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Automatic Complete Boundedness
Completely bounded linear and multilinear mappings share
strong structural properties. Thanks to the complete boundedness
they have a very specific form
(cf. the corresponding representation theorems )
whence they are much more accessible than arbitrary bounded
linear (or multilinear) mappings.57
Moreover, even in the multilinear case we have at our disposal an
extension theorem
for completely bounded mappings -
which again is in striking contrast to the situation of arbitrary bounded (multi)linear mappings.
Because of the very nice structure theory
of completely bounded linear and multilinear mappings, respectively, it is highly interesting to
decide whether or not
a given bounded (multi)linear mapping actually is completely bounded.
The most elegant way to proceed is of course to check some simple conditions concerning the spaces involved and/or
some (purely) algebraic properties of the mapping which automatically imply the complete boundedness
of the latter.
In the following we shall collect various such criteria relying
- (1)
- on the initial and/or target space
- (2)
- mainly on algebraic properties of the mapping
(being, e.g., a
-homomorphism of
-algebras or a module homomorphism).
- (1)
- ``Criterion: spaces''
- (1.1)
- The linear case
-
Smith's lemma:
If
is a matricially normed space and
, where
, is a
bounded linear operator, then we have
.
In particular,
is completely bounded if and only if
is bounded
[Smi83, Thm. 2.10].
- Let
be an operator space,
a commutative
-algebra and
a bounded linear operator.
Then
is completely bounded
and
([Loe75, Lemma 1], cf. also [Arv69, Prop. 1.2.2]).
In particular, every bounded linear functional
is automatically completely bounded
with the same cb-norm.
- The following theorem shows that, roughly speaking,
the above situations are the only ones
where every bounded linear operator between (arbitrary)
-algebras is automatically completely bounded. More precisely:
Let
and
be
-algebras. In order to have the complete boundedness of very bounded linear operator from
to
it is necessary and sufficient that either
is finite dimensional
or
is a
subalgebra of
for a compact Hausdorff space
[HT83, Cor. 4], [Smi83, Thm. 2.8], cf. also [Smi83, p. 163].
- (1.2)
- The bilinear case
- (2)
- ``Criterion: mappings''
- (2.1)
- The linear case
- Every
-homomorphism between
-algebras is
completely contractive.58
- Let
and
be
-subalgebras of
, where
is a Hilbert space.
Let further
be an
-operator module and
be a bounded
-module homomorphism.
Let
and
be
quasi-cyclic59, i.e., for every finite set of vectors
,
, there exist
such that
,
,
.
Then
is completely bounded and
([SS95, Thm. 1.6.1]; cf. also
[Smi91, Thm. 2.1] and [Smi91, Remark 2.2],
[Sat82, Satz 4.16]). - For special cases of this result obtained earlier see
[Haa80], [EK87, Thm. 2.5],
[PPS89, Cor. 3.3] and
[DP91, Thm. 2.4].
- (2.2)
- The bilinear case
Footnotes
- ... mappings.57
- For example, the representation theorems provide a very useful tool
in calculating cohomology groups
(cf. for example the monograph [SS95]).
- ... contractive.58
- This is evident because
is a
-homomorphism for every
.
- ...quasi-cyclic59
- For example, cyclic
-algebras
are quasi-cyclic.
A von Neumann algebra
is quasi-cyclic if
every normal state on the commutant
is a vector state
[Smi91, Lemma 2.3].
Next: Convexity
Up: What are operator spaces?
Previous: Completely Bounded Multilinear Mappings
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Prof. Gerd Wittstock
2001-01-07