ON MODULI OF k-CONVEXITY
TECK-CHEONG LIM
Received 11 October 1999
We establish the continuity of some moduli of k-convexity. Let X be a Banach space.
We denote by X∗ the dual space of X and by BX the unit ball of X. Several moduli of
convexity for the norm of X have been defined; the last two definitions in the following
are valid for spaces having dimension ≥ k:
x + y
δX () = inf 1 −
: x, y ∈ BX , x − y ≥ (see [2]),
2
x1 + · · · + xk+1
(k)
δX () = inf 1 −
: x1 , . . . , xk+1 ∈ BX ,A x1 , . . . , xk+1 ≥ (see [10]),
k +1
(k)
inf
sup x + y − 1 (see [9]),
X () = inf
x=1
Y ⊂X
y=1
dim(Y )=k y∈Y
(1)
where

1


 
1
f 1 x1
A x1 , . . . , xk+1 = sup .

..
k!


 f k x1
···
···
···
···



1 

f1 xk+1 ∗
:
f
,
.
.
.
,
f
∈
B
.
1
..
k
X

.



fk xk+1 (2)
Evidently, by subtracting the first column from the other columns, the determinant can
be replaced by
f1 x2 − x1 · · · f1 xk+1 − x1 ..
..
(3)
.
.
···
.
fk x2 − x1 · · · fk xk+1 − x1 Also A(x1 , . . . , xk+1 ) can be thought of as the “volume” of the convex hull of x1 , . . . ,
xk+1 since that is the case in Euclidean spaces.
Copyright © 1999 Hindawi Publishing Corporation
Abstract and Applied Analysis 4:4 (1999) 243–247
1991 Mathematics Subject Classification: 46B20, 47H10
URL: http://aaa.hindawi.com/volume-4/S1085337599000202.html
244
On moduli of k-convexity
X is called uniformly convex if δX () > 0 for > 0 and k-uniformly convex if
(k)
(1)
δX () > 0 for > 0. Note that δX () = δX (); so 1-uniform convexity coincides with
(k)
uniform convexity. Lin [8] proved that X () > 0 for > 0 is equivalent to k-uniform
convexity. Gurariı̆ [5] proved that δX () is continuous on [0, 2) and there exist spaces
(k)
of which δX () = 0 for 0 ≤ < 2 and δX (2) = 1. The continuity problem of δX was
(k)
mentioned in Kirk [6]. Let µX = sup{A(x1 , . . . , xk+1 ) : x1 , . . . , xk+1 ∈ BX }. Note that
(1)
(k)
(k) µX = 2. In this paper, we prove that δX () is continuous on 0, µX . It is quite
(k)
evident that X () satisfy the Lipschitz condition with constant 1.
Definition 1. Let k ≥ 1 and 0 ≤ a < b ≤ ∞. A function f () on (a, b) is called
k-convex if
1/k
1/k k
f λ2 + (1 − λ)1
(4)
≤ λf 2 + (1 − λ)f 1
for every 1 , 2 ∈ (a, b), 0 ≤ λ ≤ 1.
Obviously 1-convexity is simply the ordinary convexity.
Lemma 2. Let 0 ≤ a < b ≤ ∞ and let f be a nondecreasing k-convex function on
(a, b) with M = supa<x<y<b (f (y) − f (x)) < ∞. Let 1 < 2 , 1 , 2 ∈ (a, b). Then
f (c) − f 1
M
≤ 1/k
(5)
1/k 1−1/k
c − 1
k −
2
1
1
for every 1 < c < 2 .
Proof. Let z(x), 1 ≤ x ≤ 2 be the function whose graph is defined by
1/k
1/k k
x = λ2 + (1 − λ)1
y = λf 2 + (1 − λ)f 1
0 ≤ λ ≤ 1.
(6)
By direct computations, we have
f 2 − f 1
M
z (x) = 1/k
≤ 1/k
.
1/k 1−1/k
1/k 1/k
1/k k−1
k 2 − 1 1
k 2 − 1
λ2 + (1 − λ)1
If 1 < c < 2 , then by the k-convexity of f and the mean-value theorem,
z(c) − z 1
f (c) − f 1
M
≤
= z (ψ) ≤ 1/k
.
1/k 1−1/k
c − 1
c − 1
k −
2
1
1
(7)
(8)
The inequality in the following lemma is a consequence of a more general result
proved in Bernal-Sullivan [1].
Teck-Cheong Lim
245
Lemma 3. Let X be a Banach space and x1 , . . . , xk+1 ∈ X. Then
1
A x1 , . . . , xk+1 ≤ k k/2 x2 − x1 · · · xk+1 − x1 .
k!
(9)
Proof. Hadamard inequality says that if r1 , r2 , . . . , rk are the rows (or columns) of a
k × k matrix, then
(10)
det r1 , r2 , . . . , rk ≤ r1 2 r2 2 · · · rk 2 .
Here · 2 denotes the Euclidean norm in Rk . Since the Euclidean norm of the j th
column of the determinant in (3) is ≤ k 1/2 xj −1 − x1 , the inequality follows.
The inequality in the next theorem for the case k = 1 improves the one obtained
in [5]. The general idea is similar to that in Goebel [3]. However, the reader should be
aware that the assertion of Lemma 1 in that paper (that δ() is convex) is incorrect;
a counterexample can be found in [7] or [4].
Theorem 4. Let X be a Banach space. Then
(k)
(k) δX (c) − δX 1
1
≤ 1/k
1/k 1−1/k
c − 1
k −
2
1
(11)
1
(k)
for every 0 < 1 < c < 2 < µX .
Proof. For simplicity, in the following we will consider k = 2 and will indicate how
to generalize to general k. Note that if A(x1 , x2 , x3 ) > 0, then x2 − x1 and x3 − x1 are
linearly independent.
For unit vectors u, u21 , u31 , and u32 in X, with {u21 , u31 } linearly independent,
consider the set
N u, u21 , u31 , u32 ; = x1 , x2 , x3 ∈ X 3 : x1 + x2 + x3 = λu, x2 − x1 = λ21 u21 ,
x3 − x1 = λ31 u31 , x3 − x2 = λ32 u32
for some λ, λ21 , λ31 , λ32 ≥ 0 and A x1 , x2 , x3 ≥ ,
(12)
and define
x1 + x2 + x3 : x1 , x2 , x3 ∈ N u, u21 , u31 , u32 ; .
δ u, u21 , u31 , u32 ; = inf 1 −
3
(13)
Obviously, δ(u, u21 , u31 , u32 ; ) is nondecreasing and has values in [0, 1].
If (x1 , x2 , x3 ) ∈ N (u, u21 , u31 , u32 ; 1 ), (y1 , y2 , y3 ) ∈ N(u, u21 , u31 , u32 ; 2 ), and
x1 + x2 + x3 = λu,
x2 − x1 = λ21 u21 ,
x3 − x1 = λ31 u31 ,
x3 − x2 = λ32 u32 ,
y1 + y2 + y3 = αu,
y2 − y1 = α21 u21 ,
y3 − y1 = α31 u31 ,
y3 − y2 = α32 u32
(14)
On moduli of k-convexity
246
for some λ, λij , α, αij ≥ 0, then by linear independence of {u21 , u31 }, there exists c ≥ 0
such that
α21 = cλ21 ,
α31 = cλ31 ,
α32 = cλ32 .
(15)
Indeed, λ32 u32 = x3 − x2 = (x3 − x1 ) − (x2 − x1 ) = λ31 u31 − λ21 u21 and α32 u32 =
α31 u31 − α21 u21 imply
(16)
α32 λ31 − λ32 α31 u31 − α32 λ21 − λ32 α21 u21 = 0
from which it follows that α31 /λ31 = α32 /λ32 = α21 /λ21 .
Let
f1 u21 f1 u31 : f1 , f2 ∈ BX∗ .
C u21 , u31 = sup f2 u21 f2 u31
(17)
Then A(x1 , x2 , x3 ) = λ21 λ31 C(u21 , u31 ) ≥ 1 and A(y1 , y2 , y3 ) = c2 λ21 λ31 C(u21 , u31 )
≥ 2 .
For 0 ≤ ζ ≤ 1, let zi = ζ xi + (1 − ζ )yi , i = 1, 2, 3. Then
z2 − z1 = ζ λ21 + (1 − ζ )cλ21 u21 = ζ + (1 − ζ )c λ21 u21 ,
z3 − z1 = ζ + (1 − ζ )c λ31 u31 ,
z3 − z2 = ζ + (1 − ζ )c λ32 u32 ,
(18)
z1 + z2 + z3 = ζ λ + (1 − ζ )α u,
2
1/2
1/2 2
,
A z1 , z2 , z3 = ζ + (1 − ζ )c λ21 λ31 C u21 , u31 ≥ ζ 1 + (1 − ζ )2
1−
z 1 + z 2 + z 3 3
= 1−
ζ x1 + x2 + x3 + (1 − ζ ) y1 + y2 + y3 3
ζ λu + (1 − ζ )αu
= 1−
3
ζ λ + (1 − ζ )α
= 1−
(19)
3
λ
α
= ζ 1−
+ (1 − ζ ) 1 −
3
3
x 1 + x 2 + x 3 y1 + y2 + y3 = ζ 1−
+ (1 − ζ ) 1 −
.
3
3
Hence
1/2
1/2 2
δ u, u21 , u31 , u32 ; ζ 1 + (1 − ζ )2
≤ ζ δ u, u21 , u31 , u32 ; 1 + (1 − ζ )δ u, u21 , u31 , u32 ; 2 .
(20)
Since
(2)
δX () = inf δ u, u21 , u31 , u32 ; : u = u21 = u31 = u32 = 1,
u21 , u31 linearly independent ,
(21)
Teck-Cheong Lim
247
and the inequality in Lemma 2 is preserved under passing to infimum, inequality (11)
for k = 2 follows.
For general k, we have k+1
2 + 1 unit vectors u, u21 , . . . and the proof is similar to
the one above.
(k)
(k) Corollary 5. Let X be a Banach space. Then δX () is continuous on 0, µX .
Proof. Take x1 = 1 and x2 , . . . , xk+1 in a small ball centered at x1 . Then, by Lemma 3,
A(x1 , . . . , xk+1 ) is small. Since 1 − x1 + · · · + xk+1 /(k + 1) is close to 0, we see that
(k)
δX () is continuous at 0.
(k)
(k) Continuity of δX () on 0, µX follows immediately from the inequality (11). Acknowledgement
The author wishes to thank the referee for his comments that led to a better formulation
of the results in this paper.
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Teck-Cheong Lim: Department of Mathematical Sciences, George Mason University,
4400 University Drive, Fairfax, VA 22030, USA