Volume 8 (2007), Issue 2, Article 48, 4 pp.
AN IDENTITY IN REAL INNER PRODUCT SPACES
JIANGUO MA
D EPARTMENT OF M ATHEMATICS
Z HENGZHU U NIVERSITY
H ENAN , C HINA
majg@zzu.edu.cn
Received 10 March, 2007; accepted 10 May, 2007
Communicated by S.S. Dragomir
A BSTRACT. We obtain an identity in real inner product spaces that leads to the Grüss inequality
and an inequality of Ostrowski.
Key words and phrases: Real inner product spaces, Equality, Grüss inequality.
2000 Mathematics Subject Classification. Primary 26D15; Secondary 46C99.
1. I NTRODUCTION
The Grüss inequality was generalized by S.S. Dragomir to the inner product spaces in [1].
It turned out to be an inequality relative to the inner products and norms of vectors in inner
product space, that is,
“Let (H; h·, ·i) be an inner product space over K(K = C, R) and e ∈ H, kek = 1. if φ, γ, Φ, Γ
are real or complex numbers and x, y are vectors in H such that the condition
(1.1)
Re hΦe − x, x − φei ≥ 0, Re hΓe − y, y − γei ≥ 0
holds, then
(1.2)
1
| hx, yi − hx, ei he, yi | ≤ |Φ − φ||Γ − γ|.”
4
In this paper, we give an identity that yields the inequality
2 1
1
1
2
2
2
2
hx, yi −
(1.3)
hx, zi hy, zi ≤ kxk −
hx, zi
kyk −
hy, zi
kzk2
kzk2
kzk2
here x, y, z ∈ H, H is a real inner product space.
From inequality (1.3), we obtain the Grüss inequality and an inequality by A. Ostrowski.
102-07
2
J IANGUO M A
2. M AIN R ESULT
Let x, y, z be three vectors in real inner product spaces. Denote by Z := span{z} the linear
subspace spanned by z, and W := span{x, z} the linear subspace spanned by x and z, denote by dist(x, span{z}) =
inf
kx − szk for the distance between x and span{z}, and
−∞<s<+∞
dist(z, span{x, y}) =
inf
−∞<s,t<+∞
kz − (sx + ty)k. The main result of this paper is:
Theorem 2.1. Suppose x, y, z are three non-zero vectors in a real inner product space, then
2
1
2
2
dist (x, span{z}) dist (y, span{z}) − hx, yi −
hx,
zi
hy,
zi
kzk2
kyk2
=
dist 2 (x, span{y}) dist 2 (z, span{x, y}).
kzk2
Proof. Let D = dist 2 (x, span{y})kyk2 . It is easy to see that
D = kxk2 kyk2 − hx, yi2 .
(2.1)
When D 6= 0, we determine the infimum of J(s, t) = kz − (sx + ty)k2 by discovering critical
points of J(s, t). Simple calculus yields
J(s, t) = kzk2 − 2 hx, zi s − 2 hy, zi t + kxk2 s2 + 2 hx, yi st + kyk2 t2 ,
thus partial derivatives of J(s, t) are
∂J
= 2kxk2 s + 2 hx, yi t − 2 hx, zi
∂s
∂J
= 2 hx, yi s + 2kyk2 t − 2 hy, zi .
∂t
(2.2)
Let
∂J
∂s
= 0 and
∂J
∂t
= 0, we obtain
1
(kyk2 hx, zi − hy, zi hx, yi)
D
1
t = (kxk2 hy, zi − hx, zi hx, yi).
D
s=
(2.3)
Substituting for s and t in
J(s, t) = kzk2 − 2 hx, zi s − 2 hy, zi t + kxk2 s2 + 2 hx, yi st + kyk2 t2 ,
by (2.3), we obtain
(2.4)
dist 2 (z, span{x, y})
kxk2 kyk2 kzk2
=
D
hx, zi2
hy, zi2
hx, yi2
hx, zi hy, zi hx, yi
1−
−
−
+
2
kxk2 kzk2 kyk2 kzk2 kxk2 kyk2
kxk2 kyk2 kzk2
!
.
On the other hand, we have
(2.5)
2
1
dist 2 (x, span{z}) dist 2 (y, span{z}) − hx, yi −
hx,
zi
hy,
zi
kzk2
!
2
hx, zi2
hy, zi
1
2
2
= kxk −
kyk
−
−
hx,
yi
−
hx,
zi
hy,
zi
kzk2
kzk2
kzk2
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 48, 4 pp.
http://jipam.vu.edu.au/
A N I DENTITY I N R EAL I NNER P RODUCT S PACES
= kxk2 kyk2
3
hx, zi2
hy, zi2
hx, yi2
hx, zi hy, zi hx, yi
1−
−
−
+2
2
2
2
2
2
2
kxk kzk
kyk kzk
kxk kyk
kxk2 kyk2 kzk2
!
.
Comparing (2.4) and (2.5), and taking note that D = dist2 (x, span{y})kyk2 , we finish our
proof for the case D 6= 0.
When D = 0, then x and y are linearly dependent. in this case we can prove the theorem by
straightforward verification.
We point out that Theorem 2.1 is true also for complex inner product spaces.
3. A PPLICATIONS
An application of Theorem 2.1 is the well known Grüss inequality [2] (see also [3]).
Theorem 3.1 (G. Grüss). Let f and g be two Lebesque integrable functions on (a, b). m, M
and n, N are four real numbers such that
m ≤ f (x) ≤ M,
(3.1)
n ≤ g(x) ≤ N
for each x ∈ (a, b), then we have the Grüss inequality
Z b
Z b
Z b
1
1
1
≤ (M − m)(N − n).
(3.2) f (x)g(x)dx −
f
(x)dx
g(x)dx
4
b−a a
(b − a)2 a
a
Proof. We consider the Hilbert space L2 (a, b) equipped with an inner product defined by
Z b
1
(3.3)
hf, gi =
f (x)g(x)dx.
b−a a
According to Theorem 2.1, we have
1
(3.4)
hx, yi − kzk2 hx, zi hy, zi ≤ dist(x, span{z}) dist(y, span{z}).
This inequality yields inequality (1.3) by (2.1).
Let x = f, y = g and z = 1. Note that by m ≤ f (x) ≤ M and n ≤ g(x) ≤ N , it is easy to
see that
2
m+M
(M − m)2
(3.5)
f (x) −
≤
2
4
and
(3.6)
n+N
g(x) −
2
2
≤
(N − n)2
.
4
Therefore,
(3.7)
dist(f, span{1}) ≤
1
b−a
Z
a
b
M +m 2
(f (x) −
) dx
2
12
≤
M −m
.
2
An identical argument yields
N −n
.
2
Substitute x, y and z in (3.4), and by f, g and 1, we obtain (3.2).
(3.8)
dist(g, span{1}) ≤
Theorem 2.1 also contains a useful inequality of A. Ostrowski [4] (see also [3]).
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 48, 4 pp.
http://jipam.vu.edu.au/
4
J IANGUO M A
Theorem 3.2 (Ostrowski). Let a = (a1 , . . . , an ) and b = (b1 , . . . , bn ) be two linearly independent vectors. If the vector x = (x1 , . . . , xn ) satisfies
n
n
X
X
(3.9)
ai xi = 0,
bi xi = 1,
i=1
i=1
then
(3.10)
n
X
x2i
i=1
Pn
a2
≥ Pn 2 Pn i=12 i Pn
2.
( i=1 ai ) ( i=1 bi ) − ( i=1 ai bi )
The equality holds if and only if
P
P
bk ni=1 a2i − ak ni=1 ai bi
(3.11)
xk = Pn 2 Pn 2
P
2,
( i=1 ai ) ( i=1 bi ) − ( ni=1 ai bi )
k = 1, 2, . . . , n.
Proof. Substituting x, y, z in inequality (1.3), by vectors x, a, b, we have
!
2
1
ha,
bi
1
(3.12)
kxk2 −
kak2 −
≥
ha, bi2 .
2
2
2
kbk
kbk
kbk
Simple calculation shows that
kak2
,
kak2 kbk2 − ha, bi2
that is, (3.10). According to Theorem 2.1, equality in (3.13) holds if and only if x, a, b are
linearly dependent, that is, there exist constants λ, µ such that x = λa + µb. Taking the inner
product of a and b, we get kak2 λ + ha, bi µ = 0 and ha, bi λ + kbk2 µ = 1. Solutions of the last
two equations are
kxk2 ≥
(3.13)
(3.14)
λ=
− ha, bi
,
kak2 kbk2 − ha, bi2
µ=
kak2
,
kak2 kbk2 − ha, bi2
thus
(3.15)
x=
kak2 b − ha, bi a
,
kak2 kbk2 − ha, bi2
that is, (3.11).
R EFERENCES
[1] S.S. DRAGOMIR, A generalization of Grüss’inequality in inner product spaces and application, J.
Math. Anal. Appl., 237 (1999), 74–82.
Rb
1
[2] G. GRÜSS, Über das maximum des absoluten Betrages von b−a
a f (x)g(x)dx −
Rb
Rb
1
f (x)dx a g(x)dx, Math. Z., 39 (1935), 215–226.
(b−a)2 a
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ
Kluwer Academic Publisher, 1993.
AND
A.M. FINK, Classical and New Inequalities in Analysis,
[4] A. OSTROWSKI, Vorlesungen Über Differential und Integralrechnung, Vol. 2, Basel, 1951, p. 289.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 48, 4 pp.
http://jipam.vu.edu.au/