A Bohl-Bohr-Kadets type theorem characterizing Banach spaces not containing c0

0.14 MB·English

by  Bálint Farkas

A BOHL–BOHR–KADETS TYPE THEOREM CHARACTERIZING
BANACH SPACES NOT CONTAINING c
0
3
1 BA´LINTFARKAS
0
2 Abstract. WeprovethataseparableBanachspaceEdoesnotcontainacopy

n ofthespacec0ofnull-sequencesifandonlyifforeverydoublypower-bounded
a operatorT onEandforeveryvectorx∈Etherelativecompactnessofthesets
J {Tn+mx−Tnx:n∈N}(forsome/allm∈N,m≥1)and{Tnx:n∈N}are
6 equivalent. With the helpof the Jacobs–de Leeuw–Glicksberg decomposition
2 ofstronglycompactsemigroupsthecaseof(notnecessarilyinvertible)power-
bounded operatorsisalsohandled.

]
A
F This note concerns the following problem: Given a Banach space E, a bounded
h. linear operator T ∈ L(E) and a vector x ∈ E, we would like to conclude the
t relative compactness of the orbit
a
m Tnx: n∈N ⊆E
[ (cid:8) (cid:9)
from the relative compactness of the consecutive differences of the iterates

1
v Tn+1x−Tnx: n∈N ⊆E.
0 (cid:8) (cid:9)
5 This problem is a discrete, “linear operator analogue” of the classical Bohl–Bohr
2 theorem about the integration of almost periodic functions. Before going to the
6 results let us explain this connection.
.
1 Given a (Bohr) almost periodic function f : R → C with its integral F(t) =
0
t
3 f(s)ds bounded, then F is almost periodic itself. This result was extended to
0
1 RBanach space valued almost periodic functions f : R → E by M. I. Kadets [7],
: provided that E does not contain an isomorphic copy of the Banach space c of
v 0
i null-sequences. Actually, the validity of this integration result for every almost
X periodicfunctionf :R→E characterizestheabsenceofc inthe BanachspaceE.
0
r
a ThegeneralizationofKadets’result—whichexplainstheconnectiontoourproblem—
was studied by Basit for functions f : G → E defined on a group G and taking
values in the Banach space E (for simplicity suppose now G to be Abelian). In
[2] Basit proved that if F : G → E is a bounded function with almost periodic
difference functions
F(·+g)−F(·) for all g ∈G,
and E does not contain c , then F is almost periodic. The relation to Kadets’
0
result is the following: If f : R → E is almost periodic, so is F (t) := t+εf(s)ds
ε t
R
2010 Mathematics Subject Classification. 47A99,46B04,43A60.
Keywordsandphrases. Power-boundedlinearoperator,(asymptotically)almostperiodicvec-
torsanddifferences, theBanachspaceofnull-sequences,Kadets’theorem.
ThispaperwassupportedbytheJa´nosBolyaiResearchFellowshipoftheHungarianAcademy
ofSciences andbytheHungarianResearchFund(OTKA-100461).
1
2 BA´LINTFARKAS
t
for every ε > 0. So Kadets’ theorem tells that if F(t)= f(s)ds is bounded and
0
E does not contain c , then the almost periodicity of R
0
F(·+ε)−F(·) for all ε>0
implies that of F.
Now returning to our problem, suppose T ∈ L(E) is doubly power-bounded (i.e.,
T,T−1 ∈ L(E) are both power-bounded). Then by applying Basit’s result to the
function F : Z → E, F(n) := Tnx, we obtain that {Tnx : n ∈ Z} is relatively
compact in E if
Tn+mx−Tnx:n∈Z is relatively compact for all m∈Z,
(cid:8) (cid:9)
for which it suffices that
Tn+1x−Tnx:n∈Z is relatively compact.
(cid:8) (cid:9)
Let us record this latter fact in the next lemma.
Lemma 1. Let E be a Banach space and T ∈L(E) be a power-bounded operator.
If for some x∈E the set
Tn+1x−Tnx:n∈N
(cid:8) (cid:9)
is relatively compact, then so is the set
Tn+mx−Tnx:n∈N
(cid:8) (cid:9)
for all m∈N.
Proof. Denote by D the first set and by D the second one. We may suppose
1 m
m≥2. BycontinuityofT thesetsTD ,T2D ,Tm−1D areallrelativelycompact.
1 1 1
Since
Tm+nx−Tnx=Tm−1(Tnx−x)+Tm−2(Tnx−x)+···+T(Tnx−x)+(Tnx−x),
we obtain D ⊆ TD +T2D +···+Tm−1D implying the relative compactness
m 1 1 1
of D . (cid:3)
m
So our problem can be answered satisfactorily for doubly power-bounded op-
erators on Banach spaces not containing c . The situation is different if T is
0
non-invertible, or invertible but with not power-bounded inverse. To enlighten
what may be true in such a situation, let us recall a result of Ruess and Summers,
who considered the generalization of the Bohl–Bohr–Kadets result for functions
f :R →E, see [8, Thm. 2.2.2] or [9, Thm. 4.3].
+
Given an asymptotically almost periodic function f : R → E, one can find an
+
almost periodic one f : R → E and another function f : R → E vanishing at
r s +
infinity such that f =f +f . Ruess and Summers proved the following. Suppose
s r
f is asymptotically almost periodic with
t
F(t):= f(s)ds bounded,
Z
0
and the improper Riemann integral of f exists in E. If E does not contain c
s 0
then F is asymptotically almostperiodic. For details and discussionwe refer to [8,
Sec. 2.2]. As we see, a Jacobs–de Leeuw–Glicksberg type decomposition plays an
essential role here.
A BOHL–BOHR–KADETS TYPE THEOREM 3
Our main result, Corollary12, provides the solution to the very first problem con-
cerning power-bounded operators in this spirit. It contains the mentioned special
case of Bolis’ result when T is doubly power-bounded. For stating the result we
first need some preparations, explaining an analogue of the decomposition above
used by Ruess and Summers. Let E be a Banach space and let T ∈ L(E), which
is from now on always assumed to be power-bounded. A vector x ∈ E is called
asymptotically almost periodic (a.a.p. for short) with respect to T if the (forward)
orbit
Tnx:n∈N ⊆E
(cid:8) (cid:9)
is relatively compact. Denote by E the collection of a.a.p. vectors, which is a
aap
closedT-invariantsubspaceofE. WeshallneedthefollowingformoftheJacobs–de
Leeuw–Glicksberg decomposition for operators with relatively compact (forward)
orbits; see [5, Chapter 16], or [6, Thm V.2.14] where the proof is explained for
continuouslyparametrizedsemigroupsinstead ofsemigroups of the form{Tn :n∈
N} (the proof is nevertheless the same).
Theorem 2 (Jacobs–deLeeuw–Glicksberg). Let E be a Banach space and let T ∈
L(E) have relatively compact orbits (T is hence power-bounded). Then there is a
projection P ∈L(E) commuting with T such that
E :=rgP = x∈E :Tx=λx for some λ∈C, |λ|=1 ,
r
E :=rg(I−P)=kerP =(cid:8)x∈E :Tnx→0 for n→∞ . (cid:9)
s
(cid:8) (cid:9)
The restriction of T to E is a doubly power-bounded operator.
r
Notethatthe occurringsubspacesandtheprojectiondependonthelinearoper-
atorT and—forthesakeofbetterlegibility,wechosenottoreflectthisdependence
in notation. Now, we can apply this decomposition to a given power-bounded
T ∈ L(E), or more precisely to the restriction of T to E . We therefore obtain
aap
a decomposition
E =rgP ⊕kerP =E ⊕E ,
aap r s
note that E ,E ⊆ E ; E is called the stable while E the reversible subspace.
r s aap s r
On rgP the restriction of T is a doubly power-bounded, and
Tn| :n∈Z
Er
(cid:8) (cid:9)
is a strongly compact group of operators.
Remark 3. Now suppose that T ∈ L(E) is even doubly power-bounded Then
in the above decomposition E = {0} must hold. So if x ∈ E , then even the
s aap
backwardorbit is relatively compact, i.e.
Tnx:n∈Z is relatively compact.
(cid:8) (cid:9)
A vector x∈E is also called almost periodic.
r
Lemma 1 tells that for n,m ∈ N with m ≥ n we have (Tm − Tn)x ∈ E
aap
whenever (T −I)x∈E .
aap
We are interested in whether x ∈ E if (T −I)x ∈ E . The answer would be
aap aap
trivially “yes”, if we knew that Tnx−x actually converges as n → ∞. Here is a
slightly more complicated view on trivial fact:
4 BA´LINTFARKAS
Remark 4. a) Suppose we know x ∈ E . Then we can apply I −P to x and
aap
obtain
(I−P)(Tn−I)x=(I −P)(Tnx−x)=Tn(I −P)x−(I−P)x→(P −I)x
for n→∞. Hence if x∈E , then (I −P)(Tn−I)x must be convergent.
aap
b) Supposethat(I−P)(Tn−I)xconvergesforn→∞(noteagainthatTnx−x∈
E , so we can apply the projection I −P to it). If (T −I)x ∈ E belongs
aap aap
even to the stable part, then (Tn−I)x∈E , so (I−P)(Tn−I)x=(Tn−I)x.
s
Hence Tnx−x converges as n→∞ by assumption, implying x∈E .
aap
Itremainstostudy the casewhen(T−I)x∈E =rgP. The nextisapreparatory
r
lemma.
Lemma 5. Suppose x is not an a.a.p. vector but (T −I)x ∈ E . Furthermore,
aap
suppose that (I−P)(Tnx−x) converges. Then there is a δ >0 and a subsequence
(n ) of N such that
k
kP(Tnkx−Tnℓx)k≥δ for all k,ℓ∈N, k 6=ℓ.
Proof. By the non-a.a.p. assumption there is a subsequence (n ) of N and a δ >0
k
such that kTnkx − Tnℓxk > 2δ for ℓ 6= k. By the other assumption, however,
(I−P)(Tnkx−x)isaCauchy-sequencesowhenleavingoutfinitelymanymembers,
we can pass to a subsequence with k(I −P)(Tnkx−Tnℓx)k < δ for all k,ℓ ∈ N,
ℓ,k≥k . The assertion follows from this. (cid:3)
0
Note that in the situation of this lemma we necessarily have kPk>0.
Lemma 6. Let E be a Banach space, let T ∈ L(E) be power-bounded, and let
x ,...x ∈E bea.a.p. vectors. Foreverysequence(n )⊆Nthereis asubsequence
1 m k
(n′) with n′ −n′ →∞ and
k k k−1
kTn′kxi−Tn′k−1xik→0 for all i=1,...,m as k →∞.
Proof. Consider the Banach space X =Em and the diagonal operator S ∈ L(X)
defined by S(y ) = (Ty ). This is trivially power-bounded and (x ) ∈ X is an
i i i
a.a.p. vector. The assertion follows from this, since (Snk(xi)) has a Cauchy subse-
quence (Sn′k(xi)) with n′k−n′k−1 →∞ as k →∞. (cid:3)
We now come to the answer of the initial question.
Theorem 7. Let E be Banach space which does not contain an isomorphic copy
of c , and let T ∈ L(E) be a power-bounded operator. If x ∈ E and (T −I)x
0
is an a.a.p. vector with (I −P)(Tn−I)x convergent for n → ∞, then x itself is
a.a.p. vector.
Theproofisbycontradiction,i.e.,wesupposethatthereissomex∈E satisfying
the assumptionsofthe theorembut beingnotasymptoticallyalmostperiodic. The
contradiction arises then by finding a copy of c in E, for which we shall use the
0
classical result of Bessaga and Pel czyn´ski [3] in the following form, see also [4,
Thms. 6 and 8].
Theorem 8 (Bessaga–Pel czyn´ski). Let E be a Banach space and let x ∈ E be
n
N
vectors such that the partial sums are unconditionally bounded (i.e., x are
j=1 nj
uniformly bounded for all subsequences (n ) of N) and such that the sPeries x is
j i
nonconvergent. Then E contains a copy of c . P
0
A BOHL–BOHR–KADETS TYPE THEOREM 5
TheideaoftheproofisbasedonBasit’spaper,butitisnotadirectmodification,
since we do not know whether we can apply the projections P to x or to Tx.
Proof of Theorem 7. We argue indirectly. Assume that x 6∈ E , so by Lemma 5
aap
we can take a subsequence (nk) and a δ > 0 such that kP(Tnkx−Tnℓx)k > δ for
all k,ℓ∈N with k 6=ℓ. Next we construct a sequence that fulfills the conditions of
the Bessaga–PelczynskiTheorem 8, hence exhibiting a copy of c in E. First of all
0
let M := max(sup{kTn| k : n ∈ Z},sup{kTnk : n ∈ N},kPk). Take k ∈ N such
Er 1
thatkP(Tk1x−x)k>δ/M (useLemma5),andsupposethatthestrictlyincreasing
finite sequence k , i = 1,...,m is already chosen. For a subset F ⊂ {1,2,...,m}
i
denotebyΣF thesum k (ifF =∅,thenΣF =0). Eachofthefinitely many
i∈F i
vectorsTΣFx−xbelonPgstoE byLemma1. ByusingLemma6wefindk,ℓ∈N
aap
with k−ℓ>k such that
m
1
Tnk(TΣFx−x)−Tnℓ(TΣFx−x) ≤ for all F ⊆ 1,...,m .
M2m
(cid:13) (cid:13) (cid:8) (cid:9)
(cid:13) (cid:13)
By setting k :=n −n we obtain
m+1 k ℓ
1
(1) Tkm+1P(TΣFx−x)−P(TΣFx−x) ≤
2m
(cid:13) (cid:13)
(cid:13) (cid:13)
for all F ⊆{1,...,m}. We also have
M P(Tkm+1x−x) ≥ TnℓP(Tkm+1x−x) = P(Tnkx−Tnℓx) ≥δ,
(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
and hen(cid:13)ce we obtain (cid:13) (cid:13) (cid:13) (cid:13) (cid:13)
δ
P(Tkm+1x−x) ≥ .
M
(cid:13) (cid:13)
(cid:13) (cid:13)
Let xi :=P(Tkix−x). We claimthat the sequence (xi) fulfills the conditions of
Theorem8. Indeed,wehavekx k≥δ/M byconstructionsotheseries x cannot
i i
be convergent. For m∈N and 1≤i < i <···<i we have P
1 2 m
m
−Xxij = TkimP(cid:16)Tki1+···+kim−1x−x(cid:17)−P(cid:16)Tki1+···+kim−1x−x(cid:17)
j=1
+Tkim−1P Tki1+···+kim−2x−x −P Tki1+···+kim−2x−x
(cid:16) (cid:17) (cid:16) (cid:17)
.
.
.
+Tki2P Tki1x−x −P Tki1 −x
(cid:16) (cid:17) (cid:16) (cid:17)
+P(x−Tki1+···+kimx).
By (1) we obtain
m m
1
x ≤ +Mkxk+M2kxk≤M′ <+∞.
(cid:13)X ij(cid:13) X2ij−1
(cid:13)j=1 (cid:13) j=2
(cid:13) (cid:13)
It follows that E contains a copy of c , a contradiction. (cid:3)
0
If T is doubly power-bounded, then by Remark 3 we have E ={0}, and hence
s
(I−P)=0. So we obtain the following special case of Basit’s more generalresult:
6 BA´LINTFARKAS
Corollary 9 (Basit). Let E be Banach space E which does not contain a copy of
c , and let T ∈L(E) be a doubly power-bounded operator. If x∈E and (T −I)x
0
is an a.a.p. vector, then so is x itself.
The above results are certainly not valid for arbitrary Banach spaces. A coun-
terexample is actually provided by the very same one showing that the analogue
ofthe Bohl–Bohrtheoremfails for arbitraryBanach-valuedfunctions, see [7] or [9,
Sec. 2.1].
Example 10. Consider E = BUC(R;c ), and T the shift by a > 0, and x(t) :=
0
(sin2tn)n∈N. Then T ∈L(E) is doubly power-bounded, and we have
sint+h −sin t = sin h cos2t+h ≤ε
(cid:12) 2n 2n(cid:12) (cid:12) 2n+1 2n+1(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
for all n ∈ N and t (cid:12)∈ R if |h| is suffi(cid:12)cien(cid:12)tly small, theref(cid:12)ore x ∈ BUC(R;c ). On
0
the other hand x is not an a.a.p. vector since the set
(sint+2mna)n∈N :m∈N,t∈R ⊆c0
(cid:8) (cid:9)
is not relatively compact. On the other hand,
y(t):=[(T −I)x](t)=(sint2+na −sin2tn)n∈N =(sin2na+1 cos22nt++a1)n∈N
is almost periodic, because y(t) →0 uniformly in t∈R as n→∞.
n
Next we show that on c itself there is bounded linear operator satisfying the
0
assumptions of Theorem 7 but for which the conclusion of that theorem fails to
hold.
Example 11. It suffices to exhibit an example on E := c. Let (a ) ∈ c be a
n
sequencewith|an|=1foralln∈Nandan 6=b:=limn→∞an foralln∈N. Define
T(x ):=(a x ).
n n n
Then T ∈ L(E) with kTk = 1. Moreover, T is invertible and doubly power-
bounded. Sincethestandardbasisvectorsofc areeigenvectorsofT corresponding
0
to unimodular eigenvalues, they all belong to E , and hence c ⊆ E ⊆ E .
r 0 r aap
Moreover, since T is doubly power-bounded we have E = {0}, I −P = 0, and
s
the condition “(I −P)(Tnx−x) converges” is trivially satisfied for every x ∈ E.
Not all vectors are a.a.p. with respect to T. It suffices to prove this for the case
when b = limn→∞an = 1, otherwise we can pass to the operator b−1T, which has
preciselythesamea.a.p.vectorsasT. NowsupposebycontradictionthatE =E
aap
holds. ThenT ismeanergodiconE,whichisequivalenttothefactthatker(T−I)
separates ker(T′−I), see, e.g., [5, Ch. 8]. But this is false, as dimker(T −I) = 0
and dimker(T′−I)≥1. Hence E 6=E and it also follows that E =E =c .
aap r aap 0
Finally, we indeed suppose b = 1. Then, since ran(T −I) ⊆ c = E , we obtain
0 aap
that (T −I)x is a.a.p., for every x∈E, but not all x∈E belongs to E .
aap
By [1, Sec. 2.5] if c is a closed subspace in a separable Banach space, then it is
0
complemented in there. Thus Example 11 in combination with Theorem 7 yields
the following:
Corollary 12. For a separable Banach space E the following assertions are equiv-
alent:
(i) The Banach space E does no contain a copy of c .
0
A BOHL–BOHR–KADETS TYPE THEOREM 7
(ii) For every power-bounded linear operator T ∈L(E) and x∈E the orbit
{Tnx:n∈N}⊆E
is relatively compact if and only if
{Tn+1x−Tnx:n∈N}⊆E
is relatively compact and (I −P)(Tnx−x) is convergent for n→∞.
We close this paper by two consequences of the previous results, interesting in
their own right:
Corollary 13. Let E be Banach space not containing c . Then for every x ∈
0
E, T ∈ L(E) doubly power-bounded operator and m ∈ N, m ≥ 1, the relative
compactness of the two sets
D := Tn+1x−Tnx:n∈N ⊆E
1
(cid:8) (cid:9)
and
D := Tn+mx−Tnx:n∈N ⊆E
m
(cid:8) (cid:9)
are equivalent.
Proof. If D is relatively compact, then so is {Tnm+mx−Tnmx : n ∈ N} and by
m
Corollary9 even{Tnmx:n∈N}. By the continuity of T the set B :={Tnm+kx:
k
n∈N} is relatively compact for all k =0,...,m−1. Since
Tnx:n∈N =B1∪B2∪···∪Bm−1,
(cid:8) (cid:9)
the relative compactness of the (forward) orbit of x follows. But this implies the
relative compactness of D . That the relative compactness of D implies that of
1 1
D , is true without any assumption on the Banach space E, see Lemma 1. (cid:3)
m
Example 14. Let E := c and for m ∈ N, m ≥ 2 fixed let T be as in Example 10
with limn→∞an = b ∈ C an mth root of unity. Then we have Eaap = Er = c0.
Sincerg(Tm−I)⊆c andrg(T −I)⊆c +(b−1)1(1istheconstant1sequence),
0 0
we obtain that for this doubly power-bounded operator T and for every x∈E the
set D as in Corollary 13 is compact, while D is not.
m 1
Similarly as for Corollary 12, we obtain from Corollary 13 and Example 14 the
next characterization.
Corollary 15. A separable Banach space E does not contain a copy of c if and
0
only if for every x∈E, T ∈L(E) doubly power-bounded operator and m∈ N the
compactness of the two sets
Tn+1x−Tnx:n∈N
(cid:8) (cid:9)
and
Tn+mx−Tnx:n∈N
(cid:8) (cid:9)
are equivalent.
8 BA´LINTFARKAS
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Bergische Universita¨tWuppertal
Facultyof MathematicsandNaturalScience
Gaussstrasse 20,D-42119Wupertal,Germany
E-mail address: [email protected]