I ntn. J. Math. h. Sol.
Vol. 6 No. 3 (Iq83) 409-418
409
DOT PRODUCT REARRANGEMENTS
PAUL ERDOS and GARY WEISS
Mathematics Institute, Budapest; University of Cincinnati
(Received September 18, 1980)
ABSTRACT. Let a
(a), x
n
(x)
n
denote nonnegative sequences; x
the rearranged sequence determined by the permutation
Za x
n n
examine
and S(a,x) denotes {a
n,
a"
(x
(n)
denotes
x denotes the dot product
is a permuation of the positive integers}. We
x
S(a,x) as a subset of the nonnegative real line in certain special circum-
stances. The main result is that if
if and only if
an+l/an
KEY WORDS AND PHRASES.
a
n
+,
then S(a,x)
x
[a" x, ] for every
,
n
0
is uniformly bounded.
Dot product, series rearrangements, conditional convergence.
1982 MATHEMATICS SUBJECT CLASSIFICATION CODE.
40A05.
An elementary classical result of Riemann on infinite series states that a condi-
tionally convergent series that is not absolutely convergent can be rearranged to sum
any extended real number.
A slightly similar group of questions arose in connec-
tion with certain formulas in operator theory [i, p. 181]. Namely, if we le’t a
M
(x) denote any two non-negative sequences and
n
x
denote the sequence
(x
(a),
n
(n)
where n is any permutation of the positive integers, then what can be said about the
Set of non-negative real numbers
ntegers}.
S(a,x)
{a" x
n is a permutation of the positive
More specifically, which subsets of the non-negative real line can be
realized as the form
S(a,x)
Various facts about
for some such
S (a, x)
are obvious
a
and
x?
410
P. ERDOS and G. WEISS
(i)
S(a,x)
(2)
If
a
0
The values
may be obtained.
and
are strictly positive sequences or are at most finitely zero,
x
and
S(a,x)
then
(3)
[0,].
(0,].
[0,]
Not all subsets of
a cardinality argument.
If
of the class of sequences
S(a,x)
(4)
If either
[0,]
(5)
An example:
is
x
and
(0,2,0,2
a
if
[0,],
and
c
then the
but the cardinality
c
c- c
S(a,x)
is finitely non-zero then
x
2
is
This follows by
and thus the cardinality of the
c
is less than or equal to
or
a
a
set.
denotes the cardinality of
c
cardinality of the class of subsets of
subsets
S(a,x)
are realizable as an
(3
x
-n
),
is countable.
is precisely
S(a,x)
then
the Cantor set except for those non-negative real numbers whose ternary expansion consists of a tail of
O’s
2’s
or a tail of
(i.e., a subset of the
rational numbers.
For this
It seems too ambitious to consider the general question at this time.
reason we shall restrict our attention to the cases when
sequence and
If
x
and
S(a,x)
then
x
n
+
x
or
0
the problem is trivial and
0,
t
x ].
n
[0, M
{0}.
S(a,x)
{}
S(a,x)
the problem is trivial and
0,
n
is a non-increasing sequence,
x
a
is a non-decreasing
a
If
a
a
If
0
I
is bounded by
n
In any case, hereafter we shall assume
a
n
"
M,
and
unless otherwise specified.
0,
The Lemma that follows is a well-known fact, but we give a proof for complete-
ness and because the proof contains some of the ideas used in the main result.
LEMMA.
and if
x
n
a
0,
n
If
+
a
n
and
.
Define
0
n
.
+
S(a,x) [a. x,].
then
for all
n
or if
a
n
%
and
In addition,
a
> 0
n
It suffices to show that for every permutation
integers, we have
every
x
n
a. x e S(a,x),
for some
n
e S(a,x).
then
PROOF.
x
and
%
a. x
_<
anXz(n)
or, equivalently,
The rest of the lemma is clear.
nl
in terms of
n
as follows.
Set
a- x
n
of the positive
--< az(n)Xn
for
and
DOT PRODUCT REARRANGEMENTS
1
zl(n)
n= 1
(I)
-i(i)
n
(n)
It is straightforward to verify that
otherwise
z
is also a permutation of the positive
I
integers (one-to-one and onto) which fixes
To see this, note that
X1
X
-i
> 0.
z(1) > 1
411
and
-i
n
We assert that
1
x < a
a
x
1
(i)
Hence
i.
a
a
(i)
0
I
and
Therefore
(i)
[(a(n) al(n))Xn
(a
(a
n(1)
(i)
-a I (i) )Xl+ (a ( -i (i))
a
I) (xi
x
-i
a
H
(n
I
-i
)x
(I))
z
-i
(I)
(i)
0
Proceeding inductively, we obtain a sequence of permutations
for which
x < a
a
k
Hence, for every
x
we obtain
k
n=l
a
Zk (n)
x < a
z
n
k,
k
x
for which
The main question of this paper is:
S(a,x)
with
a,x
The main result of this paper gives a partial answer.
a
acterize which
x
n
+,
n
+
have the property that
it is quite rare.
be
[a- x,]
a
n
2
2
S(a,x)
[a- x, ]
for every
S(a,x)
[a" x, ]
can never be
for every
x
such that
x
n
0.
That
was first decided by an example due to Robert Young.
n
and
n
Namely, we can char-
The first result in this direction was that if
[a- x, ]
S(a,x)
then
x
x
n
2
_2n+l
Both results are unpublished.
a
n
n
or that
for every
S(a,x)
may not
Namely, let
The succeeding results
and techniques are due to the work of the authors in collaboration with Hugh
Mont gomery.
0
x
and
n
0.
On first sight, it might appear that
n,
a
x,]?
[a
such that
,k
k
n n
n=l
is
that fix 1,2
k
k-i
k
Letting
n
412
P. ERDOS and G. WEISS
(The Main Theorem)
THEOREM i.
.
a
n
Let
(a)
a
n
> 0
for every
n
and
Consider the following conditions:
is bounded.
(i)
a
(2)
For the non-negative sequence
n+i/an
and
(x
(a)
a
n
[
(b)
of
x
k
as
(x), there exist subsequences
n
x
k
and
x
and
+
n
0
(x)
n
x
that tends to
that for every positive integer
and
a ,x
n n
where
for all
0
x= (x)
hat tends to
n
0
then (2) implies (3).
n,
To prove that (i) implies that (2) holds for every strictly positive
PROOF.
sequence
k
[a- x,]
+
a
n
n
respectively such that
Then (i) implies (2)for every strictly positive sequence
Also if
(a
.
x
nk mk
S(a,x)
(3)
x
and
a
0
m
k
a
k
n_K
a
where
n
(k+l)
for which
Kn
<
a
n+ i/an
for all
M
We assert
n
there exist arbitrarily large positive integers
k,
-i
suppose
0,
< Mk
ankX
-i
If this assertion were true, then
clearly we could choose two strictly increasing subsequences of positive integers
(n
k)
k)
and (m
such that
0
ankxmk
For each fixed positive integer
x
m
(a (k+l))
e
-i
n
for which
n,m
large
M(a k)
n
x
-i
(k+l)
-i
< a x
n m
< Mk
-i
if and only if
All we need show is that there exist arbitrarily
].
[(a (k+l))
e
m
k,
to prove the assertion.
k
as
-i
M(a k)
n
n
-i
].
x
m
x
m
[(a (k+l))
U
[(a
n>N
M(a k)-l]
n
x
m
-i
n
0
n
M(a k)
-i
for every
n
(k+l))-i
M(a
n
k)-l].
as
U
[(a
For each
-i
(0,e)
N,
<
n
(k+l))-i
m > N,
Xm
let
M(a k)
n
n
m
there exists
0
e
e
for some
n
_>
(k+l))-I
0, since
such that
-1].).
denote the least positlve integer
which exists since
an
as
n
N,
Indeed, the proof below can be
However, this is not the case.
m
[(a
This would imply that
cannot contain any interval of the form
n>N
k)__
M(an+l
(Note:
m
In other words, for every
n>N
used to show that for every
(0 e)
n,m > N.
for which
N
Suppose to the contrary that there exists a positive integer
and hence
n
such that
M(an+ik)
-i
0
DOT PRODUCT REARRANGEMENTS
as
.
n
For
Also, since
a
n +i
M(a
we have
x
n +1
k)
< M(a
m--
-i
<
(a
-i
M
k)
m
M(a
(a
<
m
k
a
x
nk
n
n
k)
k
implies
m
sufficiently large,
m
n
-i]
k)
m
Therefore
(k+l))
m,
for
This implies that
m
m,
for infinitely many
or equivalently,
which contradicts our assumption
m,
Hence (2) is proved.
n.
whenever
m
k
a- x
can assume that
an x
k}
> m
n
.
n
a
+
n
x
n n
> m
n
2
k
m
k
k
2
ankXn
(n
determine subsequences of
k),
(m
for which
n
To see this, let
k.
{k
1
ankXmk +
nan x
i
k
k)
[
k
<_ mk}
x
a
keZ
2
which satisfy conditions
addition satisfy
n
k
a
> m
k
x
Then by
n n
nk
and
b
for all
Z
denote the set
1
Then
m
k
Therefore
n
i
k e Z
(mk),
and
n
a
n
1
in the 2
nd
n
k
m
k
Let
m
k
Z
k)
(x
and
of a and x
m
k
k
condition of the theorem, and in
(a
n
k.
n
k
> m
k
for all
are strictly increasing (a property of subsequences).
k
1
(in increasing order)
Next we assert that without loss of generality we may assume
To see this, note that we have
x
k
which for simplicity we again call (n
This gives us subsequences
1
a
a- x
respectively, by taking only those entries
k e Z
a
we next assert that without loss of generality we
[
keZ
Z
(a
denote the set
Z2
x
ke Z
suppose (2) holds for
0,
n
and hence (3) holds.
for every
k
and let
{}
,
.a
k
x
and
To see this suppose
S(a,x)
[kank mk
k,j.
M(a
m--
Hence, for infinitely many
].
(k+l))-i
x
we have
for all
m
k)
x
0 as
’and (x
such a x
m
n m
k
k
k k
We first assert that without loss of generality we may
Assuming that
and
n +1
m
n
> N
m--
for infinitely many
for all
the lemma we have that
k
[(a
n
,
k)
m
< M
and
n
M(a
m
M(a
n
so that there exist subsequences
assume that
{k
x
we have
M(k+l)/k
x,
k
(k+l))
n
k
To prove (2)/ (3)
and
Therefore
m
and
(k+l))
n
=.
m
m
as
0
and
nm+ i
m
an+l/an_
that
[(a
k)
n
m
-I
m,
m
an m+l/an
n
x
x
m
m, x
m
infinitely many
M(a
k)_
n +i
m
and hence
and for these
M(a
sufficiently large, we have
m
413
and that
Therefore if
n
#
k
< n
n
k
for all
m0
k
>
and < m
k>
m.
for
414
P. ERDOS and G. WEISS
some
k, j,
then
and
k
most once among the
m0
’s.
n
M
k
(nk2’mk2)
mk2+l
exists, such that
(n
S
mk3)
k
3
that
mk 3+i
e S
n
such that
1
k
n
mkl+l
where
2
j.
kl
,m
I,S 2
S
1
for some
k
<_
kl+l
e S
or
1
S
<_ k I
or
2.
and in each set, no
+
a
$2
n
S
and so
x
k
m
k
appears as an
2
<_
k <
k <
either
> m
k
Z
a
S1
a
n
a
<-- ankXmk
a"
a
m
k
0
n
as
a
x
a
m
k
Put
(nk2+l, mk2+l)
Continuing in this way, if
k3+l.
n
S
or
1
S
Othez-ise
is finite.
2
(n
when both
mj
k
n
.
ankX
x
or
m
k k
a
$2
n
k
k) (nj,mj)
(nk,mk)
m
a
m
k
n
1
Choosing
x
m
k k
x
k
S
n
M
k
for all
m0
+
m
k
or
1
(i.e.,
with the desired properties,
(nk,mk)
a"
x
(x
m
k
n
m
.
x <
k
We shall now show that for every
tive integers such that
and
k
(a
x
k
m
k
n
+ a
n
x
k
k
x
a
m
k
a
a
m n
k k
(a
0
n
n
x
k
x
k
n
n
n
and
k,j
n
(x
a
nk
mk
k,
a
(x
m
0
x
m n
k k
x
m
k
k
k
[ k an
x
k
n
n
k
k
we have
k
a
k
).
x
mk
we h ave
m
k
x
m
k
there exists a subsequence
(a
k
[ k an
amkX
0,
e
[
k e
n
Since
and so
x
x
m
n
k
k
k
Furthermore since
.
x
k(an k amk) (xmk xnk).
0 < x
and
k
k
k m m
k k
[k(ank
+ I.
is the
k.).
for every
k
kl+l
Put
k2+l.
a) and b) in Theorem 1 and also satisfying
Now consider the series
0
where
I
Therefore
m0
accordingly we produce the sequence
2
k
k
k
satisfying
n
n
k < k
can occur at
k
is a disjoint partition of the set of all
SI,S 2
Then clearly
1
n
is the least positive integer, if it
k2+l
For either case, no
are infinite.
S
k
no such least positive integer exists, then either
both
That is,
is the least positive integer, if it exists, such
k3+l
for some
nk
(n
for some
k
i
(nl,mI)
Put
least posltive integer such that
(nkl+l’ mkl+l)
for all
m.l
of posi-
(k)
n
This follows from the
nk
following more general fact.
Suppose
[d(k)
that
(d(k))
is a non-negative sequence for which
We assert that very every
e
[d(kn
).
d(k)
0
as
e > 0, there exists a subsequence
and
k
(k) such
n
The proof of this fact proceeds along the same lines as the
proof of Riemann’stheorem on rearrangments of conditionally convergent series.
Fix
DOT PRODUCT REARRANGEMENTS
e > 0
and choose
n
>_ N I
I
d(k) < e
so that
I
k=N
<_
d(k) < e
1
Choose
N
choose
n
n
2
> n
I
k=N
d(k).
1
so that
I
I
[
d(k).
n
2
[
Hence
y
d(k) < e
k=N
1
> N
k > N
p
>
n
d(k) < e
d(k)
d(k))/2
.
2
d(k)
as
0
for every
N
d(k)
k/ and
N
k
n
(n)
P
2
d(k)
k=N
d(k).
e
2
Proceeding inductively
of positive integers for which
q
d(k))/2
k=N
q=l
and then
n
k=N2
and
P
2
such that
2
[
<
k=Nl
(N)
0 < d(k) < (e
p-i
I
Hence
n2+l
I
p-i
p--
n
1
[
d(k) < (e
in this way, we obtain sequences
n
and so that
k=Nl
This can be done since
n
1
to be the largest integer greater than
2
[
such that
1
>_ N l,
k
1
k=N
n
k=N
N
nl+l
nl
for every
n
is the greatest integer greater than
415
p-I
for every
and every
p
q
and
p
n
p
n
p-i
I
q=l
d(k) < e
k=N
P
n +i
q
p
<
d(k)
k=N
d(k)
k=N
q
P
This implies that
p
n
d(k) < d(n
q=l k=N
n
p-i
q
[
0 < e
P
+ i)
I
q=l
(e
q
e/2
p-I
q
k=N
0
d(k))/2P-i
q
as
.
p
n
Therefore
q
e
d(k).
q=l k=N
Hence, if we choose (k) to be the strictly increasing
n
q
sequence of positive integers
U
{k
N
p=l
< k < n
p--
k,
we have
p
k
where
e
d(k n
Applying this result to the sequence
negative, tends to
0,
exist subsequences of
()’K
for which
e
and sums to
k)
Yk (ank
(n
and
(mk),
Let
and
e
so th at
e S(a,x).
r- a. x
Suppose
(x
x ), since it is nona
m
m
n
k
k
k
k
> 0, there
we obtain that for every
(a
n
which we shall again denote by
(n
k)
and
amk) (xmk xnk)
Now recall that we wish to show that
a- x
is taken to range over the set
a. x
S(a,x)
r <.
[a- x,
].
We already know
It suffices to show
and choose subsequences which we again call
(n
k)
r e S(a,x).
and
k)
(m
416
P. ERDOS and G. WEISS
k(an k amk) (xmk xnk)
e
We now choose
and
n
ni
k)
z(m
n
for which
for each
k
n
I
n
k,
and let
for every
nk,mk
for all
n
.
I
anX(n)
n
I
a x +
n n
Z
[
a x +
n n
a-x+
r e S(a,x)
and so
THEOREM 2.
Z
(a
(a)
a
If
a
(a
[
(a
n
n
x
+ a
m
/a
x
k
n
+ a
)(x
m
Conversely, if
/a
n+ i’
a
n
S(a,x)
[a
I
S(a,x).
letting
Indeed,
n.
x
a
a
k
anXn(n)
(x
m
k
x
m
k
n
))
k
k
a
.
n
x
an+l/a
n-
Then
S(a,x)
0
n
x
0,
n
a
for every
(x)
n
x
is
[a
+
n
(x)
n
a
and
[a
x
n
0,
1
].
x,
for which
].
x,
x
then
S(a,x)
That is,
x.
> a
z
_i
is not bounded.
an+i/an
>_
k
x
Define
[a" x,
x
(k)
Z
+
if
>
a
>_
4
k
h (k)
k
3- k
ak+l/ak
and
n
(a
n
z
h (n)
h(n)
Let
>_
4
3n) -I
n
%
0, we
-i
+iXk
> 1
3
n)-I
(k)
_<
Then
3 -n
h(k)
denote the
h (n)
Clearly
x
n
In fact, we claim that
].
anXn -7an(ah(n)
be any permuation of
I
n
Also by Theorem i, since
x,
S(a,x)
we claim that
z
n
for which
n
for which
k
a non-decreasing function of
n
for some
m
k
n
must remain bounded.
least positive integer
(x),
x
m
k
and
0
1
(x)
x
Suppose to the contrary that
x
or
k
Q.E.D.
condition (3) of the theorem is satisfied by
a
n
+ (a
x
m m
k k
k
k
k
satisfies condition (2) of the theorem.
claim that
n
is bounded, then by Theorem i, if
n+l’ n
(i.e., those
n
x
m n
k k
k
k
a
n
where
n
k
is well-defined since
which proves (3).
Let
m
r
g
bounded if and only if, for every
PROOF.
{n
k)
Then
a x +
n n
nZ
n
n(n
Let
as follows.
The permutation
denote the set
n
+
fix all other integers
k).
Z
Let
i,j.
(n)
Hence
Z
the requisite permutation on
for every
m.3
k
n
1/2
for some
ah (k) +i (ah (k) 3k)
-i
is
0. Letting
a" x
1
but
i.
Furthermore,
k,
then
DOT PRODUCT REARRANGEMENTS
-i
On the other hand, if
(k)
h(k)
lanXz(n) lan_ 1 (n) xn
[anXn(n)
In any case,
i,
for every
<
a
).3-n
x
1/2 < 1
S(a,x).
1
hence
then
k,
h (n) n
417
Q.E.D.
In the proof of Theorem i, each time we constructed a permutation
NOTE.
[anX(n)
solve the equation
that we constructed was the product of disjoint 2-cycles.
is, each such
That
it sufficed to use only disjoint 2-cycles.
r,
to
w
This
seems odd and leads us to ask if there are any circumstances in which the use of
In other words, is it always true that
infinite-cycles or n-cycles yields more.
S(a,x)
[anX(n)
is the same as
disjoint
Z
z is a permutation of
+
which is a product of
2-cycles} ?
The following question seems likely to have an affirmative answer.
would give a characterization for those sequences
and
0
x
n
which satisfy
QUESTION I.
If
S(a,x)
x
and
an ’
where
/
> 0,
al
However, it remains unsolved
[a- x,
are as above, does (3)
x
and
a
a
If so, this
=
(2) in Theorem i?
Finally, we wish to point out that Theorems i and 2 imply analogous theorems in
which
a
and
x
Indeed, the proofs of the following two corollaries
switch roles.
follow naturally along the same lines as those of Theorems 1 and 2.
Let
COROLLARY 3.
(x)
x
where
n
x
n
for all
> 0
n,
and
x
n
0
as
.
n
Consider the following conditions.
(i)
Xn/Xn+ 1
(2)
For the non-negative sequence
and
a)
b)
is bounded below.
(x
a
n
of
x
k
m
k
.aKnk
a
0
(a),
n
there exist subsequences (a
k
,
k
and
x
m
k
Then (i) implies that (2) holds for every strictly positive sequence
that tends to
n
x, respectively, such that
and
as
a
a
(a)
n
418
P. ERDOS and G. ISS
Let
COROLLARY 4.
x
(x)
n
be a non-negative sequence.
bounded below if and only if, for every
S(a,x)
[a
x,
QUESTION 2.
S(a,x)?
a
(an)
for which
Then
an
+
x /x
n
and
is
n+l
a
I
> 0,
].
Is there anything to be said about the qualitative nature of
Is it always a Borel set, measurable,
F
G
?
REFE REN CE
i.
WEISS, GARY "Commutators and operator ideals", dissertation, University of
Michigan, 19 75.