AN IDENTITY IN REAL INNER PRODUCT SPACES
JIANGUO MA
Department of Mathematics
Zhengzhu University
Henan, China
EMail: majg@zzu.edu.cn
Identity In Real Inner
Product Spaces
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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Received:
10 March, 2007
Accepted:
10 May, 2007
Communicated by:
S.S. Dragomir
2000 AMS Sub. Class.:
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Primary 26D15; Secondary 46C99.
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Key words:
Real inner product spaces, Equality, Grüss inequality.
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Abstract:
We obtain an identity in real inner product spaces that leads to the Grüss inequality and an inequality of Ostrowski.
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Contents
1
Introduction
3
2
Main Result
4
3
Applications
7
Identity In Real Inner
Product Spaces
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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1.
Introduction
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
Identity In Real Inner
Product Spaces
Jianguo Ma
Re hΦe − x, x − φei ≥ 0, Re hΓe − y, y − γei ≥ 0
(1.1)
vol. 8, iss. 2, art. 48, 2007
holds, then
(1.2)
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1
| hx, yi − hx, ei he, yi | ≤ |Φ − φ||Γ − γ|.”
4
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In this paper, we give an identity that yields the inequality
(1.3)
2
1
hx, yi −
hx,
zi
hy,
zi
kzk2
≤ kxk2 −
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1
hx, zi2
2
kzk
kyk2 −
1
hy, zi2
2
kzk
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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.
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2.
Main Result
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
−∞<s<+∞
x and span{z}, and 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
2
kzk
2
kyk
dist 2 (x, span{y}) dist 2 (z, span{x, y}).
=
kzk2
2
2
Proof. Let D = dist (x, span{y})kyk . It is easy to see that
2
2
Identity In Real Inner
Product Spaces
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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2
D = kxk kyk − hx, yi .
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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
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(2.1)
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
(2.2)
∂J
= 2kxk2 s + 2 hx, yi t − 2 hx, zi
∂s
∂J
= 2 hx, yi s + 2kyk2 t − 2 hy, zi .
∂t
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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
Substituting for s and t in
s=
(2.3)
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)
kxk2 kyk2 kzk2
D
!
2
hx, zi
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
1
2
2
(2.5) dist (x, span{z}) dist (y, span{z}) − hx, yi −
hx,
zi
hy,
zi
kzk2
!
2
hx, zi2
hy, zi
1
2
2
hx, yi −
= kxk −
kyk
−
−
hx,
zi
hy,
zi
kzk2
kzk2
kzk2
2
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
dist 2 (z, span{x, y}) =
×
Identity In Real Inner
Product Spaces
2
= kxk kyk
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hx, zi2
hy, zi2
1−
−
kxk2 kzk2 kyk2 kzk2
hx, yi2
hx, zi hy, zi hx, yi
−
+2
2
2
kxk kyk
kxk2 kyk2 kzk2
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!
.
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.
Identity In Real Inner
Product Spaces
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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3.
Applications
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
(3.1)
m ≤ f (x) ≤ M,
n ≤ g(x) ≤ N
for each x ∈ (a, b), then we have the Grüss inequality
Z b
Z b
Z b
1
1
(3.2)
f
(x)g(x)dx
−
f
(x)dx
g(x)dx
b − a
2
(b
−
a)
a
a
a
1
≤ (M − m)(N − n).
4
2
Proof. We consider the Hilbert space L (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
≤ dist(x, span{z}) dist(y, span{z}).
hx, yi −
(3.4)
hx,
zi
hy,
zi
kzk2
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
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Identity In Real Inner
Product Spaces
Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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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
Identity In Real Inner
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Jianguo Ma
An identical argument yields
vol. 8, iss. 2, art. 48, 2007
N −n
(3.8)
dist(g, span{1}) ≤
.
2
Substitute x, y and z in (3.4), and by f, g and 1, we obtain (3.2).
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Theorem 2.1 also contains a useful inequality of A. Ostrowski [4] (see also [3]).
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
X
(3.9)
i=1
ai xi = 0,
n
X
(3.10)
i=1
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bi xi = 1,
i=1
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then
n
X
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Pn
2
i=1 ai
x2i ≥ Pn 2 Pn 2
P
2.
( i=1 ai ) ( i=1 bi ) − ( ni=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.
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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
kbk
kbk
kbk2
Simple calculation shows that
kak2
kxk ≥
,
kak2 kbk2 − ha, bi2
2
(3.13)
Identity In Real Inner
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Jianguo Ma
vol. 8, iss. 2, art. 48, 2007
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
(3.14)
λ=
− ha, bi
,
2
kak kbk2 − ha, bi2
µ=
kak2
,
kak2 kbk2 − ha, bi2
thus
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2
(3.15)
x=
kak b − ha, bi a
,
kak2 kbk2 − ha, bi2
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that is, (3.11).
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References
[1] S.S. DRAGOMIR, A generalization of Grüss’inequality in inner product spaces
and application, J. Math. Anal. Appl., 237 (1999), 74–82.
[2] G. R GRÜSS, Über
b
1
f (x)g(x)dx −
b−a a
215–226.
das R maximumR des absoluten Betrages von
b
b
1
f (x)dx a g(x)dx, Math. Z., 39 (1935),
(b−a)2 a
Identity In Real Inner
Product Spaces
Jianguo Ma
[3] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publisher, 1993.
vol. 8, iss. 2, art. 48, 2007
[4] A. OSTROWSKI, Vorlesungen Über Differential und Integralrechnung, Vol. 2,
Basel, 1951, p. 289.
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