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Directionally bounded sets in c0 with equivalent norms
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J. Math. Anal. Appl. ••• (••••) •••–•••
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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Directionally bounded sets in c0 with equivalent norms Francisco Eduardo Castillo Santos a,1 , Helga Fetter a,∗,1 , Berta Gamboa de Buen a,1 , Fernando Núñez-Medina b a b
Centro de Investigación en Matemáticas (CIMAT), Apdo. Postal 402, 36000 Guanajuato, Gto., Mexico Departamento de Matemáticas, Universidad de Guanajuato, CP 36240, Guanajuato, Gto., Mexico
a r t i c l e
i n f o
Article history: Received 2 December 2013 Available online xxxx Submitted by T. Domínguez Benavides
a b s t r a c t We study the directionally bounded sets in c0 with respect to two different families of norms and conclude that in nonreflexive spaces directional boundedness is not necessarily preserved by isomorphisms. © 2014 Published by Elsevier Inc.
Keywords: Directionally bounded sets Directional sequences AFPP
1. Introduction In [3], the author defined the concept of directionally bounded set of a Banach space X. In the same article the author proved that a Banach space X is reflexive if and only if a closed convex linearly bounded set is directionally bounded. Thus if X is a reflexive space, the class of directionally bounded sets does not change with an equivalent norm. Therefore we ask if the same result holds on a nonreflexive Banach space. Explicitly, does the class of directionally bounded sets of a nonreflexive Banach space X remain the same when we equip X with an equivalent norm? In this paper we show that in c0 we can find two different families of equivalent norms, such that in one case the class of directionally bounded sets is the same as in c0 and in the other it is different. This result not only shows that directional boundedness in a nonreflexive space depends on the equivalent norm given, but also by a result of Shafrir [3] the same holds for the AFPP. Here the AFPP is defined as follows. * Corresponding author. E-mail addresses: [email protected] (F.E. Castillo Santos), [email protected] (H. Fetter), [email protected] (B. Gamboa de Buen), [email protected] (F. Núñez-Medina). 1 Fax: +52 473 7325749. http://dx.doi.org/10.1016/j.jmaa.2014.05.006 0022-247X/© 2014 Published by Elsevier Inc.
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Definition 1. Let X be a Banach space and C ⊂ X be closed and convex. We will say that C has the approximate fixed point property, AFPP for short, if for all nonexpansive maps T : C → C we have inf T (x) − x : x ∈ C = 0. The following definitions appeared in [3]. Definition 2. Let (X, · ) be a Banach space and {xn } ⊂ X. We say that the sequence {xn } is a directional sequence, if limn→∞ xn = ∞ and there exists b ≥ 0 such that for every n1 < n2 < · · · < ns , s ∈ N we have xn1 − xns ≥
s−1
xni+1 − xni − b.
i=1
Definition 3. Let X be a Banach space and C ⊂ X be a closed and convex set. We will say that C is directionally bounded if it does not contain a directional sequence. Shafrir [3] proved the following two theorems. Theorem 4. Let X be a Banach space and C ⊂ X. Then C has the AFPP if and only if it is directionally bounded. Theorem 5. Let X be a Banach space and C ⊂ X, closed, convex and unbounded. Then C is directionally bounded if for every {xn } ⊂ C with limn→∞ xn = ∞ and for every extreme point f of SX ∗ we have that
xn lim sup f n→∞ xn
< 1.
2. The space (c0 )α Recall that c0 is the space of real null sequences and · ∞ denotes its usual norm. The sequences {en }, {e∗n } will denote the canonical bases of c0 and 1 respectively. We will define the following family of norms in c0 . Definition 6. Let α > 0. For x = {x(n)} ∈ c0 define ∞ |x(n)| xα = x∞ + α . 2n n=1
It is clear that these norms are equivalent to the usual c0 norm. In [2] it is shown that, for any α > 0, the dual space of (co , · α ) is the space (1 , · α∗ ). Here the norm ∞ · α∗ is given for h = n=1 cn e∗n ∈ 1 by hα∗ = sup
n∈F
1+α
|cn |
1 n∈F 2n
:F ⊂N .
We are interested in studying the class of directionally bounded sets with respect to these norms. In order to do so, we need to look at directional sequences. However, as the next lemma shows, this is not enough to determine the directionally bounded sets.
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Lemma 7. There are nondirectional sequences in (c0 , · α ) with a nondirectionally bounded convex hull. Proof. Let xn = 2n+1 e1 + 2n en+1 . Clearly xn ∞ = 2n+1 and for any n, k ∈ N, xn − xk ∞ = |2n+1 − 2k+1 |. Thus for any n1 < n2 < · · · < ns : s−1
xni+1 − xni ∞ =
i=1
s−1
2ni+1 +1 − 2ni +1
i=1
= 2ns +1 − 2n1 +1 = xn1 − xns ∞ . Taking b = 0, we obtain that {xn } is a directional sequence in (c0 , · ∞ ). On the other hand: xn − xk α = 2n+1 − 2k+1 e1 + 2n en+1 − 2k ek+1 α
1 1 α
n+1 2 + . = 1+ − 2k+1 + α 2 2 2 From this, s−1 i=1
xni+1
s−1 ni+1 +1 α ni +1 + (s − 1)α − xn i α = 2 −2 1+ 2 i=1 ns +1 α n1 +1 = 2 + (s − 1)α −2 1+ 2 = xns − xn1 α + (s − 2)α.
For a fixed b, the inequality b ≥ (s − 2)α, cannot hold for all s ∈ N. Hence {xn } is not a directional sequence in (c0 , · α ). Now let 1 1 i+1 1 i xi = 2 e1 + 2 ei+1 . n i=1 n i=1 n i=1 n
yn =
n
n
If n > m, yn − ym =
1 1 − n m
+
m
2i+1 e1 +
i=1
1 1 − n m
m
n 1 i+1 2 e1 n i=m+1
2i ei+1 +
i=1
n 1 i 2 ei+1 . n i=m+1
Thus n 1 α yn − ym α = 1 + 2i+1 − 2 n i=1 α 4 n = 1+ 2 −1 − 2 n
and
1 i+1 2 m i=1 m
α 1 α1 − n i=1 2 m i=1 2 n
+
4 m 2 −1 m
m
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s−1
ynj+1
j=1
4 ns α 4 n1 2 −1 − 2 −1 − ynj α = 1 + 2 ns n1 = yns − yn1 α .
Therefore {yn } is a directional sequence in C = conv{xn } in (c0 , · α ). Since we also proved that {xn } ⊂ C is a directional sequence in (c0 , · ∞ ), C is the desired convex set. 2 The proof of the previous result also shows that the families of directional sequences of the spaces (c0 , · ∞ ) and (c0 , · α ) are different, however we will see later that the classes of directionally bounded sets are the same. The following three propositions are crucial for the main theorem of this section. Proposition 8. Every directional sequence in (c0 , ·α ) is also a directional sequence in (c0 , ·∞ ). Therefore if C ⊂ c0 is directionally bounded in (c0 , · ∞ ), it is also directionally bounded in (c0 , · α ). Proof. Let {xn } be a directional sequence in (c0 , · α ). Take b ≥ 0 as given by the definition of directional sequence. Let n1 < n2 < · · · < ns . Then b≥
s−1
xni − xni+1 α − xns − xn1 α
n=1
=
s−1
xni − xni+1 ∞ − xns − xn1 ∞
n=1
s−1 ∞ ∞ |xni+1 (m) − xni (m)| |xn1 (m) − xns (m)| +α − . 2m 2m m=1 i=1 m=1
By the triangle inequality: s−1 ∞
1
xn (m) − xni (m) − xn1 (m) − xns (m) ≥ 0. 2m i=1 i+1 m=1 From this: s−1
xni+1 − xni ∞ − xn1 − xns ∞ ≤ b.
2
i=1
The next proposition is similar to Lemma 2.2 of [1]. Proposition 9. Let C be a convex unbounded set of c0 . Assume that C is directionally bounded in (c0 , · α ). Let {xn } ⊂ C be a sequence such that limn→∞ xn ∞ = ∞, write xn = {xn (k)}∞ k=1 . Then for every n0 , k0 ∈ N, there exist n > n0 and k > k0 such that xn ∞ = |xn (k)|. Proof. Recall that for every x = {x(k)} ∈ c0 there exists k ∈ N such that x∞ = |x(k)|. If the proposition were false, there would exist n0 , k0 ∈ N such that for every n > n0 , if k is such that xn ∞ = |xn (k)| then k ≤ k0 . Since {xn } is a directional sequence if and only if {−xn } is a directional sequence, by passing to a subsequence, we may assume that there exists k ≤ k0 such that for all n ∈ N we have xn ∞ = xn (k). Without loss of generality we will assume that k = 1, that is, xn ∞ = xn (1).
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Let n1 = 1 and θ1 = 1. We may choose a subsequence S1 ⊂ N = S0 such that for all n ∈ S1 we have that sgn(xn (2)) = θ2 is constant. Take n2 = min(S1 \{n1 }) and a subsequence S2 ⊂ S1 such that for n ∈ S2 , sgn(xn (3)) = θ3 is constant. If for j = 1, ..., k we have nj ∈ Sj−1 , n1 < n2 < ... < nk and sequences S0 ⊃ S1 ⊃ ... ⊃ Sk so that for n ∈ Sj , we have that (xn (j + 1)) = θj+1 is constant, take nk+1 = min Sk \{n1 , ..., nk } and a subsequence Sk+1 ⊂ Sk such that for n ∈ Sk+1 , sgn(xn (k + 2)) = θk+2 is constant. By taking the subsequence {xni }, we may assume that {xn } satisfies that for n ≥ j, sgn(xn (j)) depends only on j. ∞ For x = {x(n)} ∈ c0 take f (x) = θ1 x(1) + α i=1 θi2x(i) . i Thus f α∗ = 1 and
xn f xn α
=
|xn (1)| + α
=1+
α |xn (1)|
n j=1
|xn (j)| 2j
+α
∞
j=n+1 ∞ α j=1 |xn2(j)| j
θj xn (j) 2j
|xn (1)| + ∞ θj xn (j)
∞ − |xnα(1)| j=n+1 ∞ 1 + |xn1(1)| α j=1 |xn2(j)| j
j=n+1
2j
|xn (j)| 2j
.
But since |xn (1)| ≥ |xn (j)| for all j ∈ N: 0≤
∞ ∞ ∞ α α |xn (j)| θj xn (j) 1 − ≤ 2α . j |xn (1)| j=n+1 2j |xn (1)| j=n+1 2j 2 j=n+1
Thus limn→∞ f ( xxnnα ) = 1, contradicting that C is a directionally bounded set. 2 Proposition 10. Let {xn } ⊂ c0 be a sequence such that limn→∞ xn ∞ = ∞. If for every n0 , k0 ∈ N there exist n > n0 and k > k0 such that xn ∞ = |xn (k)|, then {xn } is not a directional sequence in (c0 , · ∞ ). Proof. It is easy to see that if {zn } is a directional sequence and {yn } is a sequence such that zn − yn is s n sufficiently small then {yn } is also a directional sequence. Hence we can also assume that xn = i=1 xn (i)ei where {sn } is a strictly increasing sequence of natural numbers, with xn ∞ = xn (in ) > xn−1 (in−1 ) ≥ 0, in < sn < in+1 . Now let n > m, then xn − xm ∞ ≥ xn (in ), s−1
xn+1 − xn ∞ − xs − x1 ∞ ≥
n=1
s−1
xj+1 (ij+1 ) − xs (is ) − x1 (i1 )
j=1
=
s−2
xj+1 (ij+1 ) − x1 (i1 )
j=1
and lims→∞
s−2 j=0
xj+1 (ij+1 ) = ∞.
2
Now we present the main theorem of this section. Theorem 11. Let C be a convex, closed and unbounded subset of c0 . Then C is directionally bounded in (co , · ∞ ) if and only if it is directionally bounded in (co , · α ). Proof. By Proposition 8 it is enough to show that if C is directionally bounded in (c0 , · α ) then it is also directionally bounded in (c0 , · ∞ ).
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To prove this, let {xn } ⊂ C be a sequence such that limn→∞ xn ∞ = ∞, by Proposition 9, for every n0 , k0 ∈ N, there exist n > n0 and k > k0 such that xn ∞ = |xn (k)| and by Proposition 10, {xn } is not directional in (c0 , · ∞ ). 2 Corollary 12. Let C be a convex closed subset of c0 . Then C has the AFPP in (c0 , · α ) if and only if C has the AFPP in (c0 , · ∞ ). Proof. It is known that bounded closed convex sets have the AFPP. If C is unbounded we obtain the result by Theorem 11 and Theorem 4. 2 Corollary 13. The norm · ∞ is the starting point of a ray of equivalent norms in c0 such that the respective classes of directionally bounded sets are the same as in (c0 , · ∞ ). 3. The space (c0 , · Dα ) We start by defining the norms · Dα . Definition 14. Let α > 0. For x = {x(n)} ∈ c0 define
xDα = sup x(n) − αx(m) : n = m . It is clear that these norms are equivalent to the usual c0 norm. Observe the following: For x ∈ c0 choose j0 ∈ N such that x∞ = |x(j0 )|. Then 1. If α ≤ 1:
xDα = x∞ + α sup x(i) : sgn x(i) = − sgn x(j0 ) , where we define sup ∅ = 0. 2. If α > 1 then
xDα = αx∞ + sup x(i) : sgn x(i) = − sgn x(j0 ) . We will show that in this space the class of directionally bounded sets is different to the corresponding class in (c0 , · ∞ ). To do so we will use Theorem 5, so we need to know the dual space of (c0 , · Dα ) and the extreme points of the unit ball of its dual space. It is straightforward to prove the following. Theorem 15. The dual space of (c0 , · Dα ) is the space (1 , · Dα∗ ), where for α ≤ 1 and f = an e∗n ∈ l1 , f Dα∗ = max and for α ≥ 1 and f =
a+ i ,
a− i ,
1 |ai | , 1+α
an e∗n ∈ l1 ,
f Dα∗
1 + 1 − 1 = max ai , ai , |ai | . α α 1+α
− Here a+ i = max{ai , 0} and ai = max{−ai , 0}.
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The sets of extreme points of S(1 ,·D∗ ) are given by: α If α ≤ 1: {±e∗n n ∈ N and for m, n ∈ N, m = n, (e∗m − αe∗n )}. If α ≥ 1: {±αe∗n n ∈ N and for m, n ∈ N, m = n, ±(αe∗m − e∗n )}. Similarly to Proposition 8, we have the following result. Theorem 16. Let {xn } be a directional sequence in (c0 , · Dα ). Then {xn } contains a subsequence which is a directional sequence in (co , · ∞ ). Therefore every directionally bounded set in (co , · ∞ ) is directionally bounded in (co , · Dα ). Proof. Assume first that α ≤ 1. Let {xn } be a directional sequence in (co , · Dα ), write xn ∞ = ln and ln = |xn (in )|. Let jn = max{|xn (i)| : xn (i) = − sgn(xn (in ))}. Then xn Dα = ln + αjn . For r < n let
inr = min i : xr − xn ∞ = (xr − xn )(i) , lrn = xn − xr ∞ and
jrn = sup (xn − xr )(j) : sgn (xn − xr )(j) = − sgn (xn − xr ) inr . Let n1 = 1, it is easy to see that there exists a subsequence S1 ⊂ N\{1}, such that sgn((xn1 − xn )(inn1 )) = θ1 for every n ∈ S1 . Take n2 = min S1 and S2 ⊂ S1 \{n2 } such that sgn((xn2 − xn )(inn2 )) = θ2 is constant for n ∈ S2 . Having constructed n1 < n2 < · · · < nk and sequences S1 ⊃ S2 ⊃ · · · ⊃ Sk such that sgn((xnj − xn )(innj )) = θj is constant for n ∈ Sj , take nk+1 = min Sk and Sk+1 ⊂ Sk \{nk+1 } such that θk+1 = sgn((xnk+1 − xn )(innk+1 )) is constant for n ∈ Sk+1 . By passing to a subsequence and changing {xn } to {−xn } if necessary, we may assume that for n > m, sgn((xm −xn )(inm )) = 1. Since {xn } is a directional sequence, there exist b ≥ 0 such that for any subsequence {xnk } and for all s ∈ N: s−1
xni − xni+1 Dα − xns − xn1 Dα ≤ b.
i=1
Let x = {x(n)} ∈ c0 , define ix ∈ N and jx by x∞ = |x(ix )| and jx = sup{|x(j)| : sgn(x(j)) = − sgn x(ix )}. Then, if x(ix ), y(iy ) > 0 we get that jx+y ≤ jx + jy . To prove this, assume first that x + y∞ = −(x + y)(ix+y ). This implies that either x(ix+y ) ≤ 0 in which case −x(ix+y ) ≤ jx , or y(ix+y ) ≤ 0 in which case −y(ix+y ) ≤ jy . Also, jx+y = sup{(x + y)(i) ≥ 0}, but for all i such that (x + y)(i) ≥ 0 we have 0 ≤ (x + y)(i) ≤ jx+y ≤ −(x + y)(ix+y ) ≤ jx + jy . The case where x + y∞ = (x + y)(ix+y ) is treated similarly. Using this for x = xnk − xnk+1 and y = xnk+1 − xnk+2 we obtain by induction that: s−1 i=1
jnnii+1 ≥ jnns1 .
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Thus s−1
jnnkk+1 − jnn1s ≥ 0
k=1
s−1
s−1
nj+1 j=1 lnj
Recall that j=1 xnj − xnj+1 ∞ − xn1 − xns ∞ = for this expression as follows. s−1
lnnjj+1
−
lnn1s
≤
j=1
s−1
lnnjj+1
−
lnn1s
+α
s−1
j=1
=
s−1
− lnn1s . We can obtain an upper bound
jnnkk+1
−
jnn1s
k=1
xnj − xnj+1 Dα − xn1 − xns Dα
j=1
≤b which proves that {xn } is a directional sequence in (c0 , · ∞ ). The case for α > 1 is similar. 2 So far, the results for the · Dα norm are similar to those of the · α norm. But in contrast to this, we have the following result. Theorem 17. Let 0 < α. There exists a directionally bounded set C ⊂ c0 with respect to the norm · Dα which is not directionally bounded with respect to the norm · ∞ . Proof. Let x1 = e1 and for n ≥ 2 n xn = ne1 − 2
1 1 1 1− e2 + ... + 1 − k−1 ek + ... + 1 − n−1 en . 2 2 2
Let C = conv{xn }. Clearly C is not directionally bounded in (c0 , · ∞ ) since xn ∞ = n and lim e∗1 n Now let yn =
mn j=1
(n)
λj xj with
mn
limn→∞ yn Dα = ∞. Note that yn ∞ mn (n) jλj → ∞. norms, this implies that j=1 For mn ≥ k > 1,
xn xn ∞
= 1.
(n)
(n)
= 1, mn < mn+1 and λj ≥ 0. Assume that mn (n) = yn (1) = j=1 jλj . As · Dα and · ∞ are equivalent
j=1
λj
mn 1 1 (n) 1 − k−1 , yn (k) = − jλj 2 2 j=k
and for k > mn , yn (k) = 0. Hence, if k ≤ mn then mn k−1 (n) (n) 1 1 1 1 yn (k) 1 j=k jλj j=1 jλj 2 2 1 − = 1 − . = m − − mn (n) (n) n yn (1) 2k−1 2 2k−1 jλ jλ j=1
j
Thus for a fixed k with 1 < k ≤ mn ,
j=1
j
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1 1 −yn (k) = 1 − k−1 . lim n→∞ yn (1) 2 2 Observe that since yn is finitely supported, for every n ∈ N there exists kn ≤ mn such that for every 1 < k ≤ mn , |yn (k)| ≤ |yn (kn )|, that is: sup k>1
mn mn 1 1 1 1 (n) (n) 1 − k−1 = 1 − kn −1 . jλj jλj 2 2 2 2 j=k
j=kn
Therefore for 1 ≤ k ≤ mn , 1 1 1 ≥ 1 − kn −1 2 2 2 |yn (kn )| yn (1) k−1 (n) 1 1 1 j=1 jλj 1 − . − 2m = (n) n 2 2k−1 jλ j j=1 ≥
By passing to the limit as n → ∞, since mn → ∞, we get lim n
1 |yn (kn )| = . yn (1) 2
We know divide the proof in two cases. Case α ≤ 1. In this case yn Dα = yn (1) + α|yn (kn )|. Recall that the extreme points of the unit ball of the dual space in this case are ±e∗k and for m = k, ±(e∗m − αe∗k ).
∗ yn
= |yn (k)|
ek
yn Dα yn Dα =
1
|yn (k)| yn (1) n )| + α |yynn(k(1)
.
Hence:
yn
= lim
e∗k n yn D
1 − 2k−1 ) < 1. α 1+ 2
1 2 (1
α
e∗1
Now we look at the functionals e∗m − αe∗k . Since |e∗k (yn )| ≤ yn (1) it is enough to focus on the functionals − αe∗j with j > 1. Then:
e∗1 − αe∗k
yn yn Dα
=
=
and
yn (1) − αyn (k) yn (1) − αyn (kn ) 1 − α yynn(k) (1) n) 1 − α yynn(k(1)
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lim e∗1 − αe∗k
n
yn yn Dα
=
1 1 + α2 (1 − 2k−1 ) α 1+ 2
< 1. Thus by Theorem 5 we obtain that C is directionally bounded. Case α ≥ 1. In this case yn Dα = αyn (1) + |yn (kn )|. The extreme points in this case are ±αe∗k and for m = k, ±(αe∗m − e∗k ). Notice that
∗ α|yn (k)| yn
=
αek
yn Dα αyn (1) + |yn (kn )| =
α |yynn(k)| (1) α+
|yn (kn )| yn (1)
Thus
lim
αe∗k n
1
α 12 (1 − 2k−1 ) yn
= < 1. 1
yn Dα α+ 2
As before it is enough to check for the functionals (αe∗1 − e∗j ) with j > 1. Then
αe∗1
−
e∗k
yn yn Dα
=
=
αyn (1) − yn (k) αyn (1) − yn (kn ) α− α−
yn (k) yn (1) yn (kn ) yn (1)
and lim αe∗1 − e∗k n
yn yn Dα
=
1 α + 12 (1 − 2k−1 ). 1 α+ 2
< 1. Thus, by Theorem 5, for any α > 0, C is a directionally bounded set for the norm · Dα .
2
Corollary 18. There are spaces which are arbitrarily close to (c0 , · ∞ ) in the Banach–Mazur sense, and the respective classes of directionally bounded sets are different. Proof. d((c0 , · Dα ), (c0 , · ∞ )) ≤ 1 + α.
2
Corollary 19. There are sets in c0 which for every α > 0 have the AFPP with respect to the norm · Dα , but not with respect to · ∞ . Finally we conclude the following. Theorem 20. If X is a nonreflexive space then directionally boundedness is not necessarily preserved by isomorphisms.
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Acknowledgment This paper has been partially supported by SEP-Conacyt Grant 102380. References [1] T. Domínguez Benavides, The failure of the fixed point property for unbounded sets in c0 , Proc. Amer. Math. Soc. 140 (2012) 645–650. [2] B. Gamboa de Buen, F. Núñez Medina, The fixed point property in c0 with an equivalent norm, Abstr. Appl. Anal. 2011 (2011), http://dx.doi.org/10.1155/2011/574614, Article ID 574614, 19 pages. [3] I. Shafrir, The approximate fixed point property in Banach and hyperbolic spaces, Israel J. Math. 71 (1990) 211–223.
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Contents lists available at ScienceDirect
Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa
Directionally bounded sets in c0 with equivalent norms Francisco Eduardo Castillo Santos a,1 , Helga Fetter a,∗,1 , Berta Gamboa de Buen a,1 , Fernando Núñez-Medina b a b
Centro de Investigación en Matemáticas (CIMAT), Apdo. Postal 402, 36000 Guanajuato, Gto., Mexico Departamento de Matemáticas, Universidad de Guanajuato, CP 36240, Guanajuato, Gto., Mexico
a r t i c l e
i n f o
Article history: Received 2 December 2013 Available online xxxx Submitted by T. Domínguez Benavides
a b s t r a c t We study the directionally bounded sets in c0 with respect to two different families of norms and conclude that in nonreflexive spaces directional boundedness is not necessarily preserved by isomorphisms. © 2014 Published by Elsevier Inc.
Keywords: Directionally bounded sets Directional sequences AFPP
1. Introduction In [3], the author defined the concept of directionally bounded set of a Banach space X. In the same article the author proved that a Banach space X is reflexive if and only if a closed convex linearly bounded set is directionally bounded. Thus if X is a reflexive space, the class of directionally bounded sets does not change with an equivalent norm. Therefore we ask if the same result holds on a nonreflexive Banach space. Explicitly, does the class of directionally bounded sets of a nonreflexive Banach space X remain the same when we equip X with an equivalent norm? In this paper we show that in c0 we can find two different families of equivalent norms, such that in one case the class of directionally bounded sets is the same as in c0 and in the other it is different. This result not only shows that directional boundedness in a nonreflexive space depends on the equivalent norm given, but also by a result of Shafrir [3] the same holds for the AFPP. Here the AFPP is defined as follows. * Corresponding author. E-mail addresses: [email protected] (F.E. Castillo Santos), [email protected] (H. Fetter), [email protected] (B. Gamboa de Buen), [email protected] (F. Núñez-Medina). 1 Fax: +52 473 7325749. http://dx.doi.org/10.1016/j.jmaa.2014.05.006 0022-247X/© 2014 Published by Elsevier Inc.
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Definition 1. Let X be a Banach space and C ⊂ X be closed and convex. We will say that C has the approximate fixed point property, AFPP for short, if for all nonexpansive maps T : C → C we have inf T (x) − x : x ∈ C = 0. The following definitions appeared in [3]. Definition 2. Let (X, · ) be a Banach space and {xn } ⊂ X. We say that the sequence {xn } is a directional sequence, if limn→∞ xn = ∞ and there exists b ≥ 0 such that for every n1 < n2 < · · · < ns , s ∈ N we have xn1 − xns ≥
s−1
xni+1 − xni − b.
i=1
Definition 3. Let X be a Banach space and C ⊂ X be a closed and convex set. We will say that C is directionally bounded if it does not contain a directional sequence. Shafrir [3] proved the following two theorems. Theorem 4. Let X be a Banach space and C ⊂ X. Then C has the AFPP if and only if it is directionally bounded. Theorem 5. Let X be a Banach space and C ⊂ X, closed, convex and unbounded. Then C is directionally bounded if for every {xn } ⊂ C with limn→∞ xn = ∞ and for every extreme point f of SX ∗ we have that
xn lim sup f n→∞ xn
< 1.
2. The space (c0 )α Recall that c0 is the space of real null sequences and · ∞ denotes its usual norm. The sequences {en }, {e∗n } will denote the canonical bases of c0 and 1 respectively. We will define the following family of norms in c0 . Definition 6. Let α > 0. For x = {x(n)} ∈ c0 define ∞ |x(n)| xα = x∞ + α . 2n n=1
It is clear that these norms are equivalent to the usual c0 norm. In [2] it is shown that, for any α > 0, the dual space of (co , · α ) is the space (1 , · α∗ ). Here the norm ∞ · α∗ is given for h = n=1 cn e∗n ∈ 1 by hα∗ = sup
n∈F
1+α
|cn |
1 n∈F 2n
:F ⊂N .
We are interested in studying the class of directionally bounded sets with respect to these norms. In order to do so, we need to look at directional sequences. However, as the next lemma shows, this is not enough to determine the directionally bounded sets.
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Lemma 7. There are nondirectional sequences in (c0 , · α ) with a nondirectionally bounded convex hull. Proof. Let xn = 2n+1 e1 + 2n en+1 . Clearly xn ∞ = 2n+1 and for any n, k ∈ N, xn − xk ∞ = |2n+1 − 2k+1 |. Thus for any n1 < n2 < · · · < ns : s−1
xni+1 − xni ∞ =
i=1
s−1
2ni+1 +1 − 2ni +1
i=1
= 2ns +1 − 2n1 +1 = xn1 − xns ∞ . Taking b = 0, we obtain that {xn } is a directional sequence in (c0 , · ∞ ). On the other hand: xn − xk α = 2n+1 − 2k+1 e1 + 2n en+1 − 2k ek+1 α
1 1 α
n+1 2 + . = 1+ − 2k+1 + α 2 2 2 From this, s−1 i=1
xni+1
s−1 ni+1 +1 α ni +1 + (s − 1)α − xn i α = 2 −2 1+ 2 i=1 ns +1 α n1 +1 = 2 + (s − 1)α −2 1+ 2 = xns − xn1 α + (s − 2)α.
For a fixed b, the inequality b ≥ (s − 2)α, cannot hold for all s ∈ N. Hence {xn } is not a directional sequence in (c0 , · α ). Now let 1 1 i+1 1 i xi = 2 e1 + 2 ei+1 . n i=1 n i=1 n i=1 n
yn =
n
n
If n > m, yn − ym =
1 1 − n m
+
m
2i+1 e1 +
i=1
1 1 − n m
m
n 1 i+1 2 e1 n i=m+1
2i ei+1 +
i=1
n 1 i 2 ei+1 . n i=m+1
Thus n 1 α yn − ym α = 1 + 2i+1 − 2 n i=1 α 4 n = 1+ 2 −1 − 2 n
and
1 i+1 2 m i=1 m
α 1 α1 − n i=1 2 m i=1 2 n
+
4 m 2 −1 m
m
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s−1
ynj+1
j=1
4 ns α 4 n1 2 −1 − 2 −1 − ynj α = 1 + 2 ns n1 = yns − yn1 α .
Therefore {yn } is a directional sequence in C = conv{xn } in (c0 , · α ). Since we also proved that {xn } ⊂ C is a directional sequence in (c0 , · ∞ ), C is the desired convex set. 2 The proof of the previous result also shows that the families of directional sequences of the spaces (c0 , · ∞ ) and (c0 , · α ) are different, however we will see later that the classes of directionally bounded sets are the same. The following three propositions are crucial for the main theorem of this section. Proposition 8. Every directional sequence in (c0 , ·α ) is also a directional sequence in (c0 , ·∞ ). Therefore if C ⊂ c0 is directionally bounded in (c0 , · ∞ ), it is also directionally bounded in (c0 , · α ). Proof. Let {xn } be a directional sequence in (c0 , · α ). Take b ≥ 0 as given by the definition of directional sequence. Let n1 < n2 < · · · < ns . Then b≥
s−1
xni − xni+1 α − xns − xn1 α
n=1
=
s−1
xni − xni+1 ∞ − xns − xn1 ∞
n=1
s−1 ∞ ∞ |xni+1 (m) − xni (m)| |xn1 (m) − xns (m)| +α − . 2m 2m m=1 i=1 m=1
By the triangle inequality: s−1 ∞
1
xn (m) − xni (m) − xn1 (m) − xns (m) ≥ 0. 2m i=1 i+1 m=1 From this: s−1
xni+1 − xni ∞ − xn1 − xns ∞ ≤ b.
2
i=1
The next proposition is similar to Lemma 2.2 of [1]. Proposition 9. Let C be a convex unbounded set of c0 . Assume that C is directionally bounded in (c0 , · α ). Let {xn } ⊂ C be a sequence such that limn→∞ xn ∞ = ∞, write xn = {xn (k)}∞ k=1 . Then for every n0 , k0 ∈ N, there exist n > n0 and k > k0 such that xn ∞ = |xn (k)|. Proof. Recall that for every x = {x(k)} ∈ c0 there exists k ∈ N such that x∞ = |x(k)|. If the proposition were false, there would exist n0 , k0 ∈ N such that for every n > n0 , if k is such that xn ∞ = |xn (k)| then k ≤ k0 . Since {xn } is a directional sequence if and only if {−xn } is a directional sequence, by passing to a subsequence, we may assume that there exists k ≤ k0 such that for all n ∈ N we have xn ∞ = xn (k). Without loss of generality we will assume that k = 1, that is, xn ∞ = xn (1).
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Let n1 = 1 and θ1 = 1. We may choose a subsequence S1 ⊂ N = S0 such that for all n ∈ S1 we have that sgn(xn (2)) = θ2 is constant. Take n2 = min(S1 \{n1 }) and a subsequence S2 ⊂ S1 such that for n ∈ S2 , sgn(xn (3)) = θ3 is constant. If for j = 1, ..., k we have nj ∈ Sj−1 , n1 < n2 < ... < nk and sequences S0 ⊃ S1 ⊃ ... ⊃ Sk so that for n ∈ Sj , we have that (xn (j + 1)) = θj+1 is constant, take nk+1 = min Sk \{n1 , ..., nk } and a subsequence Sk+1 ⊂ Sk such that for n ∈ Sk+1 , sgn(xn (k + 2)) = θk+2 is constant. By taking the subsequence {xni }, we may assume that {xn } satisfies that for n ≥ j, sgn(xn (j)) depends only on j. ∞ For x = {x(n)} ∈ c0 take f (x) = θ1 x(1) + α i=1 θi2x(i) . i Thus f α∗ = 1 and
xn f xn α
=
|xn (1)| + α
=1+
α |xn (1)|
n j=1
|xn (j)| 2j
+α
∞
j=n+1 ∞ α j=1 |xn2(j)| j
θj xn (j) 2j
|xn (1)| + ∞ θj xn (j)
∞ − |xnα(1)| j=n+1 ∞ 1 + |xn1(1)| α j=1 |xn2(j)| j
j=n+1
2j
|xn (j)| 2j
.
But since |xn (1)| ≥ |xn (j)| for all j ∈ N: 0≤
∞ ∞ ∞ α α |xn (j)| θj xn (j) 1 − ≤ 2α . j |xn (1)| j=n+1 2j |xn (1)| j=n+1 2j 2 j=n+1
Thus limn→∞ f ( xxnnα ) = 1, contradicting that C is a directionally bounded set. 2 Proposition 10. Let {xn } ⊂ c0 be a sequence such that limn→∞ xn ∞ = ∞. If for every n0 , k0 ∈ N there exist n > n0 and k > k0 such that xn ∞ = |xn (k)|, then {xn } is not a directional sequence in (c0 , · ∞ ). Proof. It is easy to see that if {zn } is a directional sequence and {yn } is a sequence such that zn − yn is s n sufficiently small then {yn } is also a directional sequence. Hence we can also assume that xn = i=1 xn (i)ei where {sn } is a strictly increasing sequence of natural numbers, with xn ∞ = xn (in ) > xn−1 (in−1 ) ≥ 0, in < sn < in+1 . Now let n > m, then xn − xm ∞ ≥ xn (in ), s−1
xn+1 − xn ∞ − xs − x1 ∞ ≥
n=1
s−1
xj+1 (ij+1 ) − xs (is ) − x1 (i1 )
j=1
=
s−2
xj+1 (ij+1 ) − x1 (i1 )
j=1
and lims→∞
s−2 j=0
xj+1 (ij+1 ) = ∞.
2
Now we present the main theorem of this section. Theorem 11. Let C be a convex, closed and unbounded subset of c0 . Then C is directionally bounded in (co , · ∞ ) if and only if it is directionally bounded in (co , · α ). Proof. By Proposition 8 it is enough to show that if C is directionally bounded in (c0 , · α ) then it is also directionally bounded in (c0 , · ∞ ).
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To prove this, let {xn } ⊂ C be a sequence such that limn→∞ xn ∞ = ∞, by Proposition 9, for every n0 , k0 ∈ N, there exist n > n0 and k > k0 such that xn ∞ = |xn (k)| and by Proposition 10, {xn } is not directional in (c0 , · ∞ ). 2 Corollary 12. Let C be a convex closed subset of c0 . Then C has the AFPP in (c0 , · α ) if and only if C has the AFPP in (c0 , · ∞ ). Proof. It is known that bounded closed convex sets have the AFPP. If C is unbounded we obtain the result by Theorem 11 and Theorem 4. 2 Corollary 13. The norm · ∞ is the starting point of a ray of equivalent norms in c0 such that the respective classes of directionally bounded sets are the same as in (c0 , · ∞ ). 3. The space (c0 , · Dα ) We start by defining the norms · Dα . Definition 14. Let α > 0. For x = {x(n)} ∈ c0 define
xDα = sup x(n) − αx(m) : n = m . It is clear that these norms are equivalent to the usual c0 norm. Observe the following: For x ∈ c0 choose j0 ∈ N such that x∞ = |x(j0 )|. Then 1. If α ≤ 1:
xDα = x∞ + α sup x(i) : sgn x(i) = − sgn x(j0 ) , where we define sup ∅ = 0. 2. If α > 1 then
xDα = αx∞ + sup x(i) : sgn x(i) = − sgn x(j0 ) . We will show that in this space the class of directionally bounded sets is different to the corresponding class in (c0 , · ∞ ). To do so we will use Theorem 5, so we need to know the dual space of (c0 , · Dα ) and the extreme points of the unit ball of its dual space. It is straightforward to prove the following. Theorem 15. The dual space of (c0 , · Dα ) is the space (1 , · Dα∗ ), where for α ≤ 1 and f = an e∗n ∈ l1 , f Dα∗ = max and for α ≥ 1 and f =
a+ i ,
a− i ,
1 |ai | , 1+α
an e∗n ∈ l1 ,
f Dα∗
1 + 1 − 1 = max ai , ai , |ai | . α α 1+α
− Here a+ i = max{ai , 0} and ai = max{−ai , 0}.
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The sets of extreme points of S(1 ,·D∗ ) are given by: α If α ≤ 1: {±e∗n n ∈ N and for m, n ∈ N, m = n, (e∗m − αe∗n )}. If α ≥ 1: {±αe∗n n ∈ N and for m, n ∈ N, m = n, ±(αe∗m − e∗n )}. Similarly to Proposition 8, we have the following result. Theorem 16. Let {xn } be a directional sequence in (c0 , · Dα ). Then {xn } contains a subsequence which is a directional sequence in (co , · ∞ ). Therefore every directionally bounded set in (co , · ∞ ) is directionally bounded in (co , · Dα ). Proof. Assume first that α ≤ 1. Let {xn } be a directional sequence in (co , · Dα ), write xn ∞ = ln and ln = |xn (in )|. Let jn = max{|xn (i)| : xn (i) = − sgn(xn (in ))}. Then xn Dα = ln + αjn . For r < n let
inr = min i : xr − xn ∞ = (xr − xn )(i) , lrn = xn − xr ∞ and
jrn = sup (xn − xr )(j) : sgn (xn − xr )(j) = − sgn (xn − xr ) inr . Let n1 = 1, it is easy to see that there exists a subsequence S1 ⊂ N\{1}, such that sgn((xn1 − xn )(inn1 )) = θ1 for every n ∈ S1 . Take n2 = min S1 and S2 ⊂ S1 \{n2 } such that sgn((xn2 − xn )(inn2 )) = θ2 is constant for n ∈ S2 . Having constructed n1 < n2 < · · · < nk and sequences S1 ⊃ S2 ⊃ · · · ⊃ Sk such that sgn((xnj − xn )(innj )) = θj is constant for n ∈ Sj , take nk+1 = min Sk and Sk+1 ⊂ Sk \{nk+1 } such that θk+1 = sgn((xnk+1 − xn )(innk+1 )) is constant for n ∈ Sk+1 . By passing to a subsequence and changing {xn } to {−xn } if necessary, we may assume that for n > m, sgn((xm −xn )(inm )) = 1. Since {xn } is a directional sequence, there exist b ≥ 0 such that for any subsequence {xnk } and for all s ∈ N: s−1
xni − xni+1 Dα − xns − xn1 Dα ≤ b.
i=1
Let x = {x(n)} ∈ c0 , define ix ∈ N and jx by x∞ = |x(ix )| and jx = sup{|x(j)| : sgn(x(j)) = − sgn x(ix )}. Then, if x(ix ), y(iy ) > 0 we get that jx+y ≤ jx + jy . To prove this, assume first that x + y∞ = −(x + y)(ix+y ). This implies that either x(ix+y ) ≤ 0 in which case −x(ix+y ) ≤ jx , or y(ix+y ) ≤ 0 in which case −y(ix+y ) ≤ jy . Also, jx+y = sup{(x + y)(i) ≥ 0}, but for all i such that (x + y)(i) ≥ 0 we have 0 ≤ (x + y)(i) ≤ jx+y ≤ −(x + y)(ix+y ) ≤ jx + jy . The case where x + y∞ = (x + y)(ix+y ) is treated similarly. Using this for x = xnk − xnk+1 and y = xnk+1 − xnk+2 we obtain by induction that: s−1 i=1
jnnii+1 ≥ jnns1 .
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Thus s−1
jnnkk+1 − jnn1s ≥ 0
k=1
s−1
s−1
nj+1 j=1 lnj
Recall that j=1 xnj − xnj+1 ∞ − xn1 − xns ∞ = for this expression as follows. s−1
lnnjj+1
−
lnn1s
≤
j=1
s−1
lnnjj+1
−
lnn1s
+α
s−1
j=1
=
s−1
− lnn1s . We can obtain an upper bound
jnnkk+1
−
jnn1s
k=1
xnj − xnj+1 Dα − xn1 − xns Dα
j=1
≤b which proves that {xn } is a directional sequence in (c0 , · ∞ ). The case for α > 1 is similar. 2 So far, the results for the · Dα norm are similar to those of the · α norm. But in contrast to this, we have the following result. Theorem 17. Let 0 < α. There exists a directionally bounded set C ⊂ c0 with respect to the norm · Dα which is not directionally bounded with respect to the norm · ∞ . Proof. Let x1 = e1 and for n ≥ 2 n xn = ne1 − 2
1 1 1 1− e2 + ... + 1 − k−1 ek + ... + 1 − n−1 en . 2 2 2
Let C = conv{xn }. Clearly C is not directionally bounded in (c0 , · ∞ ) since xn ∞ = n and lim e∗1 n Now let yn =
mn j=1
(n)
λj xj with
mn
limn→∞ yn Dα = ∞. Note that yn ∞ mn (n) jλj → ∞. norms, this implies that j=1 For mn ≥ k > 1,
xn xn ∞
= 1.
(n)
(n)
= 1, mn < mn+1 and λj ≥ 0. Assume that mn (n) = yn (1) = j=1 jλj . As · Dα and · ∞ are equivalent
j=1
λj
mn 1 1 (n) 1 − k−1 , yn (k) = − jλj 2 2 j=k
and for k > mn , yn (k) = 0. Hence, if k ≤ mn then mn k−1 (n) (n) 1 1 1 1 yn (k) 1 j=k jλj j=1 jλj 2 2 1 − = 1 − . = m − − mn (n) (n) n yn (1) 2k−1 2 2k−1 jλ jλ j=1
j
Thus for a fixed k with 1 < k ≤ mn ,
j=1
j
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1 1 −yn (k) = 1 − k−1 . lim n→∞ yn (1) 2 2 Observe that since yn is finitely supported, for every n ∈ N there exists kn ≤ mn such that for every 1 < k ≤ mn , |yn (k)| ≤ |yn (kn )|, that is: sup k>1
mn mn 1 1 1 1 (n) (n) 1 − k−1 = 1 − kn −1 . jλj jλj 2 2 2 2 j=k
j=kn
Therefore for 1 ≤ k ≤ mn , 1 1 1 ≥ 1 − kn −1 2 2 2 |yn (kn )| yn (1) k−1 (n) 1 1 1 j=1 jλj 1 − . − 2m = (n) n 2 2k−1 jλ j j=1 ≥
By passing to the limit as n → ∞, since mn → ∞, we get lim n
1 |yn (kn )| = . yn (1) 2
We know divide the proof in two cases. Case α ≤ 1. In this case yn Dα = yn (1) + α|yn (kn )|. Recall that the extreme points of the unit ball of the dual space in this case are ±e∗k and for m = k, ±(e∗m − αe∗k ).
∗ yn
= |yn (k)|
ek
yn Dα yn Dα =
1
|yn (k)| yn (1) n )| + α |yynn(k(1)
.
Hence:
yn
= lim
e∗k n yn D
1 − 2k−1 ) < 1. α 1+ 2
1 2 (1
α
e∗1
Now we look at the functionals e∗m − αe∗k . Since |e∗k (yn )| ≤ yn (1) it is enough to focus on the functionals − αe∗j with j > 1. Then:
e∗1 − αe∗k
yn yn Dα
=
=
and
yn (1) − αyn (k) yn (1) − αyn (kn ) 1 − α yynn(k) (1) n) 1 − α yynn(k(1)
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lim e∗1 − αe∗k
n
yn yn Dα
=
1 1 + α2 (1 − 2k−1 ) α 1+ 2
< 1. Thus by Theorem 5 we obtain that C is directionally bounded. Case α ≥ 1. In this case yn Dα = αyn (1) + |yn (kn )|. The extreme points in this case are ±αe∗k and for m = k, ±(αe∗m − e∗k ). Notice that
∗ α|yn (k)| yn
=
αek
yn Dα αyn (1) + |yn (kn )| =
α |yynn(k)| (1) α+
|yn (kn )| yn (1)
Thus
lim
αe∗k n
1
α 12 (1 − 2k−1 ) yn
= < 1. 1
yn Dα α+ 2
As before it is enough to check for the functionals (αe∗1 − e∗j ) with j > 1. Then
αe∗1
−
e∗k
yn yn Dα
=
=
αyn (1) − yn (k) αyn (1) − yn (kn ) α− α−
yn (k) yn (1) yn (kn ) yn (1)
and lim αe∗1 − e∗k n
yn yn Dα
=
1 α + 12 (1 − 2k−1 ). 1 α+ 2
< 1. Thus, by Theorem 5, for any α > 0, C is a directionally bounded set for the norm · Dα .
2
Corollary 18. There are spaces which are arbitrarily close to (c0 , · ∞ ) in the Banach–Mazur sense, and the respective classes of directionally bounded sets are different. Proof. d((c0 , · Dα ), (c0 , · ∞ )) ≤ 1 + α.
2
Corollary 19. There are sets in c0 which for every α > 0 have the AFPP with respect to the norm · Dα , but not with respect to · ∞ . Finally we conclude the following. Theorem 20. If X is a nonreflexive space then directionally boundedness is not necessarily preserved by isomorphisms.
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Doctopic: Real Analysis
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Acknowledgment This paper has been partially supported by SEP-Conacyt Grant 102380. References [1] T. Domínguez Benavides, The failure of the fixed point property for unbounded sets in c0 , Proc. Amer. Math. Soc. 140 (2012) 645–650. [2] B. Gamboa de Buen, F. Núñez Medina, The fixed point property in c0 with an equivalent norm, Abstr. Appl. Anal. 2011 (2011), http://dx.doi.org/10.1155/2011/574614, Article ID 574614, 19 pages. [3] I. Shafrir, The approximate fixed point property in Banach and hyperbolic spaces, Israel J. Math. 71 (1990) 211–223.