# Envelopes in Banach spaces

V. Ferenczi, J. Lopez-Abad

###### Abstract.

We define the notion of isometric envelope of a subspace in a Banach space,
and relate it to a) the mean ergodic projection on the space of fixed points
of a semigroup of contractions, b) results on Korovkin sets from the 70’s, and
c) extension properties of linear isometric embeddings. We use this concept to
address the recent conjecture that the Gurarij space and the spaces $L_{p}$,
$p\notin 2{\mathbb{N}}+4$ are the only separable Approximately
Ultrahomogeneous Banach spaces (a certain multidimensional transitivity of the
action of the linear isometry group on the space considered in [21]). The
similar conjecture for Fraïssé Banach spaces (a strenghtening of the
Approximately Homogeneous Property) is also considered. We characterize the
Hilbert space as the only separable reflexive space in which any closed
subspace coincides with its envelope. We compute some envelopes in the case of
Lebesgue spaces, showing that the reflexive $L_{p}$-spaces are the only
reflexive rearrangement invariant spaces on $[0,1]$ for which all
$1$-complemented subspaces are envelopes. We also identify the isometrically
unique ”full” quotient space of $L_{p}$ by a Hilbertian subspace, for
appropriate values of $p$, as well as the associated topological group
embedding of the unitary group into the isometry group of $L_{p}$.

###### Key words and phrases:

Isometries on $L_{p}$ spaces, contractive projection, Korovkin sets,
ultrahomogeneity

###### 2010 Mathematics Subject Classification:

Primary 41A35, 46A22, 46B04; Secondary, 46E30

V.F. was supported by FAPESP, project 2016/25574-8, and by CNPq, grant
303731/2019-2. J.L.-A. was partially supported by the Ministerio de Ciencia e
Innovación grant PID2019-107701GB-I00 (Spain) and Fapesp grant 2016/25574-8
(Brazil)

##  1\. Introduction and notation

###  1.1. The motivation

An initial motivation for this work stems from the the famous rotations
problem of Mazur. Assume the linear isometry group of a separable Banach space
$X$ acts transitively on its unit sphere; must $X$ be isometric to the Hilbert
space? (see for example [15] or the recent survey [17]). One may find many
examples of non-Hilbertian spaces which are almost transitive (AT), meaning
that the orbits of the action of the isometry group on the sphere are dense.
Striking examples of this are the Lebesgue spaces $L_{p}(0,1)$ for $1\leq
p<\infty$. For $p\neq 2$ those spaces, although not transitive, are very close
to being so. Indeed they admit exactly two orbits for the action of the
isometry group, the orbit of functions with full support, resp. without full
support. In some sense this suggests that the $L_{p}$ spaces are transitive
“with respect to the supports”, that is, provided one sees each function $f$
inside its “appropriate $L_{p}$-space” $L_{p}({\rm supp\ }f)$. The concept of
isometric envelope defined in the present paper initially aimed to imitate
some aspects of the support of a function, in an abstract setting which would
not require the structure of a function space.

The correct setting for defining the isometric envelope is multidimensional,
so let us recall a few facts. A Banach space $X$ is ultrahomogeneous when any
isometric map between finite dimensional subspaces extends to a (surjective)
isometry on $X$. A multidimensional version of Mazur’s rotation problem asks:
must any separable ultrahomogeneous Banach space be isometric to the Hilbert
space ([21]).

The separable Lebesgue spaces are also very close to being ultrahomogeneous,
although one has to exclude the even values of $p\geq 4$. For $p\notin
2{\mathbb{N}}+4$, $L_{p}(0,1)$ is approximately ultrahomogeneous
$\mathrm{(AuH)}$, meaning that any isometric map between finite dimensional
subspaces may be approximated as accurately as wished by a (surjective)
isometry (essentially due to Lusky [39]). The Gurarij space is also
$\mathrm{(AuH)}$ [35]. A conjecture formulated in [21] and which is our main
interest in this paper is the following: is every separable $\mathrm{(AuH)}$
space isometric either to some $L_{p}(0,1)$ or to the Gurarij space?

Again isometric maps between subspaces of the $\mathrm{(AuH)}$ $L_{p}$ extend
to isometries between respective appropriate $L_{p}$-subspaces “enveloping”
them, a consequence of Plotkin and Rudin equimeasurability theorem [48, 52],
see also [33] (the “envelope” terminology is ours and does not appear in those
works). Based on these examples, our aim is to define envelopes of subspaces
in general Banach spaces, so that $\mathrm{(AuH)}$ spaces can be proved to be
“ultrahomogeneous with respect to envelopes”, in a precise sense to be defined
later. The notion of envelope is then expected to be useful to address the
conjecture from [21].

Let us note here another motivation for studying $\mathrm{(AuH)}$ spaces.
These are natural universal objects in the sense that two separable
$\mathrm{(AuH)}$ are isometric as soon as they have the same finite
dimensional subspaces up to isometry [21, Proposition 2.22]. In fact, for a
strengthening of the $\mathrm{(AuH)}$ property, called the Fraïssé property
(see Definition 2.43), there is a correspondence between these spaces and
classes of finite dimensional spaces with a natural amalgamation property,
and, in this way, the separable Fraïssé Banach space corresponding to one of
such classes $\mathcal{F}$ (called the Fraïssé limit of $\mathcal{F}$) is the
“generic” universal space for all those spaces having dense subspaces
consisting of an increasing union of elements in $\mathcal{F}$. Examples of
this situation are precisely the Gurarij space and the spaces $L_{p}[0,1]$,
for $p\notin 2{\mathbb{N}}+4$, that are the generic universal spaces
corresponding to the amalgamation classes of all separable spaces, and the
class of $L_{p}$-spaces, respectively (see [21]). They are the only known
examples, and it is unknown whether there exist $\mathrm{(AuH)}$ spaces which
are not Fraïssé. Moreover, the Kechris-Pestov-Todorcevic correspondence for
$\mathrm{(AuH)}$ Banach spaces characterizes a fixed point property, called
extreme amenability, of their linear isometry group in terms of an approximate
Ramsey property, of the collection of their finite dimensional subspaces. All
the known $\mathrm{(AuH)}$ Banach spaces have this property (see [27] for
$\ell_{2}$, [28] and [21] for the other $L_{p}$’s, and [7] for the Gurarij
space). New examples of $\mathrm{(AuH)}$ Banach spaces could provide new
extremely amenable groups; on the other hand if all separable $\mathrm{(AuH)}$
Banach spaces are isometric the $L_{p}$’s or the Gurarij, then, in some sense,
their isometry groups appear as canonical extremely amenable groups of
isometries of Banach spaces. Let us note finally that Fraïssé spaces must
contain an isometric copy of the Hilbert space, [21, Proposition 2.13.], and
therefore the study of the envelopes of hilbertian subspaces of Fraïssé spaces
will be particularly relevant.

Here is the definition of the isometric envelope
$\operatorname{\mathrm{Env}}(Y)$ of a subspace $Y$ of $X$: it is the subspace
of $x\in X$ such that $(T_{i}(x))_{i}$ converges to $x$ whenever the net
$(T_{i})_{i}$ in $\operatorname{\mathrm{Isom}}(X)$ converges pointwise on $Y$
to $\operatorname{\mathrm{Id}}_{Y}$. Under reflexivity we may relate the
isometric envelope to ergodic decompositions relative to semigroups of
contractions, in the so-called Jacobs-de Leeuw- Glicksberg theory, as well as
to results on Korovkin sets (see [55]). In particular, we prove the
fundamental result that if $X$ is reflexive strictly convex, then
$\operatorname{\mathrm{Env}}(Y)$ is always $1$-complemented, and if $X$ is
reflexive locally uniformly convex, then $\operatorname{\mathrm{Env}}(Y)$ is
actually the smallest superspace of $Y$ complemented by a projection in the
“isometric hull” $\overline{\rm conv}(\operatorname{\mathrm{Isom}}(X))$. As a
byproduct, the Hilbert space is the only separable reflexive space on which
all subspaces are equal to their envelopes. We conclude Section 2 with a
fundamental extension property of the envelope, which we state here for
$\mathrm{(AuH)}$ spaces $X$: any partial isometry between subspaces $Y,Z$ of
$X$ extends uniquely to a partial isometry between their envelopes
$\operatorname{\mathrm{Env}}(Y)$ and $\operatorname{\mathrm{Env}}(Z)$; if $Y$
is separable then this extension map is SOT-SOT continuous and
$\operatorname{\mathrm{Env}}(Y)$ is maximum among superspaces of $Y$
satisfying this property. Of particular interest are the “full” subspaces of
$X$, i.e. those whose envelope is equal to $X$: any partial isometry between
full subspaces extends uniquely to a surjective isometry on $X$, and the
isometry group of a full subspace embeds naturally into ${\rm Isom}(X)$. Thus
we achieve the extension properties that envelopes were designed to satisfy.

In Section 3 we use the previous characterization to prove an important
technical result: if $X$ is reflexive and separable, then the envelope
$\operatorname{\mathrm{Env}}(Y)$ of the closure $Y$ of a directed sequence
$\bigcup_{i}Y_{i}$ of subspaces coincides with the closure of the directed
sequence $\bigcup_{i}\operatorname{\mathrm{Env}}(Y_{i})$ of their envelopes.
We then investigate envelopes in $L_{p}$ spaces, and use the above to prove,
for example, that if $Y$ is unital, the envelope coincides with some classical
notions such as the sublattice generated by $Y$, the minimal $1$-complemented
subspace containing $Y$, or the range of the conditional expectation
associated to $Y$. In particular, all $1$-complemented unital subspaces are
envelopes in those spaces, and we prove that the $L_{p}$’s, $1<p<\infty$ are
the only r.i. reflexive spaces on $[0,1]$ for which this happens.

After proving that, for $p\notin 2{\mathbb{N}}+4$, the $L_{p}$’s contain a
full copy of the Hilbert space, we apply the isometric extension results of
Section 2 to study the associated exact sequence of Banach spaces and define
objects such as the group embedding of $U(\ell_{2})$ inside
$\operatorname{\mathrm{Isom}}(L_{p})$ or the associated “full quotient” of
$L_{p}$ by $\ell_{2}$.

In the final section, we establish relations between the AUH or Fraïssé
property and geometrical conditions from local theory such as type or cotype,
or complementations of euclidean or hilbertian subspaces. With these estimates
we obtain “local versions” of the known fact that $L_{p}$-spaces are Fraïssé
if and only if $p\notin 2{\mathbb{N}}+4$. With this we reinforce the
conjecture that all separable reflexive Fraïssé spaces must be $L_{p}$-spaces
as well as identify natural intermediary steps towards proving this fact. We
also answer a question of G. Godefroy by proving that any Fraïssé Banach space
with a $C_{\infty}$-norm must be isomorphic to a Hilbert space.

We shall define envelopes in arbitrary Banach spaces, but with a focus on the
reflexive case. We aimed at identifying the minimal hypotheses regarding the
norm for each of our results (strict convexity/local uniform convexity, strict
convexity of the dual, etc…). However it should be noted that $\mathrm{(AuH)}$
spaces are always uniformly convex and uniformly smooth as soon as they are
reflexive. Therefore the reader mainly interested in the concept of envelope
inside a reflexive $\mathrm{(AuH)}$ space $X$ can safely assume that the norm
on $X$ is always uniformly convex and uniformly smooth.

###  1.2. Notation

Given a Banach space $X=(X,\|\cdot\|)$ we shall write ${\rm Sph(X)}$ to denote
the unit sphere of $X$ and $B_{X}$ to denote its unit ball. We write
$\operatorname{\mathrm{GL}}(X)$ and $\operatorname{\mathrm{Isom}}(X)$ to
denote the group of surjective linear isomorphisms and surjective linear
isometries, respectively, ${\mathcal{L}}_{1}(X)$ to denote the semigroup of
contractions on $X$, and if $X$ is a Banach lattice,
${\mathcal{L}}_{1}^{+}(X)$ to denote the semigroup of positive contractions.

For the definition of classical properties of norms (strict convexity, uniform
convexity, etc…) and related notions we refer to [18] and/or [37]. For
completeness we recall the maybe slightly less classical notion of local
uniform convexity. A Banach space $(X,\|.\|)$ is locally uniformly convex if
$\forall x_{0}\in{\rm Sph(X)}$ and $\forall\varepsilon>0$, there exists
$\delta>0$ such that whenever $x\in B_{X}$,

$\|\frac{x+x_{0}}{2}\|>1-\delta\Rightarrow\|x-x_{0}\|<\varepsilon.$

Locally uniformly convex norms are also called locally uniformly rotund and
for these reasons this property is abbreviated as “LUR”.

If $T$ is an operator on $X$ we denote by ${\rm Fix}(T)$ the closed subspace
of fixed points of $T$, i.e. ${\rm Fix}(T)=\\{x\in X:Tx=x\\}$. If $S$ is a
semigroup of operators on $X$ then ${\rm Fix}(S)$ denotes the closed subspace
of fixed points of $S$, i.e. ${\rm Fix}(S):=\bigcap_{T\in S}{\rm Fix}(T)$ is
the largest subspace where all $T\in S$ act as the identity. On the other
hand, if $Y$ is a subspace of $X$, then ${\rm Stab}_{S}(Y)$ denotes the
semigroup of operators in $S$ acting as the identity on $Y$.

#### Acknowledgements

We are grateful to G. Godefroy and W. T. Johnson for helpful conversations and
remarks.

##  2\. Envelopes

An (abstract) envelope is a set-valued map $e$ to every subset $A$ of a Banach
space $X$ its envelope, i.e. a closed subspace $e(A)$ of $X$ satisfying the
following properties

  1. (a)

$A\subseteq e(A)$.

  2. (b)

If $B\subseteq A$, then $e(B)\subseteq e(A)$.

  3. (c)

$e(\overline{\rm span}(A))=e(A)$.

  4. (d)

$e(e(A))=e(A)$.

So, $e$ is a standard (algebraic) closure operator with the additional
property (c). Alternatively, any envelope map on $X$ may be fully described by
the class ${\mathcal{E}}$ of closed subspaces of $X$ which are envelopes, and
by the condition that $e(A)$ is the smallest element of ${\mathcal{E}}$
containing $A$ as a subset. It is routine to check that a necessary and
sufficient condition for a given class ${\mathcal{E}}$ to define an envelope
map in this manner is to be stable by (finite or infinite) intersections.

Perhaps the more geometrical example of envelope is the minimal envelope that
assigns to a subset, if it exists, the smallest superspace of it that is
1-complemented. This set-mapping is not always well defined: there are
examples of Banach spaces containing pairs of $1$-complemented subspaces whose
intersection is not $1$-complemented (and therefore does not have a minimal
envelope) - see W.B. Johnson’s comment in Mathoverflow [31] for a simple proof
using pushout. However in reflexive spaces with strictly convex norm we have a
positive result, which was also indicated to us by W.B. Johnson.

######  Proposition 2.1.

Let $X$ be a reflexive space with strictly convex norm. Then any intersection
of a family $(Y_{i})_{i\in I}$ of $1$-complemented subspaces of $X$ must be
$1$-complemented.

######  Proof.

Pick a family $(p_{i})_{i\in I}$ such that $p_{i}$ is a contractive projection
on $Y_{i}$ for each $i$. Given any non-empty finite $F\subseteq I$, strict
convexity implies that the average $T_{F}:=(1/\\#F)\sum_{i\in F}p_{i}$ of the
projections $p_{i}$ for $i\in F$ has $\bigcap_{i\in F}Y_{i}$ as its space
${\rm Fix}(T_{F})$ of fixed points. Consequently, the Yosida’s mean ergodic
theorem for reflexive spaces and power-bounded operators [56] implies that the
averages of powers of $T_{F}$ $(1/n\sum_{i=0}^{n-1}T_{F}^{i})_{n}$ converges
SOT to a contractive projection $p_{F}$ onto $\bigcap_{i\in F}Y_{i}$. By
reflexivity we can pick a subnet of $(p_{F})_{F\subseteq I,\text{ finite}}$
that converges WOT. The limit will be a contractive projection onto
$\bigcap_{i\in I}Y_{i}$. ∎

######  Remark 2.2.

Let us note from the proof of Proposition 2.1 that if $X$ is reflexive
strictly convex and $p_{i}$ is a contractive projection onto $Y_{i}$ for each
$i\in I$, then a contractive projection onto $\bigcap_{i\in I}Y_{i}$ may be
chosen in the WOT-closure of the convex hull of $\\{p_{i}\\}_{i\in I}$.

######  Definition 2.3 (Minimal envelope of a subspace).

Let $X$ be a Banach space, and $A$ be a subset of $X$. We let
$\operatorname{\mathrm{Env}}_{\mathrm{min}}(A)$ be the envelope defined as the
smallest $1$-complemented subspace of $X$ containing $A$, when such subspace
exists.

######  Remark 2.4.

When $\operatorname{\mathrm{Env}}_{\rm min}(A)$ exists,
$\operatorname{\mathrm{Env}}_{\rm min}(TA)=T(\operatorname{\mathrm{Env}}_{\rm
min}(A))$ for every $T\in\operatorname{\mathrm{Isom}}(X)$.

From Proposition 2.1, the minimal envelope is always well-defined in reflexive
spaces with strictly convex norm. Under the additional hypothesis that $X$ is
strictly convex, then we also have that the contractive projection on a
$1$-complemented subspace is unique [14]. In such spaces, the duality mapping
plays a central role in understanding the contractive projection onto the
minimal envelope.

######  Definition 2.5.

For a reflexive space $X$ such that $X,X^{*}$ are both strictly convex, we let
$J:\mathrm{Sph}(X)\mapsto\mathrm{Sph}({X^{*}})$ be the duality mapping, i.e.
$\langle Jx,x\rangle=1$ for every $x\in\mathrm{Sph}(X)$.

It is convenient to extend $J$ to the whole of $X$ by homogeneity. We
recommend the survey [51], and also [23, Section 4], to better understand the
relationship betweem $J$ and the isometry groups. In particular, there it is
observed that $J$ is well defined, and that it is a bijection between
$\mathrm{Sph}(X)$ and $\mathrm{Sph}(X^{*})$ whose inverse is the
corrresponding duality mapping $J_{*}$ between $\mathrm{Sph}({X^{*}})$ and
$\mathrm{Sph}(X)$; in addition, when $X$ and $X^{*}$ are assumed to be locally
uniformly rotund, then $J$ is an homeomorphism; finally, $JT={T^{*}}^{-1}J$
whenever $T\in\operatorname{\mathrm{Isom}}(X)$.

The next relations between $J$ and $1$-complemented subspaces are due to
Calvert [10], and Cohen and Sullivan [14], see [51, Thm 5.5 and Thm. 5.6.].

######  Theorem 2.6.

Let $X$ be reflexive strictly convex and with strictly convex dual. A closed
subspace $Y$ of $X$ is 1-complemented in $X$ if and only if $JY$ is a linear
subspace of $X^{*}$. In this case there is a unique contractive projection of
$X$ onto $Y$ which is associated to the decomposition $X=Y\oplus(JY)^{\perp}$,
and if $JY$ is closed then it is $1$-complemented, associated to
$X^{*}=JY\oplus(Y)^{\perp}.$

Assume that the space $X$ is reflexive and both $X$ and $X^{*}$ have strictly
convex norms. Of special interest is the minimal (contractive) projection
associated to $\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)$. Recall that a
minimal contractive projection on a Banach space is a contractive projection
which is minimal with respect to the usual order $p\leq q$ i.e. if and only if
$pq=qp=p$. As we mentioned above under the current hypothesis on $X$, the
intersection of 1-complemented subspaces is again 1-complemented, but it is
not so obvious how to find (if exists) a minimal contractive projection for a
1-complemented subspace. To this end we describe the mean ergodic projection
associated to a semigroup of contractions, which may be seen as part of the
Jacobs-de Leeuw-Glicksberg (or JdLG) theory, see [20, Chapters 8 and 16] for a
reference. Recall that $\mathrm{Fix}(S)$ denotes the subspace of points which
are fixed by all elements of a semigroup $S$ of operators on $X$. We denote by
$S^{*}$ the corresponding semigroup $\\{s^{*}:s\in S\\}$ of operators on
$X^{*}$.

Part of the following description of the decomposition of the mean ergodic
projection was obtained in [22] under the restriction that $S$ is an isometry
group and that $X^{*}$ is strictly convex. We also include a description
through the convex hull of the semigroup. Note that the description of the
second summand through duality differs from the classical JdLG description.

######  Proposition 2.7.

Assume $X$ is reflexive strictly convex, and that $S$ is a semigroup of
contractions on $X$. Then

  1. (1)

$\mathrm{Fix}(S)$ is $1$-complemented by a projection in
$\overline{\rm{conv}(S)}^{\mathrm{WOT}}$.

If furthermore $X^{*}$ is assumed strictly convex, then

  1. (2)

The contractive projection onto $\mathrm{Fix}(S)$ is the unique minimal
projection $p_{S}$ in $\overline{\rm{conv}(S)}^{\mathrm{WOT}}$, and we have
the formula

$p_{S}(x)={\rm the\ unique\ element\ of\ }\mathrm{Fix}(S)\cap\overline{\rm
conv}(Sx).$

  2. (3)

The associated decomposition of $X$ is the $S$-invariant decomposition

$X=\mathrm{Fix}(S)\oplus
J(\mathrm{Fix}(S))^{\perp}=\mathrm{Fix}(S)\oplus\mathrm{Fix}(S^{*})^{\perp}.$

######  Proof.

We assume $X$ is reflexive and strictly convex and we prove (1). For any $T\in
S$, the mean ergodic Theorem states that the averages of powers of $T$
converge to a contractive projection $p_{T}$ onto ${\rm Fix}(T)$; therefore
$p_{T}$ belongs to the SOT closure of ${\rm conv}(S)$. By Remark 2.2, there
exists a contractive projection onto $\bigcap_{T\in S}{\rm Fix}(T)={\rm
Fix}(S)$, belonging to the WOT-closure of the convex hull of the $p_{T}$’s,
and therefore to the WOT-closure of ${\rm conv}(S)$.

We now assume $X^{*}$ is strictly convex and prove (2) and (3). From the fact
that $J$ is a bijection it follows immediately that
$J(\mathrm{Fix}(S))=\mathrm{Fix}(S^{*})$ is a linear subspace. Therefore the
$1$-complementation of $\mathrm{Fix}(S)$ by a unique projection (called $q$
for now) satisfying decomposition (3) follow from previously mentioned [51,
Theorems 5.5 and 5.6.].

By the mean ergodic Theorem [20, Theorem 8.34], we know that
$\overline{S}^{\mathrm{WOT}}$ is mean ergodic, and therefore by [20, Theorem
8.33 b)] determines uniquely the so-called mean ergodic projection onto
$\rm{Fix}(S)=\rm{Fix}(\overline{S}^{\mathrm{WOT}})$ called $p$ for now, which
is characterized by $px\in\overline{\rm conv}(Sx)$ and $pT=Tp=p$ for all $T\in
S$. Strict convexity of $X^{*}$ imply by [14] that $p=q$.

The uniqueness and existence of the minimal idempotent $p_{S}$ in
$\overline{{\rm conv}(S)}^{\mathrm{WOT}}$ is in [20, Theorem 16.20]. Since
$pT=Tp=p$ for all $T\in S$ it is immediate that this also holds for all
$T\in\overline{{\rm conv}(S)}^{\mathrm{WOT}}$ and therefore for $p_{S}$, i.e.
$p\leq p_{S}$. To conclude that $p=p_{S}$ it is now enough to observe that
$p\in\overline{{\rm conv}(S)}^{\mathrm{WOT}}$ by (1). ∎

######  Remark 2.8.

The condition that $X^{*}$ strictly convex is necessary for the uniqueness in
(2) and the decomposition in (3). Indeed if $z$ is a point of $X$ of norm 1
admitting two norming functionals $\phi_{0}$ and $\phi_{1}$, then consider the
two contractive projections onto the span of $z$ defined by
$p_{i}(x)=\phi_{i}(x)z$, and let $S=\\{p_{0},p_{1}\\}$. Note that ${\rm
Fix}(S)=[z]$ and both $p_{0}$ and $p_{1}$ are projections onto it. Also for
$i=0,1$, $Fix(p_{i}^{*})=[\phi_{i}]$, therefore ${\rm Fix}(S^{*})=\\{0\\}$ and
${\rm Fix}(S^{*})^{\perp}=X$.

###  2.1. The algebraic envelope

Recall that given a semigroup $S$ and a subset $Y$ of $X$, the (point)
$S$-stabilizer $\mathrm{Stab}_{S}(Y)$ of $Y$ is the collection of $s\in S$
that acts as the identity on $Y$. Inspired by the decomposition in Proposition
2.7 we define the following.

######  Definition 2.9 ($S$-algebraic envelope).

Given a subspace $Y$ of $X$ and a semigroup $S$ of operators on $X$, the
$S$-algebraic envelope of $Y$ is defined as

$\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y)={\rm
Fix}(\mathrm{Stab}_{S}(Y)).$

In other words, the $S$-algebraic envelope of $Y$ is the subspace of points
which are fixed by all operators of $S$ fixing each point of $Y$.

It is worth noting that if $J$ is a bijection between the unit spheres of $X$
and $X^{*}$, and $S$ a semigroup of contractions, then
$(\mathrm{Stab}_{S}(Y))^{*}:=\\{{s^{*}}\,:\,{s\in\mathrm{Stab}_{S}(Y)}\\}=\mathrm{Stab}_{S^{*}}({JY})$.
Note also that
$\overline{\mathrm{Stab}_{S}(Y)}^{\mathrm{WOT}}\subseteq\mathrm{Stab}_{\overline{S}^{\mathrm{WOT}}}(Y)$,
but the equality does not seem to hold in general; unless $S$ is WOT-closed,
in which case $\mathrm{Stab}_{S}(Y)$ is also WOT-closed. The next is a
consequence of Proposition 2.7 and it relates the minimal and algebraic
envelope.

######  Proposition 2.10.

Assume $X$ is reflexive strictly convex, $S$ is a semigroup of contractions on
$X$, and $Y$ is a closed subspace of $X$. Then
$\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y)$ is $1$-complemented in $X$
by a contractive projection in $\overline{{\rm
conv}(\operatorname{\mathrm{Stab}}_{S}(Y))}^{\mathrm{WOT}}$. If furthermore
$X^{*}$ is strictly convex, then holds the
$\operatorname{\mathrm{Stab}}_{S}(Y)$-invariant decomposition

$X=\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{S}(Y)\oplus\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{S^{*}}(JY)^{\perp},$

and the projection upon the first summand is the unique minimal projection of
$\overline{{\rm conv}(\operatorname{\mathrm{Stab}}_{S}(Y))}^{\mathrm{WOT}}$.

######  Proof.

This is a consequence of Proposition 2.7 and of the fact that
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{S}(Y)=\mathrm{Fix}(\operatorname{\mathrm{Stab}}_{S}(Y))$
and that
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{S^{*}}(JY)=\mathrm{Fix}(\operatorname{\mathrm{Stab}}_{S^{*}}({JY}))=\mathrm{Fix}((\operatorname{\mathrm{Stab}}_{S}(Y))^{*})=J(\mathrm{Fix}(\operatorname{\mathrm{Stab}}_{S}(Y)))=J(\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{S}(Y))$.
∎

######  Remark 2.11.

If $T\in S$ is an automorphism then
$\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(T(Y))=T(\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y))$:
note that ${\rm Stab}_{S}{TY}=T{\rm Stab}_{S}(Y)T^{-1}$ and that ${\rm
Fix}(TST^{-1})=T({\rm Fix}(S)).$

Even though the equality ${\rm Fix}(S)={\rm Fix}({\rm conv}(S))$ always holds,
the algebraic envelopes of a semigroup $S$ and of its convex hull do not seem
to coincide in general. Indeed we have ${\rm conv}({\rm
Stab}_{S}(Y))\subseteq{\rm Stab}_{{\rm conv}(S)}(Y)$ but no reason to assume
equality, unless under additional hypotheses:

######  Proposition 2.12.

Let $X$ be a space with strictly convex norm, let $S$ be a semigroup of
contractions on $X$, and let $Y$ be a subspace of $X$. Then

$\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{conv}(S)}^{\mathrm{alg}}(Y).$

######  Proof.

It is enough to prove that ${\rm conv}({\rm Stab}_{S}(Y))={\rm Stab}_{{\rm
conv}(S)}(Y)$, the direct inclusion of which always holds. Assume
$T\in\mathrm{conv}(S)$ acts as the identity on $Y$. Writing $T$ as a convex
combination $\sum_{k}\lambda_{k}s_{k}$, $s_{k}\in S$, we see that for any
$y\in Y$, $y=\sum_{k}\lambda_{k}s_{k}(y)$. By strict convexity and the fact
that the $s_{k}$’s are contractions, we deduce $s_{k}(y)=y$ for every $k$ and
$y\in Y$, i.e. $s_{k}\in{\rm Stab}_{S}(Y)$. This means that $T$ belongs to
${\rm conv}({\rm Stab}_{S}(Y))$. We have proved that ${\rm conv}({\rm
Stab}_{S}(Y))\supseteq{\rm Stab}_{{\rm conv}(S)}(Y)$. ∎

######  Definition 2.13 (Algebraic envelope).

The algebraic envelope of a subspace $Y$ of $X$, denoted by
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)$, is the subspace
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{\operatorname{\mathrm{Isom}}(X)}(Y)$.

###  2.2. Korovkin and isometric envelopes

In some aspects the algebraic envelope seems to be “too rigid” and does not
capture topological properties of a semigroup. We introduce a (possibly)
smaller topological modification of the algebraic envelope.

######  Definition 2.14 (Korovkin envelope).

Let $X$ be a Banach space and $S$ be a bounded semigroup of operators. We
define the (Korovkin) envelope $\operatorname{\mathrm{Env}}_{S}(A)$ of a
subset $A$ of $X$ as the set of $x\in X$ such that whenever a net
$(T_{i})_{i\in I}$ in $S$ converges pointwise on $A$ to
$\operatorname{\mathrm{Id}}_{A}$, then the net $(T_{i}(x))_{i\in I}$ converges
to $x$.

It is straightforward that $Y\mapsto\operatorname{\mathrm{Env}}_{S}(Y)$ is an
envelope map. Also, we have the following inequalitiles.

$\operatorname{\mathrm{Env}}_{S}(Y)\subseteq\operatorname{\mathrm{Env}}_{\overline{S}^{\mathrm{SOT}}}^{\mathrm{alg}}(Y)\subseteq\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y).$

To see this, for suppose that $x$ belong to
$\operatorname{\mathrm{Env}}_{S}(Y)$, and assume that $T=\lim_{i}T_{i}$, with
each $T_{i}\in\overline{S}^{\mathrm{SOT}}$ and that $T\upharpoonright
Y=Id_{Y}$. Then $(T_{i}\upharpoonright Y)_{i}$ tends SOT to
$\operatorname{\mathrm{Id}}_{Y}$, and since
$x\in\operatorname{\mathrm{Env}}_{S}(Y)$, by definition of the Korovkin
envelope, we have that $(T_{i}x)_{i}$ converges in norm to $x$, and
consequently $Tx=x$.

The name we chose for this envelope has the following explanation. We recall
that we denote by ${\mathcal{L}}_{1}(X)$ the semigroup of contractions on $X$,
and if $X$ is a Banach lattice, by ${\mathcal{L}}_{1}^{+}(X)$ the semigroup of
positive contractions. When $S$ is equal to ${\mathcal{L}}_{1}(X)$ (resp.
${\mathcal{L}}_{1}^{+}(X)$), results on envelope maps have been obtained by
several authors as extensions of Korovkin theorems, as described for example
in [5], [6]. Namely, if $\mathbf{T}=(T_{i})_{i\in I}$ is a net of
contractions, the convergence set $C_{\mathbf{T}}$ is defined as

$C_{\mathbf{T}}:=\\{x\in X\,:\,(T_{i}(x))_{i\in I}{\rm\ converges\ to\ }x\\},$

and if $A$ is a subset of $X$, the shadow $\Gamma_{1}(A)$ (resp.
$\Gamma_{1}^{+}(A)$)) of $A$ is the intersection of all convergence sets
(resp. of positive contractions) containing $A$ (in the 1994 survey of
Altomare and Campiti [1, Def 3.1.1. p 142.], the term Korovkin closure is also
used). In other words, $\Gamma_{1}(A)$ coincides with the envelope
$\operatorname{\mathrm{Env}}_{{\mathcal{L}}_{1}(X)}(A)$ and
$\Gamma_{1}^{+}(A)$ with
$\operatorname{\mathrm{Env}}_{{\mathcal{L}}_{1}^{+}(X)}(A)$.

We use the terminology Korovkin set (resp. Korovkin+ set) in $X$ for a set $A$
for which $\Gamma_{1}(A)=X$ (resp. $\Gamma_{1}^{+}(A)=X$). Korovkin Theorem
[34] states that $\\{1,t,t^{2}\\}$ is a Korovkin+ set in $C(0,1)$. This was
later improved to being a Korovkin set by Wulbert [55] and independently,
Saskin [54]. Bernau [5] proves that $\\{1,t\\}$ is a Korovkin set in
$L_{p}(0,1),1<p<+\infty$, and Berens-Lorentz [6] for $p=1$. In [10], Calvert
obtains that if $X$ is a reflexive space with locally uniformly convex norm
and dual norm, then convergence sets are ranges of linear contractive
projections, and the envelope $\Gamma_{1}(A)$ is the smallest $1$-complemented
subspace containing $A$, i.e.
$\Gamma_{1}(A)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(A)$, [10] Corollary
2. We shall revisit these results in the present section.

######  Definition 2.15 (Isometric envelope).

When $G$ is the isometry group of $(X,\|\cdot\|)$ then
$\operatorname{\mathrm{Env}}_{G}(Y)$ is called the (isometric) envelope of $Y$
in $X$, and we denote it by $\operatorname{\mathrm{Env}}_{\|\cdot\|}(Y)$, or
simply $\operatorname{\mathrm{Env}}(Y)$, when the norm on $X$ is clear from
the context.

It will be relevant to give a name to those subspaces whose envelope coincides
with the whole space:

######  Definition 2.16 (Full envelope, full subspace).

A subspace $Y$ of $X$ has full envelope, or it is a full subspace of $X$, when
$\operatorname{\mathrm{Env}}(Y)=X$.

When $G$ is a bounded group of isomorphisms, we have the following
reformulation of the definition of the envelope
$\operatorname{\mathrm{Env}}_{G}(A)$. Because of item c) in the definition of
envelope, from now on we only consider envelopes of closed subspaces $Y$ of a
Banach space $X$.

######  Proposition 2.17.

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. If $G$ is a
bounded group of isomorphisms on $X$, then a point $x\in X$ belongs to
$\operatorname{\mathrm{Env}}_{G}(Y)$ if and only if whenever a net
$(T_{i})_{i\in I}$ in $G$ converges pointwise on $Y$, then $(T_{i}(x))_{i\in
I}$ also converges.

######  Proof.

Let $G$ be a bounded group of isomorphisms on $X$. Assume that $x$ satisfies
the last condition, and let us see that
$x\in\operatorname{\mathrm{Env}}_{G}(Y)$, so suppose that $(T_{i})_{i\in I}$
is a net that converges pointwise on $Y$ to $\operatorname{\mathrm{Id}}_{Y}$.
We consider the set $J:=I\times\\{0,1\\}$ lexicographically ordered, and the
net $(U_{(i,j)})_{(i,j)\in J}$, $U_{(i,0)}:=\operatorname{\mathrm{Id}}$ and
$U_{(i,1)}:=T_{i}$ for all $i$. It follows that $(U_{(i,j)})_{(i,j)}$
pointwise converges in $Y$ to $\operatorname{\mathrm{Id}}_{Y}$, so by
hypothesis, $(U_{(i,j)}(x))_{(i,j)}$ also converges, and the limit must be
$x$. This means in particular that $(U_{i}(x))_{i}$ converges to $x$.

Suppose now that $x$ does not satisfy the last condition. We can find a net
$(T_{i})_{i\in I}$ in $G$ converging pointwise on $Y$ and some $\varepsilon>0$
such that the subset $J\subseteq I$ of those $i$ such that there are
$j_{i},k_{i}\geq i$ such that $\|T_{j_{i}}(x)-T_{k_{i}}(x)\|\geq\varepsilon$.
This implies that,
$\|x-T_{j_{i}}^{-1}(T_{k_{i}}(x))\|\geq\varepsilon/\sup_{T\in G}\|T\|$ for
every $i\in J$. For each $i\in J$ let $U_{i}:=T_{j_{i}}^{-1}\circ T_{k_{i}}\in
G$. Then the net $(U_{i})_{i\in J}$ in $G$ converges pointwise on $Y$ to the
identity of $Y$ but $(U_{i}(x))_{i\in J}$ does not converge to $x$, so
$x\notin\operatorname{\mathrm{Env}}_{G}(Y)$. ∎

So, $\operatorname{\mathrm{Env}}_{G}(Y)$ is the maximum superset $Z$ of $Y$ so
that pointwise convergence on $Y$ of elements of $G$ (to a map $t:Y\rightarrow
X$ which does not necessarily extend to an element of $G$) implies pointwise
convergence of these elements on $Z$ (necessarily to an extension of $t$).

Note that when $S\subseteq S^{\prime}$ are bounded semigroups then
$\operatorname{\mathrm{Env}}_{S^{\prime}}(Y)\subseteq\operatorname{\mathrm{Env}}_{S}(Y)$,
and also that whenever $T\in S$ is an automorphism, then
$\operatorname{\mathrm{Env}}_{S}(TY)=T(\operatorname{\mathrm{Env}}_{S}(Y))$.

######  Example 2.18.

If $X=H$ is a Hilbert space and $Y$ is a closed subspace of it then
$\operatorname{\mathrm{Env}}(Y)=Y$. Indeed we can decompose $H$ as $Y\oplus
Y^{\perp}$ and if $x$ is not in $Y$, then let $z={\rm proj}_{Y^{\perp}}x\neq
0$ and define
$T_{n}=\operatorname{\mathrm{Id}}\oplus-\operatorname{\mathrm{Id}}$. This
proves that $x$ is not in $\operatorname{\mathrm{Env}}(Y)$.

A group of isomorphisms $G$ on a space $X$ is said to be trivial if all
elements of $G$ are multiples of the identity map.

######  Example 2.19.

If the isometry group $\operatorname{\mathrm{Isom}}(X)$ is trivial, then any
subspace $Y$ of $X$ of dimension at least $1$ has envelope equal to $X$: if a
net $(T_{i})_{i\in I}$ converges pointwise on $Y$, then each
$T_{i}=\lambda_{i}\operatorname{\mathrm{Id}}_{X}$ with $(\lambda_{i})_{i\in
I}$ a converging net; thus, $(T_{i})_{i\in I}$ converges pointwise on the
whole of $X$.

As in the case of the algebraic envelope, in general the Korovkin envelope of
a semigroup $S$ does not seem to need to coincide with the envelope of its
WOT-closure or of its convex hull. There is however such an identification
under mild topological conditions on the norm. The hypothesis of the norm
being locally uniformly convex (LUR) in the next proposition should be related
to the fact that the Korovkin envelope is of a topological nature.

######  Proposition 2.20.

Let $X$ be a space with a LUR norm and let $S$ be a semigroup of contractions
on $X$. Let $Y$ be a subspace of $X$. Then

$\operatorname{\mathrm{Env}}_{S}(Y)=\operatorname{\mathrm{Env}}_{\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}}(Y).$

######  Proof.

Only the direct inclusion is not trivial. Let $x$ be normalized in
$\operatorname{\mathrm{Env}}_{S}(Y)$ and fix $\varepsilon>0$. By definition of
$\operatorname{\mathrm{Env}}_{S}(Y)$, there exist normalized
$y_{1},\ldots,y_{n}$ in $Y$ and $\delta>0$ such that for any $g\in S$,

if $\|g(y_{i})-y_{i}\|\leq\delta$ for every $i=1,\ldots,n$, then
$\|gx-x\|\leq\varepsilon.$

We use the LUR property of the norm to choose for each $i=1,\ldots,n$ some
$\delta_{i}>0$ such that if $\|z\|\leq 1$ is such that $\|z+y_{i}\|\geq
2-\delta_{i}$ then $\|z-y_{i}\|\leq\delta$. Let $\beta>0$ be small enough so
that $\sqrt{2\beta}\leq\min_{i}\delta_{i}$, $\beta\leq\varepsilon$ and
$2n\sqrt{2\beta}\leq\varepsilon$. We claim that whenever
$T\in\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}$ satisfies that
$\max_{i}\|Ty_{i}-y_{i}\|\leq\beta$, then it follows that $\|Tx-x\|\leq
3\varepsilon$. This implies that whenever $(T_{i})_{i\in I}$ is a net in
$\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}$ converging pointwise to the
identity on $Y$, then $(T_{i}(x))_{i\in I}$ converges to $x$, and this means
that $x$ belongs to
$\operatorname{\mathrm{Env}}_{\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}}(Y)$.
This therefore proves the required direct inclusion.

We now give the proof of the claim. For $i=1,\ldots,n$ let $\phi_{i}$ be a
normalized functional norming $y_{i}$, and let $\phi_{0}$ be a normalized
functional norming $Tx-x$. Denote $y_{0}:=x$ and consider a convex combination
$\sum_{k\in K}\lambda_{k}g_{k}$ with each $g_{k}\in S$, such that
$|\langle\phi_{i},(T-\sum_{k\in K}\lambda_{k}g_{k})y_{i}\rangle|\leq\beta$ for
$i=0,\ldots,n$. Note that for every $i=1,\ldots,n$,

$1-2\beta\leq\phi_{i}(T(y_{i}))-\beta\leq\sum_{k\in
K}\lambda_{k}\phi_{i}(g_{k}(y_{i})).$ (1)

For each $i=1,\dots,n$, let $A_{i}$ be the set of indices $k\in K$ such that
$\phi_{i}(g_{k}(y_{i}))\geq 1-\sqrt{2\beta}$ and let $B_{i}=K\setminus A_{i}$
be the complement of $A_{i}$. Note that for $k\in A_{i}$, we have that
$\|g_{k}(y_{i})-y_{i}\|\leq\delta$. Also from the above equation (1) we have

$1-2\beta\leq\sum_{k\in A_{i}}\lambda_{k}+(1-\sqrt{2\beta})\sum_{k\in
B_{i}}\lambda_{k},$

from which it is immediate to deduce

$\sum_{k\in B_{i}}\lambda_{k}\leq\sqrt{2\beta}.$

Finally let $A=\bigcap_{i=1}^{n}A_{i}$ and $B=\bigcup_{i=1}^{n}B_{i}$.
Whenever $k\in A$, we have $\|g_{k}y_{i}-y_{i}\|\leq\delta$ for all
$i=1,\ldots,n$, and therefore $\|g_{k}x-x\|\leq\varepsilon$, by the choice of
$\delta$. Using this estimate, we compute

$\|Tx-x\|=\phi_{0}(Tx-x)\leq\beta+\sum_{k\in
K}\lambda_{k}\|g_{k}x-x\|\leq\beta+(\sum_{k\in
A}\lambda_{k})\varepsilon+2\sum_{i=1}^{n}\sum_{k\in B_{i}}\lambda_{k}\leq
2\varepsilon+2n\sqrt{\beta}\leq 3\varepsilon,$

and this concludes the proof of the claim. ∎

######  Remark 2.21.

An interesting consequence of Proposition 2.20 is that if $Y$ is a subspace of
an LUR space $X$ and $S$ a semigroup of contractions, then
$\operatorname{\mathrm{Env}}_{S}(Y)$ is contained in the range of any
contractive projection $p\in\overline{{\rm conv}(S)}^{\rm WOT}$ acting as the
identity on $Y$. Indeed from $p\upharpoonright Y=Id_{Y}$ it follows that
$px=x$ for any $x\in\operatorname{\mathrm{Env}}_{\overline{{\rm conv}(S)}^{\rm
WOT}}(Y)=\operatorname{\mathrm{Env}}_{S}(Y)$.

We now show that under natural conditions on the space (such as reflexivity),
the Korovkin envelope may be seen as an algebraic envelope. This is important
to obtain a JdLG decomposition associated to this envelope. We shall use the
following facts. If $X$ is reflexive (actually the point of continuity
property is enough) and if $G$ is a bounded group of automorphisms on $X$,
then the weak and norm topologies coincide on each $G$-orbit of a non-zero
point $x_{0}$ of $X$ [41, Theorem 2.5], applied for $G$ equipped with the
topology of weak convergence in the point $x_{0}$. Also it is a classical and
easy fact that if the norm on $X$ is locally uniformly rotund (LUR), then weak
convergence of a net on the unit ball of $X$ to a point of the sphere, implies
strong convergence to this point. See [3, Remark 2.4.] for more details on
both these facts.

######  Proposition 2.22.

Let $X$ be a reflexive space. Assume that either $S$ is a bounded group of
isomorphisms on $X$, or that $X$ is LUR and $S$ is a semigroup of
contractions. Then

$\operatorname{\mathrm{Env}}_{S}(Y)=\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{\overline{S}^{\mathrm{WOT}}}(Y).$

In particular the two envelopes coincide as soon as $S$ is WOT-closed.

######  Proof.

In the first case, we may assume that $S$ is a group of isometries for some
renorming of $X$. Therefore in either case, for any net $(T_{i})_{i}$ in $S$,
and normalized vector $x\in X$, weak convergence of $(T_{i}x)_{i}$ to $x$
implies norm convergence. Assume some $x$ does not belong to
$\operatorname{\mathrm{Env}}_{S}(Y)$ and let $(T_{i})_{i}$ converge pointwise
to $\operatorname{\mathrm{Id}}_{Y}$ on $Y$ such that $(T_{i}x)_{i}$ does not
converge to $x$ in norm and therefore does not converge weakly to $x$. Since
reflexivity of $X$ is equivalent to saying that the space of contractions with
the WOT is compact, there is convergent subnet for $(T_{i})_{i}$. W.l.o.g., we
may assume that $(T_{i})_{i}$ converges WOT to some $T$, and note that $T$ is
the identity on $Y$, but that $Tx$ is not equal to $x$. This proves that $x$
does not belong to
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{\overline{S}^{\mathrm{WOT}}}(Y)$.

Likewise, let $x\in\operatorname{\mathrm{Env}}_{S}(Y)$ and
$T\upharpoonright{Y}=\operatorname{\mathrm{Id}}_{Y}$ for
$T\in\overline{S}^{\mathrm{WOT}}$, then let $(T_{i})_{i}$ a net in $S$
converging WOT to $T$, so that $(T_{i})_{i}$ converges WOT to
$\operatorname{\mathrm{Id}}$ on $Y$. Then the convergence of $(T_{i})_{|Y}$ to
$\operatorname{\mathrm{Id}}_{Y}$ is SOT and so $(T_{i}x)_{i}$ converges to
$x$, so $Tx=x$, and consequently
$x\in\operatorname{\mathrm{Env}}^{\mathrm{alg}}_{\overline{S}^{\mathrm{WOT}}}(Y)$.
∎

The next proposition will allow us to compare the Korovkin and the minimal
envelopes. We have the following decomposition for the Korovkin envelope that
should be compared with Proposition 2.10.

######  Proposition 2.23.

Let $X$ be reflexive strictly convex, $S$ a semigroup of contractions on $X$,
and let $Y$ be a closed subspace of $X$. Assume additionally that either (i)
$S$ is a group of isometries on $X$, or that (ii) $X$ is LUR. Then

  1. (1)

$\operatorname{\mathrm{Env}}_{S}(Y)$ is a $1$-complemented subspace of $X$; in
case (ii) $\operatorname{\mathrm{Env}}_{S}(Y)$ is the smallest superspace of
$Y$ complemented by a projection in $\overline{{\rm conv}(S)}^{\mathrm{WOT}}$.

  2. (2)

If furthermore $X^{*}$ is strictly convex, then holds the ${\rm
Stab}_{\overline{\rm S}^{\mathrm{WOT}}}({Y})$-invariant decomposition

$X=\operatorname{\mathrm{Env}}_{S}(Y)\oplus\operatorname{\mathrm{Env}}_{S}(JY)^{\perp},$
(2)

and in case (ii) the projection on the first summand is the unique minimal
projection of the semigroup $\overline{{\rm conv}(S)}^{\rm{WOT}}$ acting as
the identity on $Y$.

######  Proof.

The $1$-complementation in (1) and the decomposition in (2) is a consequence
of Proposition 2.10 and the relation between the algebraic and the Korovkin
envelope in Proposition 2.22. Regarding the projection in case (1) (ii), note
that as a first step the set it belongs to is $\overline{{\rm conv}({\rm
Stab}_{\overline{S}^{\mathrm{WOT}}}(Y))}^{\mathrm{WOT}}$. Under the LUR
property and by Proposition 2.20, this set is the same as $\overline{{\rm
conv}({\rm Stab}_{\overline{{\rm
conv}(S)}^{\mathrm{WOT}}}(Y))}^{\mathrm{WOT}}$. But it is also the same as
${\rm Stab}_{\overline{{\rm conv}(S)}^{\mathrm{WOT}}}(Y)$ since this set is
convex and WOT-closed. This proves that $\operatorname{\mathrm{Env}}_{S}(Y)$
is complemented by a projection in $\overline{{\rm conv}(S)}^{\mathrm{WOT}}$,
and that it is the smallest among those containing $Y$ is Remark 2.21. The
statement about the projection in (2)(ii) follows from the one in case (1)(ii)
and strict convexity of the dual. ∎

As a consequence we spell out the following (note that (3) is an improvement
of Calvert’s result [10] on the Korovkin envelope, since there it is needed
$X^{*}$ to be LUR and we impose no requirement on it).

######  Proposition 2.24.

Let $X$ be reflexive strictly convex and let $Y$ be a closed subspace of $X$.
Then

  1. (1)

the isometric envelope $\operatorname{\mathrm{Env}}(Y)$ is $1$-complemented.

If furthermore $X$ is LUR then

  1. (2)

$\operatorname{\mathrm{Env}}(Y)$ is the smallest superspace of $Y$
complemented by a projection in $\overline{{\rm
conv}(\operatorname{\mathrm{Isom}}(X))}^{\rm{WOT}}$; if $X^{*}$ is strictly
convex, then this projection is the minimal projection in $\overline{{\rm
conv}(\operatorname{\mathrm{Isom}}(X))}^{\rm{WOT}}$ acting as the identity on
$Y$.

  2. (3)

The Korovkin envelope $\operatorname{\mathrm{Env}}_{\mathcal{L}_{1}(X)}(Y)$
coincides with the minimal envelope
$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)$.

######  Proof.

All items follow from Proposition 2.23. ∎

######  Remark 2.25.

Note that under the assumption that $X$ is reflexive strictly convex, we have

$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)\subseteq\operatorname{\mathrm{Env}}(Y)\subseteq\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)$

.

We do not know of a general characterization of situations where
$\operatorname{\mathrm{Env}}(Y)$ and
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)$ must coincide. However when
$X=L_{p}$, $p\neq 2$, and $Y$ is a unital subspace, then we will see in
Theorem 3.12 that
$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)=\operatorname{\mathrm{Env}}(Y)=L_{p}([0,1],\Sigma_{Y})$,
where $\Sigma_{Y}$ is the minimal subalgebra of the Borel algebra
$\mathcal{B}([0,1])$ making all functions in $Y$ measurable. And in this case
$\operatorname{\mathrm{Env}}(Y)=\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)$
exactly when $\Sigma_{Y}$ is a fixed-point subalgebra (see Remark 3.6).

######  Remark 2.26.

In [8] Beauzamy and Maurey consider a non-linear procedure associating to a
subset $M$ the set $M\subseteq\min M$ of its minimal points in a metric sense.
When $X$ is reflexive strictly convex and with strictly convex dual, it
follows from their results that the minimal envelope of a subspace may be
obtained through iteration of their ”min” procedure. In this way they recover
the results of Bernau [5] and Calvert [10] mentioned earlier.

######  Remark 2.27.

It may also be useful to spell out, for $X$ reflexive LUR:
$Y=\operatorname{\mathrm{Env}}(Y)$ if and only if $Y$ is complemented by a
contractive projection in $\overline{{\rm
conv}(\operatorname{\mathrm{Isom}}(X))}^{\rm WOT}$. This characterizes the
subspaces which are isometric envelopes as those which are ranges of
projections in the “isometric hull” $\overline{{\rm
conv}(\operatorname{\mathrm{Isom}}(X))}^{\rm WOT}$. This fully describes the
isometric envelope map, as observed at the beginning of this section just
after the definition of abstract envelope maps.

######  Proposition 2.28.

Assume $X$ is separable reflexive, let $G$ be a subgroup of
$\operatorname{\mathrm{Isom}}(X)$, and let $Y$ be a subspace of $X$. Then
$\operatorname{\mathrm{Env}}_{G}(Y)$ is $1$-complemented in $X$ by a
projection in $\overline{{\rm conv}(G)}^{\rm WOT}$.

######  Proof.

By Lancien [36] we may renorm $X$ with a norm $\|\cdot\|$ which is
$1+1/n$-equivalent to the original norm and so that $X$ is LUR and
$G\subseteq\operatorname{\mathrm{Isom}}(X,\|\cdot\|)$. It follows from
Proposition 2.23 (1)(ii) that $\operatorname{\mathrm{Env}}_{G}(Y)$ is
$1$-complemented with respect to $\|\cdot\|$ and therefore
$(1+1/n)^{2}$-complemented in the original norm, by a projection in
$\overline{{\rm conv}(G)}^{\rm WOT}$. A WOT-cluster point of the sequence of
associated projections $P_{n}$ is a norm $1$-projection onto
$\operatorname{\mathrm{Env}}_{G}(Y)$ belonging to $\overline{{\rm
conv}(G)}^{\rm WOT}$. ∎

######  Corollary 2.29.

The Hilbert space is the only separable reflexive space for which each closed
subspace $Y$ ($2$-dimensional subspace is enough) is equal to its envelope
$\operatorname{\mathrm{Env}}(Y)$.

######  Proof.

Indeed it is a known result by Kakutani [32] that X is Hilbertian if and only
if all subspaces are $1$-complemented (in fact it is enough to require that
all $2$-dimensional subspaces are 1-complemented); in the complex case this is
due to Bohnenblust [9]. ∎

###  2.3. Strong isometric embeddings

Closely related to the isometric envelope, as well as a motivation for its
definition, are the embeddings who are limits or restriction of (global)
isometries.

######  Definition 2.30.

Given a Banach space $X$, and a subspace $Y\subseteq X$, one defines the
collection ${\rm Emb}_{\mathrm{ext}}(Y,X)$ of extendable isometric embeddings
as

${\rm Emb}_{\mathrm{ext}}(Y,X):={\rm Isom}(X)\upharpoonright{Y}\subseteq{\rm
Emb}(Y,X),$

and the collection ${\rm Emb}_{\mathrm{s}}(Y,X)$ of strong isometric
embeddings as

${\rm Emb}_{\mathrm{s}}(Y,X):=\overline{{\rm
Isom}(X)\upharpoonright{Y}}^{\mathrm{SOT}}\subseteq{\rm Emb}(Y,X).$

Note that ${\rm Emb}_{\mathrm{s}}(Y,X)$ is by definition SOT-closed in ${\rm
Emb}(Y,X)$, hence a motivation for the name.

When $X$ is a Hilbert space, then ${\rm Emb}_{\mathrm{ext}}(F,X)={\rm
Emb}(F,X)$ for all finite dimensional subspaces $F$ of $X$. Spaces with these
properties are called ultrahomogeneous. The Hilbert space is the only known
separable example; non separable examples appear in [2] and [21]. When ${\rm
Emb}_{\mathrm{s}}(F,X)={\rm Emb}(F,X)$ for every finite dimensional $F$, the
space is called approximately ultrahomogeneous (see [21]). Separable examples
are the Hilbert spaces, all $L_{p}[0,1]$ for $p\notin 4+2{\mathbb{N}}$, or the
Gurarij space.

We shall sometimes use the word partial isometry on a space $X$ to mean a
surjective isometry $t$ between subspaces $Y$ and $Z$ of $X$. A strong partial
isometry is a partial isometry $t:Y\rightarrow Z$ such that $i_{ZX}\circ t$
belongs to ${\rm Emb}_{s}(Y,X)$, where $i_{ZX}$ is the inclusion map
$Z\rightarrow X$.

######  Lemma 2.31.

The inverse of a strong partial isometry is a strong partial isometry. When
defined, the composition of two strong partial isometries is a strong partial
isometry.

######  Proof.

Let $\gamma:Y\rightarrow Z$ be a strong partial isometry. Given
$z_{1},\dots,z_{n}\in Z$ of norm 1, let $y_{j}:=\gamma^{-1}z_{j}$ for each
$1\leq j\leq n$. Let $g\in\mathrm{Isom}(X)$ be such that
$\|g(y_{j})-\gamma(y_{j})\|\leq\varepsilon$ for every $1\leq j\leq n$. Then
for each $1\leq j\leq n$ one has that
$\|\gamma^{-1}(z_{j})-g^{-1}(z_{j})\|=\|y_{j}-g^{-1}(\gamma(y_{j}))\|=\|g^{-1}gy_{j}-g^{-1}\gamma(y_{j})\|=\|gy_{j}-\gamma
y_{j}\|\leq\varepsilon$. Similarly one proves the second part of the
statement. ∎

We may therefore define an equivalence relation between subspaces of $X$ as
follows: $Y$ and $Z$ are strongly isometric if there exists an strong partial
isometry between $Y$ and $Z$. The next proposition substantiates the idea that
isometric envelopes are well preserved under strong partial isometries.

######  Proposition 2.32.

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. Then

  1. (1)

Every $t\in{\rm Emb}_{s}(Y,X)$ extends uniquely to some $\tilde{t}\in{\rm
Emb}_{s}(\operatorname{\mathrm{Env}}(Y),X)$.

  2. (2)

The image $\tilde{t}(\operatorname{\mathrm{Env}}(Y))$ is equal to
$\operatorname{\mathrm{Env}}(tY)$.

In particular if $Y$ and $Z$ are strongly isometric subspaces of $X$, then so
are $\operatorname{\mathrm{Env}}(Y)$ and $\operatorname{\mathrm{Env}}(Z)$.
Furthermore,

  1. (3)

If $Y$ is separable, the map $t\mapsto\tilde{t}$ is SOT-SOT continuous from
${\rm Emb}_{s}(Y,X)$ to ${\rm Emb}_{s}(\operatorname{\mathrm{Env}}(Y),X)$.

######  Proof.

(1): An extension to $\operatorname{\mathrm{Env}}(Y)$ is defined by
$\tilde{t}(x)=\lim_{i\in I}T_{i}(x)$, if $t=\lim_{i}T_{i}\upharpoonright Y$
with $T_{i}\in{\rm Isom}(X)$. If $t^{\prime}$ is a strong isometric embedding
on $\operatorname{\mathrm{Env}}(Y)$ defined as pointwise limit of some
$(U_{i})_{i}$, extending $t$, then note that since the sequence
$(T_{i}^{-1}U_{i})_{i}$ is pointwise convergent to
$\operatorname{\mathrm{Id}}_{Y}$ on $Y$ it is also pointwise convergent on
$\operatorname{\mathrm{Env}}(Y)$, implying that $\tilde{t}$ and $t^{\prime}$
coincide on $\operatorname{\mathrm{Env}}(Y)$.

(2): We claim that
$\tilde{t}(\operatorname{\mathrm{Env}}(Y))=\operatorname{\mathrm{Env}}(tY)$.
We first prove that
$\tilde{t}(\operatorname{\mathrm{Env}}(Y))\subseteq\operatorname{\mathrm{Env}}(tY)$,
so fix $x\in\tilde{t}(\operatorname{\mathrm{Env}}(Y))$. Let $(T_{i})_{i\in I}$
be a net of isometries of $X$ that defines $\tilde{t}$. Suppose that
$(U_{j})_{j\in J}$ is an arbitrary sequence on
$\operatorname{\mathrm{Isom}}(X)$ such that $(U_{j}ty)_{j}$ converges for all
$y\in Y$. Without loss of generality we may assume that $I=J$: One can use the
product ordering $I\times J$ and define $T_{i,j}:=T_{i}$ and $U_{i,j}:=U_{j}$.
It follows that $(U_{i}\circ T_{i}y)_{i}$ converges for all $y\in Y$, so, by
hypothesis, $(U_{i}\circ T_{i}x)_{i}$ converges, or equivalently
$(U_{i}\tilde{t}(x))_{i}$ converges. This means that
$\tilde{t}(x)\in\operatorname{\mathrm{Env}}(tY)$. Now let $u:tY\to Y$ be the
inverse of $t$. We know that
$u\in\operatorname{\mathrm{Emb}}_{\mathrm{s}}(TY,X)$, hence,
$\tilde{u}\operatorname{\mathrm{Env}}(tY)\subseteq\operatorname{\mathrm{Env}}(utY)=\operatorname{\mathrm{Env}}(Y)$.
This means that if $x\in\operatorname{\mathrm{Env}}(tY)$, then
$\tilde{u}(x)\in\operatorname{\mathrm{Env}}(Y)$, and so
$x=\tilde{t}(\tilde{u}(x))\in\tilde{t}(\operatorname{\mathrm{Env}}(Y))$.

(3): By separability of $Y$, ${\rm Emb}(Y,X)$ is a metric space. Assume
$(t_{n})_{n}$ converges SOT to $t$ and let
$x\in\operatorname{\mathrm{Env}}(Y)$. Let $V_{n}$ be an element of ${\rm
Isom}(X)$ such that $d((V_{n})\upharpoonright{Y},t_{n})\leq 1/n$, where $d$ is
the SOT-convergence metric on ${\rm Emb}(Y,X)$, and so that
$\|V_{n}x-\tilde{t}_{n}x\|\leq 1/n$ (by definition of $\tilde{t}_{n}$). Then
note that $V_{n}$ converges SOT to $t$ on $Y$ and therefore $(V_{n}x)_{n}$
converges to $\tilde{t}x$ (by definition of $\tilde{t}$). So, $\tilde{t_{n}}x$
converges to $\tilde{t}x$ for all $x\in\operatorname{\mathrm{Env}}(Y)$. ∎

######  Corollary 2.33.

The following are equivalent for $Y\subseteq X$:

  1. (1)

$Y$ is strongly isometric to a full subspace of $X$.

  2. (2)

$\operatorname{\mathrm{Env}}(Y)$ is strongly isometric to $X$.

######  Proof.

Suppose that $t:Y\to X$ is a strong isometry such that $tY$ is a full
subspace. it follows from Proposition 2.32 (2) that
$\tilde{t}(\operatorname{\mathrm{Env}}(Y))=\operatorname{\mathrm{Env}}(tY)=X$,
hence $\tilde{t}:\operatorname{\mathrm{Env}}(Y)\to X$ is a (surjective) strong
isometry.

Suppose that $T:\operatorname{\mathrm{Env}}(Y)\to X$ is a surjective strong
isometry. Then $t:=T\upharpoonright
Y\in\operatorname{\mathrm{Emb}}_{\mathrm{s}}(Y,X)$, $\tilde{t}=T$, and it
follows from Proposition 2.32 (2) that
$X=T(\operatorname{\mathrm{Env}}(Y))=\operatorname{\mathrm{Env}}(T(Y))$, so
$TY$ is a full subspace of $X$. ∎

The envelope of $Y$ admits an important characterization when $Y$ is assumed
separable.

######  Proposition 2.34.

Assume $Y$ is a separable subspace of $X$. Then
$\operatorname{\mathrm{Env}}(Y)$ is the maximum among subspaces $Z$ of $X$
with the property that

  1. (1)

$Z$ contains $Y$.

  2. (2)

Any embedding $t$ in ${\rm Emb}_{s}(Y,X)$ extends uniquely to an embedding
$\tilde{t}$ in ${\rm Emb}_{s}(Z,X)$.

  3. (3)

The mapping $t\mapsto\tilde{t}$ is SOT-SOT continuous.

######  Proof.

The space $\operatorname{\mathrm{Env}}(Y)$ does satisfy (1), (2) and (3). Let
now $Z$ be such a subspace and $z\in Z$. If $(T_{n})_{n}$ is a sequence of
isometries on $X$ converging SOT on Y, let $t_{n}=T_{n|Y}\in{\rm
Emb}_{\mathrm{ext}}(Y,X)\subseteq{\rm Emb}_{\mathrm{s}}(Y,X)$ and
$t=\lim_{n}t_{n}$. By definition of $Z$, $(T_{n}\upharpoonright
Z)_{n}=(\tilde{t}_{n})_{n}$ tends SOT to $\tilde{t}$; therefore $(T_{n}z)_{n}$
converges. This proves that $z\in\operatorname{\mathrm{Env}}(Y)$ and therefore
that $Z\subseteq\operatorname{\mathrm{Env}}(Y)$. ∎

Propositions 2.32 and 2.34 lead us to see the class of subspaces of $X$ which
are isometric envelopes as a “strongly isometric skeleton” of $X$, in the
sense that it is a subfamily of subspaces of $X$ “containing all the
information” on strong isometric embeddings inside $X$.

Finally a relevant case of strong isometric embedding defined on a subspace
$Y$ are those whose image is equal to $Y$, and may therefore be seen as
surjective isometries on $Y$. In general, not all surjective isometries on $Y$
is of this form. However in a class of spaces that we will consider in the
next subsection 2.4 this is the case and we will see that is implies that
$Y\subseteq X$ is a $g$-embedding.

######  Definition 2.35 (Strong isometry).

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. We let
$\operatorname{\mathrm{Isom}}_{\mathrm{s}}(Y)$ be the set of strong isometries
on $Y$, i.e., of strong isometric embeddings of $Y$ whose image is $Y$.

Equivalently $t\in\operatorname{\mathrm{Isom}}(Y)$ is a strong isometry on $Y$
if and if only if it is a surjective isometry on $Y$ which belongs to ${\rm
Emb}_{s}(Y,X)$. In other words, elements of
$\operatorname{\mathrm{Isom}}_{\mathrm{s}}(Y)$ are isometries on $Y$ which, as
maps with value in $X$, belong to
$\operatorname{\mathrm{Emb}}_{\mathrm{s}}(Y,X)$. In what follows we always
consider the SOT.

######  Lemma 2.36.

Let $X$ be a Banach space and $Y$ a closed subspace of $X$. Then
$\operatorname{\mathrm{Isom}}_{\mathrm{s}}(Y)$ is a closed subgroup of
$\operatorname{\mathrm{Isom}}(Y)$.

######  Proof.

It is clearly closed. If $t,u$ are two such elements, with
$t=\lim_{i}T_{i}\upharpoonright Y$ and $u=\lim_{i}U_{i}\upharpoonright Y$,
then $({T_{i}U_{i}}\upharpoonright{Y})_{i}=(T_{i}(U_{i}\upharpoonright
Y-u)+T_{i}u)_{i}$ also tends SOT to $tu$. Likewise
$t^{-1}=\lim_{i}(T_{i}^{-1})\upharpoonright{Y}$. ∎

######  Corollary 2.37.

If $Y$ is a closed subspace of $X$, then
$t\in\operatorname{\mathrm{Isom}}_{\mathrm{s}}(Y)\mapsto\tilde{t}\in\operatorname{\mathrm{Isom}}_{\mathrm{s}}(\operatorname{\mathrm{Env}}(Y))$
is an embedding of topological groups.

######  Proof.

The group embedding property is clear. Let us denote this embedding by $e$.
Then let $t_{i}\in\operatorname{\mathrm{Isom}}_{s}(Y)$ converge to
$t\in\operatorname{\mathrm{Isom}}_{s}(Y)$, and let $T_{i}=e(t_{i})$ defined on
$\operatorname{\mathrm{Env}}(Y)$ extending $t_{i}$. Note that $T_{i}$
converges pointwise to $t$ on $Y$, so by Proposition 2.17,
$i_{\operatorname{\mathrm{Env}}(Y),X}T_{i}$ converges pointwise on
$\operatorname{\mathrm{Env}}(Y)$ to some $T\in{\rm
Emb}_{\mathrm{s}}(\operatorname{\mathrm{Env}}(Y))$ such that $T_{|Y}=t$. By
the uniqueness result (1) in Proposition 2.32, $T_{i}$ converges pointwise to
$e(t)$. ∎

###  2.4. Envelopes in AUH or Fraïssé spaces

Recall that a space $X$ is called ultrahomogeneous (UH) when
$\operatorname{\mathrm{Emb}}_{\mathrm{ext}}(Y,X)=\operatorname{\mathrm{Emb}}(Y,X)$
for every finite dimensional subspace $Y$ of $X$ ; $X$ is called approximately
ultrahomogeneous (AUH) when
$\operatorname{\mathrm{Emb}}_{\mathrm{s}}(Y,X)=\operatorname{\mathrm{Emb}}(Y,X)$
for such subspaces $Y$ of $X$.

A separable $\mathrm{(AuH)}$ space without finite cotype must be isometric to
the Gurarij space. This follows from the fact that two separable Fraïssé
spaces which are finitely representable into each other must be isometric [21,
Theorem 2.19]. In the finite cotype case let us also note that if a Banach
space $X$ is reflexive (or even just Asplund or with the RNP) and
$\mathrm{(AuH)}$, then, since $X$ is in particular almost transitive, then it
is also superreflexive and $X$ and $X^{*}$ are uniformly convex [15].

We obtain the following results regarding isometric embeddings in
$\mathrm{(AuH)}$ spaces, easy to prove directly from the corresponding results
for strong isometric embeddings in general spaces.

The next is Proposition 2.32 reformulated for $\mathrm{(AuH)}$ spaces:

######  Proposition 2.38.

Assume $X$ is $\mathrm{(AuH)}$, let $Y$ be a subspace of $X$, let $t\in{\rm
Emb}(Y,X)$ be an isometric embedding of $Y$ into $X$. Then

  1. (1)

$t$ extends uniquely to an isometric embedding $\tilde{t}\in{\rm
Emb}(\operatorname{\mathrm{Env}}(Y),X)$; and
$\tilde{t}(\operatorname{\mathrm{Env}}(Y))=\operatorname{\mathrm{Env}}(t(Y))$.

Assume that $Y$ is separable. Then,

  1. (2)

the map $t\mapsto\tilde{t}$ is SOT-SOT continuous.

  2. (3)

$\operatorname{\mathrm{Env}}(Y)$ is the maximum among spaces $Y^{\prime}$ such
that any isometric embedding $t:Y\rightarrow X$ extends uniquely to an
isometric embedding $\tilde{t}:Y^{\prime}\rightarrow X$, and such that
$t\mapsto\tilde{t}$ is SOT-SOT continuous. ∎

######  Definition 2.39 (linear $g$-embedding).

We say that a subspace $Y$ of $X$ is a linear $g$-embedding when is an
embedding of topological groups
$e:\operatorname{\mathrm{Isom}}(Y)\to\operatorname{\mathrm{Isom}}(X)$ such
that $e(h)\upharpoonright Y=h$ for every
$h\in\operatorname{\mathrm{Isom}}(Y)$.

This is the linear version of the notion of $g$-embedding of a subspace $M$ of
a metric space $L$ introduced in [53, Definition 3.1] (see also [46,
Definition 5.2.6]), that demands the existence of an embedding of topological
groups $e:\mathcal{I}so(M)\to\mathcal{I}so(L)$ such that $e(h)\upharpoonright
Y=h$ for every $h\in\mathcal{I}so(L)$, where $\mathcal{I}so(N)$ is the group
of isometries of $N$. Recall that the classical Mazur-Ulam Theorem states that
the group $\mathcal{I}so(Y)$ of surjective, not necessarily linear, isometries
on $Y$ is the group of the affine ones on $Y$. From this it follows easily
that if $e:\operatorname{\mathrm{Isom}}(Y)\to\operatorname{\mathrm{Isom}}(X)$
is an embedding of topological groups, then
$\bar{e}:\mathcal{I}so(Y)\to\mathcal{I}so(X)$ defined by
$\bar{e}(\gamma)=e(\gamma-\gamma(0))+\gamma(0)$ is also an embedding as
topological groups, and consequently every linear $g$-embedding is a
$g$-embedding.

We have the following direct consequence of Corollary 2.37.

######  Proposition 2.40.

Assume $X$ is $\mathrm{(AuH)}$ and let $Y$ be a subspace of $X$. Then

  1. (1)

any surjective isometry $t$ on $Y$ extends uniquely to a surjective isometry
$\tilde{t}$ on $\operatorname{\mathrm{Env}}(Y)$.

  2. (2)

The map $t\mapsto\tilde{t}$ defines a topological (group) embedding of
$\operatorname{\mathrm{Isom}}(Y)$ into
$\operatorname{\mathrm{Isom}}(\operatorname{\mathrm{Env}}(Y))$.

  3. (3)

this is the unique such map for which $\tilde{t}$ is an extension of $t$ for
each $t\in\operatorname{\mathrm{Isom}}(Y)$.

Consequently $Y\subseteq\operatorname{\mathrm{Env}}(X)$ is a linear
$g$-embedding. ∎

######  Corollary 2.41.

Suppose that $X$ is $\mathrm{(AuH)}$, let $Y\subseteq X$ be a subspace, and
let $\gamma\in\operatorname{\mathrm{Emb}}(Y,\operatorname{\mathrm{Env}}(Y))$.
Then $\gamma$ is such that $\operatorname{\mathrm{Env}}(\gamma
Y)=\operatorname{\mathrm{Env}}(Y)$ if and only if
$\gamma\in\operatorname{\mathrm{Emb}}_{\mathrm{ext}}(Y,\operatorname{\mathrm{Env}}(Y))$,
that is, if $\gamma$ is a restriction of a surjective isometry of
$\operatorname{\mathrm{Env}}(Y)$. In particular, any partial isometry between
full subspaces of $X$ extends to an isometry of $X$.

######  Proof.

For suppose that
$g\in\operatorname{\mathrm{Isom}}(\operatorname{\mathrm{Env}}(Y))$, and set
$\gamma:=g\upharpoonright Y$. Then, by uniqueness, $\bar{\gamma}=g$ and, by
Proposition 2.40, $\operatorname{\mathrm{Env}}(\gamma
Y)=\bar{\gamma}\operatorname{\mathrm{Env}}(Y)=g\operatorname{\mathrm{Env}}(Y)=\operatorname{\mathrm{Env}}(Y)$.
For suppose now that
$\gamma\in\operatorname{\mathrm{Emb}}(Y,\operatorname{\mathrm{Env}}(Y))$ is
such that $\operatorname{\mathrm{Env}}(\gamma
Y)=\operatorname{\mathrm{Env}}(Y)$. We claim that
$\bar{\gamma}\in\operatorname{\mathrm{Isom}}(\operatorname{\mathrm{Env}}(Y))$:
First of all,
$\bar{\gamma}(\operatorname{\mathrm{Env}}(Y))=\operatorname{\mathrm{Env}}(\gamma
Y)\subseteq\operatorname{\mathrm{Env}}(Y)$, so
$\bar{\gamma}\in\operatorname{\mathrm{Emb}}(\operatorname{\mathrm{Env}}(Y),\operatorname{\mathrm{Env}}(Y))$.
Let $\eta\in\operatorname{\mathrm{Emb}}(\gamma
Y,\operatorname{\mathrm{Env}}(Y))$ be such that $\eta(\gamma Y)=Y$ and
$\eta\circ\gamma=\operatorname{\mathrm{Id}}_{Y}$. Then, similarly we get that
$\bar{\eta}\in\operatorname{\mathrm{Emb}}(\operatorname{\mathrm{Env}}(\gamma
Y),\operatorname{\mathrm{Env}}(\gamma
Y))=\operatorname{\mathrm{Emb}}(\operatorname{\mathrm{Env}}(Y),\operatorname{\mathrm{Env}}(Y))$.
Since $\bar{\eta}\circ\bar{\gamma}\upharpoonright
Y=\eta\circ\gamma=\operatorname{\mathrm{Id}}_{Y}$, it follows by uniqueness of
the extensions that
$\bar{\eta}\circ\bar{\gamma}=\operatorname{\mathrm{Id}}_{\operatorname{\mathrm{Env}}(Y)}$,
and consequently $\bar{\eta}$ is surjective and $\bar{\gamma}$ as well. ∎

Observe that a subspace $Y$ of $X$ and $gY$ for
$g\in\operatorname{\mathrm{Isom}}(X)$ are placed similarly inside $X$ because
both $Y\equiv gY$ but also the quotient map $\widehat{g}:X/Y\to X/(gY)$,
$x+Y\mapsto g(x)+gY$ is a surjective isometry compatible with
$g\upharpoonright Y:Y\to gY$. In this way, any of the
$\operatorname{\mathrm{Isom}}(X)$-orbits of the action by composition in
$\operatorname{\mathrm{Emb}}(Y,X)$ is called an isometric position of $Y$
inside $X$. In this way, the previous Corollary says that when $X$ is
$\mathrm{(AuH)}$ the isometric position defined by the inclusion
$i:Y\to\operatorname{\mathrm{Env}}(Y)$ consists of the isometric embeddings
$\gamma:Y\to\operatorname{\mathrm{Env}}(Y)$ such that
$\operatorname{\mathrm{Env}}(\gamma Y)=\operatorname{\mathrm{Env}}(Y)$. From a
homological point of view, the exact sequences $0\rightarrow Y\rightarrow
E(Y)\rightarrow E(Y)/Y\rightarrow 0$ and $0\rightarrow\gamma Y\rightarrow
E(\gamma Y)\rightarrow E(\gamma Y)/\gamma Y\rightarrow 0$ are isometrically
equivalent (in the sense that the extension of $\gamma$ to a map between the
envelopes making the diagram commute is an isometric map), and the associated
quotient $E(Y)/Y$ is isometrically unique (in the sense that $E(\gamma
Y)/\gamma Y$ is independent of $\gamma$).

######  Definition 2.42 (position).

Given a $\mathrm{(AuH)}$ space $X$, the position of a subspace $Y$ inside its
envelope $\operatorname{\mathrm{Env}}(Y)$ is the
$\operatorname{\mathrm{Isom}}(\operatorname{\mathrm{Env}}(Y))$-orbit of the
inclusion $i:Y\to\operatorname{\mathrm{Env}}(Y)$.

The full position of a subspace $Y$ of $X$ is (if exists) the position of some
(any) full isometric copy of $X$. We call full quotient of $X$ by $Y$ the
isometrically unique associated quotient $X/Y$.

This definition was inspired by a notion of isomorphic position related to the
so-called automorphic space problem, see [13]. It is worth mentioning that it
does not seem possible to extend the result of uniqueness of full position
inside the space $L_{1}$ to the context of “almost-isometric” position: It is
proved in [25, Theorem V.1 and Remark (2) on page 284] that there is a
subspace $X$ of $L_{1}$ such that (a) every isometric embedding of $X$ into
$L_{1}$ extends uniquely to a surjective isometry on $L_{1}$, but (b) for
every $\delta>0$ there is a $\delta$-embedding $\gamma_{\delta}:X\to L_{1}$
such that
$\inf_{\delta>0}d_{\mathrm{BM}}(L_{1}/X,L_{1}/\gamma_{\varepsilon}(X))>1$,
where $d_{\mathrm{BM}}$ is the Banach-Mazur (multiplicative) distance.

We finish this subsection with some additional observations on
$\mathrm{(AuH)}$ with a stronger extension property. Recall that ${\rm
Emb}_{\delta}(E,X)$ is the collection of all linear mappings $\gamma:E\to X$
such that $(1+\delta)^{-1}\|e\|\leq\|\gamma e\|\leq(1+\delta)\|e\|$ for every
$e\in E$.

######  Definition 2.43 (Fraïssé space).

A Banach space $X$ is Fraïssé when for every $\varepsilon>0$ and every
$k\in{\mathbb{N}}$, there is a $\delta>0$ such that for every $k$-dimensional
subspace $E$ of $X$ and every $t\in{\rm Emb}_{\delta}(E,X)$, there exists
$T\in\operatorname{\mathrm{Isom}}(X)$ such that $\|T\upharpoonright
E-t\|\leq\varepsilon$.

The following is the main conjecture of [21].

######  Conjecture 2.44.

The spaces $L_{p}$ for $p\neq 4,6,..$ and the Gurarij space $\mathbb{G}$ are
the only separable Fraïssé or even $\mathrm{(AuH)}$ spaces.

It was proved in [21, Proposition 2.13.] that a Fraïssé space must contain an
isometric copy of the Hilbert space. It remains open whether any
$\mathrm{(AuH)}$ space is Fraïssé, or simply whether it contains an isometric
copy of the Hilbert. But even in the $\mathrm{(AuH)}$ case, Hilbertian
subspaces seem to be particularly relevant to the study of envelopes. It
should be noted that we do not know whether every separable Fraïssé space
contains a full copy of the Hilbert space. In Section 3 we shall give a
positive answer in the case of the Fraïssé $L_{p}$-spaces. For more general
Fraïssé spaces we have the following:

######  Proposition 2.45.

Consider $X$ an $\mathrm{(AuH)}$ Banach space and $Y$ a full Hilbertian
subspace of $X$. Then $Y$ is a minimal full subspace of $X$ and a maximal
Hilbertian subspace of $X$.

######  Proof.

Let $Y$ be some full Hilbertian subspace of $X$. Note that no proper subspace
$Z$ of $Y$ has full envelope: Indeed the identity embedding $i_{Z,X}$ of $Z$
into $X$ admits several extensions as an isometric embedding of $Y$ into $X$
and therefore several extensions as an isometric embedding of $X$ into $X$;
while the unique isometric embedding of $\operatorname{\mathrm{Env}}(Z)$
extending $i_{Z,X}$ is $i_{\operatorname{\mathrm{Env}}(Z),X}$ by Proposition
2.38 (1). On the other hand all superspaces of $Y$ have full envelope.
Therefore $Y$ is minimal with full envelope.

This also means that $Y$ is maximal Hilbertian inside $X$. Indeed a non-
trivial Hilbertian extension of $Y$ would have full envelope and would not be
a minimal full subspace, contradicting the first assertion of the proposition.
∎

##  3\. Envelopes in rearrangement invariant spaces and $L_{p}$-spaces

In this section we identify the envelope of subspaces of $L_{p},1\leq
p<+\infty,p\neq 2,4,6,\ldots$. We start with facts valid for general reflexive
spaces, and then for r.i. spaces on $[0,1]$ (also called symmetric by Peller
[45]). We start with the following result that allows to compute the Korovkin
envelope.

######  Proposition 3.1.

Assume $X$ is reflexive LUR. Let $S$ be a semigroup of contractions on $X$.
Let $Y\subseteq X$ be the closure of a directed sequence $(Y_{i})_{i\in I}$ of
subspaces of $X$. Then

$\operatorname{\mathrm{Env}}_{S}(Y)=\overline{\bigcup_{i\in
I}\operatorname{\mathrm{Env}}_{S}(Y_{i})}.$

######  Proof.

Only the direct inclusion is not trivial. For each $i\in I$ let $p_{i}$ be a
contractive projection $p_{i}\in\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}$
onto $\operatorname{\mathrm{Env}}_{S}(Y_{i})$ provided by Proposition 2.23.
Let $p$ be a WOT-cluster point of $\\{p_{i}\\}_{i\in I}$, and note that
$p\in\overline{\mathrm{conv}(S)}^{\mathrm{WOT}}$ and that $p$ is a contractive
projection onto $\overline{\bigcup_{i\in
I}\operatorname{\mathrm{Env}}_{S}(Y_{i})}$. In particular, $p\upharpoonright
Y=\operatorname{\mathrm{Id}}_{Y}$. Therefore by Remark 2.21, the range of $p$
contains $\operatorname{\mathrm{Env}}_{S}(Y)$. ∎

######  Remark 3.2.

It does not seem that the previous proposition holds for the algebraic
envelope: Assume now that $Y\subseteq X$ is the closure of a directed sequence
$(Y_{i})_{i\in I}$ of subspaces of $X$. It is clear that
$\operatorname{\mathrm{Env}}_{S}^{\rm alg}(Y)\supseteq\overline{\bigcup_{i\in
I}\operatorname{\mathrm{Env}}_{S}^{\rm alg}(Y_{i})}$. As for the direct
inclusion, assume that $d(x,\overline{\bigcup_{i\in
I}\operatorname{\mathrm{Env}}_{S}^{\rm alg}(Y_{i})})>\epsilon$. We use the
following claim.

######  Claim 3.2.1.

Assume $X$ is reflexive strictly convex. Let $S$ be a semigroup of
contractions on $X$ and let $Y\subseteq X$. If
$d(x,\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y))>\epsilon$, then there
exists $s\in S$ such that $s\upharpoonright{Y}=\operatorname{\mathrm{Id}}_{Y}$
and $d(x,sx)>\epsilon$.

To prove this, by Proposition 2.10, let $p$ be a contractive projection on
$\operatorname{\mathrm{Env}}_{S}^{\mathrm{alg}}(Y)$ belonging to the set
$\overline{{\rm conv}({\rm Stab}_{S}(Y))}^{\rm WOT}$. Note that for such $x$
one has that $\|x-px\|>\epsilon$; therefore $\phi(x-px)>\epsilon$ for some
norm $1$ functional $\phi$. Since $p\in\overline{\mathrm{conv}({\rm
Stab}_{S}(Y))}^{\mathrm{WOT}}$, there is some $T\in\mathrm{conv}({\rm
Stab}_{S}(Y))$ such that $\phi(x-Tx)>\epsilon$ and therefore some $s\in{\rm
Stab}_{S}(Y)$ such that $\phi(x-sx)>\epsilon$.

Once this is established, we can get $(s_{i})_{i\in I}$ with $s_{i}$ acting as
the identity on $Y$ and $d(x,s_{i}x)>\epsilon$. Under appropriate hypotheses
one would obtain a WOT-cluster point $T$ of $(s_{i})_{i}$ with
$T\upharpoonright{Y}=\operatorname{\mathrm{Id}}_{Y}$ and $d(x,Tx)>\epsilon$.
If $T$ were an element of $S$, this would prove that $x$ is not in
$\operatorname{\mathrm{Env}}_{S}^{\rm alg}(Y)$ ; but $T$ only belongs to
$\overline{S}^{\rm WOT}$ a priori. This perhaps explains why the result in the
desired equality was peoved for the Korovkin envelope In the case of
$X=L_{p}$, $p\neq 2$, we know the following. Suppose that
$Y=L_{p}([0,1],\Sigma)$, and $(\Sigma_{n})_{n}$ is an increasing sequence of
finite subalgebras whose union is dense in $\Sigma$. Then
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)=\overline{\bigcup_{n}\operatorname{\mathrm{Env}}^{\mathrm{alg}}(L_{p}([0,1],\Sigma_{n}))}$
only holds when $\Sigma$ is a fixed-point subalgebra (see Remark 3.6), because
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(L_{p}([0,1],\Sigma_{n}))=L_{p}([0,1],\Sigma_{n})$
(Lemma 3.5) and consequently the required equality means that
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)=Y$.

Let us note, using Lancien’s theorem about the existence of LUR renormings
preserving the isometry group in separable reflexive Banach spaces [36], we
have the following.

######  Corollary 3.3.

Assume $X$ is reflexive and either separable or LUR. Let $Y\subseteq X$ be the
closure of a directed sequence $(Y_{i})_{i\in I}$ of subspaces of $X$. Then

$\operatorname{\mathrm{Env}}(Y)=\overline{\bigcup_{i\in
I}\operatorname{\mathrm{Env}}(Y_{i})}.$

We do not know if the above is true for the algebraic envelope.

###  3.1. Envelopes in rearrangement invariant spaces

In this subsection we consider rearrangement invariant (r.i.) spaces on
$[0,1]$ as defined in [38, Definition 2.a.1.]. It is well-known that reflexive
r.i. spaces on $[0,1]$ are separable (see for example [38, pp 118-119]).

######  Definition 3.4.

For $X$ a r.i. space on $I=[0,1]$, and every $\sigma$-subalgebra $\Sigma$ of
measurable subsets of $I$, let $X_{\Sigma}$ be the subspace of $X$ consisting
of the $\Sigma$-measurable functions in $X$.

Note that $X_{\Sigma}$ is a unital space.

######  Lemma 3.5.

If $X$ is a r.i. space on $[0,1]$, and $Y$ is a finite dimensional unital
sublattice, then $\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)=Y$.

######  Proof.

The sublattice $Y$ is generated by the characteristic functions of $n$
disjoint measurable subsets $(A_{i})$ partitioning $[0,1]$. Let $f$ be such
that $T(f)=f$ for all $T\in{\rm Stab}_{\operatorname{\mathrm{Isom}}(X)}(Y)$.
We claim that $f\upharpoonright A_{i}$ is constant for each $i$, and
consequently, $f\in Y$. We prove the claim by contradiction. We can find
$c\in{\mathbb{R}}$, and $B,C$ of positive (equal) measure in some $A_{i}$ such
that $f\geq c$ on $B$ and $f<c$ on $C$. Let $\sigma$ be a measure isomorphism
such that $\sigma(B)=C$, and $\sigma(A_{j})=A_{j}$ for all $j=1,\dots,n$.
Observe that the isometry $T_{\sigma}$ determined by $\sigma$ belongs to ${\rm
Stab}_{\operatorname{\mathrm{Isom}}(X)}(Y)$, but $T_{\sigma}(f)\neq f$,
because $T_{\sigma}(f)\upharpoonright C$ is a function with value $\geq c$
while $f\upharpoonright C$ has always value $<c$. This is impossible since we
are assuming that $f\in\mathrm{Env}^{\mathrm{alg}}(Y)$. ∎

######  Remark 3.6.

Recall that a $\sigma$-subalgebra $\mathcal{B}$ of a measure algebra
$(\mathcal{A},\mu)$ is called a fixed-point subalgebra when there is a set
$\Gamma$ of measure-preserving isomorphisms of $\mathcal{A}$ such that
$\mathcal{B}=\\{{a\in A}\,:\,{\pi a=a\text{ for every $\pi\in\Gamma$}}\\}$
(see [24, Section 333]). This subalgebras are characterized in [24, Theorem
333R], where in particular is shown that fixed-point subalgebras can be
determined by a single measure-preserving isomorphism of $(\mathcal{A},\mu)$.
It is straightforward to see that A unital sublattice $Y$ of $L_{p}[0,1]$,
$p\neq 2$, satisfies that $\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)=Y$
exactly when $Y=L_{p}([0,1],\Sigma)$ for some fixed-point subalgebra of the
Borel algebra $\mathcal{B}([0,1])$. Moreover, we will see in Theorem 3.12 that
if $Y\subseteq L_{p}$, $p\neq 2$, is a unital subspace, then
$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)=\operatorname{\mathrm{Env}}(Y)=L_{p}([0,1],\Sigma_{Y})$,
and $\operatorname{\mathrm{Env}}^{\mathrm{alg}}(Y)=L_{p}([0,1],\Sigma_{Y})$
iff $\Sigma_{Y}$ is a fixed-point subalgebra.

Note that by [51, Theorem 5.25], this is a $1$-complemented subspace.

######  Proposition 3.7.

If $X$ is a reflexive r.i. space on $I=[0,1]$, then for any
$\sigma$-subalgebra $\Sigma$ one has that
$\operatorname{\mathrm{Env}}(X_{\Sigma})=X_{\Sigma}$.

######  Proof.

Each $\sigma$-algebra $\Sigma$ is separable, hence we can find an increasing
sequence $(\Delta_{n})_{n}$ of finite subalgebras of $\Sigma$ whose union is
dense (see [29]), and consequently $\bigcup_{n}X_{\Delta_{n}}$ is dense in
$X_{\Sigma}$, because $X$ is a r.i. space. Since each $X_{\Delta_{n}}$ is a
finite dimensional unital sublattice, it follows from the previous lemma that
$\operatorname{\mathrm{Env}}^{\mathrm{alg}}(X_{\Delta_{n}})=X_{\Delta_{n}}=\operatorname{\mathrm{Env}}(X_{\Delta_{n}})$.
We know that $X$ is separable, so by Corollary 3.3 we deduce that
$\operatorname{\mathrm{Env}}(X_{\Sigma})=\overline{\bigcup_{n}\operatorname{\mathrm{Env}}(X_{\Delta_{n}})}=\overline{\bigcup_{n}X_{\Delta_{n}}}=X_{\Sigma}$.
∎

Note that holds: $X_{\cap_{i}\Sigma_{i}}=\bigcap_{i}X_{\Sigma_{i}}$. Therefore
we may define the following envelope.

######  Definition 3.8 (Conditional envelope).

Suppose that $X$ is a r.i. space on $I=[0,1]$. For $Y\subseteq X$, let
$\Sigma_{Y}$ be the smallest $\sigma$-algebra making all functions of $Y$
measurable. The conditional envelope
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$ is the space $X_{\Sigma_{Y}}$ of
$\Sigma_{Y}$-measurable functions of $X$.

The name conditional we chose is due to the fact that
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=L_{p}([0,1],\Sigma_{Y})$ is the
range of the conditional expectation projection $\mathcal{E}^{\Sigma_{Y}}$.

Note that for reflexive r.i. spaces over $[0,1]$ this defines an envelope
which is $1$-complemented. Observe that
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$ is always unital. When
$X=L_{p}[0,1]$, $X_{\Sigma}$ is the Lebesgue space $L_{p}([0,1],\Sigma)$, and
consequently
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=L_{p}([0,1],\Sigma_{Y})$. We have
also the following classical result.

######  Proposition 3.9.

Suppose that $1\leq p<\infty$, $p\neq 2$, and $Y$ is a unital subspace of
$L_{p}[0,1]$. Then,

$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\mathrm{Lat}(Y),$

where ${\rm Lat}(Y)$ is the lattice generated by $Y$.

######  Proof.

We first prove that
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)$.
We know that conditional expectation operator $\mathcal{E}^{\Sigma_{Y}}$ is a
contractive projection whose range is
$L_{p}([0,1],\Sigma_{Y})=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$. On the
other hand, classical results by Douglas [19] (for $p=1$) and by Ando [4] (for
the rest of $p$’s) states that a contractive projection that fixes
$\mathbbm{1}_{[0,1]}$ must be a conditional expectation defined by some
$\sigma$-subalgebra. Hence any contractive projection $q$ whose range $R(q)$
contains $Y$ satisfies $R(q)=L_{p}([0,1],\Sigma)$ for some $\sigma$-subalgebra
$\Sigma$. Since $Y\subseteq R(q)$, it follows that
$\Sigma_{Y}\subseteq\Sigma$, so
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=L_{p}([0,1],\Sigma_{Y})\subseteq
L_{p}([0,1],\Sigma)=R(q)$. This proves that $\operatorname{\mathrm{Env}}_{\rm
c}(Y)=\operatorname{\mathrm{Env}}_{\rm min}(Y)$ (including that the latter
exists when $p=1$).

We now prove that
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\mathrm{Lat}(Y)$. We start with
the following.

######  Claim 3.9.1.

Suppose that $Z$ is a unital sublattice. Then
$\Sigma_{Z}=\\{{A}\,:\,{\mathbbm{1}_{A}\in Z}\\}$ and
$Z=L_{p}([0,1],\Sigma_{Z})$.

###### Proof of Claim:.

Set $\Delta:=\\{{A}\,:\,{\mathbbm{1}_{A}\in Z}\\}$. Obviously,
$\Delta\subseteq\Sigma_{Z}$. Observe that in general for an arbitrary subspace
$V$ one has that $\Sigma_{V}$ is the $\sigma$-subalgebra generated by the sets
$\\{f>c\\}$ for $c>0$ and $f\in V$, and when $V$ is assumed to be lattice then
those $f\in V$ can be assumed to be positive. Fix now a positive $f\in Z$,
$c>0$, and set $g:=(f-c\mathbbm{1}_{[0,1]})^{+}\in Z$. Then,

$\mathbbm{1}_{\\{f>c\\}}=\mathbbm{1}_{\\{g>0\\}}=(\sup_{n\in{\mathbb{N}}}n\cdot
g)\wedge\mathbbm{1}_{[0,1]}\in Z,$

and consequently $\\{f>c\\}\in\Delta$. By the previous remark,
$\Sigma_{Z}\subseteq\Delta$. Finally, trivially we have that $Z\subseteq
L_{p}([0,1],\Sigma_{Z})$; as for the reverse inclusion, observe that it
follows from the equality $\Sigma_{Z}=\\{{A}\,:\,{\mathbbm{1}_{A}\in Z}\\}$
that every simple function in $L_{p}([0,1],\Sigma_{Z})$ belongs to $Z$, hence
$L_{p}([0,1],\Sigma_{Z})\subseteq Z$. ∎

We have by minimality that $\mathrm{Lat}(Y)\subseteq
L_{p}([0,1],\Sigma_{Y})=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$. Since
$\mathrm{Lat}(Y)$ is 1-complemented, it follows that
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)\subseteq\mathrm{Lat}(Y)$
and we are done. ∎

######  Lemma 3.10.

If $X$ is a reflexive r.i. space on $[0,1]$, then for any unital subspace $Y$
one has that

$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)\subseteq\operatorname{\mathrm{Env}}(Y)\subseteq\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y).$

######  Proof.

From Proposition 2.24 and Proposition 2.28 we know that
$\operatorname{\mathrm{Env}}(Y)$ is 1-complemented, hence
$\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)\subseteq\operatorname{\mathrm{Env}}(Y)$.
Also, it follows from Proposition 3.7 that
$\operatorname{\mathrm{Env}}(\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y))=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$
and consequently,
$\operatorname{\mathrm{Env}}(Y)\subseteq\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)$.
∎

######  Lemma 3.11.

If $Y$ is a unital subspace of $L_{1}$, then ${\rm
Lat}(Y)\subseteq\operatorname{\mathrm{Env}}(Y)$.

######  Proof.

By the equimeasurability formula, any isometric map $t$ from $Y$ into $L_{1}$
extends uniquely to an isometric map $T$ on ${\rm Lat}(Y)$ by the formula

$T(|f|):=\frac{t(\mathbbm{1})}{|t(\mathbbm{1})|}|t(f)|.$

So, fix $f\in Y$, and assume $(t_{n})_{n}$ tends SOT to $t$ on $Y$. We claim
that $(t_{n}(|f|))_{n}$ converges to $T(|f|)$, that by the characterization of
$\operatorname{\mathrm{Env}}(Y)$ will mean that
$|f|\in\operatorname{\mathrm{Env}}(Y)$. Let us prove this claim. By composing
by an isometry acting by change of signs, it is enough to prove the claim when
$t(\mathbbm{1})$ is non-negative, and by composing with an isometric embedding
of $L_{1}(\mathrm{supp\,}(t(\mathbbm{1})))$ onto $L_{1}$ we may assume that
$t(\mathbbm{1})$ is positive, and therefore that $T(|f|)=|t(f)|$. We compute:

$T_{n}(|f|)-T(|f|)=\frac{t_{n}(\mathbbm{1})}{|t_{n}(\mathbbm{1})|}|t_{n}(f)|-|t(f)|=\frac{t_{n}(\mathbbm{1})}{|t_{n}(\mathbbm{1})|}(|t_{n}(f)|-|t(f)|)+(\frac{t_{n}(\mathbbm{1})}{|t_{n}(\mathbbm{1})|}-\mathbbm{1})|t(f)|$
(3)

The norm of the first part of the sum is at most
$\int||t_{n}(f)|-|t(f)||\leq\int|(t_{n}-t)(f)|$, and this tends to $0$. Since
$(t_{n}(\mathbbm{1}))_{n}$ tends to $t(\mathbbm{1})$ in the $L_{1}$-norm, the
measure of the sets $A_{n}:=\\{s\in[0,1]\,:\,t_{n}(1)(s)\leq 0\\}$ tend to
$0$. The norm of the second summand in (3) is controlled by
$2\int_{A_{n}}|t(f)|$ and therefore tends to $0$ as well. ∎

######  Theorem 3.12.

If $1\leq p<+\infty,p\neq 2$ and $Y$ is a unital subspace of $L_{p}$ then

$\operatorname{\mathrm{Env}}(Y)={\rm
Lat}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y).$

######  Proof.

For $p>1$, we know from Proposition 3.9 and Lemma 3.10 that

${\rm
Lat}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y)\subseteq\operatorname{\mathrm{Env}}(Y)\subseteq\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)={\rm
Lat}(Y).$

The proof for $p=1$ uses the Mazur maps. For some fixed $1<p\neq 2$ we let
$\phi:S_{L_{1}}\rightarrow S_{L_{p}}$ denote the Mazur map, extend it by
homogeneity to a map between $L_{1}$ and $L_{p}$. Observe that
$T\in\operatorname{\mathrm{Isom}}(L_{1})$ if and only if $\phi
T\phi^{-1}\in\operatorname{\mathrm{Isom}}(L_{p})$. Let us also note that if
$Y$ is a unital lattice of $L_{1}$ then $\phi(Y)=L_{p}\cap Y$ which is a
unital sublattice of $L_{p}$. As we have seen in Claim 3.9.1, there is a 1-1
correspondence between the class of unital sublattices of $L_{p}$ and the
class of sub-$\sigma$-algebras of Borel subsets of $[0,1]$,

$Y\mapsto\Sigma_{Y}=\\{A\in{\mathcal{B}}:1_{A}\in Y\\}$

and

$\Sigma\mapsto Y_{\Sigma}=L_{p}([0,1],\Sigma).$

To avoid misinterpretations, we use the terminology
$\operatorname{\mathrm{Env}}_{(L_{p})}(Z)$ to denote the isometric envelope of
some $Z\subseteq L_{p}[0,1]$ in $L_{p}[0,1]$.

######  Claim 3.12.1.

$\phi(\operatorname{\mathrm{Env}}_{(L_{1})}(Y)\subseteq\operatorname{\mathrm{Env}}_{L_{p}}(\phi(Y))$.

Admitting the claim, we deduce that if $Z\subseteq L_{1}[0,1]$ is a unital
sublattice, then so is $\phi(Z)$, hence
$\phi(\operatorname{\mathrm{Env}}_{(L_{1})}(Z))\subseteq\operatorname{\mathrm{Env}}_{(L_{p})}(\phi(Z))=\phi(Z)$,
and consequently $\operatorname{\mathrm{Env}}_{(L_{1})}(Z)=Z$, because $\phi$
is 1-1. This implies that
$\operatorname{\mathrm{Env}}(Y)\subseteq\operatorname{\mathrm{Env}}({\rm
Lat}(Y))={\rm Lat}(Y)$, and the equality $\operatorname{\mathrm{Env}}(Y)={\rm
Lat}(Y)$ follows from Lemma 3.11. The rest of the equalities are in Theorem
3.9.

To prove the claim, assume $(T_{i})_{i}$ is a net of isometries of $L_{p}$ so
that $(T_{i}(g))_{i}$ tends to $g$ for all $g\in\phi(Y)$. This is equivalent
to that $(\phi U_{i}(y))_{i}$ tends to $\phi(y)$ for all $y$ in $Y$ where each
$U_{i}:=\phi^{-1}T_{i}\phi$ is the associated isometry on $L_{1}$ to $T_{i}$.
Since the Mazur map is an homeomorphism (see [40]), this is equivalent to that
$(U_{i})_{i}$ converges to $\operatorname{\mathrm{Id}}$ pointwise on $Y$,
which, since $f\in\operatorname{\mathrm{Env}}_{(L_{1})}(Y)$, implies that
$(U_{i}(f))_{i}$ tends to $f$. Therefore $(T_{i}\phi(f))_{i}$ converges to
$\phi f$, because $T_{i}\phi(f)=\phi U_{i}(f)$. ∎

######  Remark 3.13.

It follows easily that the previous proposition is true for the Lebesgue
spaces $L_{p}(A)$ for every Borel subset $A$ of the unit interval. Hence, if
$Y\subseteq L_{p}[0,1]$, $p\neq 2$, is a subspace such that
$\mathbbm{1}_{A}\in Y$ where $A$ is the support of $Y$, then
$\operatorname{\mathrm{Env}}(Y)={\rm
Lat}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=\operatorname{\mathrm{Env}}_{\mathrm{min}}(Y).$
This is so, because all these envelopes on the space $L_{p}(A)$ are equal to
their versions in the full space $L_{p}[0,1]$.

######  Remark 3.14.

In the case $1<p<+\infty$ this result can also be deduced from Proposition
2.24(b) and the description by Peller of the WOT closure ${\mathcal{M}}$ of
the isometry group of $L_{p}$ [45], from which it follows that any contractive
projection on $L_{p}$ belongs to ${\mathcal{M}}$. There does not seem to be
such a direct proof in the case $p=1$, since our WOT characterization of the
envelope is, apparently, only valid in the reflexive case.

We have the following (partial) extension of Theorem 3.12.

######  Proposition 3.15.

Suppose that $1\leq p<+\infty$, and let $Y$ be a subspace of $L_{p}$. Then we
have that

  1. (1)

$\operatorname{\mathrm{Env}}(Y)=\operatorname{\mathrm{Env}}_{\rm min}(Y)$.

  2. (2)

If $p$ is not even, and $g\in Y$ is of full support in $Y$ then

$\operatorname{\mathrm{Env}}(Y)=g\cdot\operatorname{\mathrm{Env}}(Y/g)=g\cdot{\rm
Lat}(Y/g)=g\cdot\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y/g).$

######  Proof.

Pick $g$ of full support in $Y$. To prove (1) first assume $Y$ has full
support. Let $T$ be an isometry on $L_{p}$ sending $g$ to $1$, so that $TY$ is
unital. Then (1) follows from the equality
$\operatorname{\mathrm{Env}}(TY)=\operatorname{\mathrm{Env}}_{\rm min}(TY)$
(Proposition 3.12), and from Remarks 2.4 and before Example 2.18 about
preservation of the minimal and isometric envelopes by isometries. When $Y$
does not have full support the proof is similar using Remark 3.13.

The last two equalities in (2) for every $p\neq 2$ also follow from Remark
3.13. We have to show that
$\operatorname{\mathrm{Env}}(Y)=g\operatorname{\mathrm{Env}}(Y/g)$ for non-
even $p$’s. It is well-known that for every $1\leq p<\infty$, the envelope
$\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y)=L_{p}([0,1],\Sigma_{Y})$ is the
closure of the subspace of functions $B(f_{1},\dots,f_{n})$ where
$f_{1},\dots,f_{n}\in Y$ and $B:{\mathbb{R}}^{n}\to{\mathbb{R}}$ is Borel.
Suppose now that $p\notin 2{\mathbb{N}}$, fix $Y\subseteq L_{p}[0,1]$ and
$g\in Y$ with full support. Let $t$ be an isometry between a unital subspace
$Z$ of $L_{p}$ and $Y$ sending $\mathbbm{1}$ to $g$. Since $L_{p}[0,1]$ is an
approximately ultrahomogeneous space, it follows that there is a unique
extension $T$ of $t$ from $\operatorname{\mathrm{Env}}(Z)$ onto
$\operatorname{\mathrm{Env}}(Y)$. In fact, $T$ is defined by

$T(B(f_{1},\dots,f_{n}))=g\cdot B(tf_{1}/g,\dots,tf_{n}/g)$

for every $f_{1},\dots,f_{n}\in Z$ and every Borel function
$B:{\mathbb{R}}^{n}\to{\mathbb{R}}$ (see [33, Theorem 4]). Hence,

$\displaystyle\operatorname{\mathrm{Env}}(Y)=$ $\displaystyle
T(\operatorname{\mathrm{Env}}(Z))=\\{{T(B(f_{1},\dots,f_{n}))}\,:\,{f_{1},\dots,f_{n}\in
Z,B\text{ Borel }}\\}=$ $\displaystyle=$ $\displaystyle
g\cdot\\{{B(g_{1}/g,\dots g_{n}/g)}\,:\,{g_{1}\dots,g_{n}\in Y,B\text{
Borel}}\\}=g\cdot L_{p}(\mathrm{supp\,}g,\Sigma_{Y/g})=$ $\displaystyle=$
$\displaystyle g\cdot
L_{p}([0,1],\Sigma_{Y/g})=g\cdot\operatorname{\mathrm{Env}}_{\mathrm{c}}(Y/g)=g\cdot\operatorname{\mathrm{Env}}(Y/g).$

∎

Finally we may observe the following, as a consequence of the description by
Peller of the WOT-closure of the isometry group in separable r.i. spaces on
$[0,1]$ different from the $L_{p}$’s. Recall that a contractive map on a r.i.
space on $[0,1]$ is absolute if it defines a contraction on $L_{1}$ and on
$L_{\infty}$.

######  Theorem 3.16.

The $L_{p}$’s, $1<p<+\infty$ are the unique reflexive r.i. spaces on $[0,1]$
for which all $1$-complemented subspaces ($1$-dimensional subspaces is enough)
are envelopes.

######  Proof.

Let $X$ be a reflexive r.i. space on $[0,1]$. In particular, $X$ is separable.
The assertion that all $1$-dimensional subspaces are envelopes implies, by
Proposition 2.28, that any $1$-dimensional subspace is the range of a
contractive projection belonging to the WOT-closure of the convex hull of the
isometry group. On the other hand, it is a classical result (see [57, Theorem
2]) that if $X$ is different from one of the $L_{p}$’s, then it follows that
every isometry of $X$ is absolute. Since absolute contractions are closed
under convex combinations and WOT-limits, it follows that under our hypothesis
every 1-dimensional subspace of $X$ is 1-complemented by an absolute
contractive projection, and we are going to see (probably a well-known result)
in the next claim that this is impossible.

######  Claim 3.16.1.

Every reflexive r.i. space on $[0,1]$ admits a $1$-dimensional subspace which
is not the range of an absolute projection.

###### Proof of Claim:.

Assume otherwise. We are going to see that if $u$ and $v$ are disjointly
supported elements of $L_{\infty}$ and $\|u\|_{\infty}>\|v\|_{\infty}$, then
$\|u+v\|=\|u\|$. Since $\|v\|\leq\|u+v\|$ this implies that actually
$\|u+v\|=\max(\|u\|,\|v\|)$; by induction we would find an isometric copy of
$\ell_{\infty}$ in $X$, which is impossible by hypothesis, contradiction. Let
us see that $\|u+v\|=\|u\|$. We may assume without loss of generality that
$x_{0}:=u+v$ has norm $1$ in $X$. Let $P$ define an absolute projection with
range equal to the span of $x_{0}$. Since $P$ is a rank $1$ projection on $X$,
it must satisfy the formula $P(x)=\langle\phi_{0},x\rangle x_{0}$ where
$\phi_{0}\in X^{*}$ and $\phi_{0}(x_{0})=\|\phi_{0}\|=1$. Since $P$ is also a
contraction in $L_{\infty}$, let us define $\gamma:L_{\infty}\to{\mathbb{R}}$
by $\gamma(x):=\|x\|_{\infty}-\langle\phi_{0},x\rangle\|x_{0}\|_{\infty}$, and
let us observe that

$\gamma(x)=\|x\|_{\infty}\pm\|P(x)\|_{\infty}\geq 0.$

We see now that $\phi_{0}$ and $v$ are disjointly supported: Let
$h:=\mathbbm{1}_{A}$ where $A\subseteq\mathrm{supp\,}v$. Note that for $|t|$
small enough one has that $\|x_{0}+th\|_{\infty}=\|x_{0}\|_{\infty}$, hence

$\gamma(x_{0}+th)=\|x_{0}+th\|_{\infty}-(1+t\langle\phi_{0},h\rangle)\|x_{0}\|_{\infty}=-t\|x_{0}\|_{\infty}\langle\phi_{0},h\rangle.$

Since $\gamma$ is non-negative we deduce that $\phi_{0}(h)=0$, and this
implies that $\mathrm{supp\,}(\phi_{0})\cap\mathrm{supp\,}(v)=\emptyset$ and
therefore

$\phi_{0}(u)=\phi_{0}(x_{0}),$

so $\phi_{0}$ also norms $u$ and $\|x_{0}\|=\|u\|$. ∎

∎

We do not know whether the minimal and isometric envelopes coincide for an
arbitrary reflexive Fraïssé or even $\mathrm{(AuH)}$ Banach space. Going
further, it would be interesting to study the relationship between the
envelopes in non-reflexive Fraïssé spaces, for example in the Gurarij space,
or even in quasi-Banach spaces, such as $L_{p}[0,1]$ for $0<p<1$.

###  3.2. Full copies of $L_{q}$-spaces inside $L_{p}$

We study the envelopes of $L_{q}$-spaces inside $L_{p}$, obtaining the
following.

######  Theorem 3.17.

Assume that $1\leq p<+\infty$ is not even.

  1. (1)

the envelope of any Hilbertian subspace of $L_{p}$ of dimension at least $2$
is isometric to $L_{p}$.

  2. (2)

If $q$ satisfies $1\leq p\leq q\leq 2$, then the envelopes of $L_{q}$ and
$\ell_{q}$ inside $L_{p}$ are isometric to $L_{p}$.

Consequently,

  1. (3)

$L_{p}$ admits full copies of $\ell_{2}^{n}$, $n\geq 2$, $\ell_{2}$, and of
$L_{q}$ and $\ell_{q}$ provided that $1\leq p\leq q\leq 2$.

######  Proof.

In fact, we have the following formally stronger statement: Assume $1\leq
p<+\infty$, $p$ not even and let $Y$ be a subspace of $L_{p}$ such that one of
the two conditions hold:

  1. i)

the group $\operatorname{\mathrm{Isom}}(Y)$ acts almost transitively on
$S_{Y}$ and $Y$ does not embed isometrically into $\ell_{p}$;

  2. ii)

$Y$ has a $1$-symmetric Schauder basis $(e_{n})$ such that
$\inf_{n}\frac{\|\sum_{i=1}^{n}e_{i}\|_{Y}}{n^{1/p}}=0$

Then the envelope of $Y$ is isometric to $L_{p}$. In particular some isometric
copy of $Y$ has full envelope.

Its proof goes as follows. Composing with an isometric embedding of $Y$ into
$L_{p}$ sending some vector of full support in $Y$ to $1$, we may assume $Y$
is unital, hence $\operatorname{\mathrm{Env}}(Y)=L_{p}([0,1],\Sigma_{Y})$. The
proof is finished once we argue that $\Sigma_{Y}$ does not have atoms. Suppose
otherwise. Then
$\operatorname{\mathrm{Env}}(Y)=L_{p}(A_{1},\Sigma_{1})\oplus_{p}L_{p}[A_{2},\Sigma_{2})$
where $\Sigma_{1},\Sigma_{2}\subseteq\Sigma_{Y}$ are atomless and atomic
$\sigma$-algebras in $A_{1}$ and $A_{2}$, respectively, $A_{1}\sqcup
A_{2}=[0,1]$, and $\Sigma_{2}$ is non-empty.

We denote by $i$ the inclusion map of $Y$ into
$L_{p}(A_{1},\Sigma_{1})\oplus_{p}L_{p}(A_{2},\Sigma_{2})$, and choosing
$i_{1}=0$ if necessary, we write $i=(i_{1},i_{2})$ with respect to the
decomposition $L_{p}(A_{1},\Sigma_{1})\oplus_{p}L_{p}(A_{2},\Sigma_{2})$, that
is $i_{j}(f)=f\upharpoonright A_{j}$ for every $f\in Y$. Note that by the
Banach-Lamperti formula for isometries of $L_{p}$-spaces, all isometries on
this space can be written as $T=(T_{1},T_{2})$ where each $T_{1}$ is an
isometry on $L_{p}(A_{i},\Sigma_{i})$ for $i=1,2$. Observe that $i_{2}\neq 0$,
because otherwise $A_{2}=\emptyset$.

######  Claim 3.17.1.

The norm of $i_{2}y$ is constant on $\operatorname{\mathrm{Isom}}(Y)$-orbits.

###### Proof of Claim:.

For suppose that $t\in\operatorname{\mathrm{Isom}}(Y)$ and $y\in Y$. Since $p$
is not even, we know that $L_{p}$ is $\mathrm{(AuH)}$, and it follows from
Proposition 2.40 that there is unique extension of $t$ to
$T\in\operatorname{\mathrm{Isom}}(\operatorname{\mathrm{Env}}(Y))$. Then
$(i_{1}ty,i_{2}ty)=ity=Tiy=(T_{1}i_{1}y,T_{2}i_{2}y)$, and consequently
$\|i_{2}ty\|=\|T_{2}i_{2}y\|=\|i_{2}y\|$. ∎

Suppose that i) holds. Then it follows from the previous claim that the norm
of $i_{2}$ has constant value $k>0$ on $S_{Y}$ (because $i_{2}\neq 0$). Hence,
$i_{2}/k$ is an isometric embedding of $Y$ into $L_{p}[A_{2},\Sigma_{2})$, and
since $\Sigma_{2}$ is atomic, $L_{p}(A_{2},\Sigma_{2})$ is isometric to
$\ell_{p}(I)$, where $I$ is the (countable) set of atoms of $\Sigma_{2}$,
contradicting the hypothesis.

Suppose that ii) holds. Since $(e_{n})_{n}$ is a 1-symmetric basis of $Y$, the
isometries of $Y$ act transitively on the basis, and by the previous claim
this implies that the norm of $i_{2}e_{n}$ for $n\in{\mathbb{N}}$ is constant,
with value $c$. As before, we have that $c>0$. By $w^{*}$-compactness there
exists $u\in\ell_{p}^{**}$ and $N$ an infinite subset of ${\mathbb{N}}$ such
that $(i_{2}e_{n})_{n\in N}\rightarrow^{w^{*}}u$. We claim that $u=0$. To see
this, note that for any normalized $\phi$ in $\ell_{p}^{*}$, and any
$n\in{\mathbb{N}}$ we can find a finite $F\subseteq N$ of cardinality $n$ such
that $\phi(\sum_{j\in F}i_{2}e_{j})\geq\frac{1}{2}n\phi(u)$. It follows that
$n\phi(u)\leq 2\|i_{2}(\sum_{j\in F}e_{i})\|\leq
2\|i_{2}\|\|\sum_{j=1}^{n}e_{j}\|_{Y}$. The estimate in ii) then implies that
$\phi(u)=0$ and therefore the claim that $u=0$. Now since $(i_{2}e_{n})_{n\in
N}$ tends weakly to $0$ in $\ell_{p}$, we can, by a gliding hump argument and
passing to a subset of $N$, assume that the sequence $(i_{2}e_{n})_{n\in N}$
is almost sucessive in the sense that $\|\sum_{j\in
F}i_{2}e_{j}\|\geq\frac{1}{2}(\sum_{j\in F}\|i_{2}e_{j}\|^{p})^{1/p}$ whenever
$F\subseteq N$ is finite. Therefore $c|F|^{1/p}\leq 2\|i_{2}\|\|\sum_{j\in
F}e_{j}\|_{Y}$, contradicting ii) when $|F|$ is large enough. ∎

###  3.3. On $g$-embeddings of $\ell_{2}$ into $L_{p}$ and the spaces
$L_{p}/\ell_{2}$

Knowing that $\ell_{2}$ admits a full copy in $L_{p}$, we have an associated
exact sequence

$0\rightarrow\ell_{2}\rightarrow L_{p}\rightarrow L_{p}/\ell_{2}\rightarrow
0,$

to which we may apply Corollary 2.41 and the commentary thereafter.

######  Definition 3.18 (The full quotient space $L_{p}/\ell_{2}$).

Let $1\leq p<\infty$, $p\notin 2{\mathbb{N}}$. The quotient $L_{p}/\ell_{2}$
of $L_{p}$ by any full copy of $\ell_{2}$ is isometrically unique. We call
this space the full quotient of $L_{p}$ by $\ell_{2}$.

######  Proposition 3.19.

For each $1\leq p<\infty$, $p\notin 2{\mathbb{N}}+4$, there is a linear
$g$-embedding (see definition just before Definition 2.40) of $U(\ell_{2})$
into $\operatorname{\mathrm{Isom}}(L_{p})$. It is defined by extension of
unitaries on any fixed full copy of $\ell_{2}$ inside $L_{p}$, to isometries
on $L_{p}$.

######  Proof.

This is a consequence of Theorem 3.17 and of Proposition 2.40. ∎

Many questions seem to remain open about concrete representations of full
copies of $\ell_{2}$ inside $L_{p}$, about the exact sequence

$0\rightarrow\ell_{2}\rightarrow L_{p}\rightarrow L_{p}/\ell_{2}\rightarrow 0$

associated to the linear $g$-embedding of $\ell_{2}$ inside $L_{p}$, for
appropriate values of $p$, about the associated full quotient of $L_{p}$ by
$\ell_{2}$, or about the topological embedding of $U(H)$ into
$\operatorname{\mathrm{Isom}}(L_{p})$ from a concrete point of view, . We
shall not address them in this paper but we do make a few initial comments
below.

If $1<p<+\infty$, then the exact sequence above splits, by complementation of
$\ell_{2}$ inside $L_{p}$. This means that $L_{p}/\ell_{2}$ is isomorphic to
$L_{p}$ (since it is isomorphic to a complemented subspace of $L_{p}$, and
since $L_{p}$ is primary). On the other hand when $p=1$, then

$0\rightarrow\ell_{2}\rightarrow L_{1}\rightarrow L_{1}/\ell_{2}\rightarrow 0$

does not split, and the isomorphic type of the full quotient $L_{1}/\ell_{2}$
seems to be unknown.

Many other homological questions arise: what description can be made of the
induced action of $U(\ell_{2})$ on $L_{p}/\ell_{2}$ (which is actually an
action by bounded isomorphisms on $L_{p}$ when $p>1$)? for $1<p<+\infty$, does
$\ell_{2}$ admit an $U(\ell_{2})$-invariant summand inside $L_{p}$? if it does
not, then what can be said of the quasilinear map $\Omega$ associated to the
exact sequence and of the commutator estimates of $\Omega$ with the action of
the unitary group on $L_{p}$? How may the dual exact sequence be described?
For details we refer to [12], for the general theory of exact sequences of
Banach spaces, and to [11], for a theory of their compatibility with group
actions.

Another direction which seems worthwhile to explore is the relation between
the isometric envelopes of the Hilbert space in an $L_{p}$-space and Gaussian
Hilbert spaces. See in particular the extension Theorem 4.12 in [30]. Note
however that the theory of Gaussian Hilbert spaces seems to be limited to the
context of $L_{p}$-spaces and does not seem to generalize to the
$\mathrm{(AuH)}$ or Fraïssé situation.

##  4\. Complementations in $\mathrm{(AuH)}$ spaces: numerical estimates

The aim of this section is to establish relations between the constants of
complementation of euclidean or hilbertian copies, the type and cotype of the
space, and the AUH or Fraïssé property. Recall that a Fraïssé Banach space $X$
always contain isometric copies of the Hilbert space. We shall see that the
constant of complementation of such copies in $X$ determines for which values
of $p$ the space $X$ is possibly isometric to $L_{p}$. We shall also relate
these values to the type and cotype of the space.

In consequence we shall obtain “local versions” of the fact that
$L_{p}$-spaces are Fraïssé only if $p\notin 2{\mathbb{N}}+4$, and reinforce
the conjecture that those spaces, together with the Gurarij space, are the
only separable Fraïssé spaces.

###  4.1. Complementations in $\mathrm{(AuH)}$ spaces

Our first result relates to constants of complementation in
$\mathrm{(AuH)}$-spaces. This extends some results known for $L_{p}$-spaces,
but our point is to obtain these only through properties of the isometry
group.

######  Proposition 4.1.

Let $X$ be $\mathrm{(AuH)}$ and $1$-complemented in its bidual. Then any
isometric embedding of a subspace $Y$ of $X$ into $X$ extends to a contraction
on $X$.

######  Proof.

Let $t$ be an embedding of a subspace $Y$ of $X$. For any $F$ finite-
dimensional subspace of $Y$ and $\varepsilon>0$, there is a map
$T_{F,\varepsilon}\in\operatorname{\mathrm{Isom}}(X)$ such that
$\|(t-T_{F,\varepsilon}){\upharpoonright F}\|\leq\varepsilon$. Using a non-
trivial ultrafilter refining the natural filter on the set of
$(F,\varepsilon)$, we define a map $T:X\rightarrow X^{**}$ by
$Tx=w^{*}-\lim_{U}T_{F,\varepsilon}x$. If $P$ is a norm $1$ projection from
$X^{**}$ onto $X$ then $PT$ defines a contraction on $X$ extending $t$. ∎

######  Proposition 4.2.

Let $X$ be $\mathrm{(AuH)}$, and let $Y$ be $C$-complemented in $X$ for some
$C\geq 1$. Then any isometric copy of $Y$ inside $X$ is also $C$-complemented,
as soon as one of the following conditions holds:

  1. (1)

$X$ is $1$-complemented in its bidual, or

  2. (2)

$Y$ is finite dimensional, or

  3. (3)

$Y$ is $1$-complemented in its bidual and can be written as the closure of an
increasing union of a net $(F_{i})_{i}$ of finite dimensional subspaces such
that $F_{i}$ is $1$-complemented in $Y$ for every $i$.

######  Proof.

Let $Y$ be $C$-complemented by a projection $p$ and $t$ be an isometric
embedding of $Y$ into $X$. (1): Using Proposition 4.1, $tY$ is complemented by
$tpT^{\prime}$, if $T^{\prime}$ is a contraction on $X$ extending $t^{-1}$.
(2): for any $\varepsilon>0$ we can find, by classical perturbation arguments,
a surjective $1+\epsilon$-isometry $T_{\varepsilon}$ on $X$ extending $t$, and
then $p_{\varepsilon}:=T_{\varepsilon}pT_{\varepsilon}^{-1}$ is a projection
onto $tY$ of norm at most $C(1+\varepsilon)^{2}$. A compactness argument
provides a projection onto $tY$ of norm at most $C$. (3): Applying (2) we find
a projection $p_{i}$ onto $tF_{i}$ of norm at most $C$ for each $i$. Using a
non-trivial ultrafilter refining the net, we define a map
$p:X\rightarrow(tY)^{**}$ by $px=w^{*}-\lim_{U}p_{i}x$. If $P$ is a norm $1$
projection from $(tY)^{**}$ onto $tY$ then $Pp$ defines a projection onto $tY$
of norm at most $C$. ∎

Note that as a consequence of this and of the $\mathrm{(AuH)}$ properties of
$L_{p}$-spaces, we recover a known result in the theory of $L_{p}$-spaces
([50]), i.e. the above holds inside $L_{p}(0,1),1\leq p<+\infty,p\notin
2{\mathbb{N}}+4$. It also proved in [50] that the result does not hold for
$p\in 2{\mathbb{N}}+4$, since in this case there is a complemented subspace
admitting an isometric uncomplemented copy. One may actually prove the
following:

######  Remark 4.3.

No $\mathrm{(AuH)}$ space can contain a complemented isometric copy of any
$L_{p}$ with $p\in 2{\mathbb{N}}+4$. This follows from the existence of a
constant $K$ and a sequence of pairwise isometric finite dimensional subspaces
$F_{n}$ and $G_{n}$ of $L_{p}$ such that $F_{n}$ is $K$-complementend and
$G_{n}$ is not $n$-complemented in $L_{p}$, for each $n\in{\mathbb{N}}$. It is
easy to see that this fact, together with item (2) of Proposition 4.2, is
incompatible with complementation of a copy of $L_{p}$ inside $X$. The
existence of $K,(F_{n})$ and $(G_{n})$ is based on the unconditionality of the
subspaces used in [50] and is detailed in the proof of Proposition 2.10 of
[21].

As for the Hilbert space $H$, a classical result states that $U(H)$ is WOT-
dense in ${\mathcal{L}}_{1}(H)$ (see for example [45]). This and Proposition
4.1 gives us the following.

######  Corollary 4.4.

Assume $X$ is $\mathrm{(AuH)}$ $1$-complemented in its bidual. Then every
operator defined on a hilbertian subspace of $X$ extends to an operator of
same norm on $X$.

######  Proof.

For suppose that $Y\subseteq X$ is hilbertian and $t:Y\to Y$ be a contraction
of norm 1. We choose a net $(t_{i})_{i}$ of isometries of $Y$ converging WOT
to $T$. By Proposition 4.1, we can extend each $t_{i}$ to norm one operators
$T_{i}:X\to X$. Let $U:X\to X^{**}$ be WOT-limit of $(T_{i})_{i}$, and let
$P:X^{**}\to X$ be a contractive projection. Then $PU$ has norm 1 and extends
$t$. ∎

Note that this holds in particular for any Fraïssé space $1$-complemented in
its bidual, and therefore for all the separable Fraïssé $L_{p}$-spaces.

A similar result seems possible for spaces $X$ for which the unital algebra
${\mathcal{L}}(X)$ is unitary, that is, when the convex hull of its isometries
is norm dense in ${\mathcal{L}}_{1}(X)$. It is classical that
${\mathcal{L}}(H)$ is unitary in the complex case and the real case is due to
Navarro-Pascual and Navarro [44] who actually prove that the infinite convex
hull is equal to ${\mathcal{L}}_{1}(H)$.

######  Definition 4.5.

For $X$ a Banach space, $1\leq p\leq\infty$, and $n\in{\mathbb{N}}$, let us
denote $c_{p}^{n}(X)\in[1,+\infty]$ the infimum of constants of
complementation of an isometric copy of $\ell_{p}^{n}$ inside $X$, and
$c_{p}(X)$ the infimum constants of complementation of an isometric copy of
$L_{p}$ inside $X$. Let us also denote by $\operatorname{\mathrm{Fr}}(X)$ the
set of $p$’s such that $\ell_{p}$ is finitely representable in $X$.

######  Proposition 4.6.

Assume $X$ is Fraïssé. The following holds for every
$p\in\operatorname{\mathrm{Fr}}(X),1\leq p<+\infty$:

  1. (1)

$c_{p}^{n}(X)$ is attained.

  2. (2)

$c_{p}(X)$ is attained when it is finite.

  3. (3)

We have $c_{p}(X)=\lim_{n}c_{p}^{n}(X)$.

  4. (4)

If there is $C\geq 1$ such that for any $n$, and for any $\varepsilon>0$, $X$
contains a $C$-complemented $1+\epsilon$-copy of $\ell_{p}^{n}$, then
$c_{p}(X)\leq C$.

Furthermore

  1. (5)

for fixed $n$, the map $p\mapsto c_{p}^{n}(X)$ is continuous on
$\operatorname{\mathrm{Fr}}(X)$.

######  Proof.

Recall that $L_{p}$ embeds into $X$ as soon as
$p\in\operatorname{\mathrm{Fr}}(X)$ (see [21]). Here, except for (1), we are
using that the space $X$ is assumed to be Fraïssé and not only
$\mathrm{(AuH)}$ (although we do not know examples of $\mathrm{(AuH)}$ non
Fraïssé spaces). (1): Since $X$ is $\mathrm{(AuH)}$, any isometric map between
two copies of $\ell_{p}^{n}$ may be approximated by a global isometry,
therefore extends to a $1+\varepsilon$-isometry for $\varepsilon$ arbitrarily
small, and we deduce that any isometric copy of $\ell_{p}^{n}$ is complemented
by a projection of norm at most $c_{p}^{n}(X)+\delta$, for arbitrary
$\delta>0$. We may then define a projection as some weak limit of almost
optimal projections along an ultrafilter to see that all copies of
$\ell_{p}^{n}$ are complemented by a projection of norm $c_{p}^{n}(X)$.

Note that the sequence $c_{p}^{n}(X)$ is non-decreasing, and assume it is
bounded by $C$. Writing $L_{p}$ as the closure of the increasing union of
$\ell_{p}^{n}$’s, and again defining a projection onto $L_{p}$ as some weak
limit of projections on the $\ell_{p}^{n}$’s (using that $L_{p}$ is
$1$-complemented in its bidual), we deduce that both (2) and (3) hold.

(4): Let $\varepsilon>0$, $\eta>0$, and consider a $C$-complemented
$1+\eta$-isometric copy $F$ of $\ell_{p}^{n}$. If $\eta$ had been chosen small
enough, the Fraïssé property allows to extend a partial $1+\eta$-isometry $t$
between $F$ and an isometric copy $G$ of $\ell_{p}^{n}$ to a global
isomorphism $T$ on $X$ such that $\|T\|\|T^{-1}\|<1+\varepsilon$, guaranteeing
that $G$ is $C(1+\varepsilon)$-complemented. Then we deduce that
$c_{p}^{n}(X)\leq C$, and, by c), that $c_{p}(X)\leq C$.

(5): Let $p\in\operatorname{\mathrm{Fr}}(X),$ $\varepsilon>0$, $\eta>0$. For
$q$ close enough to $p$ in $\operatorname{\mathrm{Fr}}(X)$, there is a partial
$1+\eta$-isometry $t$ between copies of $\ell_{p}^{n}$ and $\ell_{q}^{n}$. If
$\eta$ had been chosen small enough, the Fraïssé property allows to extend $t$
to a global isomorphism $T$ on $X$ such that $\|T\|\|T^{-1}\|<1+\varepsilon$,
guaranteeing that $c_{q}^{n}(X)\leq(1+\varepsilon)c_{p}^{n}(X)$. ∎

In the case (4) we shall say that $\ell_{p}$ is uniformly complementably
finitely representable in $X$; we noted that this implies that $X$ contains a
complemented copy of $L_{p}$.

###  4.2. The hilbertian case

We now concentrate on the case of Hilbertian/euclidean copies. The exact value
of the (best) constant of complementation of isometric copies of
$\ell_{2}^{n}$ or $\ell_{2}$ inside $L_{p}$, $1\leq p\leq+\infty$ is known
(see Gordon-Lewis-Retherford [26, Theorem 6]). Let us note:

$c_{2}(L_{p})=\frac{2}{\sqrt{\pi}}\Big{(}\Gamma(\frac{p+1}{2})\Big{)}^{1/p}\Big{(}\Gamma(\frac{p^{\prime}+1}{2})\Big{)}^{1/p^{\prime}},\,1<p<+\infty$
(4)

and

$c_{2}(L_{\infty})=c_{2}(L_{1})=+\infty$

where for $p\in[1,+\infty]$ we use the notation $p^{\prime}\in[1,+\infty]$ for
the conjugate of $p$.

Recall that $p(X):=\sup\\{{p}\,:\,{X\ {\rm has\ type\ }p}\\}$ and that
$q(X):=\inf\\{{p}\,:\,{X\ {\rm has\ cotype\ }q}\\}$. We say that $X$ is near
Hilbert when $p(X)=q(X)=2$. It is well-known that $X$ does not necessarily
have type $p(X)$ or cotype $q(X)$. However, the Maurey-Pisier Theorem [42]
states that both $\ell_{p(X)}$ and $\ell_{q(X)}$ are finitely representable in
$X$. Regarding duality, it is immediate that if $X$ has type $p>1$ then
$X^{*}$ has cotype $p^{\prime}$. A profound result by Pisier [47] states
conversely that if $X$ has no-trivial type and cotype $q$, then $X^{*}$ has
type $q^{\prime}$. Another classsical result by Pisier [47] states that a
Banach space $X$ has non-trivial type exactly when $X$ is locally
$\pi$-euclidean, that is, when for some $c\geq 1$, for every $n$ and
$\epsilon>0$, there exists $N$ such that any $N$-dimensional subspace of $X$
admits an $n$-dimensional subspace which is $1+\epsilon$-isometric to an
euclidean space and which is $c$-complemented in $X$ (see also Pełczynski-
Rosenthal [49, Definition and Observation on p284], noting that the
definitions in [47] and [49] are equivalent by Dvoretsky theorem).

######  Lemma 4.7.

The following hold:

  1. (1)

$c_{2}(L_{p})=c_{2}(L_{p^{\prime}})$ for $1\leq p\leq+\infty$.

  2. (2)

If $1\leq p<\infty$ is in $\mathrm{Fr}(X)$ for some Fraïssé space $X$, then
$c_{2}(L_{p})\leq c_{2}(X).$

  3. (3)

The map $p\mapsto c_{2}(L_{p})$ is decreasing on $[1,2]$ and increasing on
$[2,\infty]$.

######  Proof.

Assertion (1) is a direct consequence of the formula (4). (2): Since $X$ is
Fraïssé and $p\in\operatorname{\mathrm{Fr}}(X)$, it follows that $X$ contains
an isometric copy of $L_{p}$. For a given $n$, if $\ell_{2}^{n}$ is
$C$-complemented in $X$, then any copy of $\ell_{2}^{n}$ inside $L_{p}$ is
also $C$-complemented in $X$ (because $X$ is Fraïssé) and consequently also in
$L_{p}$. This implies that $c_{2}^{n}(L_{p})\leq c_{2}^{n}(X)$, and by
Proposition 4.6 (3) we have that $c_{2}(L_{p})\leq c_{2}(X)$. (3): since
$L_{q}$ embeds isometrically into $L_{p}$ for $1\leq p\leq q\leq 2$, this and
assertion (2) implies that the map is decreasing on $[1,2]$, and then, by
assertion (1), increasing on $[2,+\infty]$ by (2). ∎

The next relates $p(X)$, $q(X)$, and complemented copies of $L_{p}$-spaces.

######  Proposition 4.8.

Let $X$ be a Fraïssé Banach space.

  1. (1)

If $q(X)=+\infty$ then $X$ is contains an isometric copy of the Gurarij space,
and if $X$ separable then $X$ is isometric to the Gurarij space.

  2. (2)

If $p(X)=1$ and $q(X)<+\infty$, then $X$ contains isometric copies of $L_{1}$,
$\ell_{2}$ is uncomplemented in $X$, and $c_{2}^{n}(X)\geq
n\cdot\Gamma(n/2)/(\sqrt{\pi}\cdot\Gamma(1/2+n/2))$.

  3. (3)

If $p(X)>1$ then $\ell_{2}$ is complemented in $X$. There exists a unique
$1<p\leq 2$ such that $c_{2}(X)=c_{2}(L_{p})$, and we have $p(X)\geq p$ and
$q(X)\leq p^{\prime}$.

  4. (4)

If $p(X)>1$ and $X$ has type $p(X)$ (resp. has cotype $q(X)$) then $X$
contains a complemented isometric copy of $L_{p(X)}$ (resp. of $L_{q(X)}$).

  5. (5)

If $p(X)>1$ and $q(X)\in 2{\mathbb{N}}+4$, then $X$ does not have cotype
$q(X)$.

######  Proof.

Assertion (1) was already observed in [21]. It is a consequence of the Maurey-
Pisier Theorem in [42] that $\ell_{\infty}$ is finitely representable in any
space with non non-trivial cotype, implying that a space satisfying (1) must
be isometrically universal for separable spaces, and the fact that the Gurarij
space is the unique separably universal Fraïssé space.

Assertion (2) is a consequence of Maurey-Pisier Theorem and the Fraissé
property - Lemma 4.7 (2). Since $\ell_{2}$ is uncomplemented in $L_{1}$ it has
to be uncomplemented in $X$ and this also provides a lower bound for
$c_{2}^{n}(X)$ by
$c_{2}^{n}(L_{1})=n\cdot\Gamma(n/2)/(\sqrt{\pi}\cdot\Gamma(1/2+n/2))$.

Regarding (3), since $X$ has non-trivial type, it is locally $\pi$-euclidean.
Therefore $X$ contains uniformly complemented $1+\varepsilon$-copies of
euclidean spaces of arbitrary size and therefore a complemented copy of the
Hilbert space. The formula (4) implies that $p\mapsto c_{2}(L_{p})$ is
continuous, and by Lemma 4.7 (3), is also decreasing on $[1,2]$. Since
$c_{2}(L_{1})=\infty$ and $c_{2}(L_{2})=1$, it follows that there must be a
unique $1<p\leq 2$ such that $c_{2}(X)=c_{2}(L_{p})$. By the Maurey-Pisier
Theorem on the finite representability of $L_{p(X)}$ in $X$, and by using that
$X$ is Fraïssé, $X$ contains an isometric copy of $L_{p(X)}$. It follows that
$c_{2}(L_{p(X)})\leq c_{2}(X)=c_{2}(L_{p})$, and therefore $p(X)\geq p$. The
same reasoning holds for cotype.

(4): Following [43, Proposition 13.16], if we suppose that $p(X)>1$ and $X$
has type $p(X)$, then $X$ contains for some $C\geq 1$ and for any
$\varepsilon>0$, $1+\varepsilon$-isomorphic copies of $\ell_{p}^{n}$
$C$-complemented, with $p=p(X)$, and therefore a $C$-complemented copy of
$L_{p(X)}$ by Proposition 4.6 (4).

Now suppose that $p(X)>1$ and $X$ has cotype $q(X)$. Then the Pisier duality
result [47] holds to deduce that $p(X^{*})=q(X)^{\prime}$ and $X^{*}$ has type
$p(X^{*})$. Therefore as above $X^{*}$ contains $1+\varepsilon$-isomorphic
copies of $\ell_{p(X^{*})}^{n}$’s uniformly complemented, and by duality and
local reflexivity, $X$ contains $1+\varepsilon$-isomorphic copies of
$\ell_{q(X)}^{n}$’s uniformly complemented. The rest of the reasoning is as
under the first hypothesis, to deduce that $X$ admits a complemented copy of
$L_{q(X)}$.

(5) is then a direct consequence of (4) and of Remark 4.3. ∎

Using Proposition 4.2, we have the following striking corollary of Proposition
4.8:

######  Corollary 4.9.

Any Fraïssé space $X$ satisfies exactly one of the two following properties:
either

  1. (1)

$X$ contains a complemented isometric copy of $\ell_{2}$, or

  2. (2)

$X$ contains an isometric copy of $L_{1}$.∎

Applying Proposition 4.8 to the case of $1$-complemented copies of $L_{p}$
spaces we also get:

######  Proposition 4.10.

Let $X$ be a Fraïssé Banach space. Then

  1. (1)

there are at most $2$ values of $p$ (which must be conjugate) such that $X$
contains a $1$-complemented copy of $L_{p}$.

  2. (2)

if $X$ contains a $1$-complemented copy of $L_{p}$ and $p\leq 2$ (resp. $p\geq
2$), then $p(X)=p$ (resp. $q(X)=p^{\prime}$). In particular if $X$ contains a
$1$-complemented copy of $L_{2}$ then $X$ is near Hilbert.

######  Proof.

Suppose that $X$ contains a $1$-complemented copy of $L_{p}$. Then, by Lemma
4.7 (1), (2), $c_{2}(L_{p^{\prime}})=c_{2}(L_{p})\leq c_{2}(X)$. On the other,
the 1-complementation of a copy of $L_{p}$ implies that $c_{2}(X)\leq
c_{2}(L_{p})$. This determines the set $\\{p,p^{\prime}\\}$. (2): if $p\leq 2$
then it follows that $p(X)\geq p$, but also since $X$ contains a copy of
$L_{p}$, that $p(X)\leq p$. Likewise if $p\geq 2$ then it follows that
$q(X)=p^{\prime}$.∎

We do not know if any Fraïssé contains a 1-complemented copy of some $L_{p}$
and if it contains at most such 1-complemented $L_{p}$. There are other
properties of Fraïssé $L_{p}$ spaces which may be seen as natural steps
towards the conjecture that all separable Fraïssé spaces are either
$L_{p}$-spaces or the Gurarij space. We just list two of them here: must any
Fraïssé space $X$ attain its type and cotype? if $X$ is Fraïssé and reflexive,
must ${\rm Fr}(X)$ be either an interval of the form $[p,2]$ or a singleton
$\\{q\\}$ for $2\leq q<+\infty$? The reader will easily find other natural
conjectures in this vein.

We conclude with a solution to a weak version of the Mazur rotation problem
posed by G. Godefroy (see [17, Problem 5.20, and the paragraph before it]).

######  Theorem 4.11.

A Fraïssé space admitting $C_{\infty}$ bump functions must be isomorphic to a
Hilbert space.

######  Proof.

Assume $X$ is as above and not isomorphic to a Hilbert space. Since a space
with a Fréchet-differentiable bump function is necessarily Asplund, $X$ cannot
contain a copy of $L_{1}$, whereby by the Proposition 4.8 (2), $X$ has non
trivial type. According to a Theorem of Deville, see [18, Chapter V.4] for an
exposition, this implies $q(X)$ is an even number larger than $2$ and $X$ has
cotype $q(X)$, and this contradicts Proposition 4.8 (5).∎

A Banach space $X$ is Lipschitz transitive [16] when there exists $C\geq 1$
such that for any two normalized vectors $x,y$ of $X$, there is a surjective
isometry $T$ on $X$ such that $Tx=y$ and $\|T-Id\|\leq C\|y-x\|$. According to
[16, Lemma 2.6], any Lipschitz transitive norm on a separable space is
$C_{\infty}$. We immediately deduce:

######  Proposition 4.12.

A separable Fraïssé space which is Lipschitz transitive must be isomorphic to
Hilbert space. ∎

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