# On a Goldbach-type problem for the Liouville function

Alexander P. Mangerel  Department of Mathematical Sciences, Durham University,
Stockton Road, Durham, DH1 3LE, UK<EMAIL_ADDRESS>
###### Abstract.

Let $\lambda$ denote the Liouville function. We show that for all sufficiently
large integers $N$, the (non-trivial) convolution sum bound

$\left|\sum_{n<N}\lambda(n)\lambda(N-n)\right|<N-1$

holds. This (essentially) answers a question posed at the 2018 AIM workshop on
Sarnak’s conjecture.

##  1\. Introduction

###  1.1. Main result

The following problem was posed at the 2018 AIM workshop on Sarnak’s
conjecture.

######  Problem 1.1 (Problem 5.1 of [8]).

Prove that for every111Actually, no range for $N$ was given in the problem. It
is our presumption that the bound was meant to be shown for any $N\geq 3$.
$N\geq 3$ the sum

$\mathcal{L}_{\lambda}(N):=\sum_{1\leq n<N}\lambda(n)\lambda(N-n)$

satisfies $|\mathcal{L}_{\lambda}(N)|<N-1$.

Obviously, the triangle inequality furnishes the trivial bound
$|\mathcal{L}_{\lambda}(N)|\leq N-1$. Thus, the problem is to show _any
savings_ over this bound. This should be interpreted as an analogue of the
binary Goldbach problem for the Liouville function. Indeed, if $N\geq 4$ is
even and $\lambda$ is replaced by the prime-supported von Mangoldt function
$\Lambda_{1}(n):=(\log n)1_{n\text{ prime}}$, then proving the existence of
primes $p,q$ with $p+q=N$ is equivalent to

$\mathcal{L}_{\Lambda_{1}}(N):=\sum_{n<N}\Lambda_{1}(n)\Lambda_{1}(N-n)>0,$

i.e., showing any improvement over the trivial lower bound
$\mathcal{L}_{\Lambda_{1}}(N)\geq 0$.
Problem 1.1 is far weaker than what we expect to be true regarding the
convolution sum $\mathcal{L}_{\lambda}(N)$. It is natural to compare the
problem at hand with what ought to follow from Chowla’s conjecture [2], namely
that for all (fixed) $h\geq 1$,

$\frac{1}{x}\left|\sum_{n\leq x}\lambda(n)\lambda(n+h)\right|=o(1)\text{ as
}x\rightarrow\infty.$

On the heuristic basis that the additively coupled values $\lambda(n)$ and
$\lambda(N-n)$ ought to also be “almost orthogonal” on average, Corrádi and
Kátai [3] have conjectured that $|\mathcal{L}_{\lambda}(N)|=o(N)$ as
$N\rightarrow\infty$. As far as we are aware, the only result in this
direction is a theorem of De Koninck, Germán and Kátai [4], who proved the
conjecture of Corrádi and Kátai under the assumption that there are infinitely
many Siegel zeroes. Nevertheless, because $N$ is a large shift, recent methods
that have proven effective in bounding binary correlations of multiplicative
functions seem unsuited to the estimation of the convolution sum
$\mathcal{L}_{\lambda}(N)$, and thus the problem at hand remains non-trivial
unconditionally.
Our main theorem (essentially) solves Problem 1.1.

######  Theorem 1.2.

There exists $N_{0}\in\mathbb{N}$ such that if $N\geq N_{0}$ then
$|\mathcal{L}_{\lambda}(N)|<N-1$.

######  Remark 1.

The proof of Theorem 1.2 relies on Siegel’s theorem on lower bounds for
$L(1,\chi)$, where $\chi$ is a quadratic Dirichlet character. Thus, the lower
bound $N_{0}$ is ineffective. See Remark 3 below for an indication of what
sorts of effective results (up to a possible unique exception) may be proved,
in the case of $N$ _prime_ , using the Siegel-Tatuzawa theorem [9].

###  1.2. Proof strategy

We briefly explain our strategy as follows. As we show below (see Lemma 2.1),
in order to prove Theorem 1.2 we may restrict to the case in which $N=p^{k}$
is a prime power, and it is instructive to first consider the case $k=1$.
In this case, it is readily observed that if $|\mathcal{L}_{\lambda}(p)|=p-1$
in contradiction to the claim, then $\lambda(m)\lambda(p-m)$ is constant over
all $1\leq m<p$, in fact

$\lambda(m)\lambda(p-m)=\lambda(p-1)\lambda(1)=\lambda(p-1).$

But note that if $\chi_{p}=\left(\frac{\cdot}{p}\right)$ denotes the Legendre
symbol modulo $p$ then the same is true of $\chi_{p}$:

$\chi_{p}(m)\chi_{p}(p-m)=\chi_{p}(-1)\chi_{p}(m)^{2}=\chi_{p}(p-1).$

Inspired by this comparison, we seek to show that $\lambda(m)=\chi_{p}(m)$ in
the fundamental domain $[1,p-1]$ for $\chi_{p}$. Using harmonic analysis over
$\mathbb{Z}/p\mathbb{Z}$, the problem reduces to understanding the Fourier
coefficients of $n\mapsto\lambda(n)1_{[1,p-1]}(n)$, i.e., exponential sums

$S_{\lambda}(\xi):=\sum_{1\leq n<p}\lambda(n)e(n\xi/p),\quad\xi\pmod{p},$

where as usual $e(t)=e^{2\pi it}$ for $t\in\mathbb{R}$. The corresponding sums
with $\lambda$ replaced by $\chi_{p}$ are the twisted Gauss sums

$\tau(\chi_{p},\xi):=\sum_{1\leq n\leq
p}\chi_{p}(n)e\left(\frac{n\xi}{p}\right)=\chi_{p}(\xi)\tau(\chi_{p},0),\quad\xi\pmod{p}.$

In particular, we have the _dilation property_ that for each $1\leq d<p$,

$\tau(\chi_{p},d\xi)=\chi_{p}(d)\tau(\chi_{p},\xi)\text{ for all
}\xi\pmod{p}.$

We prove below (see Proposition 2.4) that whenever
$|\mathcal{L}_{\lambda}(p)|=p-1$ a similar dilation property holds for
$S_{\lambda}$, i.e., for each $1\leq d<p$,

(1) $S_{\lambda}(d\xi)=\lambda(d)S_{\lambda}(\xi)\text{ for all }\xi\pmod{p}.$

The upshot of this is that when $d$ is a primitive root modulo $p$ we may
determine all of the sums $S_{\lambda}(\xi)$, $\xi\neq 0$, which turn out to
coincide precisely with the twisted Gauss sums given above. In this way,
verifying (1) allows us to determine that $\lambda(n)=\chi_{p}(n)$ for all
$1\leq n<p$.
It turns out that, under the assumption $|\mathcal{L}_{\lambda}(p)|=p-1$,
proving the dilation property (1) is equivalent to proving that

$\lambda(m)\lambda(m+jp)=+1\text{ whenever }1\leq m<p,\,0\leq j<d\text{ and
}m\equiv-jp\pmod{d}.$

We prove that this property holds for all pairs of $(m,j)$ in question using
an iterative argument (see Section 3). The rough idea of that argument is to

$\text{replace }\lambda(m)\lambda(m+jp)\text{ by a prescribed sign multiplied
by }\lambda(m^{\prime})\lambda(m^{\prime}+j^{\prime}p),$

in which $m^{\prime}\equiv-j^{\prime}p\pmod{d}$ and $0\leq j^{\prime}<j$;
crucially, the parameter $j$ has been decremented. Iterating this procedure
must eventually result in $j^{\prime}=0$, in which case the product on the
right-hand side is simply $+1$. We are then able to calculate the original
product $\lambda(m)\lambda(m+jp)$ to be $+1$ as well. Our argument may be seen
as extending the “periodic” behaviour imposed on $\lambda$ by the relation
$\lambda(m)\lambda(p-m)=\lambda(p-1)$ for all $m\in\\{1,\ldots,p-1\\}$, to the
larger domain $[1,dp-1]$.
Having showed that $\lambda(n)=\chi_{p}(n)$ for all $n<p$, we return to the
general prime power case $N=p^{k}$, and deduce upper bounds both for $p$ and
for $k$. First, the bound on $p$ arises from the observation that if
$\lambda(n)=\chi_{p}(n)$ for sufficiently many $n$ then $\chi_{p}$ is an
exceptional character. We obtain a bound of the shape
$L(1,\chi_{p})\ll_{\varepsilon}p^{-1/6+\varepsilon}$ for the Dirichlet
$L$-function of $\chi_{p}$ at $s=1$ (see Proposition 2.6), which is in
contradiction to Siegel’s theorem [6, Thm. 5.28(2)] when $p$ is sufficiently
large. This is the source of ineffectivity in our bounds on $N_{0}$.
Next, we note that if $p^{k}||N$ then the constraint $\lambda(n)=\chi_{p}(n)$
can be extended to all $n<p^{k}$ by an inductive argument (see Lemma 2.8
below). In particular, this implies that $\chi_{p}(q)=-1$ at all primes
$q<p^{k}$. On the other hand, using Linnik’s theorem [6, Thm. 18.1] we can
find primes $q<p^{L}$, with $L$ an effectively computable constant, for which
$\chi_{p}(q)=+1$ necessarily (namely when $q\pmod{p}$ is a quadratic residue).
This implies that $k\leq L$.
It follows that $N$ must be composed of prime powers $p^{k}$ with both $k\leq
L$ and $p\leq p_{0}$, so that $N$ itself must be bounded by some $N_{0}$, as
required.

######  Remark 2.

One may also ask222We thank Mark Shusterman for pointing out this problem to
us, which he asked in the the MathOverflow post
https://mathoverflow.net/questions/307479/goldbachs-conjecture-for-the-
liouville-function. another natural Goldbach-type problem regarding the
Liouville function: given an even integer $N\geq 4$, must there exist $1\leq
a,b\leq N$ with $a+b=N$, such that $\lambda(a)=\lambda(b)=-1$? This is
obviously implied by the binary Goldbach conjecture, and therefore a weakening
of it.
The methods of this paper appear to be far too rigid to address this problem
directly. Note however, that even a result of the form

(2) $\left|\sum_{n<N}\lambda(n)\lambda(N-n)\right|<N-g(N),$

where $g:\mathbb{R}\rightarrow\mathbb{R}$ is a (sufficiently quickly)
increasing function satisfying $g(x)=o(x)$, would suffice to prove the
existence of such a pair $(a,b)$.
Indeed, suppose otherwise. Then for any $1\leq n<N$,
$(1-\lambda(n))(1-\lambda(N-n))=0$. It follows that

$0=\sum_{1\leq
n<N}(1-\lambda(n))(1-\lambda(N-n))=N-1-2\sum_{n<N}\lambda(n)+\sum_{n<N}\lambda(n)\lambda(N-n).$

We deduce from this and the prime number theorem that, e.g.,

$\left|\sum_{n<N}\lambda(n)\lambda(N-n)\right|>N-2\left|\sum_{n<N}\lambda(n)\right|\geq
N-CNe^{-\sqrt{\log N}},$

for some absolute constant $C>0$ and all $N\geq 3$. Thus, the choice
$g(x)=Cxe^{-\sqrt{\log x}}$ would suffice to this end.
It is natural to ask to what extent the techniques in this paper may be
perturbed in order to prove a bound like (2). We plan to return to this
problem in a future paper.

## Acknowledgments

We would like to thank Oleksiy Klurman, Brad Rodgers and Mark Shusterman for
useful comments and encouragement.

##  2\. Reduction to the prime power case

Assume for the sake of contradiction that $N\geq 2$ satisfies
$|\mathcal{L}_{\lambda}(N)|\geq N-1$. By the triangle inequality, we must have
$|\mathcal{L}_{\lambda}(N)|=N-1$. Our first lemma shows that we may restrict
our attention to prime power values of $N$.

######  Lemma 2.1.

Let $N\in\mathbb{N}$ and assume that there is a divisor $d|N$ such that
$|\mathcal{L}_{\lambda}(d)|<d-1$. Then $|\mathcal{L}_{\lambda}(N)|<N-1$. In
particular, if $|\mathcal{L}_{\lambda}(N)|=N-1$ then
$|\mathcal{L}_{\lambda}(d)|=d-1$ for all $d|N$.
Moreover, if $|\mathcal{L}_{\lambda}(N)|=N-1$ then
$\mathcal{L}_{\lambda}(d)=\lambda(d-1)(d-1)$ and $\lambda(N-1)=\lambda(d-1)$
for all $d|N$.

###### Proof.

Write $N=md$. Splitting the sum over $n$ defining $\mathcal{L}_{\lambda}(N)$
according to whether $m|n$ or not, the triangle inequality implies

$\displaystyle|\mathcal{L}_{\lambda}(N)|$
$\displaystyle\leq|\sum_{\begin{subarray}{c}n<N\\\
m|n\end{subarray}}\lambda(n)\lambda(N-n)|+|\sum_{\begin{subarray}{c}n<N\\\
m\nmid n\end{subarray}}\lambda(n)\lambda(N-n)|$
$\displaystyle\leq|\sum_{\begin{subarray}{c}k<d\end{subarray}}\lambda(mk)\lambda(m(d-k))|+N-1-|\\{1\leq
n<N:m|n\\}|$
$\displaystyle=|\sum_{k<d}\lambda(k)\lambda(d-k)|+N-1-(d-1)=|\mathcal{L}_{\lambda}(d)|+N-d.$

Thus, if $|\mathcal{L}_{\lambda}(d)|<d-1$ then
$|\mathcal{L}_{\lambda}(N)|<d-1+N-d=N-1$, as required.
For the second claim, since each summand of $\mathcal{L}_{\lambda}(d)$ is $\pm
1$ and there are $d-1$ terms in its support, we have
$|\mathcal{L}_{\lambda}(d)|=d-1$ if and only if

$\lambda(n)\lambda(d-n)=\lambda(1)\lambda(d-1)=\lambda(d-1)\text{ for all
}1\leq n<d.$

It thus follows that $\mathcal{L}_{\lambda}(d)=\lambda(d-1)(d-1)$, as claimed.
Finally, since

$\lambda(n)\lambda(N-n)=\lambda(N-1)\text{ for all }n<N,$

specialising to $n=km$ for $k\in\mathbb{N}$, we find

$\lambda(N-1)=\lambda(km)\lambda(m(d-k))=\lambda(k)\lambda(d-k)=\lambda(d-1),$

as required. ∎

In view of Lemma 2.1, we may analyse the condition
$|\mathcal{L}_{\lambda}(N)|=N-1$ by considering the implied constraints
$|\mathcal{L}_{\lambda}(p^{j})|=p^{j}-1$, whenever $p^{j}|N$. In the next two
subsections we will obtain constraints both on the size of $p|N$, as well as
the multiplicity $k$ such that $p^{k}||N$.

###  2.1. Bounds on the size of prime divisors of $N$

In the sequel, write

$S_{\lambda}(\xi):=\sum_{1\leq
n<p}\lambda(n)e(n\xi/p),\quad\xi\in\mathbb{Z}/p\mathbb{Z}.$

For the purposes of illustration let us observe that the condition
$|\mathcal{L}_{\lambda}(p)|=p-1$ imposes rigid constraints on the exponential
sums $S_{\lambda}(\xi)$. Indeed, we have that

$\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(\xi)^{2}=\sum_{1\leq
n,m<p}\lambda(n)\lambda(m)\frac{1}{p}\sum_{\xi\pmod{p}}e\left(\frac{(n+m)\xi}{p}\right)=\sum_{1\leq
n<p}\lambda(n)\lambda(p-n)=\mathcal{L}_{\lambda}(p),$

since $n+m\equiv 0\pmod{p}$ with $1\leq n,m<p$ if and only if $n+m=p$. As
noted in the proof of Lemma 2.1, if $|\mathcal{L}_{\lambda}(p)|=p-1$ then

(3) $\lambda(m)\lambda(p-m)=\lambda(p-1)\lambda(1)=\lambda(p-1)\text{ for all
}1\leq m<p.$

and also $\mathcal{L}_{\lambda}(p)=\lambda(p-1)(p-1)$. Therefore,

$\frac{\lambda(p-1)}{p}\sum_{\xi\pmod{p}}S_{\lambda}(\xi)^{2}=p-1=\frac{1}{p}\sum_{\xi\pmod{p}}|S_{\lambda}(\xi)|^{2}.$

To motivate our forthcoming arguments, we explicitly observe the following
relations, which will imply further rigidity in the values of
$S_{\lambda}(\xi)$.

######  Lemma 2.2.

Let $p>2$ be prime and let $1\leq m<p$. Suppose
$|\mathcal{L}_{\lambda}(p)|=p-1$. Then:

  1. (a)

if $m$ is odd then $\lambda(p+m)=\lambda(m)$,

  2. (b)

if $m\equiv p\pmod{3}$ then $\lambda(2p+m)=\lambda(m)$,

  3. (c)

if $m\equiv 2p\pmod{3}$ then $\lambda(2p-m)=\lambda(p-1)\lambda(m)$.

###### Proof.

(a) If $1\leq m<p$ is odd then $(p\pm m)/2\in\mathbb{Z}\cap[1,p-1]$, and we
have

$p=\frac{p+m}{2}+\frac{p-m}{2}.$

As $|\mathcal{L}_{\lambda}(p)|=p-1$, using (3) with $m$ replaced by $(p-m)/2$
we get

$\lambda(p-1)=\lambda\left(\frac{p+m}{2}\right)\lambda\left(\frac{p-m}{2}\right)=\lambda(p+m)\lambda(p-m).$

Since also $\lambda(p-1)=\lambda(m)\lambda(p-m)$, the first claim follows.
(b) The argument here is similar: if $m\equiv p\pmod{3}$ then
$(2p+m)/3,(p-m)/3\in\mathbb{Z}\cap[1,p-1]$ and also

$p=\frac{2p+m}{3}+\frac{p-m}{3},$

whence we obtain by (3) that

$\lambda(p-1)=\lambda\left(\frac{2p+m}{3}\right)\lambda\left(\frac{p-m}{3}\right)=\lambda(2p+m)\lambda(p-m)=\lambda(p-1)\lambda(2p+m)\lambda(m),$

from which the claim follows.
(c) Suppose $m\equiv 2p\pmod{3}$. Again, the above idea yields

$\lambda(p-1)=\lambda\left(\frac{2p-m}{3}\right)\lambda\left(\frac{p+m}{3}\right)=\lambda(2p-m)\lambda(p+m).$

If $m$ is odd then $\lambda(p+m)=\lambda(m)$ by (a), and the claim follows
immediately. Otherwise, if $m$ is even then note that since $p>2$, $p-m$ is
odd and hence (a) again yields

$\lambda(2p-m)=\lambda(p+(p-m))=\lambda(p-m)=\lambda(p-1)\lambda(m).$

Thus, claim (c) follows in this case as well. ∎

The three relations given in Lemma 2.2 allows us to deduce the following.

######  Lemma 2.3.

Let $p>3$ and assume that $|\mathcal{L}_{\lambda}(p)|=p-1$. Then

(4) $S_{\lambda}(3\xi)=S_{\lambda}(2\xi)=-S_{\lambda}(\xi)\text{ for all
}\xi\pmod{p}.$

In particular, for any $j,k\geq 0$ we have

$S_{\lambda}(3^{j})=(-1)^{j}S_{\lambda}(1),\quad
S_{\lambda}(2^{k})=(-1)^{k}S_{\lambda}(1).$

###### Proof.

The second claim follows by induction on $j$ and $k$ from the former, so it
suffices to prove (4).
First, note that as $p>3$ the maps $\xi\mapsto 2\xi$ and $\xi\mapsto 3\xi$ are
both bijections on $\mathbb{Z}/p\mathbb{Z}$. By Plancherel’s theorem,

(5)
$\frac{1}{p}\sum_{\xi\pmod{p}}|S_{\lambda}(m\xi)|^{2}=\frac{1}{p}\sum_{\xi^{\prime}\pmod{p}}|S_{\lambda}(\xi^{\prime})|^{2}=p-1\text{
for all }m\in\\{1,2,3\\}.$

Next, note that

$\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(2\xi)\overline{S}_{\lambda}(\xi)=\sum_{m,n<p}\lambda(n)\lambda(m)\frac{1}{p}\sum_{\xi\pmod{p}}e\left(\xi\frac{2n-m}{p}\right)=\sum_{\begin{subarray}{c}m,n<p\\\
2n\equiv m\pmod{p}\end{subarray}}\lambda(m)\lambda(n).$

Among $m,n<p$ with $2n\equiv m\pmod{p}$ we have that either $2n=m$ precisely
when $m$ is even, or else $2n=m+p$ precisely when $m$ is odd. If $2n=m$ then

$\lambda(n)\lambda(m)=\lambda(m/2)\lambda(m)=-\lambda(m)^{2}=-1,$

while if $2n=p+m$ then by Lemma 2.2(a) we have

$\lambda(n)\lambda(m)=\lambda\left(\frac{p+m}{2}\right)\lambda(m)=-\lambda(p+m)\lambda(m)=-1.$

It follows that

$\displaystyle\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(2\xi)\overline{S}_{\lambda}(\xi)$
$\displaystyle=\sum_{\begin{subarray}{c}m<p\\\ m\text{
even}\end{subarray}}\lambda(m)\lambda(m/2)+\sum_{\begin{subarray}{c}m<p\\\
m\text{ odd}\end{subarray}}\lambda(m)\lambda\left(\frac{p+m}{2}\right)$
$\displaystyle=-\left(\sum_{\begin{subarray}{c}m<p\\\ m\text{
even}\end{subarray}}1+\sum_{\begin{subarray}{c}m<p\\\ m\text{
odd}\end{subarray}}1\right)=-(p-1).$

This latter sum being real-valued, it follows from this and (5) that

$\displaystyle\frac{1}{p}\sum_{\xi\pmod{p}}|S_{\lambda}(2\xi)+S_{\lambda}(\xi)|^{2}$
$\displaystyle=\frac{1}{p}\sum_{\xi\pmod{p}}(|S_{\lambda}(\xi)|^{2}+|S_{\lambda}(2\xi)|^{2}+2\text{Re}(S_{\lambda}(2\xi)\overline{S}_{\lambda}(\xi)))$
$\displaystyle=2(p-1)-2(p-1)=0.$

Therefore, $S_{\lambda}(2\xi)=-S_{\lambda}(\xi)$ for all $\xi\pmod{p}$, as
claimed.
The proof with $2$ replaced by $3$ follows similar lines, using Lemma 2.2.
Here, instead we must treat pairs $1\leq n,m<p$ with $3n\equiv m\pmod{p}$, or
equivalently $3n=m+jp$, where $0\leq j<3$ and $m+jp\equiv 0\pmod{3}$ in each
case. (Note that as $p>3$, the residue classes $0,-p,-2p\pmod{3}$ cover all
integers $1\leq m<p$.) Precisely,

$\displaystyle\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(3\xi)\overline{S}_{\lambda}(\xi)=\sum_{m,n<p}\lambda(n)\lambda(m)\frac{1}{p}\sum_{\xi\pmod{p}}e\left(\xi\frac{3n-m}{p}\right)=\sum_{\begin{subarray}{c}m,n<p\\\
3n\equiv m\pmod{p}\end{subarray}}\lambda(m)\lambda(n)$
$\displaystyle=\left(\sum_{\begin{subarray}{c}m<p\\\
3|m\end{subarray}}\lambda(m/3)\lambda(m)+\sum_{\begin{subarray}{c}m<p\\\
m\equiv-2p\pmod{3}\end{subarray}}\lambda(m)\lambda\left(\frac{2p+m}{3}\right)+\sum_{\begin{subarray}{c}m<p\\\
m\equiv-p\pmod{3}\end{subarray}}\lambda(m)\lambda\left(\frac{p+m}{3}\right)\right).$

By (b) and (c) of Lemma 2.2,

$\displaystyle\lambda(m)\lambda(m/3)=\lambda(3)\lambda(m)^{2}=-1$
$\displaystyle\text{ if }3|m,$
$\displaystyle\lambda(m)\lambda\left(\frac{2p+m}{3}\right)=\lambda(3)\lambda(m)^{2}=-1$
$\displaystyle\text{ if }m\equiv-2p\equiv p\pmod{3},$
$\displaystyle\lambda(m)\lambda\left(\frac{p+m}{3}\right)=\lambda(m)\lambda(p-1)\lambda\left(\frac{2p-m}{3}\right)=\lambda(3)\lambda(m)^{2}=-1$
$\displaystyle\text{ if }m\equiv-p\equiv 2p\pmod{3}.$

We thus conclude that

$\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(3\xi)\overline{S}_{\lambda}(\xi)=-\left(\sum_{\begin{subarray}{c}m<p\\\
3|m\end{subarray}}1+\sum_{\begin{subarray}{c}m<p\\\
m\equiv-2p\pmod{3}\end{subarray}}1+\sum_{\begin{subarray}{c}m<p\\\
m\equiv-p\pmod{3}\end{subarray}}1\right)=-(p-1).$

The claim that $S_{\lambda}(3\xi)=-S_{\lambda}(\xi)$ for all $\xi\pmod{p}$ now
follows as it did with $2\xi$. ∎

It is natural, then, to speculate that $2$ or $3$ may be replaced by other
primes as well. In fact, we will prove the following more general result in
the next section.

######  Proposition 2.4.

Let $p$ be a prime with $|\mathcal{L}_{\lambda}(p)|=p-1$, and let $1\leq d<p$.
Then we have

(6) $S_{\lambda}(d\xi)=\lambda(d)S_{\lambda}(\xi)\text{ for all }\xi\pmod{p}.$

Proposition 2.4 will be beneficial in light of the following result.

######  Lemma 2.5.

Let $p>2$ be a prime satisfying $|\mathcal{L}_{\lambda}(p)|=p-1$. Suppose
$2\leq d<p$ is a primitive root modulo $p$ such that (6) holds. Then
$\lambda(n)=\chi_{p}(n)$ for all $n<p$.

###### Proof.

As $d$ is a primitive root, every $\xi\not\equiv 0\pmod{p}$ can be written as
$\xi\equiv d^{k}\pmod{p}$ for some $1\leq k\leq p-1$, and thus
$S_{\lambda}(\xi)=S_{\lambda}(d^{k})$. By (6) and induction, it follows that
$S_{\lambda}(d^{k})=\lambda(d)^{k}S_{\lambda}(1)$ for all $k\geq 1$. In
particular, $|S_{\lambda}(\xi)|=|S_{\lambda}(1)|$ for all $\xi\not\equiv
0\pmod{p}$.
On the basis of these observations, we first verify that $\lambda(d)=-1$.
Since $p$ is odd,

$\displaystyle
0=\sum_{\xi\pmod{p}}S_{\lambda}(\xi)=S_{\lambda}(0)+\sum_{k=1}^{p-1}S_{\lambda}(d^{k})$
$\displaystyle=S_{\lambda}(0)+S_{\lambda}(1)\sum_{k=1}^{p-1}\lambda(d)^{k}$
(7) $\displaystyle=S_{\lambda}(0)+(p-1)S_{\lambda}(1)1_{\lambda(d)=+1}.$

Now supposing $\lambda(d)=+1$ then we have
$S_{\lambda}(0)=-(p-1)S_{\lambda}(1)$. On the other hand, since
$|S_{\lambda}(\xi)|=|S_{\lambda}(1)|$ for all $\xi\not\equiv 0\pmod{p}$, we
find using (5) that

$p(p-1)=\sum_{\xi\pmod{p}}|S_{\lambda}(\xi)|^{2}=|S_{\lambda}(0)|^{2}+|S_{\lambda}(1)|^{2}(p-1)=|S_{\lambda}(1)|^{2}((p-1)^{2}+p-1)=|S_{\lambda}(1)|^{2}p(p-1).$

We deduce that $|S_{\lambda}(1)|=1$, and thus $|S_{\lambda}(0)|=p-1$. But if
$p=3$ we have $|S_{\lambda}(0)|=0\neq 2$ since $\lambda(1)=-\lambda(2)$, and
for $p>3$ we have

$|S_{\lambda}(0)|=\left|\sum_{1\leq n<p}\lambda(n)\right|=\left|\sum_{3\leq
n<p}\lambda(n)\right|\leq p-3<p-1,$

a contradiction. Thus, we conclude that $\lambda(d)=-1$, and it follows
further from (2.1) that $S_{\lambda}(0)=0$.
Next, as $d$ is a primitive root and $\chi_{p}$ is a non-principal character,
$\chi_{p}(d)=-1$. Hence,

$S_{\lambda}(d^{k})=\lambda(d)^{k}S_{\lambda}(1)=(-1)^{k}S_{\lambda}(1)=\chi_{p}(d)^{k}S_{\lambda}(1)\text{
for all }k.$

Now for each $n<p$, we have using $S_{\lambda}(0)=0$ that

$\lambda(n)=\sum_{m<p}\lambda(m)1_{m\equiv
n\pmod{p}}=\frac{1}{p}\sum_{\xi\pmod{p}}e\left(-\frac{n\xi}{p}\right)\sum_{m<p}\lambda(m)e\left(\frac{m\xi}{p}\right)=\frac{1}{p}\sum_{\begin{subarray}{c}\xi\pmod{p}\\\
\xi\not\equiv
0\pmod{p}\end{subarray}}S_{\lambda}(\xi)e\left(-\frac{n\xi}{p}\right).$

Reparametrising $\xi\in(\mathbb{Z}/p\mathbb{Z})^{\times}$ as $d^{k}\pmod{p}$
for $1\leq k\leq p-1$, we get

$\lambda(n)=\frac{1}{p}\sum_{k=1}^{p-1}S_{\lambda}(d^{k})e\left(-\frac{nd^{k}}{p}\right)=\frac{S_{\lambda}(1)}{p}\sum_{k=1}^{p-1}\chi_{p}(d)^{k}e\left(-\frac{nd^{k}}{p}\right)=\frac{S_{\lambda}(1)}{p}\sum_{\xi\pmod{p}}\chi_{p}(\xi)e\left(-\frac{n\xi}{p}\right).$

Using standard relations for Gauss sums, we readily find that for any $n<p$,

$\lambda(n)=\chi_{p}(n)\cdot\frac{S_{\lambda}(1)\overline{\tau(\chi_{p})}}{p},$

where, as usual,

$\tau(\chi_{p}):=\sum_{a\pmod{p}}\chi_{p}(a)e\left(\frac{a}{p}\right).$

If we set $n=1$ then we plainly have
$S_{\lambda}(1)\overline{\tau(\chi_{p})}=p$, and the claim follows. ∎

There are clearly two ways in which Lemma 2.5 may be used. One is to derive
information about the Liouville function using corresponding behaviour of
Dirichlet characters; this appears hard to do since we only know that
$\lambda$ and $\chi_{p}$ are comparable within the fundamental domain
$[1,p-1]$.
The other way is to obtain constraints on the behaviour of Dirichlet
characters from the rigidity of the Liouville function, in particular at
primes. In this direction we deduce the following.

######  Proposition 2.6.

There are effectively computable constants $C_{1},p_{1}>0$ such that the
following holds: if $p\geq p_{1}$ is a prime for which
$|\mathcal{L}_{\lambda}(p)|=p-1$ then $L(1,\chi_{p})\leq C_{1}(\log
p)/p^{1/6}$.

######  Remark 3.

We highlight here the following (consequence of a) well-known result of
Tatuzawa [9]: with the exception of at most one prime $\tilde{p}$, given
$\delta\in(0,1/12]$ we have for all $p\geq e^{1/\delta}$, $p\neq\tilde{p}$,
that

$L(1,\chi_{p})\geq\frac{3\delta}{5p^{\delta}}.$

This allows us to give an effective range (outside of a possible lone
exception) for the possible primes $p$ for which
$|\mathcal{L}_{\lambda}(p)|=p-1$. For instance, taking $\delta=1/12$, the
above bound combines with Proposition 2.6 to show that when
$p\geq\max\\{p_{1},e^{12}\\}$ and $p\neq\tilde{p}$,

$p^{1/6}\leq 20C_{1}\log p=120C_{1}\log(p^{1/6}).$

When $p\geq e^{12}$ we have $z:=p^{1/6}>4$. Setting $R:=120C_{1}$, we find
that $z<R\log z<Rz^{1/2}$. We may thus conclude that
$p<\max\\{p_{1},e^{12},R^{12}\\}$, provided $p\neq\tilde{p}$.
A back-of-the-envelope calculation invoking explicit bounds for (a) the
Riemann zeta function on the critical line (see e.g. [11]) and (b) the
constant in the Pólya-Vinogradov inequality (see e.g. [1]) yield the constants
$C_{1}=12.5$ and $p_{1}=e^{24}$ here, which means that any $p\neq\tilde{p}$
with $p>e^{88}$ satisfies $|\mathcal{L}_{\lambda}(p)|<p-1$.

The proof of Proposition 2.6 is based on the following result, which makes
more precise an old result due to Linnik and Vinogradov [7].

######  Lemma 2.7.

There are effectively computable constants $p_{2},C_{2}>0$ such that if $p\geq
p_{2}$ then

$\left|\sum_{n<p}\left(1-\frac{n}{p}\right)\sum_{d|n}\chi_{p}(d)-\frac{p}{2}L(1,\chi_{p})\right|\leq
C_{2}p^{5/6}\log p.$

###### Proof.

We follow the strategy of [7]. Expanding the smoothed convolution sum gives

$S_{p}:=\sum_{n<p}\left(1-\frac{n}{p}\right)(1\ast\chi_{p})(n)=\sum_{d<p}\chi_{p}(d)\sum_{m<p/d}\left(1-\frac{m}{p/d}\right).$

Let $\sigma_{0}:=1+1/\log p$. Using the Mellin identity

$\sum_{n\leq y}\left(1-\frac{n}{y}\right)=\frac{1}{2\pi
i}\int_{\sigma_{0}-i\infty}^{\sigma_{0}+i\infty}\frac{y^{s}}{s(s+1)}\zeta(s)ds,$

we obtain that

$S_{p}=\frac{1}{2\pi
i}\int_{\sigma_{0}-i\infty}^{\sigma_{0}+i\infty}\frac{p^{s}}{s(s+1)}\zeta(s)\left(\sum_{d<p}\frac{\chi_{p}(d)}{d^{s}}\right)ds.$

Truncating the integral at $\text{Im}(s)=p$ yields an error term of size

$\ll
p\left(\sum_{d<p}\frac{1}{p}\right)\int_{|t|>p}|\zeta(\sigma_{0}+it)|\frac{dt}{t^{2}}\ll(\log
p)^{2}.$

We shift the contour to the line $\text{Re}(s)=1/2$, picking up a residue at
the pole $s=1$, of size

$\frac{p}{2}\sum_{d<p}\frac{\chi_{p}(d)}{d}=\frac{p}{2}L(1,\chi_{p})+O\left(\sqrt{p}\log
p\right),$

the error term arising from partial summation and the Pólya-Vinogradov
inequality. Using the standard bounds (see e.g. [10, Thms. II.3.8-3.9])

$|\zeta(\sigma+it)|=\begin{cases}O_{\varepsilon}\left((2+|t|)^{\tfrac{1}{3}(1-\sigma)+\varepsilon}\right)&\text{
for }\frac{1}{2}<\sigma\leq 1,\\\ O(\log(2+|t|))&\text{ for }\sigma\geq
1-1/\log p,1\leq|t|\leq p,\end{cases}$

the horizontal lines at height $\text{Im}(s)=\pm p$ together contribute

$\leq\frac{1}{p^{2}}\int_{1/2}^{\sigma_{0}}|\zeta(u+ip)|p^{u}\left(\sum_{d<p}\frac{1}{d^{u}}\right)du\ll_{\varepsilon}\frac{1}{p}\int_{1/2}^{1-1/\log
p}p^{\frac{1}{3}(1-u)+\varepsilon}\frac{du}{1-u}+\frac{1}{p}\int_{1-1/\log
p}^{\sigma_{0}}(\log p)^{2}du\ll\frac{1}{\sqrt{p}}.$

Summarising the above, we so far have

$S_{p}=\frac{p}{2}L(1,\chi_{p})+\frac{1}{2\pi
i}\int_{1/2-ip}^{1/2+ip}\frac{p^{s}}{s(s+1)}\zeta(s)\left(\sum_{d<p}\frac{\chi_{p}(d)}{d^{s}}\right)ds+O\left(\sqrt{p}\log
p\right).$

Next, suppose $s=1/2+it$, with $|t|\leq p$, and set

$S_{\chi_{p}}(u):=\sum_{n\leq u}\chi_{p}(n),\quad u\geq 1.$

Let $1\leq y\leq p^{3/4}$ be a parameter to be chosen shortly. We have by a
dyadic decomposition and partial summation that

$\displaystyle\left|\sum_{d<p}\frac{\chi_{p}(d)}{d^{s}}\right|$
$\displaystyle\leq\sum_{d\leq y}\frac{1}{\sqrt{d}}+\sum_{y\leq 2^{j}\leq
p}\left|\sum_{2^{j}\leq d<2^{j+1}}\frac{\chi_{p}(d)}{d^{1/2+it}}\right|$
$\displaystyle\ll y^{1/2}+\sum_{y\leq 2^{j}\leq
p}\left(\frac{|S_{\chi_{p}}(2^{j+1})|}{2^{j+1}}+\frac{|S_{\chi_{p}}(2^{j})|}{2^{j}}+(1/2+|t|)\int_{2^{j}}^{2^{j+1}}\frac{|S_{\chi_{p}}(u)|}{u^{3/2}}du\right).$

Applying the Pólya-Vinogradov inequality $\max_{1\leq u\leq
p}|S_{\chi_{p}}(u)|\ll\sqrt{p}\log p$, we obtain the upper bound

$\displaystyle\ll y^{1/2}+\sqrt{p}(\log p)\sum_{y\leq 2^{j}\leq
p}\left(\frac{1}{2^{j}}+(1/2+|t|)2^{-j/2}\right)\ll
y^{1/2}+(1+|t|)\frac{\sqrt{p}\log p}{\sqrt{y}}.$

Inserting this together with the subconvexity bound

$|\zeta(1/2+it)|\ll(|t|+2)^{1/6}\log(2+|t|),\quad|t|\leq p,$

(see e.g., [10, Cor. II.3.7]) and setting $y:=p^{2/3}(\log p)^{2}$, we obtain

$\displaystyle\left|\frac{1}{2\pi
i}\int_{1/2-ip}^{1/2+ip}\frac{p^{s}}{s(s+1)}\zeta(s)\left(\sum_{d<p}\frac{\chi_{p}(d)}{d^{s}}\right)ds\right|$
$\displaystyle\ll p^{1/2}y^{1/2}\int_{|t|\leq
p}\frac{(|t|+2)^{1/6}\log(|t|+2)}{(2+|t|)^{2}}dt+\frac{p(\log
p)}{\sqrt{y}}\int_{|t|\leq p}\frac{(2+|t|)^{7/6}\log(|t|+2)}{(2+|t|)^{2}}dt$
$\displaystyle\ll p^{5/6}\log p.$

Collecting the above estimates yields

$S_{p}=\frac{p}{2}L(1,\chi_{p})+O\left(p^{5/6}\log p\right),$

as claimed. ∎

###### Proof of Proposition 2.6.

Suppose $|\mathcal{L}_{\lambda}(p)|=p-1$. Since $p$ is prime we can find a
primitive root $2\leq d<p$ modulo $p$, and by Proposition 2.4, $d$ satisfies
(6) for all $\xi\pmod{p}$. Lemma 2.5 therefore implies that
$\lambda(n)=\chi_{p}(n)$ for all $n<p$.
Since $1\ast\lambda$ is the indicator function for the set of perfect squares,
by Lemma 2.7 we find that

$\frac{p}{2}L(1,\chi_{p})+O(p^{5/6}\log
p)=\sum_{n<p}\left(1-\frac{n}{p}\right)\sum_{d|n}\lambda(d)=\sum_{n<p}\left(1-\frac{n}{p}\right)(1\ast\lambda)(n)<\sum_{m^{2}<p}1<\sqrt{p}.$

We deduce from this that $L(1,\chi_{p})\ll p^{-1/2}+(\log p)p^{-1/6}\ll(\log
p)p^{-1/6}$, as claimed. ∎

###  2.2. Bounds on the multiplicity of prime divisors of $N$

We next extend Lemma 2.5 to higher powers of $p$.

######  Lemma 2.8.

Let $p$ be a prime and assume that $|\mathcal{L}_{\lambda}(p^{k})|=p^{k}-1$
for some $k\geq 1$. Then $\lambda(n)=\chi_{p}(n)$ for all $n<p^{k}$ with
$p\nmid n$.

###### Proof.

By Lemma 2.1, we know that $|\mathcal{L}_{\lambda}(p^{j})|=p^{j}-1$ and
$\lambda(p^{j}-1)=\lambda(p-1)$ for all $1\leq j\leq k$. In the case $j=1$,
combining Propositions 2.4 and 2.5, we see that $\lambda(n)=\chi_{p}(n)$ for
all $n<p$. We now extend this to show that $\lambda(n)=\chi_{p}(n)$ for all
$n<p^{k}$ with $p\nmid n$.
To do this, we prove by induction on $1\leq j\leq k$ that $\chi_{p}(q)=-1$ for
all primes $q<p^{j}$, $q\neq p$. The outcome of this induction is that
$\lambda(q)=\chi_{p}(q)$ for all $q<p^{k}$, $q\neq p$, so that by complete
multiplicativity we obtain $\lambda(n)=\chi_{p}(n)$ for all $n<p^{k}$ with
$p\nmid n$. The base case $j=1$ follows immediately from
$\lambda(n)=\chi_{p}(n)$ for all $1\leq n<p$, which we have just established.
Assume therefore that $\chi_{p}(q)=-1$ for all $q<p^{i}$, $q\neq p$, for all
$1\leq i<j$. We next prove that this is the case for all $q<p^{j}$, $q\neq p$,
or equivalently that $\chi_{p}(q)=-1$ for all primes $p^{j-1}<q<p^{j}$. Note
that by complete multiplicativity, the inductive assumption actually implies
that $\lambda(n)=\chi_{q}(n)$ for all $n$ that satisfy $P^{+}(n)<p^{j-1}$ with
$p\nmid n$. Now, we know that

$\lambda(n)\lambda(p^{j}-n)=\lambda(p^{j}-1)=\lambda(p-1)$

for all $n<p^{j}$ with $p\nmid n$. We observe furthermore that for any
$n<p^{j}$ with $p\nmid n$, $P^{+}(n/P^{+}(n))<p^{j-1}$, since either
$P^{+}(n)<p^{j-1}$ or else $n$ has precisely one prime factor
$p^{j-1}<q<p^{j}$, and necessarily $n/q<p$. It follows by the inductive
assumption that whenever $1\leq n<p^{j}$ with $p\nmid n$,

$\displaystyle\lambda(p-1)$
$\displaystyle=\lambda(n)\lambda(p^{j}-n)=\lambda(P^{+}(n))\lambda(P^{+}(p^{j}-n))\lambda\left(\frac{n}{P^{+}(n)}\right)\lambda\left(\frac{p^{j}-n}{P^{+}(p^{j}-n)}\right)$
$\displaystyle=\chi_{p}\left(\frac{n}{P^{+}(n)}\right)\chi_{p}\left(\frac{p^{j}-n}{P^{+}(p^{j}-n)}\right)=\chi_{p}(P^{+}(n))\chi_{p}(P^{+}(p^{j}-n))\chi_{p}\left(n\right)\chi_{p}\left(p^{j}-n\right)$
$\displaystyle=\chi_{p}(P^{+}(n))\chi_{p}(P^{+}(p^{j}-n))\chi_{p}(p-1)=\chi_{p}(P^{+}(n))\chi_{p}(P^{+}(p^{j}-n))\lambda(p-1).$

We deduce, therefore, that

(8) $\chi_{p}(P^{+}(n))=\chi_{p}(P^{+}(p^{j}-n))\text{ for all }1\leq
n<p^{j},\,p\nmid n,$

which we will presently use to show that $\chi_{p}(q)=-1$ for all
$p^{j-1}<q<p^{j}$.
Let us enumerate the primes in the interval $(p^{j-1},p^{j})$ as
$\\{q_{1},\ldots,q_{R}\\}$. We prove by a second induction on $1\leq r\leq R$
that $\chi_{p}(q_{r})=-1$. In the base case $r=1$, observe that if
$n_{1}:=\lfloor p^{j}/q_{1}\rfloor q_{1}$, which is coprime to $p$, then as
$n_{1}/q_{1}<p$ we have $P^{+}(n_{1})=q_{1}$ (recalling that $j>1$ here).
Moreover,

$p^{j}-n_{1}=p^{j}-q_{1}\left\lfloor\frac{p^{j}}{q_{1}}\right\rfloor=q_{1}\left\\{\frac{p^{j}}{q_{1}}\right\\}<q_{1}.$

Thus, $P^{+}(p^{j}-n_{1})<q_{1}$, and since $q_{1}$ is the minimal prime
$>p^{j-1}$ (and $p\nmid p^{j}-n_{1}$), we have $P^{+}(p^{j}-n_{1})<p^{j-1}$.
By our inductive assumption on $j$, we have $\chi_{p}(P^{+}(p^{j}-n_{1}))=-1$,
and hence by (8), $\chi_{p}(q_{1})=\chi_{p}(P^{+}(n_{1}))=-1$.
Now supposing that $\chi_{p}(q_{s})=-1$ for all $1\leq s\leq r$, we find by
the same construction that if $n_{r+1}:=\lfloor p^{j}/q_{r+1}\rfloor q_{r+1}$
then $P^{+}(n_{r+1})=q_{r+1}$ whereas

$P^{+}(p^{j}-n_{r+1})=P^{+}\left(q_{r+1}\left\\{\frac{p^{j}}{q_{r+1}}\right\\}\right)<q_{r+1}.$

It follows that $P^{+}(p^{j}-n_{r+1})$ is either $q_{s}$ for some $1\leq s\leq
r$, or else it is $<p^{j-1}$. Either by our inductive assumption on $r$ or on
$j$, we obtain $\chi_{p}(P^{+}(p^{j}-n_{r+1}))=-1$, and thus
$\chi_{p}(q_{r+1})=-1$ as well. By induction on $r$, we obtain
$\chi_{p}(q_{r})=-1$ for all $1\leq r\leq R$, and thus for all primes
$p^{j-1}<q<p^{j}$. By induction on $j$, the claim follows. ∎

######  Corollary 2.9.

There are effectively computable constants $p_{3}$ and $L$ such that:

  1. (1)

if $p\geq p_{3}$ and $|\mathcal{L}_{\lambda}(p^{k})|=p^{k}-1$ then $k\leq 5$,
and

  2. (2)

if $p<p_{3}$ and $|\mathcal{L}_{\lambda}(p^{k})|=p^{k}-1$ then $k\leq L$.

###### Proof.

Suppose $|\mathcal{L}_{\lambda}(p^{k})|=p^{k}-1$. By the previous lemma we
obtain that $\chi_{p}(n)=\lambda(n)$ for all $n<p^{k}$, and in particular
$\chi_{p}(q)=-1$ for all primes $q<p^{k}$.
By Heath-Brown’s version of Linnik’s theorem [5], there is an effectively
computable constant $c>0$ and a prime $q<cp^{5.5}$ such that $q\equiv
1\pmod{p}$. If $q<p^{k}$ then we obtain the contradiction
$-1=\chi_{p}(q)=\chi_{p}(1)=+1$. We therefore conclude that
$p^{k}<q<cp^{5.5}$. This implies that either $k\leq 5$, or else $p<p_{3}$,
with $p_{3}:=c^{2}$. Moreover, in the latter case $p<p_{3}$ we also have that
$2^{k-5.5}<c$, so that $k\leq L$ for $L:=5.5+\tfrac{\log c}{\log 2}$. ∎

###### Proof of Theorem 1.2.

Suppose $N\geq 2$ satisfies $|\mathcal{L}_{\lambda}(N)|=N-1$. By Lemma 2.1, we
have $|\mathcal{L}_{\lambda}(p^{k})|=p^{k}-1$ for each prime power $p^{k}||N$,
and in particular $|\mathcal{L}_{\lambda}(p)|=p-1$ for each prime $p|N$.
By Proposition 2.6, there is a(n effectively computable) constant $C>0$ such
that

$L(1,\chi_{p})\leq C(\log p)/p^{1/6}\text{ for each }p|N$

(taking $C\geq C_{1}$ as needed to cover the range $p<p_{1}$). But by Siegel’s
theorem [6, Thm. 5.28(2)], this is impossible as soon as $p\geq p_{0}$, where
$p_{0}$ is a(n ineffective) constant. It follows that all primes $p|N$ satisfy
$p<p_{0}$, as claimed.
Furthermore, by Corollary 2.9 there is a(n effectively computable) constant
$L>0$ such that if $p<p_{0}$ and $p^{k}||N$ then $k\leq L$. It follows that

$N=\prod_{p^{k}||N}p^{k}\leq\left(\prod_{p<p_{0}}p\right)^{L}=:N_{0}.$

Therefore, if $N>N_{0}$ then $|\mathcal{L}_{\lambda}(N)|<N-1$, as claimed. ∎

######  Remark 4.

The ineffectivity inherent in our main result is due to the possible presence
of Siegel zeros, which renders Siegel’s theorem ineffective. This seems
compelling in that, typically, Siegel zeros are _useful_ for establishing non-
trivial estimates about correlations of the Liouville function. For example,
in regards to the convolution sums treated in this paper, De Koninck, Germán
and Kátai proved [4] that if $(p_{k})_{k}$ is a sequence of exceptional
moduli, and if $\eta_{k}:=(1-\beta_{k})\log p_{k}$, then for any
$N\in(p_{k}^{10},p_{k}^{c\log\log(1/\eta_{k})}]$ (albeit with the condition
$(N,p_{k})=1$),

$\left|\frac{1}{N}\sum_{n<N}\lambda(n)\lambda(N-n)\right|\leq\frac{1}{\log\log(1/\eta_{k})}+o_{k\rightarrow\infty}(1).$

However, besides the restriction $(N,p_{k})=1$ not suiting our purposes here,
our results are limited by the fact that $\lambda(n)=\chi_{p_{k}}(n)$ only in
the limited range $1\leq n<p_{k}$. It is unclear whether the result of [4],
for example, which requires a broader range in which $\lambda$ behaves
periodically in some sense, may be applicable in the present circumstances in
order to render our theorem effective.

##  3\. Proof of Proposition 2.4

In this section we prove Proposition 2.4. As we show later, it suffices to
consider the case when $d$ is prime. Having shown the cases $d=2$ and $3$ in
Lemma 2.3, we focus here on $d>3$.

###  3.1. An iterative argument for $d=q$ prime

As previously, let $p>3$ be a prime with $|\mathcal{L}_{\lambda}(p)|=p-1$. Let
$3<q<p$ be an odd prime. We wish to show that

$S_{\lambda}(q\xi)=-S_{\lambda}(\xi)\text{ for all }\xi\pmod{p}.$

As in the proof of Lemma 2.3, it suffices to show that

$\displaystyle-(p-1)$
$\displaystyle=\frac{1}{p}\sum_{\xi\pmod{p}}S_{\lambda}(q\xi)\overline{S}_{\lambda}(\xi)=\sum_{m,n<p}\lambda(m)\lambda(n)\frac{1}{p}\sum_{\xi\pmod{p}}e\left(\xi\frac{qn-m}{p}\right)$
$\displaystyle=\sum_{\begin{subarray}{c}m,n<p\\\ qn\equiv
m\pmod{p}\end{subarray}}\lambda(m)\lambda(n).$

As before, we split the set of $1\leq m,n<p$ according to the choice of $0\leq
j<q$ for which $qn=m+jp$; in each case we have $m\equiv-jp\pmod{q}$, and thus
we must show that

$-(p-1)=\sum_{j=0}^{q-1}\sum_{\begin{subarray}{c}m<p\\\ m\equiv-
jp\pmod{q}\end{subarray}}\lambda(m)\lambda\left(\frac{m+jp}{q}\right)=-\sum_{j=0}^{q-1}\sum_{\begin{subarray}{c}m<p\\\
m\equiv-jp\pmod{q}\end{subarray}}\lambda(m)\lambda\left(m+jp\right).$

Equivalently, it is our goal to prove that for every $0\leq j<q$ and $1\leq
m<p$ we have

(9) $\lambda(m)\lambda(m+jp)=+1\text{ whenever }m\equiv-jp\pmod{q}.$

For $1\leq r\leq q-1$ we define the sets

$\mathcal{A}_{q,r}:=\\{(m,j)\in\\{1,\ldots,p-1\\}\times\\{0,\ldots,q-1\\}:\frac{pq}{r+1}<m+jp<\frac{pq}{r}\\},\quad\mathcal{A}_{q,q}:=\\{(m,0):1\leq
m<p\\}.$

Note that the sets $\\{\mathcal{A}_{q,r}\\}_{1\leq r\leq q}$ partition the set
of all pairs $(m,j)\in\\{1,\ldots,p-1\\}\times\\{0,\ldots,q-1\\}$, in view of
the following observations:

  1. (a)

for each such pair, $1\leq m+jp<pq$, and therefore must either satisfy $1\leq
m+jp<p$ (so $j=0$), or else $pq/(r+1)\leq m+jp<pq/r$ for some $1\leq r\leq
q-1$;

  2. (b)

as $p$ and $q$ are prime we can never have $m+jp=pq/(r+1)$ for any $1\leq
r\leq q-1$ unless $r=q-1$, but in this case $m+jp=p$ is not solvable with
$1\leq m<p-1$;

  3. (c)

if $(m,j)\notin\mathcal{A}_{q,r}$ for all $1\leq r\leq q-1$ then $1\leq
m+jp<p$, equivalently, $j=0$ and $(m,0)\in\mathcal{A}_{q,q}$.

For each $1\leq r\leq q-1$ define a map $\psi_{r}$ on pairs
$(m,j)\in\mathcal{A}_{q,r}$ via

$\psi_{r}(m,j):=\left(\left\lceil\frac{rm}{p}\right\rceil
p-rm,q-jr-\left\lceil\frac{rm}{p}\right\rceil\right),$

where, as usual, given $t\in\mathbb{R}$ we denote by $\lceil t\rceil$ the
least integer $k\geq t$.

######  Lemma 3.1.

Let $(m,j)\in\mathcal{A}_{q,r}$ for some $1\leq r\leq q-1$, and set
$(m^{\prime},j^{\prime}):=\psi_{r}(m,j)$. The following properties hold:

  1. (a)

$m^{\prime}+j^{\prime}p\equiv 0\pmod{q}$ whenever $m+jp\equiv 0\pmod{q}$,

  2. (b)

$m^{\prime}=\\{-rm/p\\}p=p\left(1-\\{rm/p\\}\right)$,

  3. (c)

$(m^{\prime},j^{\prime})\in\bigcup_{r+1\leq s\leq q}\mathcal{A}_{q,s}$, and

  4. (d)

if $m+jp\equiv 0\pmod{q}$ then
$\lambda(m+jp)=\lambda(r)\lambda(p-1)\lambda(m^{\prime}+j^{\prime}p)$.

In fact, (d) may be rewritten as

(10)
$\lambda(m)\lambda(m+jp)=\left[\lambda(rm)\lambda\left(p\left\\{\frac{rm}{p}\right\\}\right)\right]\lambda(m^{\prime})\lambda(m^{\prime}+j^{\prime}p).$

###### Proof.

(a) We observe that

(11) $m^{\prime}+j^{\prime}p=\left(\left\lceil\frac{rm}{p}\right\rceil
p-rm\right)+\left(q-jr-\left\lceil\frac{rm}{p}\right\rceil\right)p=p(q-jr)-rm=pq-r(m+jp),$

so that if $m+jp\equiv 0\pmod{q}$ then $m^{\prime}+j^{\prime}p\equiv
0\pmod{q}$ as well.
(b) Note that since $r,m<p$, $rm/p\notin\mathbb{Z}$. Whenever
$\alpha\notin\mathbb{Z}$ we have

$\left\lceil\alpha\right\rceil=\alpha+1-\\{\alpha\\}=\alpha+\\{-\alpha\\},$

so we therefore conclude that

$m^{\prime}=p\left(\frac{rm}{p}+\left\\{-\frac{rm}{p}\right\\}\right)-rm=p\left\\{-\frac{rm}{p}\right\\}=p\left(1-\left\\{\frac{rm}{p}\right\\}\right),$

as required.
(c) Using $pq/(r+1)<m+jp<pq/r$ together with (11), we deduce that

$\displaystyle m^{\prime}+j^{\prime}p$
$\displaystyle<pq-r\cdot\frac{pq}{r+1}=pq\left(1-\frac{r}{r+1}\right)=\frac{pq}{r+1}$
$\displaystyle m^{\prime}+j^{\prime}p$
$\displaystyle>pq-r\cdot\frac{pq}{r}=0.$

Together with (b), the latter bound implies that $1\leq m^{\prime}<p$.
Furthermore,

$j^{\prime}=q-\left(jr+\left\lceil\frac{rm}{p}\right\rceil\right)\geq
q-1-\frac{r}{p}(jp+m)>q-1-\frac{r}{p}\cdot\frac{pq}{r}=-1,$

so as $j^{\prime}$ is an integer with $j^{\prime}>-1$, and $\lceil
rm/p\rceil\geq 1$, we have $0\leq j^{\prime}<q$. As
$0<m^{\prime}+j^{\prime}p<pq/(r+1)$, it follows that
$(m^{\prime},j^{\prime})\in\mathcal{A}_{q,s}$ for some $r+1\leq s\leq q$, as
required.
(d) Since $(m,j)\in\mathcal{A}_{q,r}$ and $q|(m+jp)$, we see that

$1\leq\frac{m+jp}{q}<\frac{1}{q}\cdot\frac{pq}{r}=\frac{p}{r}.$

It follows that $(rm+rjp)/q\in\mathbb{Z}\cap[1,p-1]$, and so using (3),

$\displaystyle\lambda(m+jp)$
$\displaystyle=\lambda(qr)\lambda\left(\frac{rm+rjp}{q}\right)=\lambda(qr)\lambda(p-1)\lambda\left(p-\frac{rm+rjp}{q}\right)=\lambda(qr)\lambda(p-1)\lambda\left(\frac{(q-rj)p-rm}{q}\right)$
$\displaystyle=\lambda(r)\lambda(p-1)\lambda\left(\left(\left\lceil\frac{rm}{p}\right\rceil
p-rm\right)+p\left(q-rj-\left\lceil\frac{rm}{p}\right\rceil\right)\right)$
$\displaystyle=\lambda(r)\lambda(p-1)\lambda(m^{\prime}+j^{\prime}p),$

as claimed.
To prove (10), it suffices to note using (b) that

$\lambda(m^{\prime})=\lambda(p-p\\{rm/p\\})=\lambda(p-1)\lambda(p\\{rm/p\\}),$

after which the identity follows immediately from (c) upon multiplying both
sides by $\lambda(m)$. ∎

The upshot of Lemma 3.1(c) is that $\psi_{r}$ maps $\mathcal{A}_{q,r}$ to a
set of pairs $(m^{\prime},j^{\prime})$ for which $m^{\prime}+j^{\prime}p$ has
strictly decreased (and in particular $(m^{\prime},j^{\prime})$ belongs to
$\mathcal{A}_{q,r^{\prime}}$ where $r^{\prime}>r$). We see therefore that by
iteratively composing maps $\psi_{r}$, $r<q$, we must eventually find an image
pair in $\mathcal{A}_{q,q}$, i.e., where the $j$ component is $0$.
With this in mind, we introduce the following definition.

######  Definition 1.

We say that the _signature_ of a pair $(m,j)$, $1\leq m<p$ and $0\leq j<q$, is
the tuple $(r_{1},\ldots,r_{k})$ such that $1\leq r_{1}<r_{2}<\cdots<r_{k}<q$,
with

$\psi_{r_{k}}\circ\cdots\circ\psi_{r_{1}}(m,j)=(\tilde{m},0)\in\mathcal{A}_{q,q},$

for some $1\leq\tilde{m}<p$. (Here, we implicitly have that the indices
$r_{i}$ are determined such that $(m,j)\in\mathcal{A}_{q,r_{1}}$,
$\psi_{r_{1}}(m,j)\in\mathcal{A}_{q,r_{2}}$,
$\psi_{r_{2}}\circ\psi_{r_{1}}(m,j)\in\mathcal{A}_{q,r_{3}}$, and so on.)

######  Lemma 3.2.

Let $(m,j)$ have signature $(r_{1},\ldots,r_{k})$. Then

$\lambda(m)\lambda(m+jp)=\prod_{i=0}^{k-1}\lambda(m_{i}r_{i+1})\lambda\left(p\left\\{\frac{r_{i+1}m_{i}}{p}\right\\}\right),$

where we have set

$(m_{0},j_{0}):=(m,j),\quad(m_{i+1},j_{i+1}):=\psi_{r_{i+1}}(m_{i},j_{i})\text{
for }0\leq i\leq k-1.$

###### Proof.

By iteratively invoking (10), we obtain

$\displaystyle\lambda(m)\lambda(m+jp)$
$\displaystyle=\lambda(m_{0})\lambda(m_{0}+j_{0}p)=\left[\lambda(m_{0}r_{1})\lambda\left(p\left\\{\frac{r_{1}m_{0}}{p}\right\\}\right)\right]\lambda(m_{1})\lambda(m_{1}+j_{1}p)$
$\displaystyle=\prod_{i=0}^{1}\left[\lambda(m_{i}r_{i+1})\lambda\left(p\left\\{\frac{r_{i+1}m_{i}}{p}\right\\}\right)\right]\lambda(m_{2})\lambda(m_{2}+j_{2}p)$
$\displaystyle=\cdots=\prod_{i=0}^{k-1}\left[\lambda(m_{i}r_{i+1})\lambda\left(p\left\\{\frac{r_{i+1}m_{i}}{p}\right\\}\right)\right]\lambda(m_{k})\lambda(m_{k}+j_{k}p).$

By the definition of signature, we have $j_{k}=0$, and thus
$\lambda(m_{k})\lambda(m_{k}+j_{k}p)=\lambda(m_{k})^{2}=+1$. The claim
follows. ∎

###  3.2. Periodicity via dilation

In connection with Lemma 3.1 we next show the following lemma, which shows
that if $S_{\lambda}$ satisfies the dilation property in Proposition 2.4 with
$d=r<p$ then $\lambda$ exhibits mod $p$ periodicity in $[1,rp-1]$.

######  Lemma 3.3.

Assume that $1\leq r<p$ satisfies

(12) $S_{\lambda}(r\xi)=\lambda(r)S_{\lambda}(\xi)\text{ for all
}\xi\pmod{p}.$

Then for any $1\leq m<p$, $\lambda(p\\{rm/p\\})=\lambda(r)\lambda(m)$.

###### Proof.

Note that by making the bijective change of variables $\xi\mapsto
r^{-1}\xi\pmod{p}$ and rearranging, (12) yields

$S_{\lambda}(r^{-1}\xi)=\lambda(r)S_{\lambda}(\xi)\text{ for all
}\xi\pmod{p}.$

From this, we derive that

$\displaystyle 0$
$\displaystyle=S_{\lambda}(r^{-1}\xi)-\lambda(r)S_{\lambda}(\xi)=\sum_{n<p}\lambda(n)e(nr^{-1}\xi/p)-\sum_{m<p}\lambda(mr)e(m\xi/p)$
$\displaystyle=\sum_{m<p}e\left(m\xi/p\right)\left(\sum_{\begin{subarray}{c}n<p\\\
n\equiv rm\pmod{p}\end{subarray}}\lambda(n)-\lambda(mr)\right).$

Since this holds for all $\xi\pmod{p}$ we deduce that for all $1\leq m<p$,

(13) $\sum_{\begin{subarray}{c}n<p\\\ n\equiv
rm\pmod{p}\end{subarray}}\lambda(n)=\lambda(m)\lambda(r).$

On the other hand, the condition $n\equiv rm\pmod{p}$ with $1\leq n<p$ is
equivalent to $n=p\\{rm/p\\}$. Combining these two facts, we deduce that

$\lambda(p\\{rm/p\\})=\lambda(m)\lambda(r),$

as claimed. ∎

We are now in a position to prove Proposition 2.4.

###### Proof of Proposition 2.4.

We proceed by induction on $d$. When $d=1$ there is nothing to prove. Thus,
assume that for every $1\leq d<q$ the equation

(14) $S_{\lambda}(d\xi)=\lambda(d)S_{\lambda}(\xi)\text{ holds for all
}\xi\pmod{p}.$

We thus must establish (14) for $d=q$ (provided $q<p$).
First, we observe that if $q$ is composite then writing $q=ab$ with $1<a,b<q$
we have that for any $\xi\pmod{p}$,

$S_{\lambda}(q\xi)=S_{\lambda}(a(b\xi))=\lambda(a)S_{\lambda}(b\xi)=\lambda(a)\lambda(b)S_{\lambda}(\xi)=\lambda(q)S_{\lambda}(\xi),$

as required. Hence, the claim holds whenever $q$ is composite.
Thus, we may assume that $q$ is prime. Assuming the induction hypothesis, we
see that for any $1\leq r<q$ (necessarily coprime to $p$) we have

$S_{\lambda}(r\xi)=\lambda(r)S_{\lambda}(\xi)\text{ for all }\xi\pmod{p}.$

By Lemma 3.3, we see that for any $1\leq m<p$,

(15)
$\lambda\left(p\left\\{rm/p\right\\}\right)=\lambda(r)\lambda(m)=\lambda(rm),$

a fact that we will use momentarily.
As discussed above, in order to prove (14) holds with $d=q$ it suffices to
show that

$\lambda(m)\lambda(m+jp)=+1\text{ for all }0\leq j<q,\,1\leq m<p\text{ with
}m\equiv-jp\pmod{q}.$

Now given $1\leq m<p$ and $0\leq j<q$ let $(r_{1},\ldots,r_{k})$ denote the
signature of $(m,j)$, recalling that $1\leq r_{1}<\cdots<r_{k}<q$. By Lemma
3.2, we have

$\lambda(m)\lambda(m+jp)=\prod_{i=0}^{k-1}\lambda(m_{i}r_{i+1})\lambda\left(p\left\\{\frac{r_{i+1}m_{i}}{p}\right\\}\right).$

But since $r_{i+1}<q$ for all $0\leq i\leq k-1$, we know that (15) holds with
each $r=r_{i+1}$. Thus, every factor in the right-hand product is simply $+1$,
and hence $\lambda(m)\lambda(m+jp)=+1$, as required. Hence, (14) holds when
$q$ is prime as well.
The inductive claim therefore follows in all cases, and so by induction, the
proof is complete. ∎

## References

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# An abstract approach to the Crouzeix conjecture

Raphaël Clouâtre  Department of Mathematics, University of Manitoba, 186
Dysart Road, Winnipeg (Manitoba), Canada R3T 2N2.
<EMAIL_ADDRESS>,  Maëva Ostermann  Département de mathématiques
et de statistique, Université Laval, Québec City (Québec), Canada G1V 0A6.
<EMAIL_ADDRESS>and  Thomas Ransford  Département de mathématiques
et de statistique, Université Laval, Québec City (Québec), Canada G1V 0A6.
<EMAIL_ADDRESS>
###### Abstract.

Let $A$ be a uniform algebra, $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism and $\alpha:A\to A$ be an antilinear contraction such
that

$\|\theta(f)+\theta(\alpha(f))^{*}\|\leq 2\|f\|\quad(f\in A).$

We show that $\|\theta\|\leq 1+\sqrt{2}$, and that $1+\sqrt{2}$ is sharp. We
conjecture that, if further $\alpha(1)=1$, then we may conclude that
$\|\theta\|\leq 2$. This would yield a positive solution to the Crouzeix
conjecture on numerical ranges. In support of our conjecture, we prove that it
is true in two special cases. We also discuss a completely bounded version of
our conjecture that brings into play ideas from dilation theory.

###### Key words and phrases:

Crouzeix conjecture, uniform algebra, homomorphism, completely bounded map

###### 2010 Mathematics Subject Classification:

primary 46J10, secondary 47A12, 47A30

Clouâtre partially supported by an NSERC Discovery Grant.

Ostermann supported by a FRQNT doctoral scholarship

Ransford supported by grants from NSERC and the Canada Research Chairs
program.

##  1\. Introduction

In this article, we propose an approach to a conjecture about numerical
ranges, the so-called Crouzeix conjecture, through the study of certain
homomorphisms on uniform algebras. Recall that given a compact Hausdorff space
$X$, by a uniform algebra we mean a unital norm-closed subalgebra $A\subset
C(X)$. Throughout the paper, homomorphisms of are always assumed to be unital.
Further, we denote by $M_{n}({\mathbb{C}})$ the algebra of complex $n\times n$
matrices.

Our starting point is the following theorem.

######  Theorem 1.1.

Let $A$ be a uniform algebra, let $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism, and let $\alpha:A\to A$ be an antilinear contraction.
Define a linear map $\theta_{\alpha}:A\to M_{n}({\mathbb{C}})$ by

(1)
$\theta_{\alpha}(f):=\frac{1}{2}\Bigl{(}\theta(f)+\theta(\alpha(f))^{*}\Bigr{)}\quad(f\in
A).$

If $\|\theta_{\alpha}\|\leq 1$, then

(2) $\|\theta\|\leq 1+\sqrt{2}.$

Moreover, the constant $(1+\sqrt{2})$ is sharp.

We shall establish this result in §2. The example demonstrating the sharpness
of the constant $(1+\sqrt{2})$ has the feature that $\alpha(1)=-1$. We believe
that if $\alpha$ is at the other extreme, namely if it satisfies
$\alpha(1)=1$, then Theorem 1.1 may be improved as follows.

######  Conjecture 1.2.

Let $A$ be a uniform algebra, $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism, and let $\alpha:A\to A$ be a unital antilinear
contraction. If $\theta_{\alpha}$, defined by (1), satisfies
$\|\theta_{\alpha}\|\leq 1$, then

(3) $\|\theta\|\leq 2.$

Our interest in this conjecture is prompted in part by the fact that a
positive solution, even in a very special case, would lead to a proof of a
celebrated conjecture of Crouzeix [Cr04]. The latter conjecture states that,
if $T$ is an $n\times n$ matrix with numerical range $W(T)$, then, for all
polynomials $p$,

(4) $\|p(T)\|\leq 2\max_{z\in W(T)}|p(z)|.$

In other words, $W(T)$ is a $2$-spectral set for $T$. For background on the
Crouzeix conjecture and various partial results, we refer the reader to
[BCD06, BG20, CGL18, Ch13, Cr04, Cr07, Cr16, CP17, GKL18, GO18].

The link between the two conjectures is made precise by the following theorem.
We write ${\mathbb{D}}$ for the open unit disk and $A({\mathbb{D}})$ for the
disk algebra, namely the algebra of continuous functions on
$\overline{{\mathbb{D}}}$ that are holomorphic on ${\mathbb{D}}$.

######  Theorem 1.3.

If Conjecture 1.2 holds when $A=A({\mathbb{D}})$ and $n=n_{0}$, then (4) holds
for all $T\in M_{n_{0}}({\mathbb{C}})$ and all polynomials $p$.

This theorem will be proved in §3. We note in passing that a similar argument,
but using Theorem 1.1 instead of Conjecture 1.2, leads to the conclusion that
(4) holds with $(1+\sqrt{2})$ in place of the constant $2$. This was already
known: it is a result of Crouzeix and Palencia [CP17]. Indeed, the
Crouzeix–Palencia theorem was the main inspiration for the present article.

We establish two partial results in support of Conjecture 1.2. In the first of
these, we add the assumption that $\alpha$ maps $A$ to multiples of the
identity. This is an abstraction of an idea of Caldwell, Greenbaum and Li
[CGL18, §6], and, as they showed, it implies the Crouzeix conjecture for the
disk, a result originally due to Okubo and Ando [OA75].

######  Theorem 1.4.

Let $A$ be a uniform algebra, let $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism, and let $\alpha:A\to A$ be a unital antilinear
contraction such that $\alpha(A)\subset{\mathbb{C}}1$. If $\theta_{\alpha}$
defined by (1) satisfies $\|\theta_{\alpha}\|\leq 1$, then $\|\theta\|\leq 2$.

In the second result, we suppose that the uniform algebra $A$ is self-adjoint,
in other words, that $A$ is a commutative C*-algebra. The conclusion here is
even stronger than the one that we seek.

######  Theorem 1.5.

Let $A$ be a commutative C*-algebra, let $\theta:A\to M_{n}({\mathbb{C}})$ be
a continuous homomorphism, and let $\alpha:A\to A$ be a unital antilinear
contraction. If $\theta_{\alpha}$, defined by (1), satisfies
$\|\theta_{\alpha}\|\leq 1$, then $\theta$ is a $*$-homomorphism, and in
particular $\|\theta\|=1$.

Theorems 1.4 and 1.5 will be proved and further discussed in §4 and §5
respectively. Finally, in §6, we shall recast the discussion in the setting of
completely bounded maps and operator algebras, and highlight the connection
with dilation theory.

##  2\. Proof of Theorem 1.1

The proof that follows is based on an argument from [RS18].

###### Proof of Theorem 1.1.

For $f\in A$, we have

$\displaystyle\|\theta(f)\|^{4}$
$\displaystyle=\|\theta(f)\theta(f)^{*}\theta(f)\theta(f)^{*}\|$
$\displaystyle=\|\theta(f)\bigl{(}\theta(f)+\theta(\alpha(f))^{*}\bigr{)}^{*}\theta(f)\theta(f)^{*}-\theta(f)\theta(\alpha(f))\theta(f)\theta(f)^{*}\|$
$\displaystyle\leq\|\theta(f)\|\|2\theta_{\alpha}(f)^{*}\|\|\theta(f)\|\|\theta(f)^{*}\|+\|\theta(f\alpha(f)f)\|\|\theta(f)\|$
$\displaystyle\leq
2\|\theta\|^{3}\|f\|^{4}+\|\theta\|^{2}\|f\alpha(f)f\|\|f\|$
$\displaystyle\leq 2\|\theta\|^{3}\|f\|^{4}+\|\theta\|^{2}\|f\|^{4}.$

(Here the second inequality used $\|\theta_{\alpha}\|\leq 1$, and the last one
used the fact that $\alpha$ is a contraction.) Therefore $\|\theta\|^{4}\leq
2\|\theta\|^{3}+\|\theta\|^{2}$, whence $\|\theta\|\leq 1+\sqrt{2}$.

The sharpness of the constant $(1+\sqrt{2})$ follows immediately from the
example exhibited in [RS18, §3]. ∎

######  Remarks.

(i) The proof of Theorem 1.1 makes it clear that the result remains true if
$A$ is an arbitrary Banach algebra and $M_{n}({\mathbb{C}})$ is replaced by an
arbitrary C*-algebra.

(ii) As remarked in the introduction, in the example showing sharpness, we
have $\alpha(1)=-1$.

##  3\. Proof of Theorem 1.3

The proof that follows is based on arguments from [CP17].

###### Proof of Theorem 1.3.

Suppose that Conjecture 1.2 holds when $A=A({\mathbb{D}})$ and $n=n_{0}$. Let
$T\in M_{n_{0}}({\mathbb{C}})$. We shall prove that (4) holds for all
polynomials $p$.

Let $\Omega$ be a bounded convex open neighbourhood of $W(T)$ with smooth
boundary. Once again, we denote by $A(\Omega)$ the uniform algebra of
continuous functions on $\overline{\Omega}$ that are holomorphic on $\Omega$.

As is shown in [CP17, §2], the containment $W(T)\subset\Omega$ is reflected in
the fact that the operator-valued measure on $\partial\Omega$ used in defining
$h(T)$ for $h\in A(\Omega)$, namely

$\frac{1}{2\pi i}(\zeta I-T)^{-1}\,d\zeta,$

has positive real part. This quickly leads to the estimate

(5) $\|h(T)+(({\mathcal{C}}_{\Omega}\overline{h})(T))^{*}\|\leq
2\sup_{\partial\Omega}|h|\quad(h\in A(\Omega)),$

where ${\mathcal{C}}_{\Omega}\overline{h}$ denotes the Cauchy transform of
$\overline{h}$ relative to $\Omega$, defined by

$({\mathcal{C}}_{\Omega}\overline{h})(z):=\frac{1}{2\pi
i}\int_{\partial\Omega}\frac{\overline{h(\zeta)}}{\zeta-z}\,d\zeta\quad(z\in\Omega).$

We remark also that, since $\Omega$ is convex, the map
$h\mapsto{\mathcal{C}}_{\Omega}\overline{h}$ is a contraction of $A(\Omega)$
into itself (see [CP17, Lemma 2.1]).

By the Riemann mapping theorem, there exists a conformal mapping $\phi$ of
$\Omega$ onto the unit disk ${\mathbb{D}}$, and by Carathéodory’s theorem,
$\phi$ extends to a homeomorphism of $\overline{\Omega}$ onto
$\overline{{\mathbb{D}}}$. For $f\in A({\mathbb{D}})$, define

$\displaystyle\theta(f)$ $\displaystyle:=(f\circ\phi)(T),$
$\displaystyle\alpha(f)$
$\displaystyle:={\mathcal{C}}_{\Omega}(\overline{f\circ\phi})\circ\phi^{-1}.$

Then $\theta:A({\mathbb{D}})\to M_{n_{0}}({\mathbb{C}})$ is a continuous
homomorphism, and $\alpha:A({\mathbb{D}})\to A({\mathbb{D}})$ is a unital,
antilinear contraction. Further, by (5), for all $f\in A({\mathbb{D}})$,

$\|\theta(f)+(\theta(\alpha(f))^{*}\|=\|(f\circ\phi)(T)+{\mathcal{C}}_{\Omega}(\overline{f\circ\phi})(T)^{*}\|\leq
2\sup_{\partial\Omega}|f\circ\phi|=2\|f\|_{A({\mathbb{D}})},$

so $\|\theta_{\alpha}\|\leq 1$. Using the supposed truth of Conjecture 1.2, we
deduce that $\|\theta\|\leq 2$. In particular, if $p$ is a polynomial, then

$\|p(T)\|=\|\theta(p\circ\phi^{-1})\|\leq
2\|p\circ\phi^{-1}\|_{A({\mathbb{D}})}=2\sup_{\Omega}|p|.$

Finally, by taking a sequence $\Omega_{n}$ of smoothly bounded convex
neighbourhoods of $W(T)$ shrinking to $W(T)$, we deduce that

$\|p(T)\|\leq 2\sup_{W(T)}|p|,$

as required. ∎

##  4\. Proof of Theorem 1.4

Theorem 1.4 is a special case of the following, more general result.

######  Theorem 4.1.

Let $A$ be a uniform algebra and let $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism. If $\|\theta\|>1$, then, for all $\beta$ in the dual
of $A$,

$\|\theta+\beta I\|\geq\|\theta\|.$

The proof of Theorem 4.1 is based on the following lemma, due to Crouzeix,
Gilfeather and Holbrook [CGH14]. We write $\sigma(T)$ for the spectrum of $T$.

######  Lemma 4.2.

Let $T$ be an $n\times n$ matrix such that
$\sigma(T)\subset\overline{{\mathbb{D}}}$ and $\|T\|>1$. Suppose that
$\|\phi(T)\|\leq\|T\|$ for all automorphisms $\phi$ of ${\mathbb{D}}$. Let $x$
be a unit vector in ${\mathbb{C}}^{n}$ on which $T$ attains its norm. Then

(6) $\langle x,Tx\rangle=0.$

For completeness, we include a short proof.

###### Proof.

For $w\in{\mathbb{D}}$, let $\phi_{w}(z):=(z-w)/(1-\overline{w}z)$. Then

$\phi_{w}(T)=(T-wI)(I-\overline{w}T)^{-1}=T-wI+\overline{w}T^{2}+O(|w|^{2})\quad(w\to
0).$

By hypothesis $\|\phi_{w}(T)\|\leq\|T\|$, so it follows that

$\displaystyle\|T\|^{2}$
$\displaystyle\geq\langle\phi_{w}(T)x,\phi_{w}(T)x\rangle$
$\displaystyle=\|Tx\|^{2}-2\operatorname{Re}\bigl{(}w\langle
x,Tx\rangle\bigr{)}+2\operatorname{Re}\bigl{(}\overline{w}\langle
T^{2}x,Tx\rangle\bigr{)}+O(|w|^{2})$
$\displaystyle=\|T\|^{2}-2\operatorname{Re}\bigl{(}w\langle(I-T^{*}T)x,Tx\rangle\bigr{)}+O(|w|^{2}).$

Letting $w\to 0$, and noting that argument of $w$ is arbitrary, it follows
that

$\langle(I-T^{*}T)x,Tx\rangle=0.$

Since $T$ attains its norm at $x$, we have $T^{*}Tx=\|T\|^{2}x$, whence

$(1-\|T\|^{2})\langle x,Tx\rangle=0.$

Finally, since $\|T\|>1$, we conclude that (6) holds. ∎

######  Remark.

The ideas behind this lemma are developed further in [BG20, CGL18, RW20].

###### Proof of Theorem 4.1.

Let $(f_{k})$ be a sequence in $A$ such that $\|f_{k}\|=1$ and
$\|\theta(f_{k})\|\to\|\theta\|$. Replacing $(f_{k})$ by a subsequence, we can
suppose that $\theta(f_{k})\to T$ in $M_{n}({\mathbb{C}})$ and
$\beta(f_{k})\to\lambda$. We then have

$\|T\|=\lim_{k\to\infty}\|\theta(f_{k})\|=\|\theta\|,$

and

$\|T+\lambda
I\|=\lim_{k\to\infty}\|\theta(f_{k})+\beta(f_{k})I\|\leq\|\theta+\beta I\|.$

Also, since $\theta$ is a homomorphism, we have
$\sigma(\theta(f_{k}))\subset\sigma(f_{k})\subset\overline{{\mathbb{D}}}$ for
all $k$, and hence $\sigma(T)\subset\overline{{\mathbb{D}}}$.

Let $\phi$ be an automorphism of ${\mathbb{D}}$. Then $\phi\circ f_{k}\in A$
and $\|\phi\circ f_{k}\|\leq 1$, so $\|\theta(\phi\circ
f_{k})\|\leq\|\theta\|$. On the other hand, as $\theta$ is a homomorphism, we
have $\theta(\phi\circ f_{k})=\phi(\theta(f_{k}))\to\phi(T)$ as $k\to\infty$.
Therefore $\|\phi(T)\|\leq\|\theta\|=\|T\|$. Thus Lemma 4.2 applies to $T$.

Let $x\in{\mathbb{C}}^{n}$ be a vector such that $\|x\|=1$ and $\|Tx\|=\|T\|$.
By Lemma 4.2, if $\|T\|>1$, then $\langle x,Tx\rangle=0$, so

$\|T\|^{2}=\langle Tx,Tx\rangle=\langle(T+\lambda I)x,Tx\rangle\leq\|T+\lambda
I\|\|\|T\|\leq\|\theta+\beta I\|\|T\|.$

Hence $\|T\|\leq\|\theta+\beta I\|$. As $\|\theta\|=\|T\|$, we deduce that
$\|\theta\|\leq\|\theta+\beta I\|$. ∎

Let $\rho\geq 1$. We say that an operator $T$ on a Hilbert space
${\mathcal{H}}$ is in the class ${\mathcal{C}}_{\rho}$ if $T$ has a unitary
$\rho$-dilation, i.e., if there exists a Hilbert space ${\mathcal{K}}$
containing ${\mathcal{H}}$ as a subspace and a unitary operator $U$ on
${\mathcal{K}}$ such that

$T^{m}=\rho P_{\mathcal{H}}U^{m}|_{\mathcal{H}}\quad(m\geq 1),$

where $P_{\mathcal{H}}:{\mathcal{K}}\to{\mathcal{H}}$ denotes the orthogonal
projection of ${\mathcal{K}}$ onto ${\mathcal{H}}$. It is known that
${\mathcal{C}}_{1}$ is precisely the set of contractions on ${\mathcal{H}}$
[Na53], and that ${\mathcal{C}}_{2}$ is the set of operators whose numerical
range is contained in $\overline{{\mathbb{D}}}$ [NF66, Be65]. The following
result is due to Okubo and Ando [OA75].

######  Theorem 4.3.

For each $\rho\geq 1$, the closed unit disk is a $\rho$-spectral set for all
matrices in the class ${\mathcal{C}}_{\rho}$.

Theorem 4.1 permits us to give a new and simple proof of this result.

###### Proof of Theorem 4.3.

Fix $\rho\geq 1$ and let $T$ be an $n\times n$ matrix in the class
${\mathcal{C}}_{\rho}$. Then there exists a unitary $\rho$-dilation $U$ of
$T$. Let $p$ be a polynomial. Writing $q:=p-p(0)$, we then have

$\displaystyle p(T)$ $\displaystyle=q(T)+p(0)I$ $\displaystyle=\rho
P_{{\mathbb{C}}^{n}}q(U)|_{{\mathbb{C}}^{n}}+p(0)I$ $\displaystyle=\rho
P_{{\mathbb{C}}^{n}}p(U)|_{{\mathbb{C}}^{n}}+(1-\rho)p(0)I,$

whence

$\|p(T)+(\rho-1)p(0)I\|=\rho\|P_{{\mathbb{C}}^{n}}p(U)|_{{\mathbb{C}}^{n}}\|\leq\rho\|p(U)\|\leq\rho\sup_{\overline{{\mathbb{D}}}}|p|.$

Defining $\theta(p):=p(T)$ and $\beta(p):=(\rho-1)p(0)$, we see that both
$\beta$ and $\theta$ extend by continuity to the whole disk algebra
$A({\mathbb{D}})$ and that the extended maps satisfy $\|\theta+\beta
I\|\leq\rho$. By Theorem 4.1, it follows that $\|\theta\|\leq\rho$, in other
words, that $\overline{{\mathbb{D}}}$ is a $\rho$-spectral set for $T$. ∎

######  Remarks.

(i) Taking $\rho=2$ in Theorem 4.1, we recover the fact that
$\overline{{\mathbb{D}}}$ is a $2$-spectral set for all matrices $T$ whose
numerical range is contained in $\overline{{\mathbb{D}}}$, i.e., that
Crouzeix’s conjecture holds for $\overline{{\mathbb{D}}}$.

(ii) Okubo and Ando established their result for general Hilbert space
operators. Although Theorem 4.1 was proved for matrices, a standard argument
recovers the more general statement of Okubo and Ando; see for instance the
proof of Theorem 2 in [Cr07].

##  5\. Proof of Theorem 1.5

Theorem 1.5 is a consequence of another result, which may be of interest in
its own right.

######  Theorem 5.1.

Let $A$ be a uniform algebra, let $\theta:A\to M_{n}({\mathbb{C}})$ be a
continuous homomorphism, and let $\alpha:A\to A$ be a unital antilinear
contraction. Assume that $\theta_{\alpha}$, defined by (1), satisfies
$\|\theta_{\alpha}\|\leq 1$. If $p$ is a self-adjoint projection in $A$, then
$\theta(p)$ is a self-adjoint projection in $M_{n}({\mathbb{C}})$.

The proof passes via the following lemma. Recall that $\sigma(T)$ and $W(T)$
denote the spectrum and numerical range of $T$ respectively. Also, we write
$\operatorname{conv}(S)$ for the convex hull of $S$.

######  Lemma 5.2.

Under the hypotheses of Theorem 5.1,

$W(\theta_{\alpha}(f))\subset\operatorname{conv}(\sigma(f))\quad(f\in A).$

###### Proof.

Let $w\in{\mathbb{C}}\setminus\operatorname{conv}(\sigma(f))$. Then there
exists a closed disk $\overline{D}(c,r)$ such that
$\sigma(f)\subset\overline{D}(c,r)$ and $w\notin\overline{D}(c,r)$. The
condition $\sigma(f)\subset\overline{D}(c,r)$ is equivalent to $\|f-c1\|\leq
r$. The assumptions on $\theta$ and $\alpha$ imply that $\theta_{\alpha}$ is a
unital linear contraction. Therefore $\|\theta_{\alpha}(f)-cI\|\leq r$. It
follows that

$|\langle(\theta_{\alpha}(f)-cI)x,x\rangle|\leq
r\quad(x\in{\mathbb{C}}^{n},~{}\|x\|=1),$

whence $W(\theta_{\alpha}(f))\subset\overline{D}(c,r)$. In particular,
$w\notin W(\theta_{\alpha}(f))$. ∎

###### Proof of Theorem 5.1.

Since $\theta(A)$ is a commutative subalgebra of $M_{n}({\mathbb{C}})$, there
exists a unitary matrix $U\in M_{n}({\mathbb{C}})$ such that $U^{*}\theta(A)U$
is an algebra of upper triangular matrices (see e.g. [Ne67, Theorem 1]).
Defining $\widetilde{\theta}:A\to M_{n}({\mathbb{C}})$ by
$\widetilde{\theta}(f):=U^{*}\theta(f)U$ gives a continuous homomorphism that
still satisfies $\|\widetilde{\theta}_{\alpha}\|\leq 1$, and if the conclusion
of the theorem holds for $\widetilde{\theta}$, then it also holds for
$\theta$. Thus, replacing $\theta$ by $\widetilde{\theta}$ at the outset, we
may as well assume that $\theta(A)$ consists of upper triangular matrices.

Let $p$ be a self-adjoint projection in $A$ and set $u:=2p-1$. Then $u$ is
self-adjoint and $u^{2}=1$. Thus $\theta(u)$ is an upper triangular matrix
such that $\theta(u)^{2}=I$, which forces it to have the form

$\theta(u)=\begin{pmatrix}I&X\\\ 0&-I\end{pmatrix},$

for some rectangular matrix $X$. Since $u$ is self-adjoint, Lemma 5.2 implies
that $\theta_{\alpha}(u)$ is self-adjoint, which entails that

$\theta(\alpha(u))=\begin{pmatrix}D&X\\\ 0&-E\end{pmatrix},$

for some real diagonal matrices $D$ and $E$. The fact that $\theta(u)$ and
$\theta(\alpha(u))$ commute implies that $DX-X=X-XE$, whence

(7) $2X-DX-XE=0.$

Also, since $\|\theta_{\alpha}(u)\|\leq\|u\|=1$, we must have

(8) $\left\|\begin{pmatrix}(I+D)/2&X/2\\\
X^{*}/2&-(I+E)/2\end{pmatrix}\right\|\leq 1.$

Suppose, if possible, that $X$ has a non-zero entry $x_{jk}$. Condition (7)
gives $(2-d_{j}-e_{k})x_{jk}=0$, where $d_{j}$ and $e_{k}$ are the $j$-th and
$k$-th entries of $D,E$ respectively. Therefore $d_{j}+e_{k}=2$. On the other
hand, condition (8) implies that $|1+d_{j}|/2\leq 1$ and $|1+e_{k}|/2\leq 1$.
Therefore $d_{j}=e_{k}=1$. But then the matrix in the left-hand-side of (8)
contains a $2\times 2$ sub-matrix of the form

$\begin{pmatrix}1&x_{jk}/2\\\ \overline{x}_{jk}/2&-1\end{pmatrix}$

which must still have norm at most $1$. This forces $x_{jk}=0$, a
contradiction. We conclude that $X=0$. Therefore $\theta(u)$ is a diagonal
matrix. Therefore so too is $\theta(p)$. Clearly $\theta(p)^{2}=\theta(p)$, so
$\theta(p)$ is a self-adjoint projection, as claimed. ∎

###### Proof of Theorem 1.5.

Suppose first that $A$ is a commutative von Neumann algebra, so
$A=L^{\infty}(\Omega,\mu)$ for some measure space $(\Omega,\mu)$. Let $f\in A$
with $\|f\|=1$. Let $\\{D_{1},\dots,D_{r}\\}$ be a partition of
$\overline{{\mathbb{D}}}$ into Borel sets of diameter at most $\epsilon$. Pick
an element $d_{j}\in D_{j}$ for each $j$ and set $g:=\sum_{1}^{r}d_{j}p_{j}$,
where $p_{j}:=1_{f^{-1}(D_{j})}$. Then $\|f-g\|_{\infty}\leq\epsilon$. Since
the $p_{j}$ are orthogonal self-adjoint projections, so are the
$\theta(p_{j})$, by Theorem 5.1. Therefore

$\|\theta(g)\|=\Bigl{\|}\sum_{1}^{r}d_{j}\theta(p_{j})\Bigr{\|}=\max_{j}|d_{j}|\leq
1.$

It follows that $\|\theta(f)\|\leq 1+\|\theta\|\epsilon$. As $\epsilon$ is
arbitrary, $\|\theta(f)\|\leq 1$. Therefore $\|\theta\|\leq 1$. Finally, as
$\theta(1)=I$, we in fact have $\|\theta\|=1$. This implies in particular that
$\theta$ is positive [paulsen2002, Proposition 2.11], so that $\theta$ is a
$*$-homomorphism.

Now suppose that $A$ is a general commutative C*-algebra. Then its double dual
$A^{\prime\prime}$ is a commutative von Neumann algebra [BLM2004, Theorem
A.5.6]. Since $A$ is weak*-dense in $A^{\prime\prime}$ and the dualized maps
$\theta^{\prime\prime}:A^{\prime\prime}\to M_{n}({\mathbb{C}})$ and
$\alpha^{\prime\prime}:A^{\prime\prime}\to A^{\prime\prime}$ are
weak*-continuous, we have $\|\alpha^{\prime\prime}\|=\|\alpha\|=1$ and
$\|\theta^{\prime\prime}_{\alpha^{\prime\prime}}\|=\|\theta_{\alpha}\|\leq 2$.
Also $\theta^{\prime\prime}$ is still a homomorphism. By the first part of the
proof, we infer that $\theta^{\prime\prime}$ is a $*$-homomorphism, and thus
so is $\theta$. ∎

##  6\. A completely bounded version of the conjecture

In this section, we discuss a completely bounded version of the preceding
ideas, thus framing them within the context of operator algebras. As we will
see below, this allows dilation techniques to enter the picture and to shed
new light on the problem.

Throughout this section, $B({\mathcal{H}})$ denotes the $C^{*}$-algebra of
bounded linear operators on some Hilbert space ${\mathcal{H}}$. An operator
algebra is simply a _unital_ subalgebra $A\subset B({\mathcal{H}})$. While it
is possible to define these objects without referring to a particular choice
of representation on Hilbert space [paulsen2002, Corollary 16.7], for our
purposes the previous concrete description suffices. We remark that
subalgebras of $C^{*}$-algebras are operator algebras, and in particular so
are uniform algebras.

Given a natural number $n\geq 1$, we denote by $M_{n}(A)$ the algebra of
$n\times n$ matrices with entries from $A$, which we view as a subalgebra of
bounded linear operators acting on
${\mathcal{H}}^{(n)}={\mathcal{H}}\oplus{\mathcal{H}}\oplus\ldots\oplus{\mathcal{H}}$.
In particular, $M_{n}(A)$ is endowed with a norm under this identification.
Given a map $\phi:A\to B({\mathcal{K}})$, for each natural number $n\geq 1$,
we may define a map $\phi^{(n)}:M_{n}(A)\to B({\mathcal{K}}^{(n)})$ as

$\phi^{(n)}([a_{ij}])=[\phi(a_{ij})]\quad\bigl{(}[a_{ij}]\in
M_{n}(A)\bigr{)}.$

It is readily verified that $\phi^{(n)}$ is linear (respectively, a
homomorphism) whenever $\phi$ is.

If $\phi$ is linear (or anti-linear), we say that $\phi$ is _completely
bounded_ if the quantity

$\|\phi\|_{cb}=\sup_{n}\|\phi^{(n)}\|$

is finite. Furthermore, we say that $\phi$ is _completely contractive_ if
$\|\phi\|_{cb}\leq 1$.

We can now state the refined version of the main conjecture, which is the
central topic of this section.

######  Conjecture 6.1.

Let $A$ be an operator algebra, let $\theta:A\to M_{n}({\mathbb{C}})$ be a
unital completely bounded homomorphism, and let $\alpha:A\to A$ be a unital
antilinear complete contraction. Define a linear map $\theta_{\alpha}:A\to
M_{n}({\mathbb{C}})$ by (1). If $\|\theta_{\alpha}\|_{cb}\leq 1$, then
$\|\theta\|_{cb}\leq 2$.

We start by showing that the estimate $\|\theta\|_{cb}\leq 1+\sqrt{2}$ always
holds, mirroring the previous setting.

######  Proposition 6.2.

Let $A$ be an operator algebra. Let $\theta,\alpha$ and $\theta_{\alpha}$ be
as in Conjecture 6.1. Then $\|\theta\|_{cb}\leq 1+\sqrt{2}$.

###### Proof.

Fix a positive integer $n$. Remark (i) following the proof of Theorem 1.1
shows that the argument therein applies to the maps $\theta^{(n)}$ and
$\alpha^{(n)}$. Hence $\|\theta^{(n)}\|\leq 2$ for every $n\geq 1$, which is
equivalent to $\|\theta\|_{cb}\leq 2$. ∎

We note that $*$-homomorphisms on $C^{*}$-algebras are always completely
contractive, so that Theorem 1.5 shows that Conjecture 6.1 is verified for
commutative $C^{*}$-algebras. In contrast, we do not know if the proof of
Theorem 1.4 can be adapted to show that Conjecture 6.1 holds in the case of
the disk algebra when $\alpha$ takes values in the scalar multiples of the
identity.

In trying to verify Conjecture 6.1 for the disc algebra, the following well-
known basic principle is useful.

######  Lemma 6.3.

Let $\theta:A({\mathbb{D}})\to B({\mathcal{H}})$ be a unital completely
bounded homomorphism and let $\psi$ be an automorphism of ${\mathbb{D}}$.
Then, $\|\theta\|_{cb}\leq 2$ if and only if there is an invertible operator
$S\in B({\mathcal{H}})$ with the property that $S\theta(\psi)S^{-1}$ is a
contraction and that $\|S\|\|S^{-1}\|\leq 2.$

###### Proof.

Assuming that $\|\theta\|_{cb}\leq 2$, the existence of $S$ follows at once
from Paulsen’s similarity theorem [paulsen1984PAMS]. Conversely, given an
operator $S$ as in the statement, define $\theta_{S}:A({\mathbb{D}})\to
B({\mathcal{H}})$ as

$\theta_{S}(f)=S\theta(f)S^{-1}\quad(f\in A({\mathbb{D}})).$

For a matrix of polynomials $[p_{ij}]\in M_{n}(A({\mathbb{D}}))$, we see that

$\displaystyle\theta^{(n)}_{S}([p_{ij}\circ\psi])$
$\displaystyle=[S\theta(p_{ij}\circ\psi)S^{-1}]$
$\displaystyle=[p_{ij}(S\theta(\psi)S^{-1})]$

so von Neumann’s inequality implies that

$\|\theta^{(n)}_{S}([p_{ij}\circ\psi])\|\leq\|[p_{ij}]\|_{M_{n}(A({\mathbb{D}}))}=\|[p_{ij}\circ\psi]\|_{M_{n}(A({\mathbb{D}}))}.$

We conclude that $\theta_{S}$ is completely contractive on the dense
subalgebra

$\\{p\circ\psi:p\text{ polynomial}\\},$

and thus on the entire algebra $A({\mathbb{D}})$. Consequently,

$\|\theta\|_{cb}\leq\|S\|\|S^{-1}\|\|\theta_{S}\|_{cb}\leq 2.\qed$

We do not know of a good description of the class of operators $T\in
B({\mathcal{H}})$ for which there is an invertible operator $S\in
B({\mathcal{H}})$ with the property that $STS^{-1}$ is a contraction and that
$\|S\|\|S^{-1}\|\leq 2.$ A sufficient condition for this property to hold was
discovered in [OA75].

We now give an instance where the previous lemma can be exploited.

######  Proposition 6.4.

Let $\theta:A({\mathbb{D}})\to M_{2}({\mathbb{C}})$ be a unital bounded
homomorphism and let $\alpha:A({\mathbb{D}})\to A({\mathbb{D}})$ be a unital
antilinear map. Define $\theta_{\alpha}:A({\mathbb{D}})\to
M_{2}({\mathbb{C}})$ by (1). Assume that $\|\theta_{\alpha}\|\leq 1$ and that
the spectrum of $\theta(z)$ consists of a single point in ${\mathbb{D}}$. Then
$\|\theta\|_{cb}\leq 2$.

###### Proof.

By assumption, there is a $\lambda\in{\mathbb{D}}$ such that
$\sigma(\theta(z))=\\{\lambda\\}$. Let $\psi\in A({\mathbb{D}})$ be defined by

$\psi(z):=\frac{z-\lambda}{1-\overline{\lambda}z}\quad(z\in{\mathbb{D}}).$

Then the spectrum of $\theta(\psi)$ is $\\{0\\}$. Since $\theta(\psi)$ and
$\theta(\alpha(\psi))$ are commuting matrices, we can assume that they are
both upper triangular, and that the diagonal of $\theta(\psi)$ consists
entirely of zeros. Hence there is a $\delta\in{\mathbb{C}}$ along with
$a,b,c\in{\mathbb{C}}$ such that

$\theta(\psi)=\begin{pmatrix}0&\delta\\\
0&0\end{pmatrix}\quad\text{and}\quad\theta(\alpha(\psi))=\begin{pmatrix}a&b\\\
0&c\end{pmatrix}.$

Then we have

$2\theta_{\alpha}(\psi)=\begin{pmatrix}\overline{a}&\delta\\\
\overline{b}&\overline{c}\end{pmatrix}.$

Since $\|\theta_{\alpha}(\psi)\|\leq\|\theta_{\alpha}\|\leq 1$, this last
matrix has norm at most $2$, and in particular $|\delta|\leq 2$. Let

$S:=\begin{pmatrix}1/2&0\\\ 0&1\end{pmatrix}.$

Then

$S\theta(\psi)S^{-1}=\begin{pmatrix}0&\delta/2\\\ 0&0\end{pmatrix},$

so $S\theta(\psi)S^{-1}$ is a contraction and $\|S\|\|S^{-1}\|\leq 2$. Lemma
6.3 finally implies that $\|\theta\|_{cb}\leq 2$. ∎

We do not know how to handle the case where $\theta(z)$ in the previous
theorem is allowed to have arbitrary spectrum. Even in the concrete case of
the usual Crouzeix conjecture, the situation gets much more complicated; see
[BCD06, Theorem 5.2].

As mentioned earlier, the setting of Conjecture 6.1 allows for the use of
dilation techniques; this is explored in the next result.

######  Theorem 6.5.

Let $\theta:A({\mathbb{D}})\to B({\mathcal{H}})$ be a unital bounded
homomorphism and let $\alpha:A({\mathbb{D}})\to A({\mathbb{D}})$ be a unital
antilinear map. Define $\theta_{\alpha}:A({\mathbb{D}})\to B({\mathcal{H}})$
by

$\theta_{\alpha}(f):=\frac{1}{2}(\theta(f)+\theta(\alpha(f))^{*})\quad(f\in
A({\mathbb{D}})).$

Consider the following statements:

  1. (i)

The map $\theta_{\alpha}$ is completely contractive.

  2. (ii)

There is a unitary operator $U$ on a Hilbert space ${\mathcal{K}}$ containing
${\mathcal{H}}$ such that

$\theta(f)+\theta(\alpha(f))^{*}=2P_{{\mathcal{H}}}f(U)|_{{\mathcal{H}}}\quad(f\in
A({\mathbb{D}})).$

  3. (iii)

The operator $\theta(z)$ admits a unitary $2$-dilation.

  4. (iv)

The numerical range of $\theta(z)$ is contained in the closed unit disk.

  5. (v)

The estimate $\|\theta\|_{cb}\leq 2$ holds.

Then we have

$(i)\Rightarrow(ii)\quad\text{and}\quad(iii)\Leftrightarrow(iv)\Rightarrow(v).$

If we assume in addition that $z^{m}\in\ker(\theta\circ\alpha)$ for every
positive integer $m$, then we have

$(i)\Rightarrow(ii)\Rightarrow(iii)\Leftrightarrow(iv)\Rightarrow(v).$

###### Proof.

Assume that (i) holds. Recall that $A({\mathbb{D}})\subset C({\mathbb{T}})$.
By combining Arveson’s extension theorem [paulsen2002, Corollary 7.6] with
Stinespring’s dilation theorem [paulsen2002, Theorem 4.1], we obtain a Hilbert
space ${\mathcal{K}}$ containing ${\mathcal{H}}$ and unital $*$-homomorphism
$\pi:C({\mathbb{T}})\to B({\mathcal{K}})$ such that

$\frac{1}{2}\bigl{(}\theta(f)+\theta(\alpha(f))^{*}\bigr{)}=\theta_{\alpha}(f)=P_{{\mathcal{H}}}\pi(f)|_{{\mathcal{H}}}\quad(f\in
A({\mathbb{D}})).$

Choosing $U:=\pi(z)$, which is unitary, we see that (ii) holds. Furthermore,
we see that (ii) implies (iii) provided that $\theta(\alpha(z^{m}))=0$ for
every $m\geq 1$.

The equivalence of (iii) and (iv) is found in [nagy2010, Theorem I.11.2].

The fact that (iii) implies (v) follows from [OA75] along with Lemma 6.3. ∎

We conclude by remarking that, when $\alpha(f)=\overline{f(0)}I$, then (i),
(ii), (iii) and (iv) above are equivalent; see [paulsen2002, Theorem 3.15].

## References