PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 78, Number 2, February 1980
r/c0
HAS NO EQUrVALENTSTRICTLYCONVEXNORM
J. BOURGAIN
Abstract. It is shown that the quotient space l°°/c0 does not admit an
equivalent strictly convex norm.
Introduction. We say that a normed space X, || || is strictly convex provided
||x + v|| < 2 whenever ||x|| = || v|| = 1 and x ^ y. This means also that any
member x* in the dual X* of X (x* ^ 0) attains its norm in at most one point
of the unit ball of X. A space is called strictly convexifiable if there exists an
equivalent strictly convex norm on the space.
It is known that ¡°° = /°°(N) is strictly convexifiable (cf. [1], [2]). However
the purpose of this note is to prove that the quotient /oo/c0 fails this property,
a problem raised independently by J. J. Schäffer and J. Diestel [3].
We will denote infinite subsets of the integers N by letters L, M, N, P. If L
is an infinite subset of N, let PX(L) be the set of all infinite subsets of L. Our
proof is based on the following elementary result.
Lemma. Suppose x* G (/°°)*, L G F^N) and e > 0. Then there is some
M G P^L) such that \x*(x)\ < e whenever x G /", ||x|| = 1 and x„ = 0 //
n G M.
Proof. We may, of course, assume ||x*|| = 1. Take an integer d > e~l and
disjoint members M¡(\ < i < d) of F0O(L).
If each M¡ fails the property, then we find elements x(,) (1 < / < d) in /°°,
so that:
(1) ||x<'>||- 1,
(2) x« = 0 if n G M¡, and
(3) x*(xw) > e.
Consider now the vector x = x(1) + x(2) + • • • +xid\
and x*(x) > d. This is the required contradiction.
We are now ready to prove the following theorem.
Obviously ||x|| = 1
Theorem. Let \\\ \\\ be an equivalent norm on l°°/c0. Then \\\ ||| is not strictly
convex.
Proof. Let Y = /°°/c0, ||| |]| and it: /°° -+ Y the quotient map be given.
Let (t„) be a sequence of positive numbers converging to 0. We make the
following construction:
Take F, = (x G /°°; ||x|| < 1} and sx =
sup{|||77(x)|||; x G F,}. Let x(1) G F, be such that |||t7(x(1))||| > sx - e, and
Received by the editors December 21, 1978.
AMS (MOS) subject classifications (1970). Primary 46B05.
© 1980 American
Mathematical
0002-9939/80/0000-0065/$01.50
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J. BOURGAIN
consider^
G Y*, \\\y*\\\ = 1 v/ithy*w(xw) > sx - e,.
Since ir*(y*) G (/°°)*, application of the lemma provides Lx G PX(N) such
that bfw(x)| < e, if x G /", x < 1 and x„ = 0 for x G Lx. Take F2»{í6
F,; xn = x<° for « <? L,} and j2 = sup{|||ir(x)|||; x G F2}. Let x(2) G F2 be
such that |||7r(x(2))||| > s2 - e2 and consider y\ G Y*, |||^||| = 1 with
.y* "'(■*(2))> i2 - e2. Again by the lemma, we get L2 G PX(LX) such that
| v*w(x)| < e2 for x G /°°, x < 1 and x„ = 0 if n £ L2.
In general Fi+X = {x G F¡; xn = xjp for n G L,} and 5,+1 =
sup{||K*)|||;
x G F(+1}. Let x«+1> G Fí+I satisfy |||7r(x<'+1>)||| > sl+l - e(+1
and take y*+x G 7*, |||v,*+1||| = 1 with y*+xir(x<i+») > si+x - e,+1. By the
lemma, there is some Li+X G PX(L¡) so that | v,*+1tt(x)| <e,+1 if x G /°°,
x < 1 and xn = 0 for n £ L,+,.
Since for x G Fj+1 we have ||x — xw|| < 2 and xn — xjp = 0 for n & L¡, it
follows that \yfw{x - x(0)| < 2c, and thus yf-n(x) > vfir(x(/)) - 2e, > s¡ -
3c,
Clearly (/•)) is a decreasing sequence of nonvoid a(/°°, /') compact subsets
of /°°. Also (j() decreases and we let j = lim^^i,. Take some element x00 in
H ¡F¡ and let y* be a co*-cluster point of (y*). We consider the subset
S = C\ MF¡) of y. For a fixed y G S1,we find ||| v||| < s.
Since y G w(/^+1), it follows that y?(y) > s¡ — 3e, > s — 3t¡ for all i. Consequently .y*(.y) > s and thus v*(>') = s = ||| v|||. We show that 5 contains
more than one point. In particular, this will imply that s > 0 and ||| v*||| = 1.
So the proof will be complete. Because (L¡) is decreasing in PX(N), there is
some L G PX(N) with L\L¡ finite for all i. Assume x G /°°, ||x|| = 1 and
xn = x„°°for n £ L. It is possible to take 7r(x) ^ 7r(x°°) since L is infinite. We
claim that tr(x) G S. To see this, fix some / and remark that there is
x' G l°°, \\x'\\ = \, x'n = x„ for n G L\L¿ and x¿ = x„°° for « G L¡ (x' depends of course on i). Thus x¿ = x„°° = x^ if « G Lp for all/ = 1, . . . , i —
1. Now x' G F, and proceeding by induction we see that x' G Fj (1 < j < i).
Because L\L¡ is finite, ir(x) = tr(x') G tt(.FJ). Consequently w(x) G H ¡^(F,),
which is what must be obtained.
References
1. M. M. Day, Normed linear spaces, Ergebnisse der Mathematik und ihrer Grenzgebiete, vol.
21, Springer-Verlag,New York-Heidelberg, 1973.
2. J. Diestel, Geometry of Banach spaces.Selected topics, Lecture Notes in Math., vol. 485,
Springer-Verlag, Berlin and New York, 1976.
3. _,
Department
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