Mixing constructions with infinite invariant measure and spectral multiplicities

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by  Alexandre I. Danilenko

MIXING CONSTRUCTIONS WITH INFINITE INVARIANT
MEASURE AND SPECTRAL MULTIPLICITIES
0
1
Alexandre I. Danilenko and Valery V. Ryzhikov
0
2

n Abstract. Weintroducehighstaircase infinitemeasurepreservingtransformations
a
and prove that they are mixing under a restricted growth condition. This is used to
J
(i) realize each subset M ⊂N∪{∞} as the set of essential values of the multiplicity
9
function for the Koopman operator of a mixing ergodic infinite measure preserving
1
transformation, (ii) construct mixing power weakly mixing infinite measure preserv-

] ing transformations, (iii) construct mixing Poissonian automorphisms with a simple
S
spectrum, etc.
D
.
h
t
a 0. Introduction
m
[ Let T be an invertible measure preserving transformation of a σ-finite measure

2 space (X,B,µ) and µ(X) = ∞. By U we denote the Koopman unitary operator
T
v
associated with T:
3
4 UTf = f ◦T, f ∈ L2(X,µ).
6
1 The set ofessential valuesof the spectral multiplicityfunction of U will bedenoted
T
8. by M(T). We note that M(T) is a non-empty subset of N ∪ {∞}. In our previ-
0 ous paper [DaR] we showed that for each subset M ⊂ N, there exists an ergodic
9
0 conservative T such that M(T) = M. The main purpose of the present paper is
v: to sharpen (and generalize) this result. We will show that T can be chosen mixing
i or, equivalently, of zero type (see [Aa], [DaS2]). This means that Un → 0 weakly
X T
as n → ∞.
r
a
Theorem 0.1. Given any subset M ⊂ N ∪ {∞}, there exists a mixing ergodic
conservative infinite measure preserving transformation T such that M(T) = M.
We first consider the case where M does not contain {∞}. The idea of the proof
is to consider the Cartesian product S ×T, where S is a rigid transformation with
M(S) = M (constructed in [DaR]) and T is a mixing transformation such that
M(S×T) = M(S) and S×T is ergodic. We note that the Cartesian product of a
mixing system with any other (even non-ergodic) system is mixing. We construct
H(n)
T as a certain limit of a sequence of transformations (T ) satisfying U k →
n n>0 Tn
(n)
δ U weakly as k → ∞ along a subsequence H of rigidity for S, where δ is a
n Tn k n
sequence of positive reals tending to 0. The transformations T are not mixing of
n
course. However they have a mixing part which occupies more and more space as
n → ∞. In the limit it fills the entire space. We thus consider this construction
as some forcing of mixing. It remains to find a model for the mixing parts. In the
case of probability preserving systems, staircase transformations [Ad] played this
role (see [Ry1], [Ry2], [Ag], [Da4]). Therefore it seems natural to use “infinite”
Typeset by AMS-TEX
staircase systems for our purposes. Hence our first step is to show that the infinite
staircasesaremixing. Unfortunately,therestrictedgrowth condition on the sequence
of cuts essentially used by Adams in [Ad] is incompatible with the infiniteness of
the invariant measure. That is why instead of pure staircases we introduce so-
called high staircases. Geometrically this means that on each step of the inductive
cut-and-stack construction we insert a layer of spacers between the tower and the
staircase roof. If the layers are sufficiently thick, the corresponding high staircase
transformation preserves an infinite measure. At the same time an analogue of the
restricted growth condition can hold for high staircases (see Definition 2.3 below).
Modifying Adams’ argument from [Ad] we show the following theorem, which is of
independent interest.
Theorem 0.2. Under the restricted growth condition each infinite measure pre-
serving high staircase transformation is mixing.
We then utilize the transformations from Theorem 0.2 to prove Theorem 0.1 in
the aforementioned way. The case where M ∋ ∞ comes to the above by considering
a product T ×B, where B is a Bernoulli shift and T is a transformation from the
claim of Theorem 1 with M(T) = M \ {∞}. This idea ‘works’ if the maximal
spectral type of T is singular. We show that such T exists.
As a byproduct we can solve an open problem related to weak mixing for infinite
measure preserving systems. A transformation T is called power weakly mixing
if for each finite integer sequence (n ,...,n ), n 6= 0 for all i = 1,...,k, the
1 k i
product Tn1 × ···× Tnk is ergodic. A number of rank-one power weakly mixing
transformations with exotic properties are known so far [AFS], [Da1], [Da2], [DaS1]
(see also surveys [Da3], [DaS2] and reference therein). However all of them are
either rigid or partially rigid. Hence a problem arises:
is there a mixing power weakly mixing rank-one infinite measure preserving map?
We answer affirmatively by showing the following theorem.
Theorem 0.3. There is a power weakly mixing high staircase transformation sat-
isfying the restricted growth condition.
We conjecture that every high staircase transformation is power weakly mixing.
The method of forcing mixing originated in [Ry1] was used in [Ry2] and [Ag] to
construct a mixing rank-one finite measure preserving transformations T such that
the unitary
∞
exp(U ) := U⊙n
T T
Mn=0
b b
has a simple spectrum. Here U denotes the restriction of U onto the orthocom-
T T
plement to the constant functions. We establish an infinite version of this result.
b
Theorem0.4. There is a mixing rank-one conservativeinfinite measure preserving
transformation T such that the unitary operator exp(U ) has a simple spectrum.
T
This theorem has some applications to the theory of Poissonian automorphisms
(see [CFS], [Ne], [Ro1], [Ro2]).
Corollary 0.5. There is a mixing (of all orders) Poisson suspension with a simple
spectrum.
2
If a Poissonian automorphism does not have a simple spectrum then the set
of its spectral multiplicities is infinite. We construct examples of mixing Poisson
automorphisms S such that M(S) = {p,p2,p3,...} (for an arbitrary p > 1), and
also M(S) = {1,3,3·5,3·5·7,...} (for other S, of course).
1. Cut-and-stack and (C,F)-constructions
To prove Main Theorem we will use the (C,F)-construction (see [dJ], [Da1],
[Da3]). We now briefly outline its formalism. Let two sequences (C ) and
n n>0
(F ) of finite subsets in Z are given such that:
n n≥0
— F = {0,1,...,h −1}, #C > 1, 0 ∈ C ,
n n n n
— F +C ⊂ F ,
n n+1 n+1
— (F +c)∩(F +c′) = ∅ if c 6= c′, c,c′ ∈ C ,
n n n+1
— lim hn = ∞.
n→∞ #C1···#Cn
Let X := F × C × C × ···. Endow this set with the (compact Polish)
n n n+1 n+2
product topology. The following map
(f ,c ,c ,...) 7→ (f +c ,c ,...)
n n+1 n+2 n n+1 n+2
is a topological embedding of X into X . We now set X := X and endow
n n+1 n≥0 n
it with the (locally compact Polish) inductive limit topology.SGiven A ⊂ F , we
n
denote by [A] the following cylinder: {x = (f,c ,...,) ∈ X | f ∈ A}. Then
n n+1 n
{[A] | A ⊂ F ,n > 0} is the family of all compact open subsets in X. It forms a
n n
base of the topology on X.
Let R stand for the tail equivalence relation on X: two points x,x′ ∈ X are R-
equivalent if there is n > 0 such that x = (f ,c ,...), x′ = (f′,c′ ,...) ∈ X
n n+1 n n+1 n
and c = c′ for all m > n. There is a non-atomic Borel infinite σ-finite measure
m m
µ on X which is invariant (and ergodic) under R and such that µ(X ) = h . It is
0 0
unique up to scaling.
Now we define a transformation T of (X,µ) by setting
T(f ,c ,...) := (1+f ,c ,...) whenever f < h −1, n > 0.
n n+1 n n+1 n n
This formula defines T partly on X . When n → ∞, T extends to the entire X
n
(minus countably many points) as a µ-preserving invertible transformation. More-
over, the T-orbit equivalence relation coincides with R (on the subset where T
is defined). We call T the (C,F)-transformation associated with (C ,F ) .
n+1 n n≥0
Below we will often use the following simple formulae
[A] ∩[B] = [A∩B] , [A] ∪[B] = [A∪B] ,
n n n n n n
µ([A] )
n
[A] = [A+c] , µ([A+c] ) = ,
(1-1) n n+1 n+1
#C
c∈GCn+1 n+1
Tr[A] = [r +A]
n n
for all subsets A,B ⊂ F and r ∈ Z such that r +A ⊂ F .
n n
We note that in a similar way we can define (C,F)-actions of an arbitrary count-
able discrete amenable group. In that case (F ) must be a Følner sequence in
n n≥0
the group. The formulae (1-1) are all satisfied for arbitrary (C,F)-actions. In this
3
paper we are mainly interested in Z-actions. While proving Theorem 0.3 we will
also need (C,F)-actions of Zd which are Cartesian products of a fixed Z-action. A
useful observation is that if T is associated with (C ,F ) then the product
n+1 n n≥0
Zd-action (Tn1×···×Tnd)(n1,...,nd)∈Zd is associated with (Cnd+1,Fnd)n≥0, where the
upper index d denotes the Cartesian power.
Another observation is that the (C,F)-construction is equivalent to the classi-
cal cut–and-stack construction of rank-one transformations. Indeed, X can be
n
interpreted as the n-th tower consisting of the levels [f] , f ∈ F . It is cut into
n n
#C subtowers [F +c] , c ∈ C , which are then stack (with some spacers
n+1 n n+1 n+1
in-between) into a new, (n+1)-tower. C is the set of locations of these subtow-
n+1
ers inside the (n + 1)-tower. More precisely, if we order the elements of C as
n+1
follows
0 = c(0) < ··· < c(r −1)
n
then c(i) is exactly the hight of the bottom level of the (i+1)-th subtower of X
n
inside X , i = 0,...,r − 1. That is why in the following we will illustrate
n+1 n
some abstract aspects of the (C,F)-construction with more common cut-and-stack
pictures.
2. High staircase construction
In this section we prove Theorems 0.2 and 0.3. We first formulate a couple of
definitions.
Definition 2.1. Let T be a measure preserving transformation of an infinite σ-
finite measure space (X,B,µ).
(i) A sequence of positive integers a is called mixing for T if µ(TanA∩B) → 0
n
as n → ∞ for all subsets A,B ⊂ X of finite measure.
(ii) A sequence of intervals [a ,b ) ⊂ N is called mixing for T if each sequence
n n
d such that a ≤ d < b is mixing for T.
n n n n
Let (z )∞ and (r )∞ be two sequences of positive integers and r → ∞ as
n n=1 n n=0 n
n → ∞. We define (C )∞ and (F )∞ inductively by setting
n n=1 n n=0
h is arbitrary,
0
h := r (h +z )+r (r −1)/2,
n+1 n n n n n
c (0) := 0, c (i+1) := c (i)+h +z +i,
n+1 n+1 n+1 n n
C := {c (i) | i = 0,...,r −1},
n+1 n+1 n
F := {0,...,h −1}.
n+1 n+1
Definition 2.2. The(C,F)-transformationT onastandardσ-finitemeasurespace
(X,B,µ) associated with (C ,F ) will be called a high staircase (see Fig-
n+1 n n>0
ure 2.1). If z = 0 for all n, we call T a pure staircase.
n
Definition 2.3. By the restricted growth condition we mean the following:
r2
lim n = 0.
n→∞ r0r1···rn−1
4
hznn.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................Hrnn−1
n
Figure 2.1. High staircase.
We note that this condition implies
r2 r2 1
(2-1) n = n · → 0 as n → ∞.
h µ(X ) r ···r
n n 0 n−1
Moreover, the restricted growth condition is equivalent to (2-1) in the case when
µ is finite. We remark that Adams originally introduced it for the finite measure
preserving pure staircases in the form of (2-1) [Ad].
We leave to the reader the proof of the following two simple lemmata. The
second one is an L2-version of [Ad, Lemma 2.1].
Lemma 2.4. Under the restricted growth condition a high staircase is infinite mea-
∞
sure preserving if and only if z /h = ∞.
n=1 n n
P
Lemma 2.5. Given positive integers R,L,r and a measurable set B ⊂ X of finite
measure, we have
R−1 L−1
1 1 rL
U−i1 ≤ U−ir1 + µ(B).
(cid:13)R T B(cid:13) (cid:13)L T B(cid:13) R
(cid:13) Xi=0 (cid:13)2 (cid:13) Xi=0 (cid:13)2 p
(cid:13) (cid:13) (cid:13) (cid:13)
(cid:13) (cid:13) (cid:13) (cid:13)
From now on we assume that T is infinite measure preserving (see Lemma 2.4).
The following statement is a slight modification of [Ad, Lemma 2.2].
Lemma 2.6. Let [α ,β ) be a sequence of intervals in Z which is mixing for each
n n
power of T and l → ∞. Take k ∈ [α ,β ). Then 1 ln−1U−kni → 0 strongly
n n n n ln i=0 T
as n → ∞. P
5
Proof. Take a measurable subset B ⊂ X, µ(B) < 1. Then for each l > 0,
1 l−1 2 k1 k2 1 l−1
U−ikn1 = B 2 + µ(T(i−j)knB ∩B).
(cid:13)l T B(cid:13) l l2
(cid:13) Xi=0 (cid:13)2 i6=Xj=0
(cid:13) (cid:13)
(cid:13) (cid:13)
Since µ(TpknB∩B) → 0 for each p 6= 0 by the assumption of the lemma, there is l
such that
l−1 2
1
U−ikn1 < ǫ
(cid:13)l T B(cid:13)
(cid:13) Xi=0 (cid:13)2
(cid:13) (cid:13)
for all sufficiently large n. It r(cid:13)emains to apply(cid:13)Lemma 2.5. (cid:3)
We now let H := h +z . The following lemma is crucial in the proof of The-
n n n
orem 0.2.
Lemma 2.7. If the restricted growth condition is satisfied then the sequence of
intervals [h ,2H ) is mixing for each non-zero power of T.
n n
Proof. Fix j > 0. We first show that the sequence (H ) is mixing for Tj. Take
n n>0
subsets A,B ⊂ F . Since
n
j(j −1)
jH +c (i) = c (i+j)−ji− if i+j < r ,
n n+1 n+1 n
2
we have
rn−1
µ(TjHn[A] ∩[B] ) = µ(TjHn[A+c (i)] ∩[B] )
n n n+1 n+1 n
Xi=0
rn−j−1
= µ([−ij +A +c (i+j)] ∩[B] )+o¯(1)
j n+1 n+1 n
Xi=0
1 rn−j−1
= µ(T−ij[A ] ∩[B] )+o¯(1),
j n n
r
n Xi=0
where A := (A−j(j −1)/2)∩F . We note that o¯(1) here and below means “uni-
j n
formlysmall”overallA,B ⊂ F asn → ∞. Since T isergodic, r−1 rn−1U−i → 0
n n i=0 T
strongly. It follows that r −1 rn−1U−ij → 0 strongly. Hence wePare done.
n i=0 T
Now take an ∈ [hn,2Hn).PWe are going to prove that (an)n>0 is mixing for
Tj. Dropping to a subsequence we may assume without loss of generality that
ja = kH +b for some 0 ≤ b < H and 0 ≤ k ≤ 2j.
n n n n n
Suppose first that k 6= 0. Partition A into three subsets A ,A ,A such that
1 2 3
b + A ⊂ F , b + A ⊂ H + F and b + A ⊂ [h ,H ). We verify mix-
n 1 n n 2 n n n 3 n n
ing separately on each of these subsets. We note first that Tbn[A ] ∩ [B] = ∅,
3 n n
µ([B] △T−kHn[B] ) = o¯(1) and hence
n n
µ(Tjan[A ] ∩[B] ) = µ(Tbn[A ] ∩T−kHn[B] ) = o¯(1).
3 n n 3 n n
To verify mixing on on A and A it is enough to notice that
1 2
ja +A = kH +(b +A )
n 1 n n 1
ja +A = (k+1)H +(b +A −H ),
n 2 n n 2 n
6
(b + A ) ∪ (b + A − H ) ⊂ F and use the fact that the sequence (H ) is
n 1 n 2 n n n n>0
mixing for both Tk and Tk+1.
Now consider the second case when k = 0. Then b ≥ jh . Hence we can
n n
partition A into subsets A and A such that b + A ⊂ [h ,H ) and b +A ⊂
2 3 n 3 n n n 2
H +F . Therefore Tjan[A ] ∩[B] = ∅ and
n n 3 n n
µ(Tjan[A ] ∩[B] ) = µ(THn[b −H +A ] ∩[B] ) → 0
2 n n n n 2 n n
because (H )∞ is mixing for T. (cid:3)
n n=1
Proof of Theorem 0.2. It is enough to prove that any sequence (m )∞ such that
n n=1
H ≤ m < H for all n is mixing (or it contains a mixing subsequence). We
n n n+1
find integers k and t such that
n n
m = k H +t
n n n n
and 1 ≤ k ≤ r and 0 ≤ t < H . We set
n n n n
C1 := {c (i) | r −k ≤ i < r },
n+1 n+1 n n n
C2 := {c (i) | 0 ≤ i < r −k }.
n+1 n+1 n n
Mixing on F +C1 . This corresponds to the domain D on Picture 2.2. Given
n n+1 4
A,B ⊂ F , we note that for each 0 ≤ j < r −1
n n+1
m +A+C1 +c (j) = D −j +c (j +1),
n n+1 n+2 n+2
where D := t +A+C1 −H . It is important to notice that the subset D−j
n n+1 n+1
is contained essentially in F , i.e. µ([(D − j) ∩ F ] ) = µ([D] ) + o¯(1).
n+1 n+1 n+1 n+1
Hence
1 rn+1−2
µ(Tmn[A+C1 ] ∩[B] ) = µ(T−j[D] ∩[B] )+o¯(1) → 0.
n+1 n+1 n r n+1 n
n+1 Xj=0
htnn−−zztnnn..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r...............................................................................................................................................................................DDD.....................................−.......................................................................................................................132....................................................................k.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................D...........................................................................4.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
n n
Figure 2.2. Partition of the n-tower into main domains.
7
Mixing on F + C2 . Dropping to a further subsequence we can assume that
n n+1
k /r → δ for some δ < 1. Indeed, if k /r → 1 then
n n 2 2 n n
max µ([D+C2 ] )/µ([D] ) → 0
n+1 n+1 n
D⊂Fn
and mixing on F +C2 follows immediately. Now take c (i) ∈ C2 . We have
n n+1 n+1 n+1
k (k −1)
n n
m +c (i) = c (i+k )−k i− +t .
n n+1 n+1 n n n
2
Consider subsets A,B ⊂ F such that µ([A ∪ B] ) < 1. Partition A into three
n n
subsetsA ,A ,A suchthatA +t ⊂ F ,A +t ⊂ H +F andA +t ⊂ [h ,H ).
1 2 3 1 n n 2 n n n 3 n n n
As for the graphical interpretation, A is the part of A that lays inside the domain
i
D , 1 ≤ i ≤ 3, on Figure 2.2.
i
As in the proof of Lemma 2.7, we verify mixing separately on every of these
subsets. However now k may be unbounded and this fact essentially complicates
n
the proof. The restricted growth condition yields
(2-2) max max |µ([(D±k i±k (k −1)/2)∩D] )−µ([D] )| = o¯(1),
n n n n n
0≤i<rnD⊂Fn
i.e. we may neglect rotations by k i + k (k − 1)/2 inside F . Taking this into
n n n n
account we obtain mixing on A in the same way as in the proof of Lemma 2.7.
3
Next, let us verify mixing on A . (Mixing on A is verified in a similar way. We
1 2
leave that to the reader.) Making use of (2-2) again, we obtain
1 rn−kn−1
(2-3) µ(Tmn[A +C2 ] ∩[B] ) = µ(T−kni[A′] ∩[B] )+o¯(1),
1 n+1 n+1 n r 1 n n
n Xi=0
where A′ := (A +t −k (k −1)/2)∩F .
1 1 n n n n
kn...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................r..................................................................................................................................................................................................................................................................................................................................................................−.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................k............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................(rn −kn)kn
n n
Figure 2.3. Image of the bottom layer of D under Tmn.
1
8
There exists an integer p such that
n
h ≤ (r −k )k < h .
pn n n n pn+1
Since δ 6= 1 and r → ∞, it follows that p → ∞. Consider now separately three
2 n n
possible cases.
(A) Let (rn−kn)kn → ∞ and k ≤ h for all n. Then we can find k′ such that
hpn n pn n
h ≤ k k′ < 2h . It follows that rn−kn → ∞. We can choose a sequence (l )∞
pn n n pn k′ n n=1
n
so that l → ∞ and rn−kn → ∞. By Lemmata 2.6 and 2.7,
n lnkn′
1 ln−1U−iknkn′ → 0 strongly as n → ∞.
l T
n Xi=0
Now we deduce from Lemma 2.5 that
1 rn−kn−1µ(T−kni[A′] ∩[B] ) ≤ 1 ln−1U−knkn′ i1 + lnkn′ .
r 1 n n (cid:13)l T [B]n(cid:13) r −k
n Xi=0 (cid:13) n Xi=0 (cid:13)2 n n
(cid:13) (cid:13)
(cid:13) (cid:13)
Mixing on A in this case follows from the above inequality and (2-3).
1
Before we pass to the remaining two cases let us show that
(2-4) H2/h → ∞.
n n+1
Indeed,
H r +r (r −1)/2 r r2 r r2
n n n n ≤ n + n ≤ n + n → 0
H2 H H2 h h2
n n n n n
in view of (2-1).
(B) Let (rn−kn)kn → ∞ and k > h for all n. If H < k for all sufficiently
hpn n pn pn n
large n then
2δ k2 H2 H2
2 > n > pn > pn → ∞
1−δ (r −k )k (r −k )k h
2 n n n n n n pn+1
according to(2-4), a contradiction. Hence, passing toasubsequence wemayassume
that
h ≤ k ≤ H
pn n pn
for all sufficiently large n. Then by Lemmata 2.7 and 2.6 we deduce from (2-3) that
µ(Tmn[A +C2 ] ∩[B] ) → 0.
1 n+1 n+1 n
(C) Consider now the remaining third case where (rn−kn)kn does not go to
hpn
infinity. Then dropping to a subsequence we may assume that there is δ ≥ 1 with
1
(r −k )k
n n n
→ δ as n → ∞.
1
h
pn
It follows that there is δ > 0 such that
(2-5) h > δk2 for all sufficiently large n.
pn n
9
Wededucefrom(2-4)thatH2 /h → ∞. Thisplus(2-5)implyH /k → ∞.
pn−1 pn pn−1 n
Hence H > k > h for all sufficiently large n. Therefore again we can find
pn−1 n pn−1
an integer k∗ such that H ≤ k∗k < 2H . The sequence of intervals
n pn−1 n n pn−1
[H ,2H ) is mixing for all powers of T by Lemma 2.7 We also have
pn pn
r −k (r −k )k (r −k )k δ h
n n = n n n ≥ n n n ≥ 1 · pn → ∞.
k∗ k∗k 2H 4 H
n n n pn−1 pn−1
Therefore arguing as in (A) we obtain mixing in the case (C). (cid:3)
Remark 2.8. We note that in the particular case when z /h → 0 as n → ∞ the
n n
proof of Theorem 0.1 is simplified significantly. It can be carried out as a slight
modification of the reasoning of Adams in [Ad]. In particular, we then no longer
need to consider the cases (B) and (C). Furthermore, we can “reduce” the interval
[h ,2H ) in the statement of Lemma 2.7 to the interval [h ,2h ). Nevertheless,
n n n n
this particular case of Theorem 0.2 is enough to demonstrate Theorems 0.3 and
0.4. However in the proof of our main result—Theorem 0.1—we essentially use
Theorem 0.2 in the full generality. More precisely, it is utilized to satisfy (3-2)
below.
Our next purpose is to construct a power weakly mixing rank-one infinite mea-
sure preserving transformation which is mixing. For that we will use high staircases
and Theorem 0.2.
Proof of Theorem 0.3. We will construct these high staircases via an inductive pro-
cedure. Suppose that after p steps we have already defined F ,C ,F ,...,C ,F .
0 1 1 np np
Step p+1. Consider an auxiliary finite measure preserving pure staircase T
p+1
(p+1) (p+1) (p+1)
associated with a sequence (C ,F ) such that F := F . Assume
k k k≥0 0 np
that the restricted growth condition is satisfied for T . Let (X(p+1),µ ) be the
p+1 p+1
space of this action. We normalize µ in such a way that
p+1
1
µ ([0] ) = .
p+1 0
#C ···#C
1 np
Take any finite sequence l = (l ,...,l ) of non-negative integers such that klk :=
1 p+1
max |l | ≤ p+1. Since T is mixing [Ad], the transformation
0<i≤p+1 i p+1
S := Tl1 ×···×Tlp+1
l p+1 p+1
of the product space (X(p+1),µ )p+1 is ergodic. Then there are N > 0 and
p+1 p+1
M > 0 such that for all disjoint subsets A,B ⊂ (F(p+1))p+1 of equal cardinality
p+1 0
there exist subsets A′ ⊂ [A] and B′ ⊂ [B] and their partitions A′ = Mp+1 A ,
0 0 i=1 i
B′ = Mp+1 B such that F
i=1 i
F
µ×(p+1)(A′) > 0.5µ×p+1([A] ), µ×p+1(B′) > 0.5µ×p+1([B] ),
p+1 p+1 0 p+1 p+1 0
(2-6) A ,B are N -cylinders (some may be empty) and
i i p+1
SiA = B for all i = 1,...,M .
l i i p+1
It is assumed that N and M are common for all S with klk ≤ p + 1. We
p+1 p+1 l
now set
(p+1) (p+1) (p+1)
C := C ,F := F ,...,F := F .
np+1 1 np+1 1 np+Np+1 Np+1
10