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\newcommand\QQ{\mathbb{Q}}
\newcommand\RR{\mathbb{R}}
\newcommand\ZZ{\mathbb{Z}}
\newcommand\NN{\mathbb{N}}

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\title{Quasihomomorphisms from the integers into Hamming metrics}

\author[\initial{J.} \lastname{Draisma}]{\firstname{Jan} \lastname{Draisma}}
\address{Mathematical Institute\\ University of Bern\\ Sidlerstrasse 5\\ 3012 Bern\\ Switzerland; and Department of Mathematics and Computer Science\\ Eindhoven University of Technology\\ P.O. Box 513\\ 5600MB\\ Eindhoven\\ the Netherlands}
\email{jan.draisma@unibe.ch}

\author[\initial{J.} \middlename{H.}
\lastname{Eggermont}]{\firstname{Rob} \middlename{H.} \lastname{Eggermont}}
\address{Department of Mathematics and Computer Science\\ Eindhoven University of Technology\\ P.O. Box 513\\ 5600MB\\ Eindhoven\\ the Netherlands}
\email{r.h.eggermont@tue.nl}

\author[\initial{T.} \lastname{Seynnaeve}]{\firstname{Tim} \lastname{Seynnaeve}}
\address{Department of Computer Science\\ KU Leuven\\ Celestijnenlaan 200A\\ 3001 Leuven\\ Belgium}
\email{tim.seynnaeve@kuleuven.be}

\author[\initial{N.} \lastname{Tairi}]{\firstname{Nafie} \lastname{Tairi}}
\address{Mathematical Institute\\ University of Bern\\ Alpeneggstrasse 22\\ 3012 Bern\\ Switzerland}
\email{nafie.tairi@unibe.ch}

\author[\initial{E.} \lastname{Ventura}]{\firstname{Emanuele} \lastname{Ventura}}
\address{Politecnico di Torino\\ Dipartimento di Scienze Matematiche ``G.L. Lagrange''\\ Corso Duca degli Abruzzi 24 10129 Torino\\ Italy}
\email{emanuele.ventura@polito.it}

\thanks{JD,TS,NT,EV were partially or fully supported by Vici grant
639.033.514 from the Netherlands Organisation for Scientific Research
(NWO) and Swiss National Science Foundation (SNSF) project grant
200021\_191981. RE was supported by NWO Veni grant 016.Veni.192.113.
EV is a member of GNSAGA of INdAM (Italy). We thank the reviewers
for many useful suggestions.}

\CDRGrant[Vivi]{639.033.514}
\CDRGrant[SNSF]{200021\_191981}
\CDRGrant[NWO]{016.Veni.192.113}

\subjclass{11B30}
\keywords{Quasihomomorphisms, Hamming distance, Linear approximation}

\begin{document}

\begin{abstract}
A function $f: \mathbb{Z} \to \mathbb{Q}^n$ is a \emph{$c$-quasihomomorphism} if the
Hamming distance between $f(x+y)$ and $f(x)+f(y)$ is at most $c$ for all
$x,y \in \mathbb{Z}$.  We show that any $c$-quasihomomorphism has distance at
most some constant $C(c)$ to an actual group homomorphism; here $C(c)$
depends only on $c$ and not on $n$ or $f$. This gives a positive answer
to a special case of a question posed by Kazhdan and Ziegler.
\end{abstract}

\maketitle

\section{Introduction}


Let $c$ be a nonnegative real number. A $c$-quasihomomorphism from a group $G$ to a group $H$ with
a left-invariant metric $d$ is a map $f:G \to H$ such that
$d(f(xy),f(x)f(y)) \leq c$ for all $x,y$ in $G$. A central question
in geometric group theory,
raised by Ulam in
\cite[Chapter 6]{Ulam1960}, is whether there exists an actual homomorphism
$f':G \to H$ such that $d(f(x),f'(x))$ is at most some constant $C$
for all $x$. (Related questions were studied
before Ulam, e.g. by Turing in his work on approximability of groups
\cite{turing1938finite}.)
Different versions of Ulam's question are of interest:
for example, $C$ may be allowed to depend on $c,G,(H,d)$ but not on
$f$; $G,(H,d)$ may be restricted to certain classes and $C$ is only allowed to depend on $c$.

A well-known example where the answer to this question is negative is the
case where $G=H=\ZZ$ with the standard metric. Here, quasihomomorphisms
modulo bounded maps are a model of the real numbers \cite{street1985efficient,ACampo2021},
and the answer is yes only for those quasihomomorphisms that correspond to
integers. In fact, this construction can be extended to construct completions of fields in general \cite{kionke2019constructing}.

Much literature in this area focusses on {\em quasimorphisms}, which
are quasihomomorphisms into the real numbers $\RR$ with the standard
metric; we refer to \cite{Kotschick2004} for a brief
introduction. In particular, the concept
of a quasimorphism features in bounded cohomology, see \cite{monod2001continuous,calegari2009scl,frigerio2017bounded}. In another branch of the research on
quasihomomorphisms $H$ is assumed nonabelian, and
one of the first positive results on the central question
above is Kazhdan's theorem on $\varepsilon$-representations of amenable
groups \cite{Kazhdan82}. For more recent results on quasihomomorphisms
into nonabelian groups we refer to \cite{FujiwaraKapovich2016,gowers2017inverse,DeChiffreOzawaThom,brandenbursky2022non} and the
references there.

The following instance of the central question was formulated
by Kazhdan and Ziegler in their work on approximate cohomology
\cite{KazhdanZiegler2018}.

\begin{question}\label{prob}
Let $c \in \NN$. Does there exist a constant $C = C(c)$ such that the
following holds: For all $n \in \NN$ and all functions $f: \ZZ \longrightarrow \CC^{n\times n}$ such that
\[
    \text{for all } x,y \in \ZZ, \ \mathrm{rk}(f(x+y) - f(x) - f(y))\leq c,
\]
there exists a matrix $g$ such that
\[
    \text{for all } x \in \ZZ, \ \mathrm{rk}(f(x) - xg)\leq C(c)?
\]
\end{question}

Here, $G$ equals $\ZZ$ and $H$ equals $\CC^{n \times n}$, both with
addition, and the metric on $H$ is defined by $d(A,B):=\rk(A-B)$. In
\cite[p1]{KazhdanZiegler2018}, the function $R(\ZZ,c,\CC)$ denotes the
minimal possible choice of $C(c)$. Our main result is an affirmative
answer to Question~\ref{prob} in the special case where all matrices $f(x)$ are assumed to be {\em diagonal}.

\begin{definition}
Let $(Q,+)$ be an abelian group. For an element $v \in Q^n$, the \emph{Hamming weight} $w_H(v)$ is the number of nonzero entries of $v$.
For a pair of elements $u,v \in Q^n$, their \emph{Hamming distance} is
$w_H(v-u)$. This metric is clearly left-invariant, and indeed even bi-invariant. 
\end{definition}

\begin{definition}\label{def:quasimor}
Let $A$ be another abelian group. A function $f: A \to Q^n$ is called a {\it $c$-quasihomomorphism} if
\begin{equation} \label{eq:defquasimor}
    \text{for all } x,y \in A, \ w_H(f(x+y)-f(x)-f(y)) \leq c. \qedhere
\end{equation}
\end{definition}

\begin{remark}
The map $\operatorname{diag}:\CC^n \to \CC^{n \times n}$ is an isometric embedding from $\CC^n$ with the Hamming metric to $\CC^{n \times n}$ with the rank metric. This connects Definition~\ref{def:quasimor} to Question~\ref{prob}.
\end{remark}

\begin{definition}\label{def:linapprox}
Let $C\in \NN$ and let $f: A\to Q^n$ be a $c$-quasihomomorphism. A
group homomorphism $h: A\to Q^n$ is a {\it $C$-approximation of $f$} if the Hamming distance between $f$ and $h$ satisfies
\[
    \text{for all } x \in A, \ w_H(f(x)-h(x)) \leq C. \qedhere
\]
\end{definition}

We are ready to state our main result.

\begin{theorem}[Main Theorem]\label{maintheorem}
Let $c\in \NN$. Then there exists a constant $C = C(c) \in \NN$ such
that for all $n \in \NN$ and $c$-quasihomorphisms $f: \ZZ \to \QQ^n$, we have:
\[
    \text{for all } x \in \ZZ, \ w_H(f(x)-xf(1)) \leq C.
\]
Moreover, we can take $C=28c$.
\end{theorem}

\begin{remark}
The coefficient $28$ is probably not optimal. 
However, we certainly have that $C(c) \geq c$. Indeed, any map $f:\ZZ
\to \mathbb Q^n$ for which the only nonzero entries of $f(x)$ are
among the first $c$, is automatically a $c$-quasihomomorphism.
\end{remark}

\begin{corollary} \label{cor:main}
    Theorem~\ref{maintheorem} also holds with $\QQ$ replaced by any torsion-free abelian group $Q$, with the same value of $C=C(c)$.
\end{corollary}
\begin{proof}
    Suppose, for a contradiction, that we have a $c$-quasihomomorphism $f:\ZZ \to Q^n$ but $w_H(f(y)-yf(1)) > C$ for some $y \in \ZZ$. Since $Q$ is torsion-free, the natural map $\iota$ from $Q$
    into the $\QQ$-vector space $V:=\QQ \otimes_\ZZ Q$ is injective. Consequently, $g:=\iota^n \circ f$ is a $c$-quasihomomorphism $\ZZ \to V^n$ with $w_H(g(y)-yg(1))> C$. Now choose any $\QQ$-linear function $\xi:V \to \QQ$ that is nonzero on the
    nonzero entries of $g(y)-yg(1)$. Then $h:=\xi^n \circ g$ is a
    $c$-quasihomomorphism $\ZZ \to \QQ^n$ with $w_H(h(y)-yh(1))>C$, a
    contradiction to Theorem~\ref{maintheorem}.
\end{proof}

\begin{remark}
As a referee kindly pointed out to us, our result fits in 
the broader context of $\mathcal G$-{\it stability} for a family $\mathcal G$ of groups endowed with a bi-invariant metric; this was first introduced in \cite{Kazhdan82} and further studied in \cite{UlamStab}
under the name of {\it Ulam stability}. Let $\mathcal G$ be
the family of groups $\lbrace \mathrm{GL}_n(\CC)\rbrace_{n\geq 1}$
with the normalized rank metric, i.e. $d(A,B)=\frac{1}{n}\rk(A-B)$.
Let $\mathcal G_d$ be the subfamily of $\mathcal G$ consisting of
diagonal matrices. Theorem~\ref{maintheorem} shows that the abelian group $\ZZ$ is uniformly $\mathcal G_d$-stable with a linear estimate.
\end{remark}

Theorem \ref{maintheorem} shows that for a $c$-quasihomomorphism $f: \ZZ \to \QQ^n$, the group homomorphism $\tilde{f}: \ZZ\to \QQ^n$ defined by $\tilde{f}(x) = xf(1)$ gives a $C$-approximation for some constant $C\in \NN$ independent on $n$. However, $\tilde{f}$ need not be the homomorphism closest to $f$, as the next example shows.

\begin{example}
Let $c=1$ and $n\geq 3$. Define $f: \ZZ \to \QQ^n$ to be
\begin{equation}
    f(x) = \left( \bigg\lfloor\frac{2x}{5}\bigg\rceil, \bigg\lfloor\frac{x}{5}\bigg\rceil, \alpha_x, 0, \ldots, 0\right),
\end{equation}
where $\alpha_x \in \QQ$ is arbitrary if $5 \mid x$, and $\alpha_x = 0$ otherwise. Here $\lfloor \rceil$ denotes rounding to the nearest integer. To check
that $f$ is a
$1$-quasihomomorphism \eqref{eq:defquasimor} we work mod $5$. For
simplicity, restrict to the case $n=3$. Then, for $k \in \ZZ$,
\begin{align*}
f(5k) &= (2k,k,\alpha_{5k}), & f(5k+1)&=(2k,k,0),\\
f(5k+2)&=(2k+1,k,0), & f(5k+3)&=(2k+1,k+1,0),\\
f(5k+4)&=(2k+2,k+1,0). & &
\end{align*}
Let $x=5k+\ell_1$ and $y=5h+\ell_2$ with $0 \leq \ell_1 \leq \ell_2 <
5$. Then we can verify that
\[ w_H(f(5(k+h)+(\ell_1+\ell_2))-f(5k+\ell_1)-f(5h+\ell_2))\leq c=1 \] 
in all cases. Roughly speaking, the check boils down to verifying that there are no cases where both $\left\lfloor\frac{x + y}{5}\right\rceil \neq \left\lfloor\frac{x}{5}\right\rceil + \left\lfloor\frac{y}{5}\right\rceil$ and $\left\lfloor\frac{2(x+y)}{5}\right\rceil \neq \left\lfloor\frac{2x}{5}\right\rceil + \left\lfloor\frac{2y}{5}\right\rceil$, and moreover that if $5 \mid x+y$, then $f(x+y) - f(x) - f(y) = (0,0,\alpha_x)$ (because in this case, $\frac{x}{5}$ is rounded down if and only if $\frac{y}{5}$ is rounded up).
Note that $w_H(f(x)-xf(1)) \leq 3$ where equality is sometimes achieved (provided there is at least one $x \neq 0$ for which we chose $\alpha_x \neq 0$). However, there also exist $2$-approximations of $f$. For instance, letting $v=(\frac{2}{5},\frac{1}{5},0,\ldots, 0)\in \QQ^n$, one verifies that
\[
    w_H(f(x)-xv) \leq 2 \text{ for all } x \in \ZZ. 
\]
In \cite{OneQuasimorphisms}, the authors show that for
every $1$-quasihomomorphism $f: \ZZ \to \QQ^n$, and even for
every $1$-quasihomomorphism from $\ZZ$ into the space of {\em
symmetric} $n \times n$-matrices with the rank metric, there is a
$2$-approximation. 

(This result is consistent with the second paragraph of
\cite{KazhdanZiegler2018}, where a proof of the corresponding statement
for general matrices is sketched. However, the above example shows that
that proof is incomplete: viewing $f$ as a map to the diagonal matrices,
and assuming $\alpha_0 = 0$ as is done in that paragraph, we obtain a counterexample
to the statement in \cite{KazhdanZiegler2018} that there exists either a
subspace of codimension $1$ living in the kernel of all matrices $f(n+m)
- f(n) - f(m)$ or else a subspace of dimension $1$ containing all their
  images.)
\end{example}

On the other hand, the following shows that the best possible
approximation of a given quasihomomorphism $f$ is at most twice as close as the homomorphism $x \mapsto xf(1)$.

\begin{remark}
Suppose that a map $f:\ZZ \to \QQ^n$ has a $C'$-approximation $h$. Then $h(x)= xv$ for some $v \in \QQ^n$, and
\[
    w_H(f(x)-xv) \leq C' \text{ for all } x \in \NN.
\]
    Substituting $x=1$ yields $w_H(f(1)-v) \leq C'$. Thus
\[
    w_H(f(x)-xf(1)) \leq w_H(f(x)-xv) + w_H(xv -xf(1)) \leq 2C'. \qedhere
\]
\end{remark}

\begin{remark}
A result similar to Theorem \ref{maintheorem} is easily proven in
positive characteristic if we allow the constant $C$ to depend on the
characteristic. Let $K$ be a field of characteristic $p>0$, and let $f:
\ZZ \to K^n$ be a $c$-quasihomomorphism.  Then there exists a constant
$C=C(p,c)$ such that $w_H(f(x)-xf(1))\leq C$, for all $x\in \ZZ$.

To see this, we observe that for all $u,v \in \ZZ$ with $u \geq 1$, we have $$
    w_H(f(uv) - uf(v)) \leq (u-1)c.
$$
This follows by repeatedly applying the inequality $w_H(f(uv) - f((u-1)v) - f(v)) \leq c$ if $u > 1$; the case $u = 1$ is trivial.

For $x = kp+r$ with $k \in \ZZ$ and $0 \leq r \leq p-1$, we
have
$$
    w_H(f(x) - x f(1)) = w_H(f(kp+r) - r f(1));
$$ here we
have used that $pf(1)=0$. We rewrite the latter as
$$
    w_H(f(kp+r) - f(kp) - f(r) + f(kp) + f(r) - r f(1)).
$$
We
have $w_H(f(kp+r) - f(kp) - f(r)) \leq c$;
$w_H(f(kp)) \leq (p-1)c$ using our observation with $u=p, v
= k$; and also $w_H(f(r) - r f(1)) \leq (p-2)c$ (in the case $r > 0$). In total, this gives $w_H(f(x) - x f(1)) \leq 2(p-1)c$, so we can take $C = 2(p-1)c$.
\end{remark}

The remainder of this paper is organized as follows. In
Section~\ref{sec:Almost} we prove an auxiliary result of independent
interest: maps from a finite abelian group into a torsion-free
group that are almost a homomorphism, are in fact almost zero. Then,
in Section~\ref{sec:Proof}, we apply this auxiliary result to the
component functions of a $c$-quasihomomorphism $\ZZ \to \QQ^n$ to prove the
Main Theorem.





\section{Almost homomorphisms are almost zero} \label{sec:Almost}

Let $A$ be a finite abelian group and let $H$ be a torsion-free abelian group. The only homomorphism $A \to H$ is the zero map. The following proposition says that maps that are, in a suitable sense, close to being homomorphisms, are in fact also close to the zero map.

\begin{prop}\label{abeliangroup}
Let $a$ be a positive integer, $A$ an abelian group of order
$a$, $H$ a torsion-free abelian group, $q \in [0,1]$, and $f:A \to H$
a map. Suppose that the {\em zero set}
$$
    Z(f):=\{b \in A \mid f(b)=0\}
$$
has cardinality at most $qa$. Then the {\em problem set}
\[
    P(f):=\{(b,c) \in A\times A \mid f(b+c) \neq f(b)+f(c)\}
\]
has cardinality at least $\frac{(1-q)^2}{4}a^2 + \frac{(1-q)}{2}a$.
\end{prop}

The contraposition of this statement says that if $P(f)$ is a small fraction of $a^2$, so that
$f$ can be thought of as an (additive) ``almost homomorphism'' $A \to H$,
then $q$ must be close to $1$ so that $f$ is essentially zero.

\begin{proof}
Since $H$ is torsion-free, it embeds into the $\QQ$-vector space $V:=\QQ \otimes_\ZZ H$. By basic linear algebra, there exists a $\QQ$-linear function $\xi:V \to \QQ$ such that $\xi(f(b)) \neq 0$ for all $b \not \in Z(f)$, so that $Z(\xi \circ f)=Z(f)$. Since $P(\xi \circ f) \subseteq P(f)$, it suffices to prove the proposition for $\xi \circ f$ instead of $f$. In other words, we may assume from the beginning that $H=\QQ$.

Set
$$
    B:=\{b \in A \mid f(b)>0\}.
$$
Let $\lambda_1>\lambda_2>\ldots>\lambda_k>0$ be the distinct
values in $f(B)$, and for each $i=1,\ldots,k$ set
$$
    B_i:=\{b \in B \mid f(b)=\lambda_i\} \text{ and } n_i:=|B_i|;
$$
as well as $n:=n_1+\cdots+n_k=|B|$.

Now for each $c \in B_1$ and each $b \in B$ we have
$$
    f(b)+f(c)=f(b)+\lambda_1>\lambda_1
$$
so that the left-hand side is not in $f(B)$ and in particular not equal to $f(b+c)$. We have thus found $n_1(n_1+\cdots+n_k)$ pairs $(b,c) \in P(f)$ with $c \in B_1$.

Next, suppose $(b,c)$ is a pair with $c \in B_2$, $b \in B$, and $(b,c) \not \in P(f)$.
Then
$$
    f(b+c)=f(b)+f(c)>f(c)=\lambda_2
$$
and hence $b+c \in B_1$. But given $c$, there are at most $n_1$ values of $b$ with $b+c \in B_1$. (Note that here we have used that $A$ is a group.) Hence we have at least $n_2(n_2+\cdots+n_k)$ pairs $(b,c) \in P(f)$ with $c \in B_2$.

Similarly, we find at least $n_i(n_i+\cdots+n_k)$ pairs $(b,c) \in P(f)$ with $c \in B_i$. In total, we have therefore found at least
\begin{equation}\label{eq:upper_triangular}
     \sum_{i=1}^k n_i(n_i+\cdots+n_k) \geq \frac{n(n+1)}{2}
\end{equation}
pairs in $P(f)$; see Figure \ref{jans_proof_picture}.

Let $B':=\{b' \in A \mid f(b')<0\}$ and $n' := |B'|$.  Repeating
the same argument above with $B'$ and $n'$, we find at least
$n'(n'+1)/2$ further pairs in $P(f)$, disjoint from those
found above. Since $|Z(f)|\leq qa$, we have $n+n'\geq a(1-q)$.
Therefore
\[
    |P(f)|\geq \frac{n(n+1)}{2}+\frac{n'(n'+1)}{2}=\frac{n^2+n'^2}{2}+\frac{n+n'}{2}\geq \left(\frac{n+n'}{2}\right)^2+\frac{n+n'}{2},
\]
where the second inequality is the Cauchy-Schwarz inequality
\[
    \left(n^2+n'^2\right)\left(\frac{1}{2^2}+\frac{1}{2^2}\right)\geq
\left(\frac{n}{2} +\frac{n'}{2}\right)^2.
\]
Since $n+n'\geq a(1-q)$, we conclude that
\begin{equation*}
    |P(f)| \geq \left(\frac{a-qa}{2}\right)\left(\frac{a-qa}{2}+1\right). \qedhere
\end{equation*}
\end{proof}

\begin{figure}[b]
	\centering
	\includegraphics[valign=t]{figures/picture}
	\includegraphics[valign=t]{figures/picture2}
	\caption{On the left, a graphical proof of the inequality
	\eqref{eq:upper_triangular}: the left-hand side is the number
	of small squares in the shaded region, the right-hand side is
	the number of squares on or above the main diagonal.
	On the right, a proof of the inequality
	\eqref{eq:ineq2}: the two expressions on top represent the area
	of the shaded region, while the bottom expression represents the area
	enclosed by the dashed line.}

	\label{jans_proof_picture}
\end{figure}

\begin{remark}
The lower bound in Proposition \ref{abeliangroup} is sharp. Let $a=2k+1\in \ZZ$, consider $A := \ZZ/a\ZZ$ and define $f:A \to \ZZ$ as $f(x):=$the representative of $x+a\ZZ$ in $\{-k,\ldots,0,\ldots,k\}$. We take $q=\frac{Z(f)}{a} = \frac{1}{2k+1}$. Then $f(x+y)=f(x)+f(y)$ if and only if the right-hand side is still inside the interval $\{-k,\ldots,k\}$, and a straightforward count shows that this is the case for $3k^2+3k+1$ pairs $(x,y) \in A^2$. Hence $P(f)$ has size $k(k+1)$, which equals $\frac{(1-q)^2}{4}a^2 + \frac{(1-q)}{2}a$.

A similar construction for $a=2k$ yields a problem set of size $\frac{a^2}{4} = k^2$, which equals the ceiling of the lower bound $\frac{a^2}{4}-\frac{1}{4}$.
\end{remark}

Below, we will use the following
strengthening of Proposition~\ref{abeliangroup}:

\begin{prop}\label{abeliangroupprime}
Let $a,A,H,q$ and $f$ be as in Proposition~\ref{abeliangroup}.
Furthermore, let $p \in [0,\frac{1-q}{2})$ and let $S \subseteq A$ be a subset of cardinality at most $pa$.
Then the set
\[
    P_S(f):=\{(b,c) \in A\times A \mid f(b+c) \neq f(b)+f(c) \text{ and } b+c \notin S\}.
\]
has cardinality at least $\frac{(1-q-2p)^2}{4}a^2 + \frac{(1-q-2p)}{2}a$.
\end{prop}
\begin{proof}
Keep the notation from the proof of Proposition~\ref{abeliangroup}.
Recall $n=|B|$ and $n' = |B'|$. Note that for a fixed $b$,
there can be at most $pa$ choices of $c$ with $b+c \in S$.
We then find at least $n_i(n_i+\cdots+n_k-pa)$ pairs
$(b,c) \in P_S(f)$ with $b \in B_i$. Letting $k'\leq k$ be
the largest index for which the second factor
$(n_{k'}+\cdots+n_k-pa)$ is nonnegative, as in the proof of
Proposition~\ref{abeliangroup}, we find that $B$
contributes at least
\begin{align}
    \sum_{i=1}^{k'}{n_i(n_i+\cdots+n_k-pa)} &=
    \sum_{i=1}^{k'}{n_i(n-n_1-\cdots-n_{i-1}-pa)}
    \notag \\
    &\geq (n-pa)(n-pa+1)/2  \label{eq:ineq2}
\end{align}
to $P_S(f)$; see Figure~\ref{jans_proof_picture}. Similarly, $B'$ contributes at least
$(n'-pa)(n'-pa+1)/2$, and these contributions
are disjoint. The desired inequality follows as in the
proof of Proposition~\ref{abeliangroup} but with $n,n'$
replaced by $n-pa,n'-pa$. \end{proof}

The key ingredient for the proof of Theorem~\ref{maintheorem} is the following
corollary of Proposition~\ref{abeliangroupprime}. Here, and in the rest of the paper,
we write $[a]$ for the set $\{1,2,\ldots,a\}$.

\begin{corollary} \label{cor:nonzeroAndPeriodicGivesProblems}
Let $p, q \in [0,1]$ such that $p < \frac{1-q}{2}$. Let $f:[2a] \to \QQ$ such that:
\begin{enumerate}
	 \item $|Z_a(f)| \leq qa$, where $Z_a(f) \coloneqq \{x \in [a] \mid f(x) = 0\}$ is the zero set of $\restr{f}{[a]}$. 
	\item $|NP(f)| \leq pa$, where $NP(f) \coloneqq \{x \in [a] \mid f(x+a) \neq f(x)\}$ is the nonperiodicity set.
\end{enumerate}
Then
\[
|P(f)| \geq \frac{(1-q-2p)^2}{4}a^2 + \frac{(1-q-2p)}{2}a,
\]
where
\[
P(f) = \{(x,y) \in [a]\times [a] \mid f(x+y) \neq f(x)+f(y)\}.
\]
\end{corollary}

\begin{proof}
Let $\tilde{f}$ be the restriction of $f$ to the interval $[a]$, and identify $\ZZ/ {a\ZZ}$ with $[a]$ with the group operation $\star$ defined by $x\star y := x+y \ (\operatorname{mod} \ a)$.

Let $S = NP(f)$, and apply Proposition~\ref{abeliangroupprime} to $\tilde{f}$. We find that
\[
P_S(\tilde{f}) = \{(b,c) \in \ZZ/ {a\ZZ} \times \ZZ/ {a\ZZ} \mid \tilde{f}(b\star c) \neq \tilde{f}(b)+\tilde{f}(c) \text{ and } b\star c \notin S\}
\]
has cardinality at least $\frac{(1-q-2p)^2}{4}a^2 + \frac{(1-q-2p)}{2}a$. Since $b \star c \notin S$ implies that $\tilde{f}(b\star c) = f(b+c)$, this set is contained in the problem set $P(f)$.
\end{proof}

\section{Proof of the main theorem} \label{sec:Proof}

The main goal of this section is to prove Theorem \ref{maintheorem}.
We start with some definitions.

\begin{definition} Let $1<a$, and $f:[2a] \to \QQ$. We define the following \emph{problem sets of $f$}:
$$
    P(f):= \{(x,y) \in [a] \times [a] \mid f(x+y) \neq f(x) + f(y)\},
$$
and
$$
	P_1(f) := \{x \in [a] \mid f(x+1) \neq f(x) + f(1)\},
$$	
and
$$
	P_a(f) := \{x \in [a] \mid f(x+a) \neq f(x) + f(a)\}.
$$
Furthermore, we recall that 
$Z_a(f)$ denotes the zero set of $\restr{f}{[a]}$:
$$
	Z_a(f):=\{x \in [a] \mid f(x) =0\}.
$$
\end{definition}

The following proposition says that $P_1(f),P_a(f),P(f)$
cannot be simultaneously small.


\begin{prop} \label{conj:variation} Let $p,q \in (0,1)$ such
that $1-q-2p>0$, $a\in \NN$ with $1<a$, and let $f:[2a] \to \QQ$ such that $f(a) \neq af(1)$. Then at least one of the following holds:
\begin{enumerate}[label=(\roman*)]
    \item\label{it0:P1}  $|P_1(f)| > qa$,
    \item\label{it0:Pa}  $|P_a(f)| > pa$,
    \item\label{it0:P}  $|P(f)| \geq F(p,q)a^2$,
\end{enumerate}
where
\begin{equation*}
     F(p,q) \coloneqq \frac{(1-q-2p)^2}{4}. \qedhere
\end{equation*}
\end{prop}

\begin{proof}
Without loss of generality we can assume $f(a)=0$ and hence
$f(1) \neq 0$. Indeed, suppose we have shown the statement
for every $\tilde{f}$ with $\tilde{f}(a)=0$. Then for any
$f:[2a] \to \QQ$ with $f(a) \neq af(1)$, we take
$\tilde{f}:[2a] \to \QQ$ to be $\tilde{f}(x) = af(x) - x f(a)$. Now we observe that $\tilde{f}(a) = 0 \neq a\tilde{f}(1)$, and that $P(f)=P(\tilde{f})$, $P_1(f)=P_1(\tilde{f})$, $P_a(f)=P_a(\tilde{f})$.

To prove the proposition we will assume that (\ref{it0:P1}) and (\ref{it0:Pa}) are false, and prove that then (\ref{it0:P}) must hold. Write $Z_a(f)=\{x_1,\ldots,x_m\}$, where $x_1 < \cdots < x_m$. Note that for $1\leq i <m$, one of the elements $x_i,x_i+1,\ldots,x_{i+1}-1$ needs to be in $P_1(f)$ since $f(x_{i+1}) \neq f(x_i) + (x_{i+1}-x_i)f(1)$. Likewise, at least one of the elements $1, 2, \ldots, x_1-1$ needs to be in $P_1(f)$. Thus we have
$$
    |Z_a(f)| \leq |P_1(f)| \leq qa,
$$
and by assumption we have $|NP(f)| = |P_a(f)| \leq pa$. Now we can
apply Corollary~\ref{cor:nonzeroAndPeriodicGivesProblems} to conclude.
\end{proof}

We now prove Theorem~\ref{maintheorem}.

\begin{proof}[Proof of the Main Theorem.]
Consider a $c$-quasihomomorphism $f=(f_1,\ldots,f_n):\ZZ \to \QQ^n$. Our goal is to show that for every $a \in \ZZ$ we have $w_H(f(a)-af(1)) \leq C$ for some constant $C$ depending only on $c$. We start with the case $a >0$.

Write $I_a\coloneqq \{i \in [a] \mid f_i(a) \neq af_i(1)\}$, and note that $|I_a| = w_H(f(a)-af(1))$.
We will show that $|I_a| \leq C'$ for some constant $C'$ depending on $c$ only.
To this end, fix small parameters $p,q \in (0,1)$ (to be
optimized over later) and
write $f_i^a \coloneqq \restr{f_i}{[2a]}$ for the restriction of $f_i$ to $[2a]$.
By Proposition~\ref{conj:variation}, for every $i \in I_a$, we have 
\begin{enumerate}[label=(\roman*)]
	\item\label{it:P}  $|P_1(f_i^a)| > qa$, or
	\item\label{it:P1} $|P_a(f_i^a)| > pa$, or
	\item\label{it:Pa} $|P(f_i^a)| \geq F(p,q)a^2$.
\end{enumerate}
Let $m_0$ be the number of coordinates $i \in I_a$ such that \ref{it:Pa} holds. We define $m_1$ and $m_2$ analogously, for \ref{it:P} and \ref{it:P1} respectively.
	
By counting the number of triples $(i,x,y) \in [n] \times [a] \times [a]$ such that $f_i(x+y)-f_i(x)-f_i(y) \neq 0$ in two ways, we see that
$$
	\sum_{x=1}^{a}\sum_{y=1}^{a}{w_H\left(f(x+y)-f(x)-f(y)\right)} = \sum_{i=1}^{n}{|P(f^a_i)|} \geq \sum_{i \in I_a}{|P(f^a_i)|}.
$$
Because $f$ is a $c$-quasihomomorphism, the very left-hand side
is at most $a^2c$. On the other hand, the very right-hand side is
at least $m_0F(p,q)a^2$, so
$$
    a^2c \geq
    \sum_{x=1}^{a}\sum_{y=1}^{a}{w_H\left(f(x+y)-f(x)-f(y)\right)}
    \geq \sum_{i \in I_a}{|P(f^a_i)|}  \geq  m_0F(p,q)a^2.
$$
So we obtain $m_0 \leq \frac{c}{F(p,q)}$. Similarly we find
$$
	ac \geq
	\sum_{x=1}^{a}{w_H\left(f(x+1)-f(x)-f(1)\right)} = \sum_{i=1}^n |P_1(f^a_i)| \geq
	\sum_{i \in I_a}{|P_1(f^a_i)|}  > m_1qa,
$$
so that $m_1 < \frac{c}{q}$. Finally,
$$
	ac \geq \sum_{x=1}^{a}{w_H\left(f(x+a)-f(x)-f(a)\right)} = \sum_{i=1}^n |P_a(f^a_i)| \geq \sum_{i \in I_a}{|P_a(f^a_i)|} > m_2pa.
$$	
So $m_2 < \frac{c}{p}$. But now $|I_a| \leq m_0 + m_1 + m_2 < c
(\frac{1}{F(p,q)}+\frac{1}{q}+\frac{1}{p})=:C'$.

The case $a=0$ is easy: we have
\[ w_H(f(0))=w_H(f(0)-f(0)-f(0)) \leq c.\]

Finally, let us consider the case $a<0$. Then
\begin{align*}
    w_H(f(a)-af(1)) \leq& w_H(f(a)+f(-a)-f(0))+w_H(f(0)) \\
    &+w_H(f(-a)-(-a)f(1)) 
\leq 2c+C'=:C.
\end{align*}
This completes the proof of the qualitative part of the Main Theorem.
To obtain the explicit bound $28c$, we
minimize the function
\[
2+\frac{1}{q}+\frac{1}{p}+\frac{1}{F(p,q)}=
2+\frac{1}{q}+\frac{1}{p}+\frac{4}{(1-q-2p)^2}.
\]
This function is strictly convex for $(p,q) \in \RR_{>0}^2$,
so it has at most one minimum in the positive orthant. We find
this by setting the partial derivatives to zero and solving for
$p,q$. The minimum is $\approx 27.6817$ and attained at $(p,q) \approx
(0.1167, 0.16500)$.
\end{proof}

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