Filomat 30:8 (2016), 2139–2145
DOI 10.2298/FIL1608139S
Published by Faculty of Sciences and Mathematics,
University of Niš, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
A Numerical Radius Version of
the Arithmetic-Geometric Mean of Operators
Alemeh Sheikhhosseinia
a Department
of Pure Mathematics, Faculty of Mathematics and Computer, Shahid Bahonar University of Kerman, Kerman, Iran
Abstract. In this paper, we obtain some numerical radius inequalities for operators, in particular for
positive definite operators A, B a numerical radius and some operator norm versions for arithmeticgeometric mean inequality are obtained, respectively as
!
A2 + B2
1
ω2 (A]B) 6 ω
− inf δ(x),
2
2 kxk=1
where δ(x) = h(A − B)x, xi2 , and
kAkkBk 6
where, δ(x, y) =
1
1
(kA2 k + kB2 k) −
inf δ(x, y),
2
2 kxk=kyk=1
2
Ay, y − hBx, xi .
1. Introduction
Let H be a complex Hilbert space with inner product h., .i and let B(H) denote the algebra of all bounded
linear operators on H. Let |||.||| denote any unitarily invariant norm, i.e., a norm with the property that
|||UAV||| = |||A||| , for all A ∈ B(H) and for all unitary U, V ∈ B(H).
For A ∈ B(H), the spectral norm of A is defined by
kAk = sup{|hAx, yi| : kxk = kyk = 1, x, y ∈ H}.
It is evident that this norm is unitary invariant.
The numerical range of a A ∈ B(H) is defined as
W(A) = sup{hAx, xi : kxk = 1, x ∈ H}.
For any A ∈ B(H), W(A) is a convex subset of the complex plane containing the spectrum of A. See
[5, Chapter 2] for this topic.
The numerical radius of A ∈ B(H) is defined by
ω(A) = sup{|λ| : λ ∈ W(A)}.
We recall the following results that were proved in [6].
2010 Mathematics Subject Classification. Primary 47A30; Secondary 47A12
Keywords. Geometric mean, Inequalities, Numerical radius, Operator norm
Received: 15 May 2014; Accepted: 24 July 2014
Communicated by Mohammad Sal Moslehian
Email address: [email protected] (Alemeh Sheikhhosseini)
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Lemma 1.1. Let A ∈ B(H) and let ω(.) be the numerical radius. Then
(i) ω(.) is a norm on B(H),
(ii) ω(UAU∗ ) = ω(A), for all unitary operators U,
(iii) ω(A) = kAk if ( but not only if) A is normal,
(iv) 21 kAk ≤ ω(A) ≤ kAk.
Moreover, ω(.) is not a unitarily invariant norm and is not submultiplicative.
For positive real numbers a and b, the most familiar form of the Young inequality is the following:
ab ≤
ap bq
+ ,
p
q
where p, q > 1 such that
(1)
1 1
+ = 1, or equivalently
p q
aν b1−ν 6 νa + (1 − ν)b,
with ν ∈ [0, 1]. Recently, Kittaneh and Manasrah [8] obtained a refinement of (1)
ab + r0 (ap/2 − bq/2 )2 ≤
ap bq
+ ,
p
q
(2)
where r0 = min{ 1p , 1q }.
For positive definite operators A, B ∈ B(H), the operator geometric mean is defined by
A]B ≡ A1/2 (A−1/2 BA−1/2 )1/2 A1/2 .
The operator geometric mean has the symmetric property (A]B = B]A). If AB = BA, then A]B = (AB)1/2 .
In this paper we obtain some inequalities (upper bound) for ω((A]B)X), where X ∈ B(H) is arbitrary.
Throughout the paper we use the notation A > 0 to mean that A is positive definite and Mn the space of all
n × n matrices.
2. Main Results
Bhatia and Kittaneh in 1990 [3] established a matrix mean inequality as follows:
|||A∗ B||| ≤
1 ∗
|||A A + B∗ B||| ,
2
(3)
for matrices A, B ∈ Mn .
In [2] a generalization of (3) was proved, for all X ∈ Mn ,
|||A∗ XB||| ≤
1
|||AA∗ X + XBB∗ ||| .
2
Ando in 1995 [1] established a matrix Young inequality:
p
Bq A
+ |||AB||| ≤ p
q
(4)
for p, q > 1 with 1/p + 1/q = 1 and positive matrices A, B. In [9] we considered the inequalities (3) and (4)
with the numerical radius norm as follows:
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Proposition 2.1. [9, Proposition 1] If A, B are n × n matrices, then
1
ω(A∗ A + B∗ B).
2
Also if A and B are positive matrices and p, q > 1 with 1/p + 1/q = 1, then
ω(A∗ B) ≤
ω(AB) ≤ ω(
Ap Bq
+ ).
p
q
Moreover, the authors, in [9, Theorem 2 ] and [10, Theorem 2.3], showed that the inequality
p
Bq A
|||AXB||| ≤ X + X p
q
does not holds for numerical radius and spectral norm for all X ∈ Mn and positive matrices A, B.
The following lemma is a consequence of the spectral theorem for positive operators and Jensen’s inequality
(see, e.g., [7]).
Lemma 2.2. Let A be a positive semidefinite operator in B(H) and let x ∈ H be any unit vector. Then for all r > 1
hAx, xir ≤ hAr x, xi,
(5)
and for all 0 6 r 6 1
hAr x, xi ≤ hAx, xir .
Theorem 2.3. Let A, B, X ∈ B(H), such that A, B > 0 and p > q > 1 where 1/p + 1/q = 1. Then for all r >
!
Arp/2 (X∗ BX)rq/2
1
r
ω ((A]B)X) 6 ω
+
− inf δ(x),
p
q
p kxk=1
2
where δ(x) = hAx, xirp/4 − hX∗ BXx, xirq/4 .
Proof. Let x ∈ H, with kxk = 1. By the Schwarz inequality in the Hilbert space (H; h., .i), we have
Er
r D
(A]B)Xx, x = A1/2 (A−1/2 BA−1/2 )1/2 A1/2 Xx, x D
Er
= (A−1/2 BA−1/2 )1/2 A1/2 Xx, A1/2 x 6 k(A−1/2 BA−1/2 )1/2 A1/2 Xxkr .kA1/2 xkr
D
Er/2
= (A−1/2 BA−1/2 )1/2 A1/2 Xx, (A−1/2 BA−1/2 )1/2 A1/2 Xx
D
Er/2
× A1/2 x, A1/2 x
= hAx, xir/2 hX∗ BXx, xir/2 .
Now, by Young’s inequality and (2) we have
hAx, xir/2 hX∗ BXx, xir/2
2
1
1
1
6 hAx, xirp/2 + hX∗ BXx, xirq/2 −
hAx, xirp/4 − hX∗ BXx, xirq/4
p
q
p
and by (5) we have
2
1
1
1
hAx, xirp/2 + hX∗ BXx, xirq/2 −
hAx, xirp/4 − hX∗ BXx, xirq/4
p
q
p
E 1
2
1 D rp/2 E 1 D ∗
rq/2
A x, x +
(X BX) x, x −
6
hAx, xirp/4 − hX∗ BXx, xirq/4
q
p
*p rp/2
!
+
2
(X∗ BX)rq/2
A
1
=
+
x, x −
hAx, xirp/4 − hX∗ BXx, xirq/4 .
p
q
p
Now, the result follows by taking the supremum over all unit vectors in H.
2
q
(6)
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Remark 2.4. Let r = p = q = 2. Then δ(x) ≡ 0 if and only if A − X∗ BX = 0. In general, δ(x) = 0 if and only if
hAx, xirp/4 = hX∗ BXx, xirq/4 .
The following example shows that, inequality (6) does not hold in general for spectral norm.
"
#
"
#
1 0
0 1
Example 2.5. If we take p = q = 2, r = 1, A =
, B = I2 and X =
, then
0 0
0 14
Arp/2 (X∗ BX)rq/2 5
r
1 = k(A]B)Xk > +
= .
p
q
8
Put X = I in Theorem 2.3, we obtain the following corollary.
Corollary 2.6. Let A, B ∈ B(H), be positive definite and p > q > 1 such that 1/p + 1/q = 1. Then for all r >
2
q
!
Arp/2 Brq/2
1
ω (A]B) 6 ω
+
− inf δ(x),
p
q
p kxk=1
r
2
where δ(x) = hAx, xirp/4 − hBx, xirq/4 .
Note that it is enough to replace r with 2r and X = I in the statement of Theorem 2.3 to obtain the following
corollary.
Corollary 2.7. Let A, B ∈ B(H) be positive definite operators and p > q > 1 such that 1/p + 1/q = 1. Then for all
r > 1q ,
ω2r (A]B) 6 ω(
1
Arp Brq
+
) − inf δ(x),
p
q
p kxk=1
(7)
2
where δ(x) = hAx, xirp/2 − hBx, xirq/2 .
Remark 2.8. Note that, if we set r = 1 and p = q = 2 in (7), then we have
!
1
A2 + B2
2
− inf δ(x),
ω (A]B) 6 ω
2
2 kxk=1
(8)
where δ(x) = h(A − B)x, xi2 . Notice that (8) is an operator numerical radius version for arithmetic-geometric mean
and moreover if, 0 < W(A − B), then infkxk=1 δ(x) > 0.
In the proof of Theorem 2.3, if we put r = 2 and X = I, then we have the following corollary.
Corollary 2.9. Let A, B ∈ B(H), be positive definite operators. Then
kA]Bk2 6 kAkkBk.
Let T, U ∈ B(H). The Euclidean radius(see [4]) is defined by
1/2
ωe (T, U) = sup |hTx, xi|2 + |hUx, xi|2
.
kxk=1
Corollary 2.10. Let A, B ∈ B(H), be positive definite operators. Then
√
2kA]Bk 6 ωe (A, B) 6 kA2 + B2 k1/2 ,
inparticular,
√
2ω(A]B) 6 ωe (A, B) 6 ω1/2 (A2 + B2 ).
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Proof. Same as, in the proof of Theorem 2.3, if we set r = p = q = 2, then we have
2 1
(A]B)x, x 6 (hAx, xi2 + hBx, xi2 )
2
(9)
and by Lemma 2.2,
E
E D
E
1
1 D
1D 2
(A + B2 )x, x .
(hAx, xi2 + hBx, xi2 ) 6 ( A2 x, x + B2 x, x ) =
2
2
2
(10)
Now, the result follows by taking the supremum in (9) and (10) over all unit vectors in H.
3. Additional Results
Proposition 3.1. Let A, B, X ∈ B(H) such that A, B > 0 and p > q > 1 such that 1/p + 1/q = 1. Then for all r >
k(A]B)Xkr 6 k
where δ(x, y) =
2
q
(X∗ BX)rq/2
1
Arp/2
k+k
k−
inf δ(x, y),
p
q
p kxk=kyk=1
2
rp/4
Ay, y
− hX∗ BXx, xirq/4 .
Proof. Let x, y ∈ H, such that kxk = kyk = 1. By the Schwarz inequality in the Hilbert space (H; h., .i), we have
Er
r D
(A]B)Xx, y = A1/2 (A−1/2 BA−1/2 )1/2 A1/2 Xx, y D
Er
= (A−1/2 BA−1/2 )1/2 A1/2 Xx, A1/2 y 6 k(A−1/2 BA−1/2 )1/2 A1/2 Xxkr .kA1/2 ykr
D
Er/2
= (A−1/2 BA−1/2 )1/2 A1/2 Xx, (A−1/2 BA−1/2 )1/2 A1/2 Xx
D
Er/2
× A1/2 y, A1/2 y
r/2 ∗
= Ay, y
hX BXx, xir/2 .
Now, by Young’s inequality and (2) we have
r/2 ∗
Ay, y
hX BXx, xir/2
2
rp/2 1 ∗
rp/4
1 1
Ay, y
+ hX BXx, xirq/2 −
Ay, y
− hX∗ BXx, xirq/4
6
p
q
p
and by (5) we have
2
rp/2 1 ∗
rp/4
1
1 + hX BXx, xirq/2 −
− hX∗ BXx, xirq/4
Ay, y
Ay, y
p
q
p
E 1D
E 1 2
rp/4
1 D rp/2
∗
rq/2
6
A y, y +
(X BX) x, x −
Ay, y
− hX∗ BXx, xirq/4 .
p
q
p
Now, the result follows by taking the supremum over all unit vectors x, y ∈ H.
Corollary 3.2. Let A, B, X ∈ B(H) be such that A, B > 0. Then for all r > 1
2k(A]B)Xkr 6 kAr k + k(X∗ BX)r k −
where δ(x, y) =
inf
kxk=kyk=1
δ(x, y),
2
r/2
Ay, y
− hX∗ BXx, xir/2 .
If in relation (11) we set X = I we obtain the following corollary.
(11)
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Corollary 3.3. Let A, B ∈ B(H) be positive definite operators. Then for all r > 1
2kA]Bkr 6 kAr k + kBr k −
where δ(x, y) =
inf
kxk=kyk=1
δ(x, y),
2
r/2
Ay, y
− hBx, xir/2 .
Proposition 3.4. Let A, B, X ∈ B(H) such that A, B > 0 and p > q > 1 such that 1/p + 1/q = 1. Then for all r > 2/q
(X∗ BX)rq/2
1
Arp/2
k+k
k−
inf δ(x, y),
p
q
p kxk=kyk=1
2
rp/4
where δ(x, y) = Ay, y
− hX∗ BXx, xirq/4 .
(kAkkX∗ BXk)r/2 6 k
(12)
Proof. Let x, y ∈ H with kxk = kyk = 1. By the inequality (2), we have
r/2 ∗
Ay, y
hX BXx, xir/2
2
rp/2 1 ∗
rp/4
1 1
Ay, y
Ay, y
+ hX BXx, xirq/2 −
− hX∗ BXx, xirq/4
6
p
q
p
and by (5) we have
2
rp/2 1 ∗
rp/4
1
1 Ay, y
Ay, y
+ hX BXx, xirq/2 −
− hX∗ BXx, xirq/4
p
q
p
E 1D
E 1 2
rp/4
1 D rp/2
∗
rq/2
A y, y +
(X BX) x, x −
Ay, y
− hX∗ BXx, xirq/4 .
6
p
q
p
Now, the result follows by taking the supremum over all unit vectors x, y ∈ H.
If in relation (12), we set X = I and r = 2, then we obtain the following corollary.
Corollary 3.5. Let A, B ∈ B(H) be positive definite operators and p > q > 1 such that 1/p + 1/q = 1. Then
Ap
Bq
1
k+k k−
inf δ(x, y),
p
q
p kxk=kyk=1
2
p/2
where δ(x, y) = Ay, y
− hBx, xiq/2 .
kAkkBk 6 k
(13)
Remark 3.6. Note that, if we set p = q = 2 in (13), then we have
1
1
(kA2 k + kB2 k) −
inf δ(x, y),
(14)
2
2 kxk=kyk=1
2
where δ(x, y) = Ay, y − hBx, xi . Notice that (14) is an operator norm version for arithmetic-geometric mean and
moreover if, W(A) and W(B) are separated, then infkxk=kyk=1 δ(x, y) > 0.
kAkkBk 6
Example 3.7. Let p = q = 2 and A = dia1(1, 2), B = dia1(5, 6) in the inequality (13). Then infkxk=kyk=1 δ(x, y) =
9 > 0 and hence,
1
1
31
12 = kAkkBk 6 (kA2 k + kB2 k) −
inf δ(x, y) =
.
2
2 kxk=kyk=1
2
Whereas, if we set this values in the inequality (4), with the spectral norm, then we obtain
12 = kABk 6 k
Ap B q
+ k = 20.
p
q
Thus, in this case, we have
kABk = kAkkBk < k
Ap
Bq
1
Ap Bq
k+k k−
inf δ(x, y) < k
+ k.
p
q
p kxk=kyk=1
p
q
Acknowledgement: The author would like to thank the anonymous referee for careful reading and the
helpful comments improving this paper.
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