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¤ ( P± + Pβ ) x ’ xγ < 2 µ.

If x ∈ E is arbitrary, then P± (x) ∈ P± (E), hence by (1)

Pβ (x) = Pβ (P± (x)) ’ P± (P± (x)) = P± (x) for β ±.

(8) ⇐ (1) & (6) Let µ > 0. Then the set {β : β < ±, Rβ (x) ≥ µ} is ¬nite,
since otherwise there would be an increasing sequence (βn ) such that Rβn (x) ≥ µ
and since P±+1 ’ P± = (1 ’ P± )P±+1 ≥ 1 also (Pβn +1 ’ Pβn )(x) ≥ µ. Let
β∞ := supn βn . Then β∞ ¤ ± is a limit ordinal and Pβ∞ (x) = limβ<β∞ Pβ (x)
according to (6), a contradiction.
(9) ⇐ (6) We prove by trans¬nite induction that P± (x) is in the closure of the
linear span of {Rβ (x) : ω ¤ β < ±} ∪ Pω (x).

53.13
53.13 53. Appendix: Projective resolutions of identity on Banach spaces 589

For ± = ω this is obviously true. Let now ± = β + 1 and assume Pβ (x) is in
the closure of the linear span of {Rγ (x) : ω ¤ γ < β} ∪ Pω (x). Since P± (x) =
Pβ (x) + P± ’ Pβ Rβ (x) we get that P± (x) is in the closure of the linear span of
{Rγ (x) : ω ¤ γ < ±} ∪ Pω (x).
Let now ± be a limit ordinal and let Pβ (x) be in the closure of the linear span of
{Rγ (x) : ω ¤ γ < ±} ∪ Pω (x) for all β < ±. Then by (6) we get that P± (x) =
limβ<± Pβ (x) is in this closure as well.

Proposition. Suppose all complemented subspaces of a Banach space E have PRI
then E has a SPRI.

Proof. We proceed by induction on µ := dens E. For µ = ω nothing is to be
shown. Now let (P± )0¤±¤µ be a PRI of E. For every ± < µ we have ± + 1 < µ and
so µ± := dens((P±+1 ’ P± )(E)) ¤ dens(P±+1 (E)) ¤ ± < µ, hence there is a SPRI
±
(Pβ )0¤β<µ± of (P±+1 ’ P± )(E). Now consider

± ±
P±,β := P± + Pβ (P±+1 ’ P± ) = (P± + Pβ (1 ’ P± ))P±+1

for ω ¤ ± < µ and ω ¤ β ¤ µ± with the lexicographical ordering. This is a
well-ordering and since the cardinality of µ2 is µ and µ± < µ it corresponds to the
ordinal segment [ω, µ). In fact for any limit ordinal ± > ω we have

|[ω, µ± )| ¤ |[ω, ±)|2 ¤ |[ω, ±)|.
|[ω, ±)| = 1¤
ω¤β<± ω¤β<±


Obviously the P±,β are projections that satisfy (1) and (3).
±
(1) For P±,β with the same ± this follows from (1) for Pβ : R± (E) ’ R± (E): In
fact

± ±
P±,β P±,β := P± + Pβ (P±+1 ’ P± ) P± + Pβ (P±+1 ’ P± )
2 ± ±
= P± + Pβ (P±+1 ’ P± )P± + P± Pβ (P±+1 ’ P± )
± ±
+ Pβ (P±+1 ’ P± )Pβ (P±+1 ’ P± )
2 ±
= P± + 0 + 0 + Pmin{β,β } (P±+1 ’ P± )

For di¬erent ± this follows, since P±1 ,β E ⊆ P±1 +1 E ⊆ P±2 and

P± E ⊆ P±,β ⊆ P±+1
±
ker P± ⊇ ker P±,β = ker(P± + Pβ (1 ’ P± ))P±+1 ⊇ ker P±+1


(3) The density of P±,β E is less or equal to ± + 1.
±
And clearly they satisfy (7) as well, since R±,β = Rβ (P±+1 ’ P± ).
β
(9) Since this is true for the P± and the P± it follows for P±,β as well.
In fact P±,β (x) belongs to the closure of the linear span of P± (x) and the R±,β =
± ±
Rβ (P±+1 ’P± )(x) for β < β by the property of the Pβ . Furthermore P± (x) belongs

53.13
590 53. Appendix: Projective resolutions of identity on Banach spaces 53.14

to the closure of the linear span of R± (x) for ± < ± and Pω (x) by the property of
±
the P± and R± (x) belongs to the closure of the linear span of all Rβ (R± x) for all
β < dens R± E.
(8) For x in the linear span of all R±,β E we obviously have that (R±,β (x))±,β ∈ c0 .
n i i
In fact for x := i=1 » R±i ,βi (xi ), we have that R±i ,βi (x) = » R±i ,βi (xi ) and
R±,β (x) = 0 for all (±, β) ∈ {(±1 , β1 ), . . . , (±n , βn )}.
/

R± Rβ = (P±+1 ’ P± )(Pβ+1 ’ Pβ ) = (1 ’ P± )P±+1 Pβ+1 (1 ’ Pβ ) = 0,

if ± + 1 ¤ β or β + 1 ¤ ±, since the factors commute. For general x we ¬nd by (9)
a point x in the linear span of the R±,β x with x ’ x < µ. Then
˜ ˜

{(±, β) : R±,β (x) ≥ µ} ⊆ {(±1 , β1 ), . . . , (±n , βn )}.


Note however that we don™t have P±,β = 1.

53.14. Theorem. Let E be a Banach space with projective generator •. Then E
admits a PRI (P± )± , where each P± is based on a norming pair (A± , B± ) with
(1) |A± | ¤ ±, |B± | ¤ ± for all ±;
(2) Aβ ⊆ A± and Bβ ⊆ B± for all β ¤ ±;
(3) ω¤β<± Aβ = A± for all limit ordinals ±;
(4) ω¤β<± Bβ = B± for all limit ordinals ±;

Proof. Choose a dense subset {x± : ± < dens E}. We construct by trans¬nite
recursion for every ordinal ± ¤ dens E a norming pair (A± , B± ) with

A± ⊇ {xβ : β < ±}, |A± | ¤ ±, |B± | ¤ ±, •(B± ) ⊆ A±
Aβ ⊆ A± and Bβ ⊆ A± for β ¤ ±.

For the ordinal ω let A0 := {x± : ± < ω} and let B0 be a countable subset of E
such that
x = sup{| x, x— : x— ∈ B0 } for all x ∈ A0 .
By (53.7) there is a norming pair (Aω , Bω ) with |Aω |, |Bω | ¤ ω, Aω ⊇ A0 , Bω ⊇ B0
and •(Bω ) ⊆ Aω .
If ± is a successor ordinal, i.e. ± = β + 1, then let A0 := Aβ ∪ {xβ } and B0 := Bβ .
Again by (53.7) we get a norming pair (A± , B± ), such that

A0 ⊆ A± , B 0 ⊆ B± ⊆ E , |A± | ¤ ±, |B± | ¤ ±, •(B± ) ⊆ A±

If ± is a limit ordinal, we set

A± := Aβ
β<±

Bβ ⊆ E .
B± :=
β<±


53.14
53.17 53. Appendix: Projective resolutions of identity on Banach spaces 591

Then obviously (A± , B± ) is a norming pair with •(B± ) ⊆ A± .
Now using the property of the projective generator • we have that there are norm-1
projections P± ∈ L(E) with P± (E) = A± and ker P± = (B± )o = (B± )o . Hence

(53.13.1) P± Pβ = Pβ = Pβ P± for β ¤ ±

dens P± E ¤ ±, dens P± (E )σ ¤ ±,
(53.13.3)

(53.13.5) P± (E) = A± = Pβ E
β<±
(53.13.4) P± = 1

and since {x± : ± < dens E} is dense in E we also have (53.13.2). Furthermore we
have that B± is weak— -dense in P± E .



53.15. Corollary. WCD and duals of Asplund spaces have SPRI.

53.16. De¬nition. A compact set K is called Valdivia compact if there exists
some set “ with K ⊆ R“ and {x ∈ K : carr(x) is countable} being dense in K.

53.17. Lemma. For a Valdivia compact set K ⊆ R“ we consider the set E :=
{x ∈ R“ : carr(x) is countable}. Let µ be the density number of K © E. Then there
exists an increasing long sequence of subsets “± ⊆ “ for ω ¤ ± ¤ µ satisfying:
(i) |“± | ¤ ±;
(ii) β<± “β = “± for limit ordinals ±;
(iii) “µ = cup{carr(x) : x ∈ K};
and such that K± := Q“± (K) ⊆ K, where Q“ : R“ ’ R“ ’ R“ , i.e.

for γ ∈ “

Q“ (x)γ := .
for γ ∈ “ \ “
0 /

Thus K± ⊆ K is a retract via Q“± .

Note that for any Valdivia compact set K ⊆ R“ we may always replace “ by
{carr(x) : x ∈ K} = {carr(x) : x ∈ K © E}, and then (iii) says “µ = “.

Proof. The proof is based on the following claim: Let ∆ ⊆ “ be a in¬nite subset.
˜ ˜ ˜
Then there exists some subset ∆ with ∆ ⊆ ∆ ⊆ “ and |∆| = |∆| and Q∆ (K) ⊆ K.
˜
By induction we construct a sequence ∆ =: ∆0 ⊆ ∆1 ⊆ · · · ⊆ ∆k ⊆ · · · ⊆ “ with
|∆k | = |∆0 | and Q∆k ({x ∈ K © E : carr(x) ⊆ ∆k+1 }) being dense in Q∆k (K):
(k+1) Since K © E is dense in K, we have that Q∆k (K © E) is dense in Q∆k (K) ⊆
R∆k — {0} ⊆ R“ . And since the topology of R∆k has a basis of cardinality |∆k |,
there is a subset D ⊆ K © E with |D| ¤ |∆k | and Q∆k (D) dense in Q∆k (K). Let
∆k+1 := ∆k ∪ x∈D carr(x) then ∆k+1 ⊇ ∆k and |∆k+1 | = |∆k |. Furthermore
Q∆k ({x ∈ K © E : carr(x) ⊆ ∆k+1 }) ⊇ Q∆k (D) is dense in Q∆k (K).
˜
Now ∆ := k ∆k is the required set. In order to show that Q∆ (K) ⊆ K let x ∈ K
˜
be arbitrary. Since Q∆k (x) is contained in the closure of Q∆k ({xk ∈ K © E :

53.17
592 53. Appendix: Projective resolutions of identity on Banach spaces 53.19

˜
carr(xk ) ⊆ ∆k+1 }) and hence in the closed set Q∆k ({xk ∈ K : carr(xk ) ⊆ “}).
˜
Thus there is an xk ∈ K with carr(xk ) ⊆ “ and such that x agrees with xk on ∆k .
˜
Thus K xk ’ Q∆ (x), since every ¬nite subset of ∆ is contained in some ∆k and
˜
˜
outside ∆ all xk and Q∆ (x) are zero. Since K is closed we get Q∆ (x) ∈ K.
˜ ˜

Without loss of generality we may assume that µ > ω. Let {x± : ω ¤ ± < µ} be a
dense subset of K © E. Let “ω := carr(xω ). By trans¬nite induction we de¬ne

(“β ∪ carr(xβ ))∼ for ± = β + 1,
“± :=
“β for limit ordinals ±.
β<±

Then the “± satisfy all the requirements.

53.18. Corollary. Let K be Valdivia compact. Then C(K) has a PRI.

Proof. We choose “± as in (53.17) and set K± := Q“± (K). Let Q± := Q“± |K .
Then Q± is a continuous retraction.

wK

! IdK±
g
e
g
ee


±

 eeeQ

incl ±

K



uU
R
U
Id
R    u

C(K± ) C(K± )
R
incl R  Q
£
 


±

C(K) u |E ±

We have dens(C(R“± )) = |±|, since we have a base of the topology of this space of
that cardinality. Hence dens(C(K± )) ¤ |±|. Let E± := (Q± )— (C(K± )). Then E± is
a closed subspace of C(K) and (53.13.3) holds. Furthermore P± := Q± —¦ incl— ± is
K
a norm-1 projection from C(K) to E± . The inclusion “± ⊆ “β for ± ¤ β implies
(53.13.1). To see (53.13.6) and (53.13.5) let µ > 0 and choose a ¬nite covering of
K± by sets
Uj := {x ∈ R“± : |xγ ’ xj | < δj for all γ ∈ ∆j },
γ

where xj ∈ R“± , δj > 0 and ∆j ⊆ “± is ¬nite and such that for x , x ∈ Uj © K
we have |f (x ) ’ f (x )| < µ. Now choose ±0 < ± such that “±0 ⊇ ∆j for all of the
¬nitely many j. Since the Uj cover K± , we have x ∈ K± © Uj for some j and hence
Qβ (x) ∈ K± © Uj for all ±0 ¤ β < ±. Hence |f (x) ’ f (Qβ (x))| < µ for all x ∈ K±
and so P± (f ) ’ Pβ (f ) = (1 ’ Pβ )P± (f ) ¤ µ. Thus we have shown that E has
a PRI (P± )± , with all E± ∼ C(K± ) and dens(K± ) ¤ |“± | ¤ ±.
=

53.19. Remark. The space C([0, ±]) has a PRI given by

for µ ¤ β
f (µ)
Pβ (f )(µ) := .
for µ ≥ β
f (β)

53.19
53.20 53. Appendix: Projective resolutions of identity on Banach spaces 593

However, there is no PRI on the hyperplane E := {f ∈ C([0, ω1 ]) : f (ω1 ) = 0} of
the space C[0, ω1 ]. And, in particular, C[0, ω1 ] is not WCD.

Proof. Assume {P± : ω ¤ ± ¤ ω1 } is a PRI on E. Put ±0 := ω0 . We may ¬nd
β0 < ω1 with
P±0 E ⊆ Eβ0 := {f ∈ E : f (±) = 0 for ± > β0 },
because for each f in dense countable subset D ⊆ P±0 E we ¬nd a βf with f (±) = 0
for ± ≥ βf . Since Eβ0 is separable, there is an ±0 < ±1 < ω1 such that

Eβ0 ⊆ P±1 E,

in fact D ⊆ Eβ0 is dense and hence for each f ∈ D and n ∈ N there exists an
˜ ˜
±f,n < ω1 and f ∈ P±f,n E such that f ’ f ¤ 1/n. Then ±1 := sup{±f,n : n ∈
N, f ∈ D} ful¬lls the requirements.
Now we proceed by induction. Let ±∞ := supn ±n and β∞ := supn βn . Then

P±n E = Fβ∞ := {f ∈ E : f (±) = 0 for ± ≥ β∞ }.
P±∞ E =
n

But Fβ∞ is not the image of a norm-1 projection: Suppose P were a norm-1 pro-
jection on Fβ∞ . Let π : E ’ C(X) be the restriction map, where X := [0, β∞ ].
˜ ˜
It is left inverse to the inclusion ι given by f ’ f with f (γ) = 0 for γ ≥ β∞ .
˜ ˜
Let P := π —¦ P —¦ ι ∈ L(C(X)). Then P is a norm-1 projection with image
˜
Cβ∞ (X) := {f ∈ C[0, β∞ ] : f (β∞ ) = 0}. Then C(X) = ker(P ) • Cβ∞ (X).
˜ /˜
We pick 0 = f0 ∈ ker(P ). Since f0 ∈ P (C(X)) = Cβ∞ (X) = ker(evβ∞ ), we
have f0 (β∞ ) = 0, and without loss of generality we may assume that f0 (β∞ ) = 1.
˜ ˜
For f ∈ C(X) we have that f ’ P (f ) ∈ ker P and hence there is a »f ∈ R
˜
with f ’ P (f ) = »f f0 . In fact evaluating at β∞ gives f (β∞ ) ’ 0 = »f 1, hence
˜
P (f ) = f ’ f (β∞ ) f0 . Since β∞ is a limit point, there is for each µ > 0 a xµ < β∞
with f0 (xµ ) > 1 ’ µ. Now choose fµ ∈ C(X) with fµ = 1 = ’fµ (β∞ ) = fµ (xµ ).
Then

= fµ ’ fµ (β∞ ) f0
P fµ ∞ ∞

≥ |fµ (xµ ) ’ fµ (β∞ ) f0 (xµ )|
≥ 1 + 1(1 ’ µ) = 2 ’ µ.

˜
Hence P ≥ 2, a contradiction.
Note however that every separable subspace is contained in a 1-complemented sep-
arable subspace.

53.20. Theorem. [Bistr¨m, 1993, 3.16] If E is a realcompact (i.e. non-measurab-
o
le) Banach space admitting a SPRI, then there is a non-measurable set “ and a
injective continuous linear operator T : E ’ c0 (“).

Proof. We proof by trans¬nite induction that for every ordinal ± with ± ¤ µ :=
dens(E) there is a non-measurable set “± and an injective linear operator T± :

53.20
594 53. Appendix: Projective resolutions of identity on Banach spaces 53.21

E± := P± (E) ’ c0 (“± ) with T± ¤ 1.
Note that if E is separable, then there are x— ∈ E with x— ¤ 1, and which are
n n
σ(E , E) dense in the unit-ball of E . Then T : E ’ c0 (N), de¬ned by T (x)n :=
1—
n xn (x), satis¬es the requirements: It is obviously a continuous linear mapping into
c0 , and it remains to show that it is injective. So let x = 0. By Hahn-Banach
there is a x— ∈ E with x— (x) = x and x— ¤ 1. Hence there is some n with
|(x— ’ x— )(x)| < x and hence x— (x) = 0.
n n
In particular we have Tω0 : Eω0 ’ c0 (“ω0 ).
For successor ordinals ± + 1 we have E±+1 ∼ E± — (E±+1 /E± ) = E± — (P±+1 ’
=
P± )(E). Let R± := (P±+1 ’ P± )/ P±+1 ’ P± , let F := (P±+1 ’ P± )(E) and let
T : F ’ c0 be the continuous injection for the, by (53.13.7), separable space F
with T ¤ 1. Then we de¬ne “±+1 := “± N and T±+1 : E±+1 ’ c0 (“±+1 ) by

T± ( P± (x) )γ for γ ∈ G±

T±+1 (x)γ := .
for γ ∈ N
T (R± (x))γ
Now let ± be a limit ordinal. We set

“± := “ω “β+1 ,
ω¤β<±

and de¬ne T± : E± := P± (E) ’ c0 (“± ) by

Tω ( Pω (x) ) for γ ∈ “ω

T± (x)γ :=
for γ ∈ “β+1
Tβ+1 (Rβ (x))γ
We show ¬rst that T± (x) ∈ c0 (“± ) for all x ∈ E. So let µ > 0. Then the set
{β : Rβ (x) ≥ µ, β < ±} is ¬nite by (53.13.8).
Obviously T± is linear and T± ¤ 1. It is also injective: In fact let T± (x) = 0
for some x ∈ E± . Then Rβ (x) = 0 for all β < ± and Pω (x) = 0, hence by
x = P± (x) = 0.
As card(E) is non-measurable, also the smaller cardinal dens(E) is non-measur-
able. Thus the union “± of non-measurable sets over a non-measurable index set
is non-measurable.

53.21. Corollary. The WCD Banach spaces and the duals of Asplund spaces
continuously and linearly inject into some c0 (“). The same is true for C(K), where
K is Valdivia compact.

For WCG spaces this is due to [Amir, Lindenstrauss, 1968] and for C(K) with K
Valdivia compact it is due to [Argyros, Mercourakis, Negrepontis, 1988.]

Proof. For WCD and duals of Asplund spaces this follows using (53.15). For
Valdivia compact spaces K one proceeds by induction on dens(K) and uses the
PRI constructed in (53.18). The continuous linear injection C(K) ’ c0 (“) is then
given as in (53.20) for ± := dens(K), where Tβ exists for β < ±, since Eβ ∼ C(Kβ )
=
with Kβ Valdivia compact and dens(Kβ ) ¤ β < ±.


53.21

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