Ann. Funct. Anal. 3 (2012), no. 2, 128–134
A nnals of F unctional A nalysis
ISSN: 2008-8752 (electronic)
URL: www.emis.de/journals/AFA/
HARDY–HILBERT TYPE INEQUALITIES FOR HILBERT
SPACE OPERATORS
MOHSEN KIAN
Communicated by L. Székelyhidi
Abstract. Some Hardy–Hilbert type inequalities for Hilbert space operators
are established. Several particular cases of interest are given as well.
1. Introduction and preliminaries
One of the applicable inequalities in analysis and differential equations is the
Hardy inequality
says that if p > 1 and {an }∞
n=1 are positive real numbers
P∞which
p
such that 0 < n=1 an < ∞, then
!p p X
∞
n
∞
X
1X
p
ak
≤
apn .
(1.1)
n
p
−
1
n=1
n=1
k=1
p
p
The inequality is sharp, i.e., the constant p−1
is the smallest number such
that the inequality holds. A continuous form of inequality (1.1) is as follows:
If p > 1 and f is a non-negative p-integrable function on (0, ∞), then
p
p Z ∞
Z ∞ Z x
p
1
f (t)dt dx ≤
f (x)p dx.
x
p
−
1
0
0
0
(1.2)
This inequality has been studied by many mathematicians [1, 3, 6]. A weighted
version of inequality (1.2) was given in [2]. A developed inequality, the so-called
Hardy–Hilbert inequality reads as follows:
Date: Received: 1 March 2012; Accepted: 20 April 2012.
2010 Mathematics Subject Classification. Primary 47A63; Secondary 47A64, 26D15.
Key words and phrases. Hardy–Hilbert inequality, self-adjoint operator, operator inequality.
128
HARDY–HILBERT TYPE INEQUALITIES
If p > 1, p1 + 1q = 1, an , bn ≥ 0 such that 0 <
then
∞ X
∞
X
an b m
π
<
n+m
sin(π/p)
n=1 m=1
P∞
n=1
∞
X
129
apn < ∞ and 0 <
! p1
apn
n=1
∞
X
P∞
q
n=1 bn
< ∞,
! 1q
bqn
.
(1.3)
n=1
An integral form of inequality (1.3) can
R ∞ be stated as the following:
R∞
If p > 1, p1 + 1q = 1, f, g ≥ 0 with 0 < 0 f (x)p dx < ∞ and 0 < 0 g(x)q dx < ∞,
then
Z ∞
p1 Z ∞
1q
Z ∞Z ∞
f (x)g(y)
π
p
q
dxdy <
f (x) dx
g(x) dx .
x+y
sin(π/p)
0
0
0
0
There are many refinements and reformulations of the above inequality. Yang
[7] proved the following generalization of (1.3):
! p1 ∞
! 1q
∞ X
∞
∞
X
X
X
an b m
m1−s apm
n1−s bqn
< L1
s
(n
+
m)
n=1 m=1
m=1
n=1
and
∞
X
n=1
(s−1)(p−1)
n
∞
X
am
(n + m)s
m=1
!p
< L1
∞
X
m1−s apm
m=1
p+s−2 q+s−2
B( p , q ),
where B(u, v) is the
in which 2 − min{p, q} < s ≤ 2 and L1 =
β-function. Also Yang [8] presented some reverse Hardy integral inequalities.
Hansen [4] gave an operator version of inequality (1.1) in the C ∗ -algebra B(H )
of all bounded linear operators on a complex Hilbert space H , in the case when
1 ≤ p ≤ 2:
Theorem.[4] Let 1 < p ≤ 2 be a real number and let f from (0, ∞) to the set
B(H )+ of all positive operators in B(H ), be a weakly measurable map such that
the integral
Z
∞
f (x)p dx
0
defines a bounded linear operator on H . Then the inequality
p
p Z ∞
Z ∞ Z x
1
p
f (t)dt dx ≤
f (x)p dx
x
p
−
1
0
0
0
p
p
holds, and the constant p−1
is the best possible. In the same paper Hansen
proved a similar trace inequality in the case where p > 1.
An operator version of inequality (1.3) was also given in [5]
In this paper, we give some inequalities analogue to (1.3) for operators in the real
space B(H )h of all self-adjoint operators on H .
2. Main results
We start this section with an analogous inequality to (1.3) for operators acting
on a Hilbert space H .
130
M. KIAN
Theorem 2.1. Let f, g be continuous functions defined on an interval J and
f, g ≥ 0. If p > 1, p1 + 1q = 1, then
1
1
hf (A)g(A)x, xi + hf (A)x, xihg(B)y, yi
2
3
1
1
+ hf (A)y, yihg(B)x, xi + hf (B)g(B)y, yi
3
4
E
D
1
1
π
p
≤
(f (A) + f (B)p ) p (g(A)q + g(B)q ) q x, x
sin(π/p)
for all operators A, B ∈ B(H )h with spectra contained in J and all unit vectors
x, y ∈ H .
Proof. Let a1 , a2 , b1 , b2 be positive scalars. Using (1.3) we have
1
1
a1 b 1 a1 b 2 a2 b 1 a2 b 2
π
+
+
+
≤
(ap1 + ap2 ) p (bq1 + bq2 ) q .
2
3
3
4
sin(π/p)
(2.1)
Let t, s ∈ J. Noticing that f (t) ≥ 0 and g(t) ≥ 0 for all t ∈ J and putting
a1 = f (t), a2 = f (s), b1 = g(t) and b2 = g(s) in (2.1) we obtain
f (t)g(t) f (t)g(s) f (s)g(t) f (s)g(s)
+
+
+
2
3
3
4
1
1
π
p
(f (t) + f (s)p ) p (g(t)q + g(s)q ) q
≤
sin(π/p)
(2.2)
for all s, t ∈ J. Using the functional calculus for A to inequality (2.2) we get
f (A)g(A) f (A)g(s) f (s)g(A) f (s)g(s)
+
+
+
2
3
3
4
1
1
π
p
≤
(f (A) + f (s)p ) p (g(A)q + g(s)q ) q ,
sin(π/p)
whence
1
1
1
f (s)g(s)
hf (A)g(A)x, xi + g(s)hf (A)x, xi + f (s)hg(A)x, xi +
2
3
3
4
D
E
1
1
π
≤
(f (A)p + f (s)p ) p (g(A)q + g(s)q ) q x, x
sin(π/p)
for any unit vector x ∈ H and any s ∈ J. Applying the functional calculus once
more to the self-adjoint operator B we get
1
1
1
f (B)g(B)
hf (A)g(A)x, xi + g(B)hf (A)x, xi + f (B)hg(A)x, xi +
2
3
3
4
D
E
1
1
π
p
p p
q
q q
≤
(f (A) + f (B) ) (g(A) + g(B) ) x, x . (2.3)
sin(π/p)
HARDY–HILBERT TYPE INEQUALITIES
131
If y ∈ H is a unit vector, then it follows from inequality (2.3) that
1
1
hf (A)g(A)x, xi + hg(B)y, yihf (A)x, xi
2
3
1
1
+ hf (B)y, yihg(A)x, xi + hf (B)g(B)y, yi
3
4
D
E
1
1
π
≤
(f (A)p + f (B)p ) p (g(A)q + g(B)q ) q x, x .
sin(π/p)
Replacing B by A and y by x in Theorem 2.1 we get:
Corollary 2.2. If f, g are continuous functions defined on an interval J and
f, g ≥ 0, then
3
3
hf (A)x, xihg(A)x, xi ≤
2π −
hf (A)g(A)x, xi
(2.4)
2
4
for any self-adjoint operator A and any unit vector x ∈ H .
With p = q = 2 in Theorem 2.1 we obtain
Corollary 2.3. If f, g are continuous functions defined on an interval J and
f, g ≥ 0, then
1
1
hf (A)g(A)x, xi + hf (A)x, xihg(B)y, yi
2
3
1
1
+ hf (A)y, yihg(B)x, xi + hf (B)g(B)y, yi
3D
4
E
1
1
2
2 2
2
2 2
≤ π f (A) + f (B)
g(A) + g(B)
x, x
for all operators A, B ∈ B(H )h with spectra contained in J and all unit vectors
x, y ∈ H .
Another version of inequality (1.3) is given in the next theorem.
Theorem 2.4. Let f, g be continuous functions defined on an interval J and
f, g ≥ 0. If p > 1 and p1 + 1q = 1, then
1
1
hf (B)y, yihf (A)x, xi + hg(B)y, yihf (A)x, xi
2
3
1
1
+ hf (B)y, yihg(A)x, xi + hg(B)y, yihg(A)x, xi
3
4
ED
E
1
1
π D
q
q q
p
p p
≤
(f (B) + g(B) ) y, y (f (A) + g(A) ) x, x
sin π/p
(2.5)
for all operators A, B ∈ B(H )h with spectra contained in J and all unit vectors
x, y ∈ H .
132
M. KIAN
Proof. Let s, t ∈ J. We use inequality (2.1) with a1 = f (t), a2 = g(t), b1 = f (s)
and b2 = g(s) to get
f (t)f (s) f (t)g(s) g(t)f (s) g(t)g(s)
+
+
+
2
3
3
4
1
1
π
p
≤
(f (t) + g(t)p ) p (f (s)q + g(s)q ) q .
sin π/p
Applying the functional calculus for A to the above inequality we get
f (A)f (s) f (A)g(s) g(A)f (s) g(A)g(s)
+
+
+
2
3
3
4
1
1
π
p
≤
(f (A) + g(A)p ) p (f (s)q + g(s)q ) q ,
sin π/p
whence
g(s)
f (s)
g(s)
f (s)
hf (A)x, xi +
hf (A)x, xi +
hg(A)x, xi +
hg(A)x, xi
2
3
3
4
D
E
1
1
π
(f (s)q + g(s)q ) q (f (A)p + g(A)p ) p x, x
≤
sin π/p
for any unit vector x ∈ H . Using the functional calculus for B to the last
inequality we obtain
1
1
hf (B)y, yihf (A)x, xi + hg(B)y, yihf (A)x, xi
2
3
1
1
+ hf (B)y, yihg(A)x, xi + hg(B)y, yihg(A)x, xi
3
4
ED
E
1
1
π D
q
q q
≤
(f (B) + g(B) ) y, y (f (A)p + g(A)p ) p x, x
sin π/p
for any unit vector y ∈ H .
With A = B and x = y, inequality (2.5) gives rise to
1
2
1
hf (A)x, xi2 + hg(A)x, xihf (A)x, xi + hg(A)x, xi2
2
3
4
ED
E
1
1
π D
q
q q
≤
(f (A) + g(A) ) x, x (f (A)p + g(A)p ) p x, x .
sin π/p
Putting p = q = 2 in the above inequality we obtain
1
2
1
hf (A)x, xi2 + hg(A)x, xihf (A)x, xi + hg(A)x, xi2
2
3
4
D
E2
1
≤ π f (A)2 + g(A)2 2 x, x
≤ π f (A)2 + g(A)2 x, x .
In the case where the functions f and g are convex, we reach to the next result:
HARDY–HILBERT TYPE INEQUALITIES
133
Theorem 2.5. Let f, g : J → [0, ∞) be convex functions and let p, q > 1 with
1
+ 1q = 1. Then
p
1
1
f (hAx, xi)g(hAx, xi) + f (hAx, xi)g(hBy, yi)
2
3
1
1
+ g(hAx, xi)f (hBy, yi) + hf (B)g(B)y, yi
3
4
1
1
π
p
p
q
q
(hf (A) x, xi + hf (B) y, yi) + (hg(A) x, xi + hg(B) y, yi)
≤
sin(π/p) p
q
for all A, B ∈ B(H )h with spectra contained in J and all unit vectors x, y.
Proof. Put t = hAx, xi in (2.2) to get
1
1
1
1
f (hAx, xi)g(hAx, xi) + f (hAx, xi)g(s) + f (s)g(hAx, xi) + f (s)g(s)
2
3
3
4
1
1
π
p
p p
q
q q
≤
(f (hAx, xi) + f (s) ) (g(hAx, xi) + g(s) ) .
sin(π/p)
A use of the functional calculus for B to the above inequality yields that
1
1
f (hAx, xi)g(hAx, xi) + f (hAx, xi)hg(B)y, yi
2
3
1
1
+ hf (B)y, yig(hAx, xi) + hf (B)g(B)y, yi
3
4
D
E
1
1
π
p
p p
q
q q
(f (hAx, xi) + f (B) ) (g(hAx, xi) + g(B) ) y, y .
≤
sin(π/p)
(2.6)
It follows from the convexity of f and g that f (hBy, yi) ≤ hf (B)y, yi and
g(hBy, yi) ≤ hg(B)y, yi. Therefore
1
1
f (hAx, xi)g(hAx, xi) + f (hAx, xi)hg(B)y, yi)
2
3
1
1
+ g(hAx, xi)hf (B)y, yi + hf (B)g(B)y, yi
3
4
1
1
≥ f (hAx, xi)g(hAx, xi) + f (hAx, xi)g(hBy, yi)
2
3
1
1
+ g(hAx, xi)f (hBy, yi) + hf (B)g(B)y, yi. (2.7)
3
4
The convexity of f and g and the power functions tr (r ≥ 1) follow that
f (hAx, xi)p ≤ hf (A)x, xip ≤ hf (A)p x, xi,
g(hAx, xi)q ≤ hg(A)x, xiq ≤ hg(A)q x, xi.
Therefore
1
1
(f (hAx, xi)p + f (B)p ) p (g(hAx, xi)q + g(B)q ) q
1
1
≤ (hf (A)p x, xi + f (B)p ) p (hg(A)q x, xi + g(B)q ) q . (2.8)
134
M. KIAN
Since the operators hf (A)p x, xi + f (B)p and hg(A)p x, xi + g(B)q commute, we
infer from the arithmetic-geometric mean inequality that
1
1
1
(hf (A)p x, xi + f (B)p ) p (hg(A)q x, xi + g(B)q ) q ≤ (hf (A)p x, xi + f (B)p )
p
1
+ (hg(A)q x, xi + g(B)q ).
q
(2.9)
Combining (2.8) and (2.9) we obtain
D
E
p
p p1
q
q 1q
(f (hAx, xi) + f (B) ) (g(hAx, xi) + g(B) ) y, y
1
1
≤ (hf (A)p x, xi + hf (B)p y, yi) + (hg(A)q x, xi + hg(B)q y, yi).
p
q
The result now follows by combining (2.6), (2.7) and (2.10).
(2.10)
An application of Corollary 2.5 with A = B yields that:
Corollary 2.6. Let f, g : J → [0, ∞) be convex functions and let If p, q > 1 with
1
+ 1q = 1. Then
p
12
2
π
2
p
q
f (hAx, xi)g(hAx, xi) ≤
hf (A) x, xi + hg(A) x, xi
17 sin(π/p) p
q
for any A ∈ B(H )h and any unit vector x ∈ H . In particular if f = g we get
12
2
π
2
2
p
q
f (hAx, xi) ≤
hf (A) x, xi + hf (A) x, xi .
17 sin(p/π) p
q
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Department of Pure Mathematics, Ferdowsi University of Mashhad, P.O. Box
1159, Mashhad 91775, Iran.
E-mail address: kian− tak@yahoo.com