 1 p ( a 1 ) a r b 1 p ( a 1 ) ( 1 b p ) 2 b p + r a r } 2 r q ( a p + r q a r q ) ,
and hence
a r b r ( 1 b p ) 2 ( b p + r q a r q ) ( b p + r q a p + r q ) ( b p + r a r ) 2 b p + r q 1 q { ( a 1 ) b + p ( b 1 ) ( b a ) b 1 p ( a 1 ) a r b 1 p ( a 1 ) ( 1 b p ) 2 b p + r a r } a p + r q a r q a 1 .
Letting a 1 + 0 and applying l’Hopital’s rule, we have
b r ( 1 b p ) 2 ( b p + r q 1 ) 2 ( b p + r 1 ) 2 b p + r q 1 p 2 q 2 ( b 1 ) 2 .
This implies that, for arbitrary 1 < b ,
b 1 + r p + r q 2 ( b p 1 ) ( b p + r q 1 ) p q ( b p + r 1 ) ( b 1 ) .
(7)

For arbitrary 0 < x < 1 , substitute 1 x for b in (7) and multiply by x, x p , x p + r , x p + r q both sides. It is easy to see that x itself satisfies (7). This completes the proof of Theorem 1.1.

Declarations

Acknowledgements

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

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Niigata University

References

  1. Löwner K: Über monotone Matrixfunktionen. Math. Z. 1934, 38: 177–216. 10.1007/BF01170633MathSciNetView ArticleGoogle Scholar
  2. Heinz E: Beiträge zur Störungstheorie der Spektralzerlegung. Math. Ann. 1951, 123: 415–438. 10.1007/BF02054965MathSciNetView ArticleGoogle Scholar
  3. Furuta T: A B 0 assures ( B r A p B r ) 1 / q B ( p + 2 r ) / q for r 0 , p 0 , q 1 with ( 1 + 2 r ) q p + 2 r . Proc. Am. Math. Soc. 1987, 101(1):85–88.MathSciNetGoogle Scholar
  4. Tanahashi K: Best possibility of the Furuta inequality. Proc. Am. Math. Soc. 1996, 124: 141–146. 10.1090/S0002-9939-96-03055-9MathSciNetView ArticleGoogle Scholar

Copyright

© Watanabe; licensee Springer 2012

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