Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 3, Issue 4, Article 55, 2002
ON MODULI OF EXPANSION OF THE DUALITY MAPPING OF SMOOTH
BANACH SPACES
PAVLE M. MILIČIĆ
FACULTY OF M ATHEMATICS
U NIVERSITY OF B ELGRADE
YU-11000, Y UGOSLAVIA .
pmilicic@hotmail.com
Received 30 March, 2002; accepted 30 April, 2002.
Communicated by S.S. Dragomir
A BSTRACT. Let X be a Banach space which is uniformly convex and uniformly smooth. We
introduce the lower and upper moduli of expansion of the dual mapping J of the space X.
Some estimation of certain well-known moduli (convexity, smoothness and flatness) and two
new moduli introduced in [5] are described with this new moduli of expansion.
Key words and phrases: Uniformly convex (smooth) Banach space, Angle of modulus convexity (smoothness), Lower (upper)
modulus of expansion.
2000 Mathematics Subject Classification. 46B20, 46C15, 51K05.
Let (X, k·k) be a real normed space, X ∗ its conjugate space, X ∗∗ the second conjugate of X
and S (X) the unit sphere in X (S (X) = {x ∈ X| kxk = 1}) .
Moreover, we shall use the following definitions and notations.
The sign (S) denotes that X is smooth, (R) that X is reflexive, (U S) that X is uniformly
smooth, (SC) that X is strictly convex, and (U C) that X is uniformly convex.
∗
The map J : X → 2X is called the dual map if J (0) = 0 and for x ∈ X, x 6= 0,
J (x) = {f ∈ X ∗ |f (x) = kf k kxk , kf k = kxk} .
∗∗
The dual map of X ∗ into 2X we denote by J ∗ . The map τ is canonical linear isometry of X
into X ∗∗ .
It is well known that functional
kx + tyk − kxk
kxk
kx + tyk − kxk
(1)
g (x, y) :=
lim
+ lim
t→−0
t→+0
2
t
t
always exists on X 2 . If X is (S) , then (1) reduces to
kx + tyk − kxk
;
t→0
t
g (x, y) = kxk lim
ISSN (electronic): 1443-5756
c 2002 Victoria University. All rights reserved.
050-02
2
PAVLE M. M ILI ČI Ć
the functional g is linear in the second argument, J (x) is a singleton and g (x, ·) ∈ J (x) . In
this case we shall write J (x) = Jx = fx . Then [y, x] := g (x, y) , definesa so called semi-inner
product [·, ·] (s.i.p) on X 2 which generates the norm of X, [x, x] = kxk2 , (see [1]). If X is an
inner-product space (i.p. space) then g (x, y) is the usual i.p. of the vector x and the vector y.
By the use of functional g we define the angle between vector x and vector y (x 6= 0, y 6= 0)
as
g (x, y) + g (y, x)
(2)
cos (x, y) :=
2 kxk kyk
(see [3]). If (X, (·, ·)) is an i.p. space, then (2) reduces to
cos (x, y) =
(x, y)
.
kxk kyk
We say that X is a quasi-inner product space (q.i.p space) if the following equality holds
(3)
kx + yk4 − kx − yk4 = 8 kxk2 g (x, y) + kyk2 g (y, x) , (x, y ∈ X)1)
The equality (3) holds in the space l4 , but does not hold in the space l1 . A q.i.p. space X is
(SC) and (U S) (see [6] and [4]).
Alongside the modulus of convexity of X, δX , and the modulus of smoothness of X, ρX ,
defined by
x + y δX (ε) = inf 1 − 2 x, y ∈ S (X) ; kx − yk ≥ ε ;
x + y ρX (ε) = sup 1 − 2 x, y ∈ S (X) ; kx − yk ≤ ε ;
0
we have defined in [5] the angle modulus of convexity of X, δX
, and the angle modulus of
0
smoothness of X, ρX by:
1 − cos (x, y) 0
δX (ε) = inf
x, y ∈ S (X) ; kx − yk ≥ ε ;
2
1 − cos (x, y) 0
ρX (ε) = sup
x, y ∈ S (X) ; kx − yk ≤ ε .
2
We also recall the known definition of modulus of flatness of X, ηX (Day’s modulus):
2 − kx + yk ηX (ε) = sup
x, y ∈ S (X) ; kx − yk ≤ ε .
kx − yk We now quote three known results.
Lemma 1. (Theorem 6 in [7] and Theorem 6 in [1]). Let X be a real normed space which is
(S) , (SC) and (R) . Then for all f ∈ X ∗ there exists a unique x ∈ X such that
f (y) = g (x, y) , (y ∈ X) .
Lemma 2. (Theorem 7 in [1]). Let X be a Banach space which is (U S) and (U C) and let [·, ·]
be an s.i.p. on X 2 which generates the norm on X (see [1]). Then the dual space X ∗ is (U S)
and (U C) and the functional
hJx, Jyi := [y, x] , (x, y ∈ X) ,
is an s.i.p on (X ∗ )2 .
1) If (·, ·) is an i.p. on X 2 then g (x, y) = (x, y) and the equality (3) is the parallelogram equality.
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D UALITY M APPING OF S MOOTH BANACH S PACES
3
Lemma 3. (Proposition 3 in [2]). Let X be a real normed space. Then for J, J ∗ and τ on their
respective domains we have
J −1 = τ −1 J ∗ and J = J ∗−1 τ.
Remark 4. Under the hypothesis of Lemma 2, the mappings J, J ∗ and τ are bijective mappings.
Then, by Lemma 3, Lemma 2 and Lemma 1, in this case, we have
hJx, Jyi = g (x, y) = g (fy , fx ) , (x, y ∈ X) .
Lemma 5. Let X be a real normed space which is (S) , (SC) and (R) . Then for x, y ∈ S (X)
we have
x + y 1 − cos (x, y)
kx − yk kfx − fy k
≤
(4)
1−
≤
.
2
2
4
Proof. Under the hypothesis of Lemma 5, using Lemma 1, we have fx = g (x, ·) (x ∈ X) .
Consequently,
kfx − fy k = sup {|g (x, t) − g (y, t)| | t ∈ S (X)}
≥ g (x, t) − g (y, t) (t ∈ S (X)) .
For t =
(5)
x−y
,
kx−yk
(x 6= y) , we obtain
x−y
x−y
g x,
− g y,
≤ kfx − fy k .
kx − yk
kx − yk
Since X is (S) , the functional g is linear in the second argument. Hence, from (5) we get
(6)
1 − g (x, y) − g (y, x) + 1 ≤ kx − yk kfx − fy k .
Using the inequality
x + y 1 − cos (x, y)
kx − yk
≤
1−
≤
2
2
2
(see Lemma 1 in [5]) and the inequality (6) we obtain the inequality (4).
Lemma 6. Let X be a Banach space which is (U S) and (U C) . Let δX ∗ be the modulus of
convexity of X ∗ . Then for each ε > 0 and for all x, y ∈ S (X) the following implications hold
(7)
(8)
kx − yk ≤ 2δX ∗ (ε) =⇒ kfx − fy k ≤ ε,
kfx − fy k ≥ ε =⇒ kx − yk ≥ 2δX ∗ (ε) .
Proof. By Lemma 2, X ∗ is a Banach space which is (U C) and (U S) . Since X ∗ is (U C) , for
each ε > 0, we have δX ∗ (ε) > 0 and, for all x, y ∈ S (X) ,
(9)
kfx + fy k > 2 − 2δX ∗ (ε) =⇒ kfx − fy k < ε.
Under the hypothesis of Lemma 6, by Remark 4, we have g (x, y) = g (fy , fx ) . Hence, by
inequality
1 − kx − yk ≤ g (x, y) ≤ kx + yk − 1
(see Lemma 1 in [6]), we obtain
(10)
1 − kx − yk ≤ g (x, y) = g (fy , fx ) ≤ kfx + fy k − 1,
so that we have
(11)
kx − yk + kfx + fy k ≥ 2.
Now, let x, y ∈ S (X) and kx − yk < 2δX ∗ (ε) . Then, by (11) we obtain
kfx + fy k > 2 − 2δX ∗ (ε) .
J. Inequal. Pure and Appl. Math., 3(4) Art. 55, 2002
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PAVLE M. M ILI ČI Ć
Thus, by (9), we conclude that
kx − yk < 2δX ∗ (ε) =⇒ kfx − fy k < ε.
(12)
On the other hand if kx − yk = 2δX ∗ (ε) and kfx − fy k > ε, by (9), it follows
kx − yk + kfx + fy k ≤ 2.
So, by (11), we get
kx − yk + kfx + fy k = 2.
Hence, using (10), we conclude that g (x, y) = 1 − kx − yk , i.e., g (x, x − y) = kxk kx − yk .
Thus, since X is (SC) , using Lemma 5 in [1], we get x = x − y, which is impossible. So, the
implication (7) is correct. The implication (8) follows from the implication (12).
We now introduce a new definition.
According to the inequality (4), to make further progress in the estimates of the moduli
0
δX , δX
, ρX , ρ0X , it is convenient to introduce
Definition 1. Let X be (S) and x, y ∈ S (X) . The function eJ : [0, 2] → [0, 2] , defined by
eJ (ε) := inf {kfx − fy k | kx − yk ≥ ε}
will be called the lower modulus of expansion of the dual mapping J.
The function eJ : [0, 2] → [0, 2] , defined as
eJ (ε) := sup {kfx − fy k | kx − yk ≤ ε}
is the upper modulus of expansion of the dual mapping J.
Now, we quote our new results. Firstly, we note some elementary properties of the moduli eJ
and eJ .
Theorem 7. Let X be (S). Then the following assertions are valid.
a) The function eJ is nondecreasing on [0, 2] .
b) The function eJ is nondecreasing on [0, 2] .
c) eJ (ε) ≤ eJ (ε) (ε ∈ [0, 2]) .
d) If X is a Hilbert space, then eJ (ε) = eJ (ε) .
Proof. The assertions a) and b) follow from the implications
ε1 < ε2 =⇒ {(x, y) | kx − yk ≥ ε1 } ⊃ {(x, y) | kx − yk ≥ ε2 } (x, y ∈ S (X)) ,
ε1 < ε2 =⇒ {(x, y) | kx − yk ≤ ε1 } ⊂ {(x, y) | kx − yk ≤ ε2 } (x, y ∈ S (X)) .
c) Assume, to the contrary, i.e., that there is an ε ∈ [0, 2] such that eJ (ε) > eJ (ε) . Then
inf {kfx − fy k | kx − yk = ε} ≥ inf {kfx − fy k | kx − yk ≥ ε}
> sup {kfx − fy k | kx − yk ≤ ε}
≥ sup {kfx − fy k | kx − yk = ε} ,
which is not possible.
d) In a Hilbert space, we have
kfx − fy k = sup {|(x, t) − (y, t)| | t ∈ S (X)} ≤ kx − yk .
x−y
On the other hand, the functional fx − fy attains its maximum in t = kx−yk
∈ S (X) .
Hence kx − yk = kfx − fy k . Because of that, we have eJ (ε) = eJ (ε) = ε.
0
In the next theorems some relation between moduli δX
, ρ0X ,eJ , eJ are given.
Theorem 8. Let X be (S) , (SC) and (R) . Then, for ε ∈ (0, 2] we have
J. Inequal. Pure and Appl. Math., 3(4) Art. 55, 2002
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D UALITY M APPING OF S MOOTH BANACH S PACES
5
1
0
a) δX
(ε) ≤ eJ (ε)
2
ε
b) ρ0X (ε) ≤ eJ (ε) ,
4
1
2
c) ρX (ε) ≤ ηX (ε) ≤ eJ (ε) .
ε
2
Proof. The proof of the assertions a) and b) follows immediately using the definitions of the
0
functions δX
and ρ0X and the inequality (4).
c) Let x, y ∈ S (X) , x 6= y. By Lemma 5, we have
2
kx + yk
2 − kx + yk
=
1−
kx − yk
kx − yk
2
1 − cos (x, y)
≤
kx − yk
kx − yk kfx − fy k
≤
2 kx − yk
kfx − fy k
=
.
2
So
kfx − fy k
2 − kx + yk
≤
.
kx − yk
2
Using the definition of ηX and eJ , we obtain
1
ηX (ε) ≤ eJ (ε) .
2
On the other hand
1
1
2 − kx + yk
2
kx + yk
(0 < kx − yk ≤ ε) =⇒
≥
=⇒
≥
1−
.
kx − yk
ε
kx − yk
ε
2
Because of that we have
2
ηX (ε) ≥ ρX (ε) .
ε
Remark 9. The last inequality is true for an arbitrary space X.
Corollary 10. For a q.i.p. space, it holds that
ε 4
(13)
eJ (ε) ≥
(ε ∈ [0, 2]) .
2
Proof. By a) of Theorem 8 and the inequality
(13).
ε4
32
0
≤ δX
(ε) (see Corollary 2 in [5]), we get
Corollary 11. If X is (S) , (SC) and (R) then
1
0
a) δX
eJ (ε) ,
∗ (ε) ≤
2
1
b) ρ0X ∗ ≤ eJ ∗ (ε) ,
2
2
1
c) ρX ∗ (ε) ≤ ηX ∗ (ε) ≤ eJ ∗ (ε) .
3
2
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PAVLE M. M ILI ČI Ć
Proof. It is well-known that if X is (S) , (SC) and (R) then X ∗ is (S) , (SC) and (R) . Hence
Theorem 8 is valid for X ∗ .
Theorem 12. Let X be a Banach space which is (U C) and (U S) . Then, for all ε > 0, we have
the following estimations:
εδX ∗ (ε)
a) ρ0X (2δX ∗ (ε)) ≤
,
2
εδX (ε)
,
b) ρ0X ∗ (2δX (ε)) ≤
2
c) eJ ∗ (ε) ≥ 2δX ∗ (ε) ,
d) eJ (2δX ∗ (ε)) ≤ ε,
Proof.
(eJ ∗ (2δX (ε)) ≤ ε) .
a) Using, in succession, the definition of the function ρ0X , the inequality (4) in
Lemma 2 and the implication (7), we obtain:
1 − cos (x, y) 0
ρX (2δX ∗ (ε)) = sup
kx − yk ≤ 2δX ∗ (ε)
2
1
sup {kx − yk kfx − fy k | kx − yk ≤ 2δX ∗ (ε)}
4
1
≤ 2εδX ∗ (ε)
4
εδX ∗ (ε)
=
.
2
b) If, in a), we set X ∗ instead of X (X ∗∗ instead of X ∗ ), we get
≤
(14)
εδX ∗∗ (ε)
.
2
Let F, G ∈ S (X ∗∗ ) . Under the hypothesis of Theorem 12, we have
kF + Gk δX ∗∗ (ε) = inf 1 −
kF − Gk ≥ ε
2
kτ x + τ yk = inf 1 −
kτ x − τ yk ≥ ε
2
kτ (x + y)k = inf 1 −
kτ (x − y)k ≥ ε
2
kx + yk = inf 1 −
kx − yk ≥ ε
2 = δX (ε) .
ρ0X ∗ (2δX ∗∗ (ε)) ≤
Consequently the inequality (14) is equivalent to the inequality b).
c) Using, in succession, the definition of eJ , Lemma 3, and the implication (8), we get
eJ ∗ (ε) = inf {kJ ∗ fx − J ∗ fy k | kfx − fy k ≥ ε}
= inf {kτ x − τ yk | kfx − fy k ≥ ε}
≥ 2δX ∗ (ε) .
d) Using the definition of eJ and the implication (7), we get
eJ (2δX ∗ (ε)) = sup {kfx − fy k | kx − yk ≤ 2δX ∗ (ε)} ≤ ε.
Replacing, here, X ∗ with X ∗∗ and J with J ∗ , we get the second inequality.
J. Inequal. Pure and Appl. Math., 3(4) Art. 55, 2002
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D UALITY M APPING OF S MOOTH BANACH S PACES
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Since in a Banach space X we have
r
ε2
0
and δX (ε) ≤ δX
(ε)
4
(see Theorem 1 in [5]), using b) and a) of Theorem 12, we obtain
Corollary 13. Under the hypothesis of Theorem 12, we have
2
2 0
a) ρ0X ∗ (2δX (ε)) ≤ δX (ε) ≤ δX
(ε) ,
ε
!
rε
2
ε
ε
.
b) ρ0X (2δX ∗ (ε)) ≤
1− 1−
4
2
δX (ε) ≤ 1 −
1−
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J. Inequal. Pure and Appl. Math., 3(4) Art. 55, 2002
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