# A series whose sum range is an arbitrary finite set

Jakub Onufry Wojtaszczyk
Department of Mathematics, Computer Science and Mechanics
University of Warsaw
ul. Banacha 2, 02-097 Warsaw, Poland
email<EMAIL_ADDRESS>
(March 21, 2004)

###### Abstract

In finitely-dimensional spaces the sum range of a series has to be an affine
subspace. It is long known this is not the case in infinitely dimensional
Banach spaces. In particular in 1984 M.I. Kadets and K. Woz̀niakowski obtained
an example of a series the sum range of which consisted of two points, and
asked whether it is possible to obtain more than two, but finitely many
points. This paper answers the question positively, by showing how to obtain
an arbitrary finite set as the sum range of a series in any infinitely
dimensional Banach space.

##  1 Introduction

For a finitely-dimensional linear space $X$ the well-known Steinitz theorem
states that for any conditionally convergent series the set of all possible
limits of the series (called the sum range) is a affine subspace of $X$. In
the ”Scottish Book” S. Banach posed the problem whether the same holds for
infinitely dimensional Banach spaces. The problem was solved negatively in the
same book by J. Marcinkiewicz. In his example the sum range is the set $M$ of
all integer-valued functions in $L_{2}[0,1]$. The next example, due to M. I.
Ostrovskii, showed that the sum range does not have to be a closed set - the
sum range of Ostrovskii’s series was of the form $M+\sqrt{2}M$. Finally M. I.
Kadets constructed an example in which the sum range consisted of two points,
disproving, in particular, H. Hadwiger’s conjecture that the sum range has to
be the coset of some additive subgroup of $X$. The justification of the
example was obtained independently by K. Woz̀niakowski and P. A. Kornilov in
1986.

It is still unknown what sets can be the sum ranges of series. In this paper
it is shown that any finite subset of $X$ can be the sum range of a
conditionally convergent series, which solves the problem posed by M. I.
Kadets along with his two-point example (the problem is stated in [S91] in the
general case, and in [U02] for $X=C(\Delta)$ and $n=3$). The example is an
extension of the 2-point example of M. I. Kadets as given in [S91]. As far as
possible I shall try to keep the notation consistent with the notation given
there, although the lack of suitable letters in the latin alphabet will force
me to abandon the notation in a few places.

Everywhere all spaces are considered with the $L_{1}$ norm, i.e.
$||f||_{X}=\int_{X}|f(x)|dx$. Frequently it is assumed it is obvious on which
space the norm is taken, and only $||f||$ is written.

##  2 The results of K. Woz̀niakowski

The work in this paper is strongly inspired by the 2-point example of M. I.
Kadets and the proof by K. Woz̀niakowski. In this paper not only the final
result of Woz̀niakowski’s work will be used, but also multiple technical facts
than can be found in the proof. Rather than force the reader to search for
those in the original paper, I shall reiterate here Woz̀niakowski’s work, at
times formulating the results in a way that will make them easier to use in
the subsequent sections. The whole content of this section in based on [S91],
and a reader familiar with this work may probably skip to the next section.

Let $Q=[0,1]^{\omega}$ be the infinite dimensional cube, i.e. the product of a
countable number of unit segments, equipped with the standard product topology
and measure. By $x=(x_{1},x_{2},\ldots)$ we shall denote the variable on $Q$.
Suppose we have two sequences of functions on the cube: $a_{m}^{n}$ and
$b_{m,j}^{n}$, where $n\in{\mathbb{N}}$ and for a given $n$ both $m$ and $j$
belong to some finite sets $M_{n}$ and $J_{n}=M_{n+1}$ respectively. By
$A_{n}$ we shall denote the set $\\{a_{m}^{n}:m\in M_{n}\\}$, and by $B_{n}$
the set $\\{b_{m,j}^{n}:m\in M_{n},j\in J_{n}\\}$. For convenience if $X$ is a
set of functions, by $\tilde{X}$ we shall denote the sum of the functions from
$X$. We shall assume the following properties of the functions $a_{m}^{n}$ and
$b_{m,j}^{n}$:

$\displaystyle\tilde{A_{n}}(x)$ $\displaystyle=$ $\displaystyle
1\qquad\forall_{n\in{\mathbb{N}}}\forall_{x\in Q}$ (1)
$\displaystyle||a_{m}^{n}||$ $\displaystyle=$ $\displaystyle\frac{1}{|M_{n}|}$
(2) $\displaystyle\lim_{n{\rightarrow}\infty}|M_{n}|$ $\displaystyle=$
$\displaystyle\infty$ (5) The function $a_{m}^{n}$ depends only on the
variable $x_{n}$ The functions $a_{m}^{n}$ assume only values 0 and 1
$\displaystyle b_{m,j}^{n}$ $\displaystyle=$ $\displaystyle-a_{m}^{n}\cdot
a_{j}^{n+1}$ (6)

We shall this collection of properties the Kadets properties on the cube $Q$.
These properties mean that for each $n$ the interval $[0,1]$ is divided into
$|M_{n}|$ sets $V_{m}^{n}$ of equal measure, and
$a_{m}^{n}(x_{1},x_{2},\ldots)=1$ iff $x_{n}\in V_{m}$. The functions
$b_{m,j}^{n}$ are negative, and are supported on rectangles
$(x_{n},x_{n+1})\in V_{m}^{n}\times V_{j}^{n+1}$.

From the Kadets properties we can easily deduce another few properties, mainly
about the behaviour of $b_{m,j}^{n}$ based on properties 1 and 6:

$\displaystyle a_{m}^{n}$ $\displaystyle=$ $\displaystyle-\sum_{j\in
J_{n}}b_{m,j}^{n},$ (7) $\displaystyle a_{j}^{n+1}$ $\displaystyle=$
$\displaystyle-\sum_{m\in M_{n}}b_{m,j}^{n}$ (8)
$\displaystyle\tilde{B_{n}}(x)$ $\displaystyle=$
$\displaystyle-{\mathbf{1}}\forall_{n\in{\mathbb{N}}}$ (9)
$\displaystyle||b_{m,j}^{n}||$ $\displaystyle=$
$\displaystyle\frac{1}{|M_{n}\times J_{n}|}$ (14) The function $b_{m,j}^{n}$
depends only on the variables $x_{n}$ and $x_{n+1}$ The functions
$b_{m,j}^{n}$ assume only values 0 and -1 $a_{m}^{n}$ and $a_{m^{\prime}}^{n}$
have almost disjoint supports for $m\neq m^{\prime}$

These properties follow easily from the Kadets properties. In property 14 by
almost disjoint supports we mean that the intersection of two supports is of
measure zero, we can obviously modify $a_{m}^{n}$ so that the Kadets
properties still hold and the sets $\\{x:a_{m}^{n}(x)>0\\}$ are disjoint for
any constant $n$ and any two different values of $m$.

Let $c_{k},k\in{\mathbb{N}}$ be any ordering of all the functions $a_{m}^{n}$
and $b_{m,j}^{n}$. Following Woz̀niakowski we shall investigate the
convergence of any reordering $c_{\sigma(k)}$ of $c_{k}$.

######  Proposition 2.1.

For any family of functions $c_{k}$ having the Kadets properties there exist
such two permutations $\sigma$ and $\tau$ of ${\mathbb{N}}$ that $\sum
c_{\sigma(k)}{\rightarrow}{\mathbf{0}}$ and $\sum
c_{\tau(k)}{\rightarrow}{\mathbf{1}}$.

###### Proof.

For $\sigma$ it is enough to order the functions $a_{m}^{n}$
lexicographically, i.e. $a_{m}^{n}$ appears before
$a_{m^{\prime}}^{n^{\prime}}$ iff $n<n^{\prime}$ or $n=n^{\prime}$ and
$m<m^{\prime}$, and then immediately after each $a_{m}^{n}$ to put the whole
set $\\{b_{m,j}^{n}:j\in J_{n}\\}$. Then the sum of each block consisting of a
single function $a_{m}^{n}$ and the functions $b_{m,j}^{n}$ following it sums
up to zero due to property 7, so the norm of each partial sum is the norm of
the currently open block, which converges to zero due to properties 2, 14 and
5.

To get $\tau$ we order the functions $a_{m}^{n}$ in the same way, but we
follow each function $a_{m}^{n}$ for $n>1$ by the set $\\{b_{l,m}^{n-1}:l\in
M_{n-1}\\}$, the functions $a_{m}^{1}$ are not followed by anything (as there
are no functions $b_{m,j}^{0}$). Then the functions $a_{m}^{1}$ sum up to the
constant function 1 due to property 1. The following blocks again sum up to
zero, this time due to property 8, so the norm of the difference between 1 and
a particular partial sum is equal to the norm of the currently open block,
which again converges to zero due to properties 2, 14 and 5.∎

######  Remark 1.

The series of functions from Proposition 2.1 converge not only in the $L_{1}$
norm, but also in any $L_{p}$ norm for any $p<\infty$.

###### Proof.

Again it is only a question of investigating the norm of any given block, as
the sum of the previous blocks is zero. Functions $a_{m}^{n}$ assume only
values 0 and 1 and have disjoint supports for a set $n$ from properties 5 and
14. Functions $b_{m,j}^{n}$ for a given $n$ have disjoint supports (this
follows from properties 6 and 14) and assume values $0$ and $-1$ (from 14).
Thus for any $f$ being a sum of any set of functions $a_{m}^{n}$ and
$b_{m,j}^{n}$ for a fixed $n$ (or $a_{m}^{n}$ and $b_{m,j}^{n-1}$ for a fixed
$n$ in the case of $\tau$) we have $\|f\|_{\infty}\leq 1$. This implies for
any $1\leq p<\infty$

$\|f\|_{p}=(\int|f|^{p})^{1/p}=(\int|f|\cdot|f|^{p-1})^{1/p}\leq(\|f\|_{1}\cdot\|f^{p-1}\|_{\infty})^{1/p}\leq\|f\|_{1}^{1/p}\cdot
1=\|f\|_{1}^{1/p}.$

Thus if the sum of the series tended to zero in the $L_{1}$ norm with $n$
tending to infinity, it also tends to zero in any $L_{p}$ norm for
$p<\infty$.∎

######  Proposition 2.2.

If a reordering $c_{\sigma(k)}$ of a family $c_{k}$ having the Kadets
properties converges, it converges to a constant integer function.

###### Proof.

Due to properties 5 and 14 and the finiteness of the sets $M_{n}$ and $J_{n}$
only finitely many of the functions $c_{\sigma(k)}$ depend on a given variable
$x_{l}$ \- precisely the functions belonging to $A_{l}$, $B_{l}$ and
$B_{l-1}$. Moreover the sum of all these functions equals to the constant
function $-1$ due to properties 1 and 9. Thus for some integer $K_{0}$ the
function $\sum_{k=1}^{K}c_{\sigma(k)}$ is constant with regard to $x_{l}$ for
$K\geq K_{0}$, and thus the limit of the series also has to be constant with
regard to $x_{l}$. As this applied to an arbitrary $l$, the limit simply has
to be constant.

As the functions $c_{k}$ are integer-valued (properties 5 and 14), their sums
also have to be integer-valued. Thus all the partial sums of the series are
integer-valued, and so the limit is also integer-valued, which ends the
proof.∎

The next step will be to show that 0 and 1 are the only possible limits of a
rearrangement of a family of functions with the Kadets property. We shall set
a fixed rearrangement $c_{\sigma(k)}$ of a given Kadets family, and we shall
assume that the sum $\sum_{k}c_{\sigma(k)}$ converges to some constant integer
$C\neq 1$ (we know $C=1$ can be achieved, it remains to prove that under these
assumptions $C=0$).

Take an arbitrary $\delta>0$ and fix $K_{0}=K_{0}(\delta)$ such that for any
$K>K_{0}$,

$\big{|\big{|}}C-\sum_{k=1}^{K}c_{\sigma(k)}\big{|\big{|}}\leq\delta$ (15)

and for any $m>l>K_{0}$ the Cauchy condition holds, i.e.

$\big{|\big{|}}\sum_{k=l}^{m}c_{\sigma(k)}\big{|\big{|}}\leq\delta.$ (16)

In addition to the sets $A_{n}$ and $B_{n}$ introduced earlier we shall also
consider $V_{n}=\bigcup_{k=1}^{n}(A_{k}\cup B_{k})$. Let $M$ be any integer
such that

$c_{\sigma(k)}\in V_{M}\cup A_{M+1}\qquad\mbox{for any $k\leq K$.}$ (17)

Let $c_{k}^{*}=c_{\sigma(k)}$ if $c_{\sigma(k)}\in V_{M}\cup A_{M+1}$, 0
otherwise. Similarly let $\bar{c}_{k}=c_{\sigma(k)}$ if $c_{\sigma(k)}\in
B_{M+1}$, 0 otherwise. By $c^{*}$ we shall denote
$\sum_{k=K_{0}+1}^{\infty}c_{k}^{*}$, while by $c$ we shall denote
$\sum_{k=1}^{K_{0}}c_{\sigma(k)}$. The sum $c+c^{*}$ is equal to
$\tilde{V}_{M}+\tilde{A}_{M+1}=0+1=1$. Hence
$||c^{*}||=||1-c||\geq||1-C||-||C-c||\geq 1-\delta$. Let $k_{0}=K_{0}$ and

$k_{j+1}=\min\big{\\{}k:\frac{1}{4}-\frac{5\delta}{4}\leq{\big{|}\big{|}}\sum_{i=k_{j}+1}^{k}c_{k}^{*}{\big{|}\big{|}}\leq\frac{1}{4}-\frac{\delta}{4}\big{\\}}.$
(18)

The indices $k_{j}$ are well defined for $j$ from 1 to 4 because the total
norm of the sum $c^{*}$ is at least $1-\delta$ and each single $c_{k}^{*}$ has
norm $\leq\delta$ due to the Cauchy condition (16). For $j=0,1,2,3$ define the
following functions:

$c_{j+1}^{**}=\sum_{k=k_{j}+1}^{k_{j+1}}c_{k}^{*},\qquad\bar{\bar{c}}_{j+1}=\sum_{k=k_{j}+1}^{k_{j+1}}\bar{c}_{k},\qquad\hat{c}_{j+1}=\sum_{k=k_{j}+1}^{k_{j+1}}c_{\sigma(k)},$

and for $j=1,2,3,4$ set $r_{j}=\hat{c}_{j}-\bar{c}_{j}-c^{**}_{j}$.

In plain words this means that we divide the functions $c_{k}$ for
$k_{j}<k\leq k_{j+1}$ into three sets - those from $A_{n}$ for $n\leq M+1$ or
$B_{n}$ for $n\leq M$ (these add up to $c_{j}^{**}$), those from $B_{M+1}$
(these add up to $\bar{\bar{c}}_{j}$) and the rest (these add up to $r_{j}$).
We will show that the functions from $B_{M+1}$ are placed in $c_{k}$ in
similar proportions as the functions from $V_{M}\cup A_{M+1}$ — if, say, about
a half of the functions from $V_{M}\cup A_{M+1}$ appeared in $c_{k}$ (that
happens at $k_{2}$) then about a half of the functions from $B_{M+1}$ must
have appeared, too.

We shall need to estimate the norm of two sums, which we would like to be
negligible: $||r_{j}||$ and $||\sum_{k=k_{4}+1}^{\infty}c_{k}^{*}||$. We know
that the sum of all $c_{k}$ up to $k_{j}$ is negligible, thus if the high-$n$
functions ($r_{j}$) are negligible, the functions from $V_{M}\cup A_{M+1}$ and
$B_{M+1}$ have to approximately cancel each other out. This motivates the
following proposition:

######  Proposition 2.3.

For a Kadets family of functions $c_{k}$, its rearrangement $c_{\sigma(k)}$
converging to some $C\neq 1$, an arbitrary $\delta$ and an arbitrary
$M>K_{0}(\delta)$ as above, with the notation as above we have
$\sum_{j=1}^{4}||r_{j}||\leq 18\delta$.

###### Proof.

As $c_{j}^{**}$ is integer-valued (being a sum of some functions from a Kadets
family), the condition $||c_{j}^{**}||\leq\frac{1}{4}$ implies $|{\rm
supp}c_{j}^{**}|\leq\frac{1}{4}$. Thus we can use lemma 1 (from the section
”Auxiliary lemmas”) to get

$||c_{j}^{**}+r_{j}||\geq||c_{j}^{**}||+(1-2|{\rm
supp}c_{j}^{**}|)||r_{j}||=||c_{j}^{**}||+\frac{1}{2}r_{j}.$

Of course $||\hat{c}_{j}||\leq\delta$ from the Cauchy condition (16). We thus
have

$1\geq\sum_{j=1}^{4}||\bar{\bar{c}}_{j}||=\sum_{j=1}^{4}||\hat{c}_{j}-c_{j}^{**}-r_{j}||\geq\sum_{j=1}^{4}||c^{**}_{j}+r_{j}||-\sum_{j=1}^{4}||\hat{c}_{j}||\geq$
$\geq\sum_{j=1}^{4}(||c^{**}_{j}||+\frac{1}{2}||r_{j}||)-4\delta\geq
1-5\delta+\frac{1}{2}\sum_{j=1}^{4}||r_{j}||-4\delta,$

which gives us the sought estimate upon $||r_{j}||$, namely
$\sum_{j=1}^{4}||r_{j}||\leq 18\delta$. In particular, of course, each
$||r_{j}||$ is bounded by $18\delta$. ∎

######  Corollary 2.4.

With the notation and assumptions as above,
$||\bar{\bar{c}}_{j}+c^{**}_{j}||\leq 19\delta$

###### Proof.

$||\bar{\bar{c}}_{j}+c^{**}_{j}||=||\hat{c}_{j}-r_{j}||\leq||\hat{c}_{j}||+||r_{j}||\leq\delta+18\delta=19\delta$.∎

######  Proposition 2.5.

For a Kadets family of functions $c_{k}$, its rearrangement $c_{\sigma(k)}$
converging to some $C\neq 1$, an arbitrary $\delta$ and an arbitrary
$M>K_{0}(\delta)$ as above, with the notation as above we have
$||\sum_{k=k_{4}+1}^{\infty}c_{k}^{*}||\leq 11\delta$.

###### Proof.

We have

$||\bar{\bar{c}}_{j}||=||\hat{c}_{j}-c_{j}^{**}-r_{j}||\geq||c_{j}^{**}+r_{j}||-||\hat{c}_{j}||\geq||c_{j}^{**}||+\frac{1}{2}||r_{j}||-||\hat{c}_{j}||\geq||c_{j}^{**}||-\delta\geq\frac{1}{4}-\frac{9\delta}{4}.$

Take any index $k^{\prime}>k_{4}$. If the norm
$||\sum_{k=k_{4}+1}^{k^{\prime}}c_{k}^{*}||$ were greater then $11\delta$,
then there would exist some $k_{5}\in(k_{4},k^{\prime}]$ such that
$12\delta\geq||\sum_{k=k_{4}+1}^{k_{5}}c_{k}^{*}||>11\delta$. Then by a
similar argument
($||\bar{\bar{c}}_{5}||\geq||c_{5}^{**}||+(1-24\delta)||r_{5}||-||\hat{c}_{5}||\geq
11\delta-\delta$) the norm of $\sum_{k=k_{4}+1}^{k_{5}}\bar{c}_{k}$ would be
larger then $10\delta$ — but all the functions $\bar{c}_{k}$ are negative, so
$||\sum\bar{c}_{k}||=\sum||\bar{c}_{k}||$, which in this case gives
$1\geq||\sum_{k=k_{0}}^{k_{5}}\bar{c}_{k}||=\sum_{j=1}^{4}||\bar{c}_{k}||+||\sum_{k=k_{4}+1}^{k_{5}}\bar{c}_{k}||>1-9\delta+10\delta$,
a contradiction. Thus the norm $||\sum_{k=k_{4}+1}^{\infty}c_{k}^{*}||$ has to
be no greater than $11\delta$ (the sum is convergent, as it is in fact the sum
of a finite number of functions, all coming from $V_{M+1}$). Let us denote
this sum by $c_{5}^{**}$.∎

Now we can prove the main theorem of Woz̀niakowski’s work:

######  Theorem 2.6.

For a Kadets family of functions $c_{k}$ and some rearrangement
$c_{\sigma(k)}$ converging to $C\neq 1$ we have
$|C-\frac{1}{2}|\leq\frac{1}{2}$, which (due to lemma 2.2) implies $C=0$.

###### Proof.

Consider any $\delta$, and the partial sum $S=\sum_{k=1}^{k_{4}}c_{\sigma(k)}$
with the notation as above. As $k_{4}>K_{0}$, from assumption 15 we know that
$||S-C||\leq\delta$, so it will suffice to estimate $||S-\frac{1}{2}||$. We
have

$||S-\frac{1}{2}||={\big{|}\big{|}}c+\sum_{j=1}^{4}c_{j}^{**}+\sum_{j=1}^{4}\bar{\bar{c}}_{j}+\sum_{j=1}^{4}r_{j}+c_{5}^{**}-c_{5}^{**}-\frac{1}{2}{\big{|}\big{|}}=$
$={\big{|}\big{|}}c+c^{*}-\frac{1}{2}+\sum_{j=1}^{4}\bar{\bar{c}}_{j}+\sum_{j=1}^{4}r_{j}-c_{5}^{**}{\big{|}\big{|}}\leq{\big{|}\big{|}}\frac{1}{2}+\sum_{j=1}^{4}\bar{\bar{c}}_{j}{\big{|}\big{|}}+{\big{|}\big{|}}\sum_{j=1}^{4}r_{j}{\big{|}\big{|}}+{\big{|}\big{|}}c_{5}^{**}{\big{|}\big{|}}.$

The function $\sum_{j=1}^{4}\bar{\bar{c}}_{j}$ is a sum of functions from
$B_{M+1}$, which means assumes only the values 0 and $-1$, thus
$|\frac{1}{2}+\sum_{j=1}^{4}\bar{\bar{c}}_{j}|$ is always equal to
$\frac{1}{2}$. Inserting this and the bounds upon $r_{j}$ and $c_{5}^{**}$ we
get

$||S-\frac{1}{2}||\leq\frac{1}{2}+18\delta+11\delta=\frac{1}{2}+29\delta.$

As $||S-C||\leq\delta$ we get $||C-\frac{1}{2}||\leq\frac{1}{2}+30\delta$. As
$\delta$ was chosen arbitrarily, we get the thesis.∎

######  Corollary 2.7.

The sum range of any Kadets family consists of two points, the constant
functions 0 and 1, in any $L_{p}$ norm for $1\leq p<\infty$

###### Proof.

From Proposition 2.1 and Remark 1 we know that the two constant functions
belong to the sum range. From the Proposition 2.2 we know that all functions
in the sum range in the $L_{1}$ norm are constant integer functions, and from
Theorem 2.6 we know that only the two functions $0$ and $1$ are eligible. If
any permutation of the series converged to some function $g$ in some $L_{p}$
norm, then $\|S_{n}-g\|_{p}$ would tend to zero. But from the Hölder
inequality we know that $\|S_{n}-g\|_{p}\geq\|S_{n}-g\|_{1}$ (as the measure
of the whole space is 1), which would imply that the series $S_{n}$ converges
also in the $L_{1}$ norm, contradicting Theorem 2.6.∎

##  3 The 3-point series

Denote by $Q_{i}=[0,1]^{\omega},i=1,2,3$ the infinite dimensional cube, i.e.,
the product of a countable number of unit segments equipped with the standard
product probability measure. The example will be constructed in
$L_{1}(Q_{1}\cup Q_{2}\cup Q_{3})$. In the whole paper
$t=(t_{1},t_{2},\ldots)$ will denote the variable on $Q_{1}$,
$u=(u_{1},u_{2},\ldots)$ will denote the variable on $Q_{2}$ and
$v=(v_{1},v_{2},\ldots)$ will denote the variable on $Q_{3}$.

Our series will consist of functions of three kinds. The functions of the
first kind are defined as follows:

$f_{m}^{n}(t)=\begin{cases}1&\mbox{if $\frac{m-1}{n}<t_{n}<\frac{m}{n}$}\\\
0&\mbox{otherwise.}\end{cases}$ $f_{m}^{n}(u)=f_{m}^{n}(v)=0$

for $n\in{\mathbb{N}},m\in\\{1,2,\ldots,n\\}$.

The second kind of functions is defined on all three cubes:

$\displaystyle g_{m,j}^{n}(t)$ $\displaystyle=$
$\displaystyle\begin{cases}-1&\mbox{if $\frac{m-1}{n}<t_{n}<\frac{m}{n}$ and
$\frac{j-1}{n+1}<t_{n+1}<\frac{j}{n+1}$}\\\ 0&\mbox{otherwise}\end{cases}$
$\displaystyle g_{m,j}^{n}(u)$ $\displaystyle=$
$\displaystyle\begin{cases}\frac{1}{n+1}&\mbox{if
$\frac{m-1}{n}<u_{n}<\frac{m}{n}$}\\\ 0&\mbox{otherwise}\end{cases}$
$\displaystyle g_{m,j}^{n}(v)$ $\displaystyle=$
$\displaystyle\begin{cases}1&\mbox{if
$\frac{(m-1)(n+1)+j-1}{n(n+1)}<v_{n}<\frac{(m-1)(n+1)+j}{n(n+1)}$}\\\
0&\mbox{otherwise}\end{cases}$

for $n\in{\mathbb{N}},m\in\\{1,2,\ldots,n\\},j\in\\{1,2,\ldots,n+1\\}$.

The functions of the third kind are defined on $Q_{2}$ and $Q_{3}$:

$\displaystyle h_{m,j,k}^{n}(t)$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle h_{m,j.k}^{n}(u)$ $\displaystyle=$
$\displaystyle\begin{cases}-\frac{1}{(n+1)^{2}(n+2)}&\mbox{if
$\frac{m-1}{n}<u_{n}<\frac{m}{n}$}\\\ 0&\mbox{otherwise}\end{cases}$
$\displaystyle h_{m,j,k}^{n}(v)$ $\displaystyle=$
$\displaystyle\begin{cases}-1&\mbox{if
$\frac{(m-1)(n+1)+j-1}{n(n+1)}<v_{n}<\frac{(m-1)(n+1)+j}{n(n+1)}$ and
$\frac{k-1}{(n+1)(n+2)}<v_{n+1}<\frac{k}{(n+1)(n+2)}$}\\\
0&\mbox{otherwise}\end{cases}$

for
$n\in{\mathbb{N}},m\in\\{1,2,\ldots,n\\},j\in\\{1,2,\ldots,n+1\\},k\in\\{1,2,\ldots,(n+1)(n+2)\\}$.

These functions have properties we want to generalize. Suppose we have three
families of indices: $M_{n}$, $J_{n}$ and $K_{n}$, with $J_{n}=M_{n+1}$ and
$K_{n}=M_{n+1}\times J_{n+1}$ (here $M_{n}=\\{1,2,\ldots,n\\}$ and the mapping
between $\\{1,2,\ldots,n\\}\times\\{1,2,\ldots,n+1\\}$ and
$\\{1,2,\ldots,n(n+1)\\}$ is given by $(m,j)\mapsto(m-1)(n+1)+j$). We have
three families of functions: the first kind
$\\{f_{m}^{n}:n\in{\mathbb{N}},m\in M_{n}\\}$, the second kind
$\\{g_{m,j}^{n}:n\in{\mathbb{N}},m\in M_{n},j\in J_{n}\\}$ and the third kind
$\\{h_{m,j,k}^{n}:n\in{\mathbb{N}},m\in M_{n},j\in J_{n},k\in K_{n}\\}$
defined on the union $Q_{1}\cup Q_{2}\cup Q_{3}$ of Hilbert cubes. The
families $f$ and $g$ form a Kadets family on $Q_{1}$, while the functions $h$
disappear on $Q_{1}$. On $Q_{3}$ the functions $g$ and $h$ form a Kadets
family (with $M_{n}\times J_{n}$ being the first index set and $K_{n}$ the
second), while functions $f$ disappear. The properties of the functions on
$Q_{2}$ are different, as follows:

$\displaystyle\sum_{m\in M_{n}}\sum_{j\in J_{n}}g_{m,j}^{n}$ $\displaystyle=$
$\displaystyle{\mathbf{1}}$ (19) $\displaystyle\sum_{m\in M_{n}}\sum_{j\in
J_{n}}\sum_{k\in K_{n}}h_{m,j,k}^{n}$ $\displaystyle=$
$\displaystyle-{\mathbf{1}},$ (20) $\displaystyle g_{m,j}^{n}$
$\displaystyle=$ $\displaystyle-\sum_{k\in K_{n}}h_{m,j,k}^{n},$ (21)
$\displaystyle\sum_{m^{\prime}\in M_{n+1}}g_{m^{\prime},j^{\prime}}^{n+1}$
$\displaystyle=$ $\displaystyle-\sum_{m\in M_{n}}\sum_{j\in
J_{n}}\sum_{m^{\prime}\in M_{n+1}}h_{m,j,(m^{\prime},j^{\prime})}^{n}.$ (22)
$\displaystyle\sum_{j\in J_{n}}g_{m,j}$ assumes only values 0 and 1 (23)
$\displaystyle\int_{Q_{2}}g_{m,j}^{n}$ $\displaystyle=$
$\displaystyle\int_{Q_{3}}g_{m,j}^{n}$ (24)
$\displaystyle\int_{Q_{2}}h_{m,j,k}^{n}$ $\displaystyle=$
$\displaystyle\int_{Q_{3}}h_{m,j,k}^{n}$ (25) $\displaystyle||g_{m,j}^{n}||$
$\displaystyle=$ $\displaystyle\frac{1}{|M_{n}\times J_{n}|}$ (26)
$\displaystyle||h_{m,j,k}^{n}||$ $\displaystyle=$
$\displaystyle\frac{1}{|M_{n}\times J_{n}\times K_{n}|}$ (28) The functions
$g_{m,j}^{n}$ and $h_{m,j,k}^{n}$ on $Q_{2}$ depend only on $u_{n}$

Such a family of functions will be called a 3-Kadets family. It is easy
(although maybe a bit tedious) to check that the family defined at the
beginning of the section is a 3-Kadets family.

We shall denote by $F_{n}$ the set $\\{f_{m}^{n}:m\in M_{n}\\}$, by $G_{n}$
the set $\\{g_{m,j}^{n}:m\in M_{n};j\in J_{n}\\}$ and by $H_{n}$ the set
$\\{h_{m,j,k}^{n}:m\in M_{n},j\in J_{n};k\in K_{n}\\}$. Also, by $V_{M}$ we
shall denote $\bigcup_{k=1}^{M}F_{k}\cup G_{k}\cup H_{k}$. Denote by $d_{n}$
any set enumeration of the whole 3-Kadets family. We are investigating the
possible limits of $\sum_{n=1}^{\infty}d_{\sigma(n)}$ for all permutations
$\sigma$ of ${\mathbb{N}}$.

If a given rearrangement $d_{\sigma(n)}$ of a 3-Kadets family converges, it
converges on each of the cubes separately. On $Q_{1}$ and $Q_{3}$ we have
Kadets families of functions, so the series on each of these cubes converges
either to ${\mathbf{0}}$ or to ${\mathbf{1}}$ due to theorem 2.6. The new part
is the behaviour on $Q_{2}$. Same as in the first part of Proposition 2.2 only
finitely many functions depend on a given variable $u_{n}$ – the functions
$g_{m,j}^{n}$ and $h_{m,j,k}^{n}$ – and their sum is constant, equal to zero
due to property (21) applied to each $j$ separately. Thus again the limit of
the series $\sum d_{\sigma(n)}$ on $Q_{2}$ has to be a constant function.

As $\int_{Q_{2}}d_{n}=\int_{Q_{3}}d_{n}$ for any $d_{n}$ (it is 0 for
functions of the first kind and follows from properties 24 and 25 for the
second and third kind), we get
$\int_{Q_{2}}\sum_{n=1}^{N}d_{\sigma(n)}=\int_{Q_{3}}\sum_{n=1}^{N}d_{\sigma(n)}$.
As the integral is a continuous functional on $L_{1}(Q_{2})$ and
$L_{1}(Q_{3})$ we get that the integrals of the limits have to be equal – but
we know that the limit of $\sum d_{\sigma(n)}$ on both $Q_{2}$ and $Q_{3}$ is
a constant function, so the equality of integrals implies the equality of the
limits. Thus the limit of the whole series is described by a pair of integers
- the value on $Q_{1}$ and the value on $Q_{3}$. Let us denote the limit
function by $d_{\infty}$.

We are to show that it is possible to obtain exactly three different sums –
precisely we can obtain
$({\mathbf{0}},{\mathbf{0}}),({\mathbf{1}},{\mathbf{0}})$ and
$({\mathbf{1}},{\mathbf{1}})$. To obtain any of these limits we first arrange
the functions $f$ and $g$ as by Proposition 2.1 for a Kadets family on
$Q_{1}$, and then after each $g$ we put the $h$ functions as by Proposition
2.1 for the cube $Q_{3}$. It remains to be seen if we get convergence on
$Q_{2}$.

In the case of $({\mathbf{0}},{\mathbf{0}})$ after a given $f_{m}^{n}$ there
appear the all functions $g_{m,j}^{n}$ and $h_{m,j,k}^{n}$ with the same $m$
and $n$. The sum of all these functions on $Q_{2}$ is equal to ${\mathbf{0}}$
due to property (21) for each $j$ separately. Thus the norm of the partial sum
on $Q_{2}$ is equal to the norm of the functions appearing after the last $f$,
and this tends to zero due to properties 26, 28 and 5 (all the functions have
the same index $m$, so the sum of their norms is equal to
$\frac{2}{|M_{n}|}{\rightarrow}0$).

In the case of $({\mathbf{1}},{\mathbf{0}})$ after a given $f_{m}^{n}$ we get
the functions $g_{l,m}^{n-1}$ and $h_{l,m,k}^{n-1}$. The sum of all these
functions on $Q_{2}$ is again ${\mathbf{0}}$ due to property 21, this time
applied to each $l$ separately. Again the norm of the difference between the
partial sum and $({\mathbf{1}},{\mathbf{0}})$ is the norm of the part after
the last $f$, and that again tends to 0.

In the case of $({\mathbf{1}},{\mathbf{1}})$ after a given $f_{m}^{n}$ we get
the functions $g_{l,m}^{n-1}$ and $h_{l^{\prime},m^{\prime},(l,m)}^{n-2}$.
Their sum is ${\mathbf{0}}$ due to property 22 applied to them all. Again the
norm of the difference between the partial sum and ${\mathbf{1}}$ tends to 0.

Again it is easy to check that the convergence occurs not only in the $L_{1}$
norm, but also in any $L_{p}$ norm for $p<\infty$ in the same way as in Remark
1 — on each of the cubes the $L_{\infty}$ norm of the partial sums is bounded
by 1.

One may wonder why the same arguments will not imply the convergence of the
series arranged by rows in $G_{n}$ and columns in $H_{n-1}$ to
$({\mathbf{0}},{\mathbf{1}})$. The answer is we lack the equivalent of
property 22 for this arrangement. To illustrate this let us look at the
3-Kadets family given at the beginning of the section arranged in this natural
way. The sum $\sum_{j=1}^{n+1}g_{m,j}^{n}$ on $Q_{2}$ is equal to $1$ on
$\frac{m-1}{n}<u_{n}<\frac{m}{n}$, while the sum of the appropriate column of
$H_{n-1}$,
$\sum_{j=1}^{n+1}\sum_{m^{\prime}=1}^{n-1}\sum_{j^{\prime}=1}^{n}h_{m^{\prime},j^{\prime},(m-1)(n+1)+j}^{n-1}$
is equal to $-\frac{1}{n}$ on the whole cube $Q_{2}$. Thus the partial sums
before each function of the first kind do not disappear as they did in the
previous three cases, and when half of these functions from a given $F_{n}$
have appeared, the norm of the partial sum on $Q_{2}$ is $\frac{1}{2}$
regardless of $n$ – thus this particular series does not converge. Of course
we still have to prove this is true for any rearrangement – but this example
shows the nature of the reason why only three and not four possible limits
exist.

##  4 Auxiliary lemmas

Before we begin the main part of this paper – i.e. the proof that our series
cannot converge to $({\mathbf{0}},{\mathbf{1}})$ – we shall need three
auxiliary lemmas:

######  Lemma 1.

(Lemma given without proof in [O89]) Let $(X,\mu)$ and $(Y,\nu)$ be measure
spaces with probability measures. Let $f(x,y)$ and $g(x,y)$ be functions in
$L_{1}(X\times Y)$, each of which depends on only one variable:
$f(x,y)=\tilde{f}(x),g(x,y)=\tilde{g}(y)$. Then

$||f+g||\geq||f||+||g||[1-2\mu({\rm supp}\tilde{f})].$

###### Proof.

$||f+g||=\int_{X\times Y}|f+g|=\int_{{\rm supp}\tilde{f}\times
Y}|f+g|+\int_{(X\setminus{\rm supp}\tilde{f})\times Y}|g|\geq\int_{{\rm
supp}(\tilde{f})\times Y}|f|-\int_{{\rm supp}(\tilde{f})\times
Y}|g|+(1-\mu({\rm supp}\tilde{f}))||g||=||f||-\mu({\rm
supp}\tilde{f})||g||+(1-\mu({\rm supp}\tilde{f}))||g||=||f||+||g||[1-2\mu({\rm
supp}\tilde{f})].$∎

######  Lemma 2.

Let $A,B,C$ be arbitrary spaces equipped with probabilistic measures and let
$X=A\times B\times C$ be equipped with the standard product measure. Suppose
$f,g$ are bounded functions defined on $X$ of the form
$f(a,b,c)=\tilde{f}(a,b)=\sum_{k=1}^{N}s_{k}\chi_{A_{k}\times B_{k}}$ and
$g(a,b,c)=\tilde{g}(b,c)=\sum_{l=1}^{N}t_{l}\chi_{B_{l}\times C_{l}}$, and
$\|f-g\|\leq{\varepsilon}$. Then there exists a function
$h(a,b,c)=\tilde{h}(b)$ such that $\|h-g\|\leq 2{\varepsilon}$ and
$\|h-f\|\leq 2{\varepsilon}$. Moreover if $f$ is integer-valued then $h$ can
also be chosen to be integer-valued, and if for a family of sets $B_{\alpha}$
we have $\forall_{\alpha}\forall_{b_{1},b_{2}\in B_{\alpha}}\forall_{a\in
A}f(a,b_{1},c)=f(a,b_{2},c)$, then we can choose a function $h$ constant on
any set $B_{\alpha}$.

###### Proof.

For any given $b\in B$ we take $\tilde{h}(b)$ such that

$\int_{A}|\tilde{f}(a,b)-\tilde{h}(b)|da=\inf_{x\in{\mathbb{R}}}\\{\int_{A}|\tilde{f}(a,b)-x|da\\}.$

This is well defined, as $f$ is bounded, and thus in fact the $\inf$ is taken
over a bounded, and thus compact set. For such an $h$ we have

$\|h-f\|=\int_{X}|f(a,b,c)-\tilde{h}(b)|=\int_{C}\int_{B}\int_{A}|\tilde{f}(a,b)-\tilde{h}(b)|=\int_{C}\int_{B}\inf\\{\int_{A}|\tilde{f}(a,b)-x(b)|\\}\leq$
$\leq\int_{C}\int_{B}\int_{A}|\tilde{f}(a,b)-\tilde{g}(b,c)|\leq\int_{C}\int_{B}\int_{A}|f(a,b,c)-g(a,b,c)|=\|f-g\|\leq{\varepsilon}.$

As $\|h-f\|\leq{\varepsilon}$ and $\|f-g\|\leq{\varepsilon}$, we immediately
have $\|g-h\|\leq 2{\varepsilon}$. As for the additional assumptions, if $f$
and $g$ are integer-valued, we can take the $\inf$ in the definition of
$\tilde{h}$ to be taken only over integers, with the same result. Regardless
of that which option we choose, if $f$ is constant with regard to $b$ on any
$B_{\alpha}$, then from the definition $h$ also can be chosen to be constant
on that set.∎

######  Lemma 3.

Let $A,B$ be arbitrary spaces equipped with probabilistic measures and
$X=A\times B$ equipped with the standard product measure. Suppose $f,g,h$ are
integer-valued functions defined on $X$ fulfilling $f(a,b)=\tilde{f}(a)$ and
$h(a,b)=\tilde{h}(b)$ for some $\tilde{f},\tilde{h}$. Suppose too that the
function $g$ assumes only two adjacent values (i.e. $k$ and $k+1$ for some
$k$) . Finally suppose that $\|f+g+h\|<\delta<\frac{1}{9}$. Then either $f$ or
$h$ is a constant function equal some integer $c$ on a set of measure $\geq
1-2\sqrt{\delta}$. Furthermore the function satisfies $\|f-c\|<3\sqrt{\delta}$
(or $\|h-c\|<3\sqrt{\delta}$, respectively).

###### Proof.

The sets $F_{n}=\tilde{f}^{-1}((-\infty,n])$ and
$H_{n}=\tilde{h}^{-1}((-\infty,n])$ form two increasing families, the sum of
each is the whole space $X$ and the intersection of each is empty. The
measures $|F_{n}|$ thus form an ascending sequence with elements arbitrarily
close to 0 when $n{\rightarrow}-\infty$ and arbitrarily close to 1 when
$n{\rightarrow}\infty$. As $F_{n}\setminus F_{n-1}=\tilde{f}^{-1}(n)$, if
$\tilde{f}$ is not constant on any set of measure $\geq 1-2\sqrt{\delta}$,
then at least one element of the sequence $|F_{n}|$, say $F_{n_{f}}$, has to
fall into the interval $[\sqrt{\delta},1-\sqrt{\delta}]$. Similarly if
$\tilde{h}$ is constant on no set of measure $\geq 1-2\sqrt{\delta}$, then for
some $n_{h}$ we have $\sqrt{\delta}\geq|H_{n_{h}}|\geq 1-\sqrt{\delta}$. Then
on the set $X_{1}=F_{n_{f}}\times H_{n_{h}}$ we have $f(a,b)+h(a,b)\leq
n_{h}+n_{f}$, while on $X_{2}=(A\setminus F_{n_{f}})\times(B\setminus
H_{n_{h}})$ we have $f(a,b)+h(a,b)\geq n_{h}+n_{f}+2$. As $g$ assumes two
adjacent values, it is either $\leq-(n_{h}+n_{f}+1)$ or $\geq-(n_{h}+n_{f}+1)$
on the whole space $X$. Thus on one of the sets $X_{1},X_{2}$ we have
$|f+g+h|\geq 1$, call it $X_{i}$. As both $X_{1}$ and $X_{2}$ are products of
two sets of measure $\geq\sqrt{\delta}$, we have
$\|f+g+h\|=\int_{X}|f(a,b)+g(a,b)+h(a,b)|\geq\int_{X_{i}}|f(a,b)+g(a,b)+h(a,b)|\geq|X_{i}|\geq\delta$,
which contradicts the assumptions of the lemma.

Thus one of the functions has to be constant on a large set. Without the loss
of generality we may assume it is $h$, and that it is equal to some integer
$c$. Let us examine the function $f$, taking into account that all the
functions are integer-valued, and thus if their sum is non-zero, it is at
least one :

$\delta>\|f+g+h\|\geq\|f+g+c\|_{A\times
h^{-1}(c)}\geq|\\{\tilde{f}(a)\not\in\\{-k-c,-k-c-1\\}\\}\times h^{-1}(c)|=$
$=|\\{\tilde{f}(a)\not\in\\{-k-c,-k-c-1\\}\\}|\cdot(1-2\sqrt{\delta}),$

which implies $\tilde{f}(a)\in\\{-k-c,-k-c-1\\}$ on a set of measure at least
$1-\frac{\delta}{1-2\sqrt{\delta}}$. Denote this set by $A^{\prime}$. Now we
return to the function $h$:

$\|h-c\|_{X}\leq\frac{1}{1-2\sqrt{\delta}}\|h-c\|_{A^{\prime}\times
B}=\frac{1}{1-2\sqrt{\delta}}\|h-c\|_{A^{\prime}\times(B\setminus
h^{-1}(c))}.$

On the set $A^{\prime}$ the function $f+g+c$ assumes values of absolute value
$\leq 1$, so by substituting $f+g$ for $-c$ we shall decrease the norm at most
by

$1\cdot|A^{\prime}\times(B\setminus
h^{-1}(c))|\leq(1-\frac{\delta}{1-2\sqrt{\delta}})(2\sqrt{\delta})\leq
2\sqrt{\delta},$

thus giving the inequality

$\|h-c\|_{X}\leq\frac{1}{1-2\sqrt{\delta}}\|h+f+g\|_{A^{\prime}\times(B\setminus
h^{-1}(c))}+2\sqrt{\delta}\leq\frac{1}{1-2\sqrt{\delta}}\|f+g+h\|_{X}+2\sqrt{\delta}\leq\frac{\delta}{1-2\sqrt{\delta}}+2\sqrt{\delta}.$

As $\delta\leq\frac{1}{9}$, we have
$\frac{\delta}{1-2\sqrt{\delta}}\leq\sqrt{\delta}$, and thus $||h-c||\leq
3\sqrt{\delta}$.

∎

##  5 The fourth point

Now we can begin to prove the main theorem of the paper:

######  Theorem 5.1.

The function $d_{\infty}=({\mathbf{0}},{\mathbf{1}})$ does not belong to the
sum range of any 3-Kadets family series.

###### Proof.

Suppose we have a rearrangement of some 3-Kadets family $d_{\sigma(n)}$ the
sum of which converges to $d_{\infty}$. Again, take an arbitrarily small
$\delta>0$ (we shall need $927\sqrt{\delta}<\frac{1}{4}$, i.e.
$\delta<\frac{1}{13749264}$) and an integer $K$ satisfying inequalities (15)
and (16), i.e. the tails and Cauchy sums are smaller than $\delta$ for $N>K$.
Then, again, we take any $M$ satisfying (17), i.e. such that $V_{M}$ contains
the first $K$ elements of our series. Then we take an $N_{0}$ such that

$V_{M}\subset\\{d_{\sigma(1)},d_{\sigma(2)},\ldots,d_{\sigma(N_{0})}\\}.$ (29)

Consider any fixed $N>N_{0}$. We will prove that

$\int_{Q_{3}}\sum_{n=1}^{N}d_{\sigma(n)}<\frac{1}{4}.$

Of course this suffices to prove that our series does not converge to $1$ on
$Q_{3}$, which contradicts the assumption the rearrangement converged to
$({\mathbf{0}},{\mathbf{1}})$.

Denote for any $L,k\in{\mathbb{Z}}$ by $D_{k}$ the set
$\\{d_{\sigma(1)},\ldots,d_{\sigma(k)}\\}$, and by
$F^{k}_{L},G^{k}_{L},H^{k}_{L}$ and $V^{k}_{L}$ the intersections of sets
$F_{L},G_{L},H_{L}$ or $V_{L}$, respectively, with the set $D_{k}$. First we
shall prove the following lemma:

######  Lemma 4.

If functions $f^{n}_{m},g_{m,j}^{n}$ and $h_{m,j,k}^{n}$ are a 3-Kadets family
on the cubes $Q_{1},Q_{2}$ and $Q_{3}$ and their set permutation
$d_{\sigma(n)}$ tends to $0$ on $Q_{1}$ and $1$ on $Q_{2}$ and $Q_{3}$, and
for a given $L$ we have
$\int_{Q_{3}}\tilde{G}^{N}_{L}\geq\frac{1}{2}+38\delta$, where $N>N_{0}$ as
above, then there exists a $P\subset[0,1]$ such that $|P|=\frac{1}{2}$ and
$[(\tilde{H}_{L}^{N})^{-1}(0)]\cap\\{v:v_{L}\in P\\}\subset Q_{3}$ has measure
$\leq 450\delta$.

######  Remark 2.

What this lemma really tells us is: if up to the $N$th element of the series
at least half plus something $(38\delta)$ of the $G_{L}$ functions have
appeared, then at least half minus something $(450\delta)$ of the $H_{L}$
functions had to appear. Moreover, the $H_{L}$ functions do not appear in a
haphazard fashion - we know that at least half minus something rows had to
appear (a row is the set of the functions $h_{m,j,k}^{L}$ with fixed $m$ and
$j$ and varying $k$).

###### Proof.

If $L\leq M$ then our thesis is automatically fulfilled – all functions from
$H_{L}$ belong to the set $D_{N}$, thus we can take any set of measure
$\frac{1}{2}$ for $P$ and the set $(\tilde{H}_{L}^{N})^{-1}(0)$ will be empty,
so $P$ will satisfy the required conditions.

Now consider the case $L>M$. The numbers $K$ and $L-1$ satisfy the conditions
(15), (16) and (17) (as $L>M$ and $M$ satisfied (17)). Thus we know there
exist numbers $n_{i}$ satisfying (18). We shall prove that $N\geq n_{2}$.

We know that $\int_{Q_{3}}\tilde{G}^{N}_{L}=-\int_{Q_{1}}\tilde{G}^{N}_{L}$
(as all $g_{m,j}^{n}$ are of the same constant sign on each cube, the absolute
value of the integral is equal to the norm, and the norms on each cube are
equal) . If $N<n_{2}$, then

$\|\tilde{G}_{L}^{N}\|_{Q_{1}}\leq\|\tilde{G}_{L}^{n_{2}}\|_{Q_{1}}=\|\bar{\bar{d}}_{1}+\bar{\bar{d}}_{2}\|\leq\|d^{**}_{1}\|+19\delta+\|d^{**}_{2}\|+19\delta<\frac{1}{2}+38\delta,$

which contradicts our assumption (the first inequality follows from the fact,
that $g_{m,j}^{n}$ are all non-positive functions on $Q_{1}$, the second
inequality from corollary 2.4).

Thus $N>n_{2}$. Consider $\tilde{V}_{L-1}^{n_{2}}+\tilde{F}_{L}^{n_{2}}$ on
$Q_{1}$. This function is dependent on variables $t_{1},t_{2},\ldots,t_{L}$,
while $\tilde{G}_{L}^{n_{2}}=\bar{\bar{d}}_{1}+\bar{\bar{d}}_{2}$ on $Q_{1}$
depends on $t_{L}$ and $t_{L+1}$. From property (15) and Corollary 2.4 we get

$\|\tilde{V}_{L-1}^{n_{2}}+\tilde{F}_{L}^{n_{2}}+\tilde{G}_{L}^{n_{2}}\|\leq\|\tilde{D}_{k}\|+\|d^{**}_{1}+\bar{\bar{d}}_{1}\|+\|d^{**}_{2}+\bar{\bar{d}}_{2}\|\leq\delta+19\delta+19\delta=39\delta.$

We can thus use lemma 2 for functions
$-\tilde{V}_{L-1}^{n_{2}}-\tilde{F}_{L}^{n_{2}}$ and $\tilde{G}_{L}^{n_{2}}$
to get that on $Q_{1}$ both these functions are closer than $39\delta$ to some
integer-valued function $\tilde{A}$ depending only on $t_{L}$.

Each function $f_{m}^{n}$ depends only on $t_{n}$ and assumes values $0$ and
$1$ only (properties 5 and 5), so it is in fact the characteristic function of
a set $\\{t:t_{n}\in S_{m}^{n}\\}$ for some $S_{m}^{n}\subset[0,1]$. As the
$f_{m}^{n}$ functions have disjoint support for a fixed $n$, they are all
constant on any given $S_{m}^{n}$. The $g$ functions are also constant with
regard to $t_{n}$ on the $S_{m}^{n}$ due to property 6, and all the other
functions are constant with regard to $t_{n}$ on the whole interval. Thus the
functions $-\tilde{V}_{L-1}^{n_{2}}-\tilde{F}_{L}^{n_{2}}$ and
$\tilde{G}_{L}^{n_{2}}$ are constant with respect to $t_{L}$ on sets
$\\{t_{L}\in S_{m}^{L}\\}$ we can choose $\tilde{A}$ to be constant on those
sets. Thus $\tilde{A}$ coincides on $Q_{1}$ with the sum of some of the rows
of $G_{L}$, i.e. $\tilde{A}$ corresponds to some subset $A$ of $G_{L}$ such
that for a fixed $m$ either all or none of the functions $g_{m,j}^{L}$ belong
to $A$. Define $\tilde{A}$ on $Q_{2}$ and $Q_{3}$ as the sum of all the
elements of $A$ as well, which agrees with our notation that $\tilde{U}$ is
the sum of all the elements of $U$ for an arbitrary set of functions.

We know from (18) and Proposition 2.5 that
$\|\sum_{n=n_{2}+1}^{\infty}d_{n}^{*}\|_{Q_{1}}\leq\frac{1-\delta}{4}+\frac{1-\delta}{4}+11\delta\leq\frac{1}{2}+11\delta$.
Remark that
$(\tilde{V}^{n_{2}}_{L-1}+\tilde{F}_{L}^{n_{2}}+\sum_{n=n_{2}+1}^{\infty}d^{*}_{n})|_{Q_{1}}=(\tilde{V}_{L-1}+F_{L})|_{Q_{1}}={\mathbf{1}}|_{Q_{1}}$,
so
$\|\tilde{V}_{L-1}^{n_{2}}+\tilde{F}_{L}^{n_{2}}\|_{Q_{1}}\geq\frac{1}{2}-11\delta$.
On the other hand
$\|\tilde{V}_{L-1}^{n_{2}}+\tilde{F}_{L}^{n_{2}}\|_{Q_{1}}=\|\tilde{D}_{K}+d_{1}^{**}+d^{**}_{2}\|_{Q_{1}}\leq\delta_{\frac{1-\delta}{4}}+\frac{1-\delta}{4}\leq\frac{1}{2}+\delta$.
As $\|\tilde{V}_{L-1}^{n_{2}}+\tilde{F}_{L}^{n_{2}}-\tilde{A}\|_{Q_{1}}\leq
39\delta$, taking into account the equality
$\|\tilde{A}\|_{Q_{1}}=\|\tilde{A}\|_{Q_{2}}$ we can estimate that

$\frac{1}{2}-50\delta\leq\|\tilde{A}\|_{Q_{2}}\leq\frac{1}{2}+40\delta.$ (30)

Distinct functions from $G_{L}$ have disjoint supports on $Q_{1}$ (this
follows from the properties 14 and 6 of Kadets families), and each has the
same norm $\psi=\frac{1}{|M_{L}\times J_{L}|}$. Thus if the distance between
two functions corresponding to two subsets of $G_{L}$ on $Q_{1}$ is smaller
than $n\psi$, then at most $n$ functions belong to the symmetric difference of
those two subsets. If at most $n$ functions belong to the symmetric
difference, then the distance between the two functions on $Q_{2}$ is at most
$n\psi$ (as on $Q_{2}$ the norm of a single function is also equal $\psi$ by
property 26). Thus, in general, if $B,C\subset G_{L}$, then
$\|\tilde{B}-\tilde{C}\|_{Q_{1}}\geq\|\tilde{B}-\tilde{C}\|_{Q_{2}}$. In
particular $\tilde{G}_{L}^{n_{2}}$ is at most $39\delta$ distant from
$\tilde{A}$ on $Q_{2}$.

Now consider what happens on $Q_{2}$. From (23) the restriction of $\tilde{A}$
to $Q_{2}$ is equal to $1$ on some set (on intervals
$t_{L}\in[\frac{m-1}{L},\frac{m}{L}]$ for $m$ such that $g^{L}_{m,j}\in A$)
and 0 on the rest. From (15), as $n_{2}>K$, we have
$\|\tilde{D}_{n_{2}}-1\|_{Q_{2}}\leq\delta$. If we substitute $\tilde{A}$ for
$\tilde{G}_{L}^{n_{2}}$, we will be at most $40\delta$ distant from zero,
precisely

$\|\tilde{D}_{n_{2}}-1-\tilde{G}^{n_{2}}_{L}+\tilde{A}\|_{Q_{2}}\leq
40\delta.$

However as only $G_{L}$ and $H_{L}$ depend on $u_{L}$, this sum is composed of
two parts - the part $\tilde{A}+\tilde{H}_{L}^{n_{2}}$ dependent on $u_{L}$
and the whole rest (i.e.
$\tilde{D}_{n_{2}}-(\tilde{G}_{L}^{n_{2}}+\tilde{H}_{L}^{n_{2}})$) dependent
on other variables. Thus we can apply a simplified version of lemma 2, with
$f=\tilde{A}+\tilde{H}_{L}^{n_{2}}$, $g=-(\tilde{D}_{n_{2}}-V_{L}^{n_{2}})$,
and a trivial one-point space as $B$. We learn that both our functions are
within $80\delta$ from a function $c$ dependent on $b$ – but as $B$ was a one-
point space, $c$ is a constant function. As $\tilde{A}$ assumes values $0$ and
$1$, and $\tilde{H}_{L}^{n_{2}}\in[-1,0]$, their sum is non-negative on ${\rm
supp}\tilde{A}$ and non-positive on the remainder of $Q_{2}$.

From (30) we know that $|{\rm supp}\tilde{A}|\geq\frac{1}{2}-50\delta$, thus
$\tilde{A}+\tilde{H}_{L}^{n_{2}}$ is non-negative on a set of measure
$\geq\frac{1}{2}-50\delta$. If $c$ is positive, then (as
$\delta<\frac{1}{200}$)

$80\delta\geq\|\tilde{A}+\tilde{H}_{L}^{n_{2}}-c\|\geq
c(\frac{1}{2}-50\delta)\geq\frac{c}{4},$

which implies $c\leq 320\delta$. Similarly if $c$ is negative, we know from
(30) that $|Q_{2}\setminus{\rm supp}\tilde{A}|\geq\frac{1}{2}-40\delta$,
yielding again $c>-\frac{800}{3}\delta$. Thus $|c|<320\delta$, so
$\|\tilde{A}+\tilde{H}_{L}^{n_{2}}\|\leq\|\tilde{A}+\tilde{H}_{L}^{n_{2}}-c\|+|c|\leq
80\delta+320\delta=400\delta$.

Thus $\tilde{H}_{L}^{n_{2}}$ is within $400\delta$ of a function with values 0
and -1 on $Q_{2}$ – the function $-\tilde{A}$. Remark, that
$-\tilde{A}=-\tilde{A}^{\prime}$ on $Q_{2}$ for a subset $A^{\prime}$ of
$H_{L}$ with the property that for a given $m$ either all of the functions
$h_{m,j,k}^{L}$ belong to $A^{\prime}$, or none of the functions belongs to
$A^{\prime}$ (if a given $g_{m,j}^{L}$ belongs to $A$, then all
$h_{m,j,k}^{L}$ belong to $A^{\prime}$) . If $\tilde{A^{\prime}}$, where
$A^{\prime}\subset H_{L}$, is a function assuming only values $0$ and $1$ on
$Q_{2}$ and $B\subset H_{L}$, then

$\displaystyle\|\tilde{A^{\prime}}-\tilde{B}\|_{Q_{2}}$ $\displaystyle=$
$\displaystyle\|\tilde{A^{\prime}}-\tilde{B}\|_{{\rm
supp}\tilde{A^{\prime}}}+\|\tilde{A^{\prime}}-\tilde{B}\|_{{\mathbb{Q}}_{2}\setminus{\rm
supp}\tilde{A^{\prime}}}$ $\displaystyle=$ $\displaystyle\frac{1}{|M_{L}\times
J_{L}\times K_{L}|}|\\{h:h\in A^{\prime}\wedge h\not\in
B\\}|+\frac{1}{|M_{L}\times J_{L}\times K_{L}|}|\\{h:h\not\in A^{\prime}\wedge
h\in B\\}|$ $\displaystyle=$ $\displaystyle\frac{1}{|M_{L}\times J_{L}\times
K_{L}|}|A\bigtriangleup B|=\|\tilde{A^{\prime}}-\tilde{B}|_{Q_{3}}.$

Let us take any subset $A^{\prime\prime}$ of $H_{L}$ depending only on $m$ and
$j$ with exactly half of the elements of $H_{L}$ and containing $A^{\prime}$
or contained in $A^{\prime}$. If $B\subset C\subset H_{L}$ or $C\subset
B\subset H_{L}$, then $\|\tilde{C}-\tilde{B}\|=|\|\tilde{C}\|-\|\tilde{B}\||$,
because all the the functions in $H_{L}$ are non-positive. As
$\tilde{A^{\prime}}=-\tilde{A}$ on $Q_{2}$ and from (30)
$|\|\tilde{A}\|_{Q_{2}}-\frac{1}{2}|\leq 50\delta$, we get
$\|\tilde{A^{\prime}}-\tilde{A^{\prime\prime}}\|_{Q_{2}}\leq 50\delta$, and
thus
$\|H_{L}^{n_{2}}-\tilde{A^{\prime\prime}}\|_{Q_{3}}=\|H_{L}^{n_{2}}-\tilde{A^{\prime\prime}}\|_{Q_{2}}\leq
450\delta$.

Now consider what happens on $Q_{3}$. As $\tilde{H}_{L}^{n_{2}}$ and
$\tilde{A^{\prime\prime}}$ are both integer-valued on $Q_{3}$, this means they
differ on a set of measure at most $450\delta$, and thus their difference can
be positive on a set of measure at most $450\delta$. When we increase $n$ from
$n_{2}$ to $N$ the set where the difference is positive can only decrease.
Thus $|\\{H_{L}^{N}-\tilde{A^{\prime\prime}}>0\\}|\leq 450\delta$. Now for $P$
we take ${\rm supp}\tilde{A}^{\prime\prime}$. The set
$[(\tilde{H}_{N}^{L})^{-1}(0)]\cap\\{v:v_{L}\in P\\}$ is the set where
$H_{N}^{L}$ is equal to zero and $\tilde{A}^{\prime\prime}$ is negative — thus
their difference is positive, so the set has to have measure smaller than
$450\delta$, which is what we had to prove. ∎

Now the main proof. Assume $d_{\infty}=({\mathbf{0}},{\mathbf{1}})$, i.e. our
series converges to ${\mathbf{1}}$ on $Q_{2}$ and $Q_{3}$ and to
${\mathbf{0}}$ on $Q_{1}$. We shall prove by induction upon $L$ that
$\int_{Q_{3}}\tilde{V}_{L}^{N}\leq\frac{1}{4}$. As
$\sum_{n=1}^{N}d_{\sigma(n)}$ is finite, its elements are contained in some
$V_{L}$, thus if the thesis is true, we get
$\int_{Q_{3}}\sum_{n=1}^{N}d_{\sigma(n)}\leq\frac{1}{4}$, which is what we had
to prove. For $L<M$ we have $V_{L}\subset D_{N}$ and from property (7)
$\int_{Q_{3}}\tilde{V}_{L}^{N}=0\leq\frac{1}{4}$. Now suppose we have the
thesis for $L-1$ and attempt to prove it for $L$. Denote by $P_{1}$ the
function $(\tilde{V}_{L-1}^{N}+\tilde{G}_{L}^{N})|_{Q_{3}}$ and by $P_{2}$ the
function $\sum_{n>L}\tilde{G}_{n}^{N}+\tilde{H}_{n}^{N}|_{Q_{3}}$. Consider
the function $\tilde{H}_{L}^{N}|_{Q_{3}}$. It depends on variables $v_{L}$ and
$v_{L+1}$. The function $P_{1}$ depends on $v_{1},\ldots,v_{L}$, while $P_{2}$
depends on $v_{L+1},\ldots,v_{Z}$ for some $Z\in{\mathbb{Z}}$. The function
$H_{L}^{N}|_{Q_{3}}$ assumes only values 0 and -1, all three functions –
$H_{L}^{N}|_{Q_{3}}$, $P_{1}$ and $P_{2}$ are integer-valued, and from (15)
their sum is less then $\delta$ distant from ${\mathbf{1}}$ on $Q_{3}$. Thus
by taking $P_{1}^{\prime}=P_{1}-1$ we have three functions fulfilling the
assumptions of lemma 3. Thus either $P_{1}$ or $P_{2}$ is within
$3\sqrt{\delta}$ of a constant function. In each of these cases the proof will
also depend on whether $\int_{Q_{3}}\tilde{G}_{L}^{N}\leq\frac{1}{2}+38\delta$
or $\int_{Q_{3}}\tilde{G}_{L}^{N}>\frac{1}{2}+38\delta$. Thus we have in total
four cases to consider.

Suppose first that $P_{2}$ is within $3\sqrt{\delta}$ of a constant function.
As $\|P_{1}+P_{2}+\tilde{H}_{L}^{N}-1\|\leq\delta$, this means that
$P_{1}+\tilde{H}_{L}^{N}$ is within $3\sqrt{\delta}+\delta\leq 4\sqrt{\delta}$
of a constant function. If
$\int_{Q_{3}}\tilde{G}_{L}^{N}\leq\frac{1}{2}+38\delta$, then
$\int_{Q_{3}}\tilde{V}_{L}^{N}=\int_{Q_{3}}\tilde{V}_{L-1}^{N}+\tilde{G}_{L}^{N}+\tilde{H}_{L}^{N}\leq\frac{1}{4}+(\frac{1}{2}+38\delta)+0=\frac{3}{4}+38\delta$.
But this function is equal $P_{1}+\tilde{H}_{L}^{N}$, and so is within
$4\sqrt{\delta}$ of some constant integer $c$ and its integral also has to be
within $4\sqrt{\delta}$ of $c$. As $4\sqrt{\delta}+38\delta<\frac{1}{4}$, we
get $c\leq 0$, thus $\int_{Q_{3}}\tilde{V}_{L}^{N}\leq
c+4\sqrt{\delta}\leq\frac{1}{4}$.

If $P_{2}$ is within $3\sqrt{\delta}$ of a constant function, and
$\int_{Q_{3}}\tilde{G}_{L}^{N}>\frac{1}{2}+38\delta$, then again
$P_{1}+\tilde{H}_{L}^{N}$ is within $4\sqrt{\delta}$ from a constant integer
$c$. From lemma 4 we have in particular that
$\int_{Q_{3}}\tilde{H}_{L}^{N}\leq-\frac{1}{2}+450\delta$. Obviously
$\int_{Q_{3}}\tilde{G}_{L}^{N}\leq 1$, thus
$\int_{Q_{3}}V_{L}^{N}=\int_{Q_{3}}V_{L-1}^{N}+\tilde{G}_{L}^{N}+\tilde{H}_{L}^{N}\leq\frac{1}{4}+1-\frac{1}{2}+450\delta=\frac{3}{4}+450\delta$.
As this is supposed again to within $4\sqrt{\delta}$ of $c$, we have $c\leq 0$
as $450\delta+4\sqrt{\delta}\leq\frac{1}{4}$. Again thus
$\int_{Q_{3}}\tilde{V}_{L}^{N}\leq c+4\sqrt{\delta}\leq\frac{1}{4}$.

In the third case we suppose that $P_{1}^{\prime}$, and thus also $P_{1}$ is
within $3\sqrt{\delta}$ of a constant function and
$\int_{Q_{3}}\tilde{G}_{L}^{N}\leq\frac{1}{2}+38\delta$. As
$\int_{Q_{3}}\tilde{V}_{L-1}^{N}\leq\frac{1}{4}$ from the inductive
assumption, we have $\int_{Q_{3}}P_{1}\leq\frac{3}{4}+38\delta$. As $P_{1}$ is
supposed to be within $3\sqrt{\delta}$ of some constant integer $c$, its
integral also has to be within $3\sqrt{\delta}$ of $c$, which again implies
$c\leq 0$ and $\int_{Q_{3}}P_{1}\leq 3\sqrt{\delta}$. As
$\tilde{V}_{L}^{N}=P_{1}+\tilde{H}_{L}^{N}$ and $\tilde{H}_{L}^{N}\leq 0$, we
get $\int_{Q_{3}}\tilde{V}_{L}^{N}\leq 3\sqrt{\delta}\leq\frac{1}{4}$.

The last case is when $P_{1}$ is within $3\sqrt{\delta}$ of a constant integer
$c$ and $\int_{Q_{3}}\tilde{G}_{L}^{N}>\frac{1}{2}+38\delta$. In this case
from lemma 4 we know there exists a set $P^{\prime}\subset Q_{3}$ dependent
only on $v_{L}$ such that $|P^{\prime}|=\frac{1}{2}$ and
$\int_{P^{\prime}}\tilde{H}_{L}^{N}\leq-\frac{1}{2}+450\delta$. If $P_{1}$ is
within $3\sqrt{\delta}$ of a constant integer function and
$P_{1}+P_{2}+\tilde{H}_{L}^{N}$ is within $\delta$ of $1$ (from 15) then
$P_{2}+\tilde{H}_{L}^{N}$ is within $3\sqrt{\delta}+\delta\leq 4\sqrt{\delta}$
of some constant integer function $C$. Taking $P_{2}^{\prime}=P_{2}-C$ we
arrrive in the situation of lemma 2: $\tilde{H}_{L}^{N}$ depends on $v_{L}$
and $v_{L+1}$ while $P_{2}^{\prime}$ depends on
$v_{L+1},v_{L+2},\ldots,v_{Z}$. This means that each of them is within
$8\sqrt{\delta}$ of some integer function $P_{3}$ dependent only on $v_{L+1}$.
As $\int_{P^{\prime}}\tilde{H}_{L}^{N}\leq-\frac{1}{2}+450\delta$ and
$\|\tilde{H}_{L}^{N}-P_{3}\|\leq 8\sqrt{\delta}$, we gather that
$\int_{P^{\prime}}P_{3}\leq-\frac{1}{2}+450\delta+8\sqrt{\delta}\leq-\frac{1}{2}+458\sqrt{\delta}$.
As $P^{\prime}$ depends only on $v_{L}$ and $P_{3}$ only on $v_{L+1}$ and
$|P^{\prime}|=|Q_{3}\setminus P^{\prime}|$ ,

$\int_{Q_{3}}P_{3}=\int_{P^{\prime}}P_{3}+\int_{Q_{3}\setminus
P^{\prime}}P_{3}=2\int_{P^{\prime}}P_{3}\leq-1+916\sqrt{\delta}.$

Returning to $\tilde{H}_{L}^{N}$ we get
$\int_{Q_{3}}\tilde{H}_{L}^{N}\leq\int_{Q_{3}}P_{3}+8\sqrt{\delta}\leq-1+924\sqrt{\delta}$.

As $\int_{Q_{3}}\tilde{G}_{L}^{N}\leq 1$ and
$\int_{Q_{3}}\tilde{V}_{L-1}^{N}\leq\frac{1}{4}$ we get
$\int_{Q_{3}}P_{1}\leq\frac{5}{4}$. As before, $\int_{Q_{3}}P_{1}$ has to be
within $3\sqrt{\delta}$ of the integer $c$, implying $c\leq 1$ and
$\int_{Q_{3}}P_{1}\leq 1+3\sqrt{\delta}$. We have
$\int_{Q_{3}}\tilde{V}_{L}^{N}=\int_{Q_{3}}P_{1}+\tilde{H}_{L}^{N}\leq
1+3\sqrt{\delta}-1+924\sqrt{\delta}\leq 927\sqrt{\delta}\leq\frac{1}{4}$.

Thus in all four cases we have completed the induction step, which proves in a
finite number of steps that $\int_{Q_{3}}\tilde{D}_{N}\leq\frac{1}{4}$. This
holds for an arbitrary $N>N_{0}$, and would thus have to hold for the limit
function, $\int_{Q_{3}}d_{\infty}\leq\frac{1}{4}$, which obviously contradicts
the assumption that $d_{\infty}|_{Q_{3}}={\mathbf{1}}$.∎

######  Corollary 5.2.

A 3-Kadets series has a 3-point sum range, consisting of the functions
$({\mathbf{0}},{\mathbf{0}})$, $({\mathbf{1}},{\mathbf{0}})$ and
$({\mathbf{1}},{\mathbf{1}})$. As previously, this holds for any $L_{p}$ with
$1\leq p<\infty$

##  6 More points

From the previous section we know how to make 3 points out of 2. The same
mechanism can be applied to make $r+1$ points out of $r$.

######  Theorem 6.1.

For any $r>1$ there exist a family $d_{k}$ of functions defined on a union of
cubes $Q_{1},\ldots,Q_{N}$ with an $r$-point sum range. Additionally we can
distinguish two disjoint subsets ${\mathcal{F}}$ and ${\mathcal{G}}$ of
$\\{d_{k}:k\in{\mathbb{N}}\\}$ which form a Kadets family on $Q_{N}$, while
all other functions $d_{k}$ disappear on $Q_{N}$. Moreover one function in the
sum range of $d_{k}$ is equal to ${\mathbf{1}}$ on $Q_{N}$ and all the other
functions from the sum range disappear on $Q_{N}$. Finally there exist
rearrangements convergent to any point of the sum range in which the sets
${\mathcal{F}}$ and ${\mathcal{G}}$ are arranged as in Proposition 2.1.

###### Proof.

We shall prove the thesis by induction upon $r$. For $r=2$ the original Kadets
example with $N=1$ satisfies the given conditions.

Suppose we have an appropriate family for $r-1$. We add two cubes to the
domain of $d_{k}$: $Q_{N+1}$ and $Q_{N+2}$. Denote by $x=(x_{1},x_{2},\ldots)$
the variable on $Q_{N+1}$ and by $y=(y_{1},y_{2},\ldots)$ the variable on
$Q_{N+2}$. All the functions except ${\mathcal{G}}$ will disappear on these
cubes. For each $n$ we divide the unit interval $[0,1]$ into $|M_{n}|$ sets
$S_{m}^{n},m\in M_{n}$ of measure $\frac{1}{|M_{n}|}$ each. We define
$g_{m,j}^{n}$ to be equal $\frac{1}{|J_{n}|}$ if $x_{n}\in S_{m}^{n}$, 0
otherwise. Next we define $K_{n}=M_{n+1}\times J_{n+1}$ and divide the unit
interval $[0,1]$ into $|K_{n}|$ sets $T_{k}^{n}$ of equal measure, and on
$Q_{N+2}$ define $g_{m,j}^{n}$ to be equal to 1 if $y_{n}\in T_{(m,j)}^{n-1}$,
0 otherwise. Finally to the functions $d_{k}$ we add a set of functions
${\mathcal{H}}=\\{h_{m,j,k}^{n}\\}$ which disappear on the cubes $Q_{1}$ to
$Q_{N}$, and satisfy $h_{m,j,k}^{n}=-\frac{1}{|K_{n}|}g_{m,j}^{n}$ on
$Q_{N+1}$ and $h_{m,j,k}^{n}=-g_{m,j}^{n}\cdot g_{k}^{n+1}$ on $Q_{M+2}$.

It is again easy, although tedious, to check that ${\mathcal{F}}$,
${\mathcal{G}}$ and the new functions ${\mathcal{H}}$ form a 3-Kadets family
on $Q_{N},Q_{N+1},Q_{N+2}$. We claim that the set
$\\{d_{k}\\}\cup{\mathcal{H}}$ satisfies the conditions given in the theorem.
The sets ${\mathcal{G}}$ and ${\mathcal{H}}$ form a Kadets family on
$Q_{N+2}$, all other functions disappear on $Q_{N+2}$. We have to check the
sum ranges. Let us fix any convergent rearrangement $e_{k}$ of
$\\{d_{k}\\}\cup{\mathcal{H}}$. From the properties of 3-Kadets families given
in section 3 we know that the limit on $Q_{N+1}$ and $Q_{N+2}$ is going to be
the same, and equal either ${\mathbf{0}}$ or ${\mathbf{1}}$. From theorem 5.1
we know that if the series converges to ${\mathbf{0}}$ on $Q_{M}$, it has to
converge to ${\mathbf{0}}$ on $Q_{N+1}$ and $Q_{N+2}$. Thus at most $r+1$
limits can be achieved - the functions with ${\mathbf{0}}$ on $Q_{N}$ generate
one each (by the 0-extension onto $Q_{N+1}\cup Q_{N+2}$), while the single
function with ${\mathbf{1}}$ on $Q_{N}$ can be extended by either
${\mathbf{0}}$ or ${\mathbf{1}}$ to $Q_{N+1}\cup Q_{N+2}$. This also satisfies
the condition that only one of the points in the sum range is ${\mathbf{1}}$
on $Q_{N+2}$, while the other points disappear on $Q_{2}$.

We can of course attain all the desired points in the sum range with
${\mathcal{G}}$ and ${\mathcal{H}}$ ordered as in Proposition 2.1 by taking
the rearrangements with ${\mathcal{F}}$ and ${\mathcal{G}}$ ordered as in the
proposition and inserting ${\mathcal{H}}$ as in section 3. ∎

Thus it is possible to attain a affine-independent finite set of any size $r$
as a sum range of a conditionally convergent series. Again, this works for any
$L_{p}$, $1\leq p<\infty$.

To attain full generality on $L_{p}$ we would attain arbitrary sum ranges, and
not only the affine-independent sum range given above. We will do that
according to the scheme from [K90], as follows:

######  Lemma 5.

Let $\Omega$ be an arbitrary probability space, $c_{n}\in{\mathbb{R}}$,
$c_{n}{\rightarrow}0$ and let $f_{n}\in L_{2}(\Omega)$ be a sequence of
integer-valued functions. Then the series $\sum_{n=1}^{\infty}(f_{n}+c_{n})$
converges if and only if both $\sum_{n=1}^{\infty}f_{n}$ and
$\sum_{n=1}^{\infty}c_{n}$ converge.

###### Proof.

The “if” part is obvious. For the “only if” part it is enough to prove that if
$\sum c_{n}$ diverges, then $\sum(f_{n}+c_{n})$ has to diverge as well. In
fact if $\sum c_{n}$ diverges then there exists an ${\varepsilon}\in(0,1/4)$
such that for any $N\in{\mathbb{N}}$ we have a large Cauchy sum above $N$,
i.e. for some $l>k>N$ we have $|\sum_{n=k}^{l}c_{n}|>{\varepsilon}$. As
$c_{n}{\rightarrow}0$ we can take $N$ large enough to ensure
$|c_{j}|<{\varepsilon}$ for $j>N$. Thus we can select $l=l(k)$ such that
${\varepsilon}<\sum_{n=k}^{l(k)}c_{n}<2{\varepsilon}<\frac{1}{2}$. But then
$\|\sum_{n=k}^{l(k)}(f_{n}+c_{n})\|\geq{\varepsilon}$ as a sum of an integer-
valued function and a constant $c\in({\varepsilon},1/2)$, which ensures the
divergence of $\sum(f_{n}+c_{n})$.∎

Now let us apply this lemma to our example from Theorem 6.1. We have a series
$d_{k}$ with an $r+2$-point sum range $D$ defined on
$\Omega=\bigcup_{i=1}^{2r+1}Q_{i}$ of cubes. We consider it as a series
defined on $L_{2}(\Omega)$. Let us denote $X={\rm
lin}\\{\chi_{Q_{1}},\chi_{Q_{2}},\ldots,\chi_{Q_{2r+1}}\\}$, i.e. the subspace
of the piece-wise constant functions on $\Omega$. Let
$P:L_{2}(\Omega){\rightarrow}X$ be the orthogonal projection onto $X$. Denote
by $Y$ the subspace of $X$ consisting of those piecewise constant functions
$(f_{i})_{i=1}^{2r+1}$, where $f_{i}$ is the value of $f$ on $Q_{i}$, that
$f_{2j}=f_{2j+1}$ for $j=1,2,\ldots,r$.

Recall that $\int_{Q_{2}j}d_{k}d\mu=\int_{Q_{2j+1}}d_{k}d\mu$ for
$j=1,2,\ldots,r$. Thus for any $d_{k}$ we have $P(d_{k})\in Y$, and thus
$P(D)$ is in fact a subset of $Y$. Recall also that for odd indices $j$ the
functions $d_{k}$ are integer-valued. Let $T:Y{\rightarrow}Y$ be an arbitrary
linear operator. Put $d_{k}^{\prime}=d_{k}+TP(d_{k})$.

######  Theorem 6.2.

The sum range $D^{\prime}$ of the series $\sum d_{k}^{\prime}$ equal
$(I+T)(D)$.

###### Proof.

The inclusion $(I+T)(D)\subset D^{\prime}$ is evident. To prove the inverse
inclusion consider an arbitrary arrangement $(b_{k}^{\prime})$ of
$(d_{k}^{\prime})$ and the corresponding rearrangement $(b_{k})$ of $(d_{k})$.
If $(b_{k}^{\prime})$ converges to some point $b^{\prime}\in D^{\prime}$, then
its restrictions to $Q_{j}$ for odd indices $j$ satisfy the conditions of the
lemma. Thus the restrictions to $Q_{j}$ for odd $j$ of $TP(b_{k})$ converge.
Now the restrictions of $TP(b_{k})$ to $Q_{j-1}$ are equal to the
corresponding restrictions to $Q_{j}$, so the whole series $TP(b_{k})$
converges. Then $\sum b_{k}=\sum(b_{k}^{\prime}-TP(b_{k}))$ also has to
converge. The sum of this series $b$ belongs to $D$, hence
$b^{\prime}=b+TP(b)$ belongs to $(I+T)(D)$.∎

This example can be transferred to any infinite-dimensional Banach space $Y$
using the results of V.M. Kadets. In [S91], Theorem 7.2.2 states: Let $X$ and
$Y$ be Banach spaces, $X\stackrel{{\scriptstyle f}}{{\Rightarrow}}Y$. Suppose
that $X$ has a basis $\\{e_{k}\\}_{k=1}^{\infty}$ and let
$\sum_{k=1}^{\infty}x_{k}$ be a series in $X$ such that
SR$(\sum_{k=1}^{\infty}x_{k})$ is not a linear set. Then for any monotone
sequence of positive numbers $\\{a_{k}\\}_{k=1}^{\infty}$ with
$a_{k}{\rightarrow}\infty,k{\rightarrow}\infty$, there exists a series
$\sum_{k=1}^{\infty}y_{k}$ in $Y$ such that SR$(\sum_{k=1}^{\infty}y_{k})$ is
not a linear set and $\|y_{k}\|\leq a_{k}\|x_{k}\|$ for all
$k\in{\mathbb{N}}$, Corollary 7.2.1 points out that if $X$ is $l_{2}$ then by
Dvoretzky’s theorem $X\stackrel{{\scriptstyle f}}{{\Rightarrow}}Y$, and
Corollary 7.2.2 states that In any infinite-dimensional Banach space there are
series whose sum range consists of two points. This is achieved by applying
the two-point example in $L_{2}$ to Corollary 7.2.1 and following the proof of
Theorem 7.2.2 to see that no new points appear and all the old ones are
transferred to the space $Y$. We have an $n$-point example in $L_{2}$ which
can be in the same manner, through obvious modifications in the proof of
Theorem 7.2.2 transferred to any Banach space $Y$. Finally for any finite-
dimensional subspaces $H_{1},H_{2}$ of a infinitely dimensional Banach space
$Y$ and any isomorphism $f:H_{1}{\rightarrow}H_{2}$ there exists an
isomorphism $\tilde{f}:Y{\rightarrow}Y$ extending $f$. Thus having any $n$
points satisfying some linear equations as a sum range of $y_{k}$ in $Y$ we
can take an $f$ transferring them to any other $n$ points satisfying the same
linear equations and then transfer the whole series by $\tilde{f}$.

## References

  * [S91] M. I. Kadets and V. M. Kadets, Series in Banach Spaces, Conditional and Unconditional Convergence, Birkhäuser Verlag, 1991.
  * [O89] M. I. Kadets and K. Woz̀niakowski, On series whose permutations have only two sums, Bull. Polish Acad. Sci. Math. 37 (1989), no. 1–6, 15–21.
  * [U02] P. L. Ulyanov, On interconnections between the research of russian and polish mathematicians in the theory of functions, Banach Center Publ, Polish Acad. Sci. Math 56 (2002), p. 122–128
  * [K90] V. M. Kadets, How many points can the sum-set of a series in a Banach space contain?, Teor. Funktsij, Funktsional. Anal. i Prilozhen. 54(1990), p. 54–57. (Russian)


# Towards Quantum Software Requirements Engineering

Tao Yue  Simula Research Laboratory
Oslo, Norway
<EMAIL_ADDRESS>Shaukat Ali  Simula Research Laboratory and
Oslo Metropolitan University
Oslo, Norway
<EMAIL_ADDRESS>Paolo Arcaini  National Institute of Informatics
Tokyo, Japan
<EMAIL_ADDRESS>
###### Abstract

Quantum software engineering (QSE) is receiving increasing attention, as
evidenced by increasing publications on topics, e.g., quantum software
modeling, testing, and debugging. However, in the literature, quantum software
requirements engineering (QSRE) is still a software engineering area that is
relatively less investigated. To this end, in this paper, we provide an
initial set of thoughts about how requirements engineering for quantum
software might differ from that for classical software after making an effort
to map classical requirements classifications (e.g., functional and extra-
functional requirements) into the context of quantum software. Moreover, we
provide discussions on various aspects of QSRE that deserve attention from the
quantum software engineering community.

###### Index Terms:

quantum software engineering, requirements engineering, requirements

##  I Introduction

Quantum software engineering (QSE) [1, 2], as classical software engineering,
is expected to focus on various phases of quantum software development,
including requirements engineering, design and modeling, testing, and
debugging. Various studies have been conducted in the literature regarding
most of these phases. However, as reported in [3, 1], the requirements
engineering phase remains relatively untouched. Only a few preliminary works
exist on requirements engineering [4, 5].

Requirements engineering, like in the classical context, if not conducted
properly, will build incorrect quantum software and cause high costs in fixing
it once problems are discovered in later phases of quantum software
development. Thus, this paper focuses on quantum software requirements
engineering (QSRE). In particular, we highlight the key aspects of QSRE that
differentiate itself from the classical domain. To illustrate differences, we
also present a motivating example of financial risk management. Moreover, we
shed light on how typical requirements engineering will be impacted due to the
quantum context and suggest key following activities.

##  II Motivating Example

We will use the motivating example of credit risk analysis with quantum
algorithms from Qiskit [6]. Detailed information about the algorithm is
published in [7, 8]. The proposed quantum algorithm is more efficient than its
equivalent classical implementations, such as using Monte Carlo simulations on
classical computers. We calculate two key risk measures, i.e., Value at Risk
(VaR) and Conditional Value at Risk (CVaR). Key requirements of the risk
estimation, including the calculation of these two risk measures (i.e.,
functional requirements), are shown in Figure 1. In addition, we present
extra-functional requirements specific to quantum computing, e.g., estimating
the number of gates and (ancilla) qubits. Moreover, we show hardware
constraints such as the limited number of qubits and limited depth of
circuits.

Figure 2 (a) present a use case diagram including actor Credit Analyst
responsible for managing risk in finance, as illustrated with use case Manage
risk in finance with quantum. This use case includes use cases Determine Var
and Determine CVaR. Also, for calculating VaR or CVaR, a credit analyst needs
to define the confidence level, captured with use case Define the confidence
level. In Figure 2 (b), we use the use case diagram notation to illustrate the
main functionalities of a quantum expert applying the Amplitude Estimation [7,
8] algorithm for calculating VaR and CVaR.

##  III Quantum Software Requirements Engineering

###  III-A Stakeholders

The ISO/IEC/IEEE 15288 standard defines stakeholders as: “Individual or
organization having a right, share, claim, or interest in a system or in its
possession of characteristics that meet their needs and expectations” [9].
Identifying stakeholders and their requirements is a crucial activity in
requirements engineering. When building quantum software systems, stakeholders
are the same as in the classical context. For example, in our example,
stakeholders related to the development of the quantum risk management system
include credit analysts (domain experts), borrowers (customers), banks, and
software developers, all having different concerns on various aspects,
including functionality, ease of use, price, and performance.

###  III-B Requirements classifications

Requirements are commonly classified into functional and extra-functional
(Section III-B1). A further classification specific to QSRE is whether
requirements are related to the quantum or the classical part (Section III-B2)
of the system.

####  III-B1 Functional requirements and extra-functional requirements

_Functional requirements_ are related to the functionality that a quantum
software system is expected to provide. For instance, the functional
requirements of our example are indicated with «Functional Requirement», such
as Determine Value at Risk (VaR) with the 95% of confidence level (see Figure
1).

Figure 1: Finance application for credit risk analysis – key requirements, in
the SysML requirements diagram notation. Stereotypes «qReq», «cReq», and
«hReq» are applied to distinguish quantum requirements, classical requirements
and the hybrid of both, respectively. Stereotypes «Functional Requirement» and
«Extra-functional Requirement» distinguish functional and extra-functional
requirements.

Identifying functional requirements for quantum software shall be the same as
for the classical one.

SEBoK defines _non-functional requirements_ (also commonly named _extra-
functional_) as “Quality attributes or characteristics that are desired in a
system, that define how a system is supposed to be” [10]. These attributes
vary from one system to another. For instance, safety requirements (i.e., one
type of extra-functional requirements) will only apply to a safety-critical
system. All the relevant extra-functional requirements from classical software
systems generally apply to quantum software systems. However, there are
additional requirements. For instance, Figure 1 shows three extra-functional
requirements: Estimation accuracy shall be a quadratic speed-up over classical
methods (e.g., Monte Carlo), which is further decomposed into another two
extra-functional requirements on estimating the required numbers of gates and
(ancilla) qubits. These two requirements relate to the hardware constraints:
Limited number of qubits and Limited depth of quantum circuits. Identifying
and realizing these extra-functional requirements require knowledge of quantum
computing. The good news is that such requirements are common across various
quantum software applications, implying that they can be reused and that
common solutions can be proposed to address them. We would also like to point
it out that Saraiva et al. [5] have already identified such five common extra-
functional requirements. Moreover, such extra-functional requirements might
need to be elicited step-wise, as their elicitation depends on identifying
other requirements. Ideally, when available in the future, an actionable
requirements elicitation process could clearly guide users through all
required activities.

####  III-B2 Quantum requirements vs. classical requirements

It is crucial to distinguish requirements that should be addressed with
classical computers and those to be addressed with quantum computers.
Moreover, there should be high-level requirements that are hybrid. For
instance, Figure 1 defines three stereotypes «qReq», «cReq», and «hReq» to
distinguish requirements that need to be addressed in the classical, quantum,
or hybrid manner, respectively. Doing so is essential, as mentioned by Weder
et al. [10, 11]; indeed, the first phase of their proposed quantum software
lifecycle is about performing quantum-classical splitting. Requirements
engineering, especially requirements analysis, and optimization, is typically
performed in this phase to decide which parts of a targeted problem need to be
solved on a quantum computer and which parts go to classical hardware.
Consequently, requirements specification and modeling solutions should provide
mechanisms to support the problem separation into classical and quantum parts.
We explain this idea by applying three stereotypes to use case modeling (see
Figure 2).

Figure 2: (a) Application for credit risk analysis – key use cases. (b) Key
functionalities of realizing Determine VaR and Determine CVaR in (a).

###  III-C Specific extra-functional concerns

####  III-C1 Portability

Near future quantum computers will be built with different hardware
technologies; thus, portability will remain a key requirement to be captured.
For example, a quantum software system for our example (i.e., credit risk
analysis) may need to be deployed to different quantum computers. Moreover, in
the near future, various quantum computing resources and classical computing
will be pooled so that more computations can be performed jointly. Thus,
converting a problem into a set of requirements, where requirements shall be
addressed with different types of quantum computers and classical computers,
is needed.

####  III-C2 Performance

Performance requirements over classical implementation are essential to
justify the need for quantum computing. For example, our example requires that
the estimation accuracy be a quadratic speed-up over classical methods (see
Figure 1). Such requirements may consider other requirements, e.g., the
estimation accuracy depends on the number of gates and the number of (ancilla)
qubits that need to be estimated at the requirements level to check whether or
not the expected quadratic speed-up on the estimation accuracy can be
achieved. These are two additional requirements. Such requirements are common
across quantum software, as, currently and in the near future, the
capabilities of quantum computers are limited. Thus, deciding early on whether
available resources can achieve the expected performance requirements and, if
yes, with which margin is important.

####  III-C3 Reliability

Currently, hardware errors affect the reliability of their computations and
consequently constrain how quantum computers should be used and how quantum
software should be designed. For instance, performing a reset after running
several calculations for a period of time might be needed; this means that a
quantum algorithm might not be run for a long time [12]. Thus, when
identifying requirements of quantum software, it is essential to identify
reliability requirements and associated constraints, especially considering
the impact of hardware errors on the reliability of quantum software systems.
Decisions such as introducing Quantum Error Correction [13] (which requires
additional quantum resources) or other fault tolerance mechanisms might be
needed early in the quantum software development lifecycle.

####  III-C4 Scalability

Current quantum computers support a limited number of qubits, i.e., resources
are scarce and expensive. Therefore, scalability requirements are carefully
considered while designing quantum software. For instance, as discussed in
[8], in the context of quantum risk analysis (our motivating example), based
on the results of the authors’ investigation, more qubits are needed to model
more realistic scenarios, thereby achieving practically meaningful advantages
over Monte Carlo simulations, which represent state of the art in risk
management. Moreover, scalability requirements (e.g., on the number of
parameters and constraints expected to be handled in the risk analysis) should
be carefully defined such that they can be satisfied with a limited depth of
the quantum circuit to mitigate the impact of decoherence, with limited use of
two-qubit gates (e.g., CNOT gates) to reduce the effect of crosstalk, and so
on, which can be ensured with more powerful quantum computers, dedicated error
mitigation mechanisms, and even carefully-designed quantum algorithms.

####  III-C5 Maintainability

Like classical software, quantum software will require maintainability. Given
that, as expected, quantum hardware will continue to evolve, existing quantum
software needs to be updated (in some cases) to deal with the hardware
changes. For example, with the decreased hardware error rates provided by the
latest technological advancements, error handing mechanisms in quantum
software systems must be updated to improve performance and reduce the cost of
additional error correction. Thus, quantum software systems shall identify and
capture such maintainability requirements.

####  III-C6 Reusability

Like classical software, the reusability of quantum software is essential to
be easily reused across different systems. Thus, such requirements shall be
captured during requirements engineering. However, some specific requirements
related to quantum software shall be explicitly captured. For instance,
quantum software is often built as hybrid software. Therefore, having tight
coupling between the two parts would reduce the reusability of quantum
software. Instead, the high cohesion of the quantum software part is expected
to enable more reusability.

##  IV Discussions and Suggestions

Requirements elicitation elicits software requirements, i.e., quantum software
in our context. Given that, in this phase, we investigate what problem a
quantum software should solve rather than how the software should be
implemented to address this problem, the requirements engineering for quantum
software shall remain similar to the classical one. For instance, identifying
stakeholders and defining system boundaries remain the same. However, one
difference might be in checking whether it is needed to solve a problem that
has been solved in the classical world with quantum, especially considering
the known limitations of quantum computing. For example, in our running
example (see Figure 1), we need to consider requirements specific to the
quantum domain, such as the required number of qubits (i.e., a hardware
constraint). Regarding stakeholders, there remain similarities between the
classical and quantum requirements elicitation. For instance, a possible
stakeholder in our example is the credit analyst, which would remain the same
as in the classical domain. Existing methods, such as interviews and
prototyping for requirements elicitation, are also expected to be largely
similar.

Functional and non-functional requirements are typically specified during
requirements specification at various formalization degrees, ranging from
informal natural language specifications to fully formal specifications.
Examples include semi-formal notations such as use cases and entity-relations
diagrams or formal notations such as Hoare logic [14]. Requirements
specifications for quantum software will be changed to accommodate concepts
related to quantum software. For example, when using use case diagrams, as
shown in our example, it is helpful to distinguish use cases from the
classical world, the quantum world, and the mix of the two. Moreover, when
specifying requirements with modeling notations (e.g., SysML requirements
diagram), they need to be extended to capture novel concepts from quantum
software. Finally, formal methods are also relevant to investigate for
specifying quantum software requirements as surveyed in [15]. Nonetheless,
such methods are also quite early in their stage of development [15].

Requirements verification of quantum software has received less attention;
when considering formal methods, only preliminary tools and methods are
available as discussed in [15]. Moreover, the survey discusses the need for
new methods for formal verification for complex quantum software. Requirements
validation via automated testing is getting popular in the software
engineering community, with several new works being published (e.g., [16, 17,
18, 19, 20, 21]). Nonetheless, as discussed in [22], many testing challenges
remain unaddressed. Finally, the classical verification and validation
methods, e.g., inspection and walk-through, apply to some extent to quantum
software requirements.

Based on our investigation, we recommend the following:

(1) Carefully consider separating parts of the problem that should be
addressed in the classical world and those on quantum computers; (2) Identify
and specify requirements related to various constraints, especially those
about quantum hardware. Realizing these requirements depends on available and
realistic quantum computing resources and explicitly specifying such
requirements support requirements analysis on the feasibility of the
realization; (3) Identify existing quantum algorithms that could be
incorporated. Selecting which quantum algorithms to use is a decision that
might need to be made at the early stage, as the availability and capability
of such quantum algorithms have an impact on the quantum part of the
realization of certain extra-functional requirements; (4) Based on the
identified and specified requirements, requirements analysis might be needed
to identify key factors (e.g., selection of quantum algorithms, determining
quantum hardware resources, assessing the feasibility of satisfying extra-
functional requirements) that have a significant impact on the development of
quantum software, and potential trade-offs among these factors. Doing so is
expected to effectively support decision-making on selecting quantum hardware
resources; (5) Identify requirements whose realization strongly depends on
constantly emerging quantum algorithms and advanced quantum computers. Doing
so is necessary because as soon as more advanced quantum algorithms or quantum
computers are available, such requirements could be realized (if not possible
before) or realized better. Also, decisions made regarding the satisfaction of
certain requirements (e.g., the required number of gates) and rationales
behind these decisions are highly recommended to be recorded.

##  V Conclusions and Future Work

Requirements engineering (RE) for quantum software has gotten less attention
than other phases, such as quantum software testing. Thus, we present some
ideas on how RE for quantum software will differ from the classical
counterpart. For instance, what will be the key differences for extra-
functional requirements? Finally, we discussed how various steps in RE, such
as requirements elicitation, specification, verification, and validation, will
be impacted, including developing requirements specification/modeling,
analyses, verification, and validation methods, with tool support, for
supporting quantum software development at the RE phase.

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