Bohnenblust theorem
Consider the space ,
, and a measure space
. Since the norm
is
-additive, it is easily seen that the following condition is satisfied: For all
satisfying
,
,
and
(in the sense of disjoint support), one has
.
In [a2] H.F. Bohnenblust showed that the spaces are the only Banach lattices (cf. also Banach lattice) possessing this property; more precisely, he proved the following theorem, now known as the Bohnenblust theorem: Let
be a Banach lattice of dimension
satisfying
for all
such that
,
,
, and
. Then there exists a
,
, such that the norm on
is
-additive.
Here, for , a norm is said to be
-additive if
for all
with
; a norm is said to be
-additive, or, equivalently,
is said to be an
-space, if
for all
with
.
It should be noted that when , every Banach lattice with a
-additive norm is isometrically isomorphic to
, with
a suitable measure space. This representation theorem is essentially due to S. Kakutani [a3], who considered the case
; the proof of the more general result follows almost the same lines. For
the situation is not so clear: there exist many
-spaces that are not isomorphic to any concrete
-space, for instance
.
In the proof of his theorem, H.F. Bohnenblust introduced an interesting and tricky method to construct the such that the norm is
-additive. A similar method was used later by M. Zippin [a5] to characterize
-spaces in terms of bases. Since the proof given by Bohnenblust is interesting in itself, the main ideas are sketched below.
By hypothesis, there exists a function defined by
![]() |
whenever and
are disjoint vectors of norm one. It can easily be verified that the function
has the following properties:
1) ;
ii) ;
iii) ;
iv) ;
v) if
,
. The only non-trivial inclusion iv) follows from
![]() |
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for all disjoint of norm one and
with
.
One defines and
for all
. Property v) implies that the sequence
is increasing. By induction one obtains
and
. If
, then properties i)–v) easily imply
for all
. Hence,
is an
-space.
Assume now that and let
. For all
there exists a
such that
. Since
is an increasing sequence, one concludes from
that
or, equivalently, that
. This yields
![]() |
Letting ,
![]() |
It is clear that does not depend on the special choice of
. Moreover,
for all
. Since
, it follows that
for all
. Consequently,
![]() |
From for all
it follows that
. This completes the proof.
Bohnenblust's theorem has some interesting consequences. For instance, T. Ando [a1] used it to prove that a Banach lattice is isometrically isomorphic to
for some measure space
, or to some
, if and only if every closed sublattice of
is the range of a positive contractive projection.
References
[a1] | T. Ando, "Banachverbände und positive Projektionen" Math. Z. , 109 (1969) pp. 121–130 |
[a2] | H.F. Bohnenblust, "An axiomatic characterization of ![]() |
[a3] | S. Kakutani, "Concrete representation of abstract ![]() |
[a4] | P. Meyer-Nieberg, "Banach lattices" , Springer (1991) |
[a5] | M. Zippin, "On perfectly homogeneous bases in Banach spaces" Israel J. Math. , 4 A (1966) pp. 265–272 |
Bohnenblust theorem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Bohnenblust_theorem&oldid=50317