*0.5in0.5in plain[]0 mystyle[] [0][][Pólya’s conjecture for Euclidean balls]N. Filonov, M. Levitin, I. Polterovich, and D. A. Sher0

# Pólya’s conjecture for Euclidean balls111A Mathematica script used for a
computer-assisted part of the paper is available for download at
https://michaellevitin.net/polya.html 222MSC(2020): Primary 35P15. Secondary
35P20, 33C10, 11P21.333Keywords:  Laplacian, eigenvalues, Weyl’s law, lattice
points, Bessel functions, Bessel phase functions, zeros of Bessel functions
and their derivaives

Nikolay Filonov N. F.: St. Petersburg Department of Steklov Institute of
Mathematics of RAS, Fontanka 27, 191023, St.Petersburg, Russia; St. Petersburg
State University, University emb. 7/9, 199034, St.Petersburg, Russia;
<EMAIL_ADDRESS>Michael Levitin M. L.: Department of Mathematics and
Statistics, University of Reading, Pepper Lane, Whiteknights, Reading RG6 6AX,
UK<EMAIL_ADDRESS>https://www.michaellevitin.net Iosif Polterovich
I. P.: Département de mathématiques et de statistique, Université de Montréal,
CP 6128 succ Centre-Ville, Montréal QC H3C 3J7, Canada;
<EMAIL_ADDRESS>https://www.dms.umontreal.ca/ĩossif David A. Sher D.
A. S.: Department of Mathematical Sciences, DePaul University, 2320 N. Kenmore
Ave, 60614, Chicago, IL, USA<EMAIL_ADDRESS>
(Revised version, 2 May 2023
to appear in Invent. Math.)

###### Abstract

The celebrated Pólya’s conjecture (1954) in spectral geometry states that the
eigenvalue counting functions of the Dirichlet and Neumann Laplacian on a
bounded Euclidean domain can be estimated from above and below, respectively,
by the leading term of Weyl’s asymptotics. Pólya’s conjecture is known to be
true for domains which tile Euclidean space, and, in addition, for some
special domains in higher dimensions. In this paper, we prove Pólya’s
conjecture for the disk, making it the first non-tiling planar domain for
which the conjecture is verified. We also confirm Pólya’s conjecture for
arbitrary planar sectors, and, in the Dirichlet case, for balls of any
dimension. Along the way, we develop the known links between the spectral
problems in the disk and certain lattice counting problems. A key novel
ingredient is the observation, made in recent work of the last named author,
that the corresponding eigenvalue and lattice counting functions are related
not only asymptotically, but in fact satisfy certain uniform bounds. Our
proofs are purely analytic, except for a rigorous computer-assisted argument
needed to cover the short interval of values of the spectral parameter in the
case of the Neumann problem in the disk.

###### Contents

  1. 1 Weyl’s law and Pólya’s conjecture
  2. 2 Dirichlet and Neumann eigenvalues of the ball and lattice counting problems
  3. 3 Proof of Theorem 2.3
  4. 4 From the weighted lattice point count towards Pólya’s conjecture
  5. 5 Proof of Theorem 4.1
  6. 6 Proof of Theorem 4.3
  7. 7 Proof of Theorem 4.2
  8. 8 Closing the gap: the proof of Theorem 4.4
  9. 9 Proof of Theorem 1.12
  10. References

## §1. Weyl’s law and Pólya’s conjecture

Let $\Omega\subset\mathbb{R}^{d}$ be a bounded domain. Consider the Dirichlet
eigenvalue problem for the Laplacian

$-\Delta:=-\sum_{j=1}^{d}\frac{\partial^{2}}{\partial x_{j}^{2}}$

in $\Omega$:

$\begin{split}-\Delta u=\lambda u\qquad&\text{in }\Omega,\\\
u=0\qquad&\text{on }\partial\Omega.\end{split}$ (1.1)

It is well known that the spectrum of (1.1) is discrete and consists of
isolated eigenvalues of finite multiplicity accumulating to $+\infty$,

$0<\lambda_{1}(\Omega)\leq\lambda_{2}(\Omega)\leq\dots\leq\lambda_{n}(\Omega)\leq\dots,$

which we enumerate with account of multiplicities.

Similarly, assuming additionally that $\partial\Omega$ is Lipschitz, consider
the Neumann eigenvalue problem

$\begin{split}-\Delta u=\mu u\qquad&\text{in }\Omega,\\\
\partial_{n}u=0\qquad&\text{on }\partial\Omega,\end{split}$ (1.2)

where $\partial_{n}u=\left.\langle\nabla u,n\rangle\right|_{\partial\Omega}$
denotes the normal derivative of $u$ with respect to the exterior unit normal
$n$ on the boundary. The spectrum of (1.2) again consists of isolated
eigenvalues of finite multiplicity accumulating to $+\infty$,

$0=\mu_{1}(\Omega)\leq\mu_{2}(\Omega)\leq\dots\leq\mu_{n}(\Omega)\leq\dots,$

enumerated with account of multiplicities.

Let, for $\lambda\in\mathbb{R}$,

$\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda):=\\#\left\\{n:\lambda_{n}(\Omega)\leq\lambda^{2}\right\\}\qquad\text{and}\qquad\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda):=\\#\left\\{n:\mu_{n}(\Omega)\leq\lambda^{2}\right\\}$

denote the _counting functions_ 444Strictly speaking, we are counting the
number of eigenvalues less than or equal to a given $\lambda^{2}$, but such
normalisation will be convenient to us throughout. of the Dirichlet and
Neumann eigenvalue problems on $\Omega$.555One can also define the counting
functions using strict inequalities; this does not affect any of the results
below. It follows from the variational principles for (1.1) and (1.2) that

$\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)\leq\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda)$

for any $\lambda\geq 0$.666This in fact can be improved to
$\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)+1\leq\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda)$,
see [Fri91] and [Fil04].

Under the assumptions stated above, the leading term asymptotics of the
counting functions is given by Weyl’s law [Wey11],

$\mathcal{N}_{\Omega}(\lambda)=C_{d}|\Omega|_{d}\lambda^{d}+R(\lambda),$ (1.3)

where $\mathcal{N}_{\Omega}(\lambda)$ denotes either
$\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)$ or
$\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda)$, $|\cdot|_{d}$ denotes the
$d$-dimensional volume, $R(\lambda)=o\left(\lambda^{d}\right)$ as
$\lambda\to+\infty$, and

$C_{d}:=\frac{1}{(4\pi)^{\frac{d}{2}}\Gamma\left(\frac{d}{2}+1\right)}$

is the so-called _Weyl constant_. We refer to [SafVas97] for a historical
review, as well as numerous generalisations and improvements.

H. Weyl himself conjectured [Wey12] a sharper version of (1.3) taking into
account the boundary conditions: for $\Omega\subset\mathbb{R}^{d}$ with a
piecewise smooth boundary,

$\mathcal{N}_{\Omega}(\lambda)=C_{d}|\Omega|_{d}\lambda^{d}\pm
C_{\mathrm{b},d}|\partial\Omega|_{d-1}\lambda^{d-1}+o\left(\lambda^{d-1}\right)\qquad\text{as
}\lambda\to+\infty,$ (1.4)

where the minus sign is taken for the Dirichlet boundary conditions and the
plus sign for the Neumann ones, and

$C_{\mathrm{b},d}:=\frac{1}{2^{d+1}\pi^{\frac{d-1}{2}}\Gamma\left(\frac{d+1}{2}\right)}.$

We note that for planar domains (1.4) takes the particularly simple form

$\mathcal{N}_{\Omega}(\lambda)=\frac{\operatorname{Area}(\Omega)}{4\pi}\lambda^{2}\pm\frac{\operatorname{Length}(\partial\Omega)}{4\pi}\lambda+o\left(\lambda\right).$
(1.5)

The two-term Weyl’s law (1.4) remains open in full generality. It has been
proved by V. Ivrii [Ivr80] under the condition that the set of periodic
billiard trajectories in $\Omega$ has measure zero. While this condition is
conjectured to be satisfied for all Euclidean domains, it has been verified
only for a few classes, such as convex analytic domains and polygons, see
[SafVas97] and references therein. Specifically for a disk, it was proved by
N. Kuznetsov and B. Fedosov in [KuzFed65].

Assuming that the two-term Weyl’s asymptotics (1.4) holds for a domain
$\Omega\subset\mathbb{R}^{d}$, we immediately obtain that for $\lambda$ above
some _sufficiently large but unspecified_ value $\Lambda_{1}$ we have

$\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)\leq
C_{d}|\Omega|_{d}\lambda^{d}\leq\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda).$
(1.6)

We refer also to [Mel80] for results of the same kind in the Riemannian
setting.

In 1954, G. Pólya [Pól54] conjectured that the inequalities (1.6) hold for
_all_ $\lambda\geq 0$.777In fact, Pólya’s original conjecture was only for
planar domains, and in a slightly different form. He later proved this
conjecture in [Pól61] for _tiling domains_ $\Omega$: that is, domains such
that $\mathbb{R}^{d}$ can be covered, up to a set of measure zero, by a
disjoint union of copies of $\Omega$. In fact, in the Neumann case, some
additional assumptions were imposed in [Pól61] that have been removed in
[Kel66]. It has been also shown that Pólya’s conjecture in the Dirichlet case
holds for a Cartesian product
$\Omega=\Omega_{1}\times\Omega_{2}\subset\mathbb{R}^{d_{1}+d_{2}}$ if it holds
for $\Omega_{1}\subset\mathbb{R}^{d_{1}}$ with $d_{1}\geq 2$, and
$\Omega_{2}\subset\mathbb{R}^{d_{2}}$ is bounded, see [Lap97, Theorem 2.8].
For general domains, somewhat weakened versions of (1.6) are known to hold as
a consequence of the so-called _Berezin–Li–Yau_ inequalities: we have

$\left(\frac{d}{d+2}\right)^{d/2}\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)\leq
C_{d}|\Omega|_{d}\lambda^{d}\leq\frac{d+2}{2}\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda)$

for all $\lambda\geq 0$, see [LiYau83], [Krö92], and [Lap97]. We refer also to
[Lin17], [KLS19], [FLP21], and [FreSal22] for some recent results on Pólya’s
conjecture and further interesting links to other problems in spectral
geometry.

######  Remark 1.1.

Pólya’s conjecture (1.6) can be equivalently restated as the inequalities for
the eigenvalues (instead of the counting functions),

$\mu_{n+1}(\Omega)\leq\left(C_{d}|\Omega|_{d}\right)^{-\frac{2}{d}}n^{\frac{2}{d}}\leq\lambda_{n}(\Omega)$
(1.7)

for all $n\geq 1$. It is known that inequalities (1.7) hold for any domain in
any dimension for $n=1,2$. In particular, for $n=1$ this follows from the
celebrated Faber–Krahn and Szegő–Weinberger inequalities, and for $n=2$ in the
Dirichlet case from the Krahn–Szego inequality, see [Hen06]. For $n=2$ in the
Neumann case, we refer to [GNP09], [BucHen19]. These are the only eigenvalues
for which it is known in full generality. We refer also to [Fre19] for further
results on the validity of the Dirichlet Pólya’s conjecture for low
eigenvalues in higher dimensions. ∎

Remarkably, since balls do not tile the space, Pólya’s conjecture has so far
remained open for Euclidean balls, including planar disks.888As stated in
[Lap12, p. 638]: “Remarkably this conjecture still remains open even for such
a simple domain as the disc, where the eigenvalues of the Dirichlet Laplacians
could be calculated via the roots of Bessel functions.” See also [FLW09, p.
1366] and [Lau12, p. 66]. Although all the eigenvalues of the Dirichlet and
Neumann Laplacians on the unit disk are explicitly known in terms of zeros of
the Bessel functions or their derivatives, see §2 below, in each case the
spectrum is given by a _two-parametric_ family, and rearranging it into a
single monotone sequence appears to be an unfeasible task.

Let $\mathbb{B}^{d}\subset\mathbb{R}^{d}$ be the $d$-dimensional unit ball.
Then
$|\mathbb{B}^{d}|_{d}=\frac{\pi^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$.
Therefore the leading Weyl’s term in (1.3) for $\mathbb{B}^{d}$ becomes

$W_{d}(\lambda):=C_{d}|\mathbb{B}^{d}|_{d}\lambda^{d}=w_{d}\lambda^{d},\qquad
w_{d}=\frac{1}{2^{d}\left(\Gamma\left(\frac{d}{2}+1\right)\right)^{2}},$ (1.8)

in particular

$W_{2}(\lambda)=\frac{\lambda^{2}}{4}\qquad\text{and}\qquad
W_{3}(\lambda)=\frac{2\lambda^{3}}{9\pi}.$

The main results of this paper address the validity of Pólya’s conjecture for
disks and balls. Namely, we prove the following results.

######  Theorem 1.2.

The Dirichlet Pólya’s conjecture for the unit ball holds in any dimension
$d\geq 2$, that is we have

$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)<W_{d}(\lambda)$

for all $\lambda>0$.

Our results in the Neumann case are restricted to the case $d=2$. Higher-
dimensional Neumann problems are harder, and we intend to treat them in a
subsequent paper.

We first state

######  Lemma 1.3.

The Neumann Pólya’s conjecture for $\mathbb{D}=\mathbb{B}^{2}$ is valid for
all $\lambda\in\left[0,\Lambda_{0}\right]$, where

$\Lambda_{0}:=2\sqrt{3}.$

###### Proof.

Taking the span of $\\{1,x,y\\}$ as a test space in the Rayleigh quotient for
the Neumann Laplacian on $\mathbb{D}$ gives $\mu_{3}(\mathbb{D})\leq 4$.
Therefore,

$\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda)\quad\geq\quad\begin{cases}1,&\quad\lambda\in[0,2),\\\
3,&\quad\lambda\geq
2,\end{cases}\quad\geq\quad\frac{\lambda^{2}}{4}\qquad\text{for
}\lambda\in[0,\Lambda_{0}].$

∎

We then prove

######  Theorem 1.4.

The Neumann Pólya’s conjecture for the unit disk holds for all

$\lambda\geq\Lambda_{1}:=\frac{6\pi}{3\pi-8}.$ (1.9)

We note that $\Lambda_{0}>3$ and $\Lambda_{1}<14$, so we already have the
validity of the Neumann Pólya’s conjecture for the disk for all $\lambda$
outside the interval $\left(3,14\right)$.

######  Theorem 1.5.

The Neumann Pólya’s conjecture for the unit disk holds for all
$\lambda\in[3,14]$.

The proof of Theorem 1.5 is _rigorous_ but _computer-assisted_. More
specifically, it is based on a realisation of an algorithm which satisfies two
fundamental principles.

Principle 1.

The algorithm should complete in a finite number of steps.

Principle 2.

The algorithm should operate only with integer or rational numbers, thus
avoiding _any_ use of floating-point arithmetic and _any_ rounding errors.

The combination of Lemma 1.3 and Theorems 1.4 and 1.5 ensures that the Neumann
Pólya conjecture for the disk is valid for all $\lambda>0$, that is we have

######  Corollary 1.6.

$\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda)>\frac{\lambda^{2}}{4}$ for all
$\lambda>0$.

######  Remark 1.7.

Since Pólya’s conjecture is scale-invariant, its validity for a unit ball
immediately implies that it is valid for any ball of the same dimension. ∎

We additionally have the following generalisation of Pólya’s result for tiling
domains: we show that Pólya’s conjecture holds not only for domains which tile
Euclidean space, but also for domains which tile another domain for which it
is known to be true.

######  Theorem 1.8.

Let $\Omega\subset\mathbb{R}^{d}$ be a domain for which either the Dirichlet
or the Neumann Pólya’s conjecture holds, and let $\Omega^{\prime}$ be a domain
which tiles $\Omega$. Then the same Pólya’s conjecture also holds for
$\Omega^{\prime}$.

###### Proof.

Assume that $\Omega$ can be tiled by $\ell\geq 2$ congruent copies of
$\Omega^{\prime}$, so that $|\Omega|_{d}=\ell|\Omega^{\prime}|_{d}$. We have,
by bracketing and since the eigenvalues of all the congruent copies coincide
with those of $\Omega^{\prime}$,

$\ell\mathcal{N}^{\mathrm{D}}_{\Omega^{\prime}}(\lambda)\leq\mathcal{N}^{\mathrm{D}}_{\Omega}(\lambda)<\mathcal{N}^{\mathrm{N}}_{\Omega}(\lambda)\leq\ell\mathcal{N}^{\mathrm{N}}_{\Omega^{\prime}}(\lambda).$

Assuming now (1.6) for all $\lambda\geq 0$, we get

$\ell\mathcal{N}^{\mathrm{D}}_{\Omega^{\prime}}(\lambda)\leq
C_{d}|\Omega|_{d}\lambda^{d}=C_{d}\ell|\Omega^{\prime}|_{d}\lambda^{d}\leq\ell\mathcal{N}^{\mathrm{N}}_{\Omega^{\prime}}(\lambda),$

and the result follows by cancelling $\ell$. ∎

######  Remark 1.9.

If the inequalities in Pólya’s conjecture (1.6) for $\Omega$ are strict, they
are also strict for $\Omega^{\prime}$. ∎

Theorem 1.8 immediately implies the following

######  Corollary 1.10.

Let $\widetilde{\Omega}\subset\mathbb{S}^{d-1}$ be a spherical domain which
tiles $\mathbb{S}^{d-1}$. Then the Dirichlet Pólya’s conjecture holds for the
spherical cone in $\mathbb{R}^{d}$ with the base $\widetilde{\Omega}$ and the
vertex at the origin.

In the planar case we get

######  Corollary 1.11.

Pólya’s conjecture holds for any circular sector $S_{\alpha}$ with an aperture
$\alpha=\frac{2\pi}{\ell}$, where $\ell\in\\{2,3,\dots\\}$.

We refer also to [FreSal22] for an alternative proof of Corollary 1.11 for
sufficiently large (but unspecified) $\ell$.

We can in fact extend the result of Corollary 1.11 to arbitrary sectors.

######  Theorem 1.12.

Pólya’s conjecture holds for any circular sector $S_{\alpha}$ with an aperture
$\alpha\in(0,2\pi]$, that is

$\mathcal{N}^{\mathrm{D}}_{S_{\alpha}}(\lambda)<\frac{\alpha\lambda^{2}}{8\pi}<\mathcal{N}^{\mathrm{N}}_{S_{\alpha}}(\lambda)$

for all $\lambda>0$.

######  Remark 1.13.

The result of Theorem 1.12 in the case $\alpha=2\pi$ (the disk with the radial
slit) follows immediately from Theorems 1.2 ($d=2$) and 1.4 by
Dirichlet–Neumann bracketing. ∎

### Plan of the paper

In the next section we describe two lattice counting problems (2.6) and (2.7),
variants of which were originally introduced by N. Kuznetsov and B. Fedosov in
[KuzFed65], and which are closely linked to the Dirichlet and Neumann
eigenvalue counting problems in the ball. The key novel tool is Theorem 2.3,
originally obtained in part in [She22], which gives a uniform bound between
the eigenvalue and the lattice counting functions, as opposed to asymptotic
relations that were previously known. We provide an independent proof of this
result in §3. In §4 we state the results on the lattice counting functions
which are sufficient for proving Pólya’s conjecture for balls. The bulk of the
paper, §§5–8, is devoted to the proofs of these results. Theorem 1.12 is
proved in §9.

### Acknowledgements

The authors would like to thank L. Friedlander, R. Laugesen, Z. Rudnick, and
I. Wigman for useful discussions. We are also very grateful to the anonymous
referee for numerous helpful suggestions. Research of NF was supported by the
grant No. 22-11-00092 of the Russian Science Foundation. Research of ML was
partially supported by the EPSRC grants EP/W006898/1 and EP/V051881/1, and by
the University of Reading RETF Open Fund. Research of IP was partially
supported by NSERC and FRQNT. Research of DS was partially supported by an
FSRG from DePaul University.

## §2. Dirichlet and Neumann eigenvalues of the ball and lattice counting
problems

Throughout this paper, with $\nu\geq 0$, and $k\in\mathbb{N}$, let
$J_{\nu}(z)$ be the Bessel functions of order $\nu$, let $j_{\nu,k}$ be the
$k$th positive zero of $J_{\nu}$, and let $j^{\prime}_{\nu,k}$ be the $k$th
positive zero of its derivative $J^{\prime}_{\nu}$, with the exception of
$J^{\prime}_{0}$ for which $j^{\prime}_{0,1}=0$.

It is well known that the eigenvalues of the Dirichlet Laplacian in the unit
ball are given by the squares of the zeros of the cylindrical Bessel
functions. Namely, considering the Dirichlet Laplacian in $\mathbb{B}^{d}$, we
have the simple eigenvalues

$\lambda_{d,0,k}=\left(j_{d/2-1,k}\right)^{2},\qquad k\in\mathbb{N},$

that is of multiplicity

$\kappa_{d,0}:=1,$

and the eigenvalues

$\lambda_{d,m,k}=\left(j_{m+d/2-1,k}\right)^{2},\qquad m,k\in\mathbb{N},$

of multiplicity

$\kappa_{d,m}:=\binom{m+d-1}{d-1}-\binom{m+d-3}{d-1}.$

######  Remark 2.1.

We note that the numbers $\kappa_{d,m}$, $d\geq 2$,
$m\in\mathbb{N}_{0}=\mathbb{N}\cup\\{0\\}$, coincide with the multiplicity of
the eigenvalue $m(m+d-2)$ of the Laplace–Beltrami operator on the unit sphere
$\mathbb{S}^{d-1}$, or alternatively with the dimension of the space of
homogeneous harmonic polynomials of degree $m$ in $\mathbb{R}^{d}$. In the
planar case we have

$\kappa_{2,m}=2\qquad\text{for }m\in\mathbb{N}.$

∎

We therefore have

$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)=\sum_{m=0}^{\infty}\kappa_{d,m}\\#\left\\{k\in\mathbb{N}:j_{m+d/2-1,k}\leq\lambda\right\\}.$
(2.1)

######  Remark 2.2.

The sum in (2.1) is in fact finite: we have999Here and further on,
$\left\lfloor x\right\rfloor=\max\\{k\in\mathbb{Z}:k\leq x\\}$ denotes the
integer part of $x\in\mathbb{R}$, and $\left\lceil
x\right\rceil=\min\\{k\in\mathbb{Z}:k\geq x\\}$ denotes its ceiling.

$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda-d/2+1\right\rfloor}\kappa_{d,m}\\#\left\\{k\in\mathbb{N}:j_{m+d/2-1,k}\leq\lambda\right\\},$
(2.2)

This is due to the fact that $j_{\nu,1}>\nu$ [DLMF, Eq. 10.21.3]. Note that in
(2.2) and further on, any sum in which the lower limit exceeds the upper limit
is assumed to be zero, which immediately gives
$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)=0$ for
$\lambda<\frac{d}{2}-1$. ∎

In the planar case the expression (2.1) simplifies to

$\mathcal{N}^{\mathrm{D}}_{\mathbb{D}}(\lambda)=\\#\left\\{k\in\mathbb{N}:j_{0,k}\leq\lambda\right\\}+2\sum_{m=1}^{\left\lfloor\lambda-d/2+1\right\rfloor}\\#\left\\{(m,k)\in\mathbb{N}^{2}:j_{m,k}\leq\lambda\right\\}.$

Similarly, the eigenvalues of the Neumann Laplacian in the unit disk
$\mathbb{D}$ are given by the squares of the zeros of the derivatives of
Bessel functions. We have the simple eigenvalues

$\mu_{0,k}=\left(j^{\prime}_{0,k}\right)^{2},\qquad k\in\mathbb{N},$

and the double eigenvalues

$\mu_{m,k}=\left(j^{\prime}_{m,k}\right)^{2},\qquad m,k\in\mathbb{N}.$

We therefore have

$\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda)=\\#\left\\{k\in\mathbb{N}:j^{\prime}_{0,k}\leq\lambda\right\\}+2\sum_{m=1}^{\left\lfloor\lambda-d/2+1\right\rfloor}\\#\left\\{(m,k)\in\mathbb{N}^{2}:j^{\prime}_{m,k}\leq\lambda\right\\},$
(2.3)

where the sum is again finite since $j^{\prime}_{\nu,1}\geq\nu$ [DLMF, Eq.
10.21.3].

For illustrative purposes only, we show the graphs of the Dirichlet and
Neumann eigenvalue counting functions for the disk in Figure 1.

Figure 1: The Dirichlet eigenvalue counting function
$\mathcal{N}^{\mathrm{D}}_{\mathbb{D}}(\lambda)$ (blue), the Neumann
eigenvalue counting function $\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda)$
(red), and the leading Weyl’s term $W_{d}(\lambda)=\frac{\lambda^{2}}{4}$
(black) in dimension $d=2$. The plot is produced using the floating-point
evaluation of zeros of the Bessel functions and their derivatives. If we were
to assume (contrary to the philosophy of this paper) the validity of floating-
point arithmetic, this plot would have presented a _numerically assisted_ (as
opposed to _computer-assisted)_ “proof” of Pólya’s conjecture for the disk for
$\lambda\lessapprox 15$.

We will be comparing the counting functions
$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)$ and
$\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda)$ with some weighted lattice
counting functions. Let

$h(x):=\frac{1}{\pi}\left(\sqrt{1-x^{2}}-x\arccos x\right),\qquad x\in[0,1],$

and let $\mathbb{P}$ be a planar region under the graph of $h(x)$,

$\mathbb{P}:=\left\\{(x,y):x\in[0,1],y\in[0,h(x)]\right\\}.$

Let, for $\lambda>0$,

$G_{\lambda}(z):=\lambda
h\left(\frac{z}{\lambda}\right)=\frac{1}{\pi}\left(\sqrt{\lambda^{2}-z^{2}}-z\arccos\frac{z}{\lambda}\right),\qquad
z\in[0,\lambda],$ (2.4)

and let $\mathbb{P}_{\lambda}$ be a dilation of $\mathbb{P}$ with coefficient
$\lambda$ with respect to the origin,

$\mathbb{P}_{\lambda}=\left\\{(z,y):0\leq z\leq\lambda,0\leq y\leq
G_{\lambda}(z)\right\\},$ (2.5)

that is, the region under the graph of $G_{\lambda}(z)$.

Let

$Q^{\mathrm{D}}_{d}(\lambda):=\left\\{(m,k)+\left(\frac{d}{2}-1,-\frac{1}{4}\right)\in\mathbb{P}_{\lambda}:(m,k)\in\mathbb{N}_{0}\times\mathbb{N}\right\\}$

and

$Q^{\mathrm{N}}_{2}(\lambda):=\left\\{(m,k)+\left(0,-\frac{3}{4}\right)\in\mathbb{P}_{\lambda}:(m,k)\in\mathbb{N}_{0}\times\mathbb{N}\right\\}$

be the sets of shifted integer lattice points which lie in
$\mathbb{P}_{\lambda}$, see Figure 2. The definitions of the two sets for
$d=2$ differ by a vertical shift. The reason for choosing this particular
notation will become evident later.

We now introduce the _weighted lattice point counting functions_

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda):=\sum_{\left(m+\frac{d}{2}-1,k-\frac{1}{4}\right)\in
Q^{\mathrm{D}}_{d}(\lambda)}\kappa_{d,m}$ (2.6)

and

$\mathcal{P}^{\mathrm{N}}_{2}(\lambda):=\sum_{\left(m,k-\frac{3}{4}\right)\in
Q^{\mathrm{N}}_{2}(\lambda)}\kappa_{2,m}.$ (2.7)

It is immediately seen from the definitions (2.5)–(2.7) that with
$\aleph\in\\{{\mathrm{D}},{\mathrm{N}}\\}$ we have

$\mathcal{P}^{\aleph}_{d}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda-d/2+1\right\rfloor}\kappa_{d,m}\left\lfloor
G_{\lambda}\left(m+\frac{d}{2}-1\right)+s^{\aleph}\right\rfloor,$ (2.8)

where

$s^{\mathrm{D}}:=\frac{1}{4},\qquad s^{\mathrm{N}}:=\frac{3}{4}.$ (2.9)

Figure 2: The region $\mathbb{P}_{\lambda}$, and the sets of shifted lattice
points $Q^{\mathrm{D}}_{d}(\lambda)$ (blue disks) and
$Q^{\mathrm{N}}_{d}(\lambda)$ (red diamonds), shown here for $d=2$ and
$\lambda=23$.

It is well known that as $\lambda\to+\infty$, the asymptotics of the lattice
point counting function $\mathcal{P}_{d}^{\mathrm{D}}(\lambda)$ is intricately
linked to the asymptotics of the eigenvalue counting function
$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)$. This was first shown in
the planar case in [KuzFed65] and later re-discovered in [CdV10], see also
[Gra07]. Namely, in some appropriate sense,

$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)\sim\mathcal{P}^{\mathrm{D}}_{d}(\lambda)\qquad\text{as
}\lambda\to+\infty.$

This observation, together with asymptotic bounds on the difference between
the two functions, has been used to great effect to estimate the remainder in
Weyl’s law for the unit ball. In particular, for the Dirichlet problem in the
disk the two-term Weyl asymptotics (1.5) holds with an improved remainder
estimate

$O\left(\lambda^{131/208}(\log\lambda)^{18627/8320}\right),$

see [GMWW21] (the remainder estimate $O\left(\lambda^{2/3}\right)$ was already
obtained in [KuzFed65], [CdV10]). Similar improved remainder estimates are
also known in the Dirichlet case for higher-dimensional balls [Guo21] and in
the planar Neumann case [GWW19].

As has been recently found in [She22] in the Dirichlet case, there is a
further simple _non-asymptotic_ relation between the lattice point and the
eigenvalue counting functions, which lies at the cornerstone of our proofs of
Theorems 1.2 and 1.4.

######  Theorem 2.3.

For any $d\geq 2$ and any $\lambda\geq 0$, we have

$\mathcal{N}^{\mathrm{D}}_{\mathbb{B}^{d}}(\lambda)\leq\mathcal{P}_{d}^{\mathrm{D}}(\lambda).$

We also have, for any $\lambda\geq 0$,

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)\leq\mathcal{N}^{\mathrm{N}}_{\mathbb{D}}(\lambda).$

## §3. Proof of Theorem 2.3

We start by introducing some additional notation. Set, for $\nu\geq 0$,
$\lambda\geq 0$, and $\aleph\in\\{{\mathrm{D}},{\mathrm{N}}\\}$,

$A_{\nu}^{\aleph}(\lambda):=\begin{cases}G_{\lambda}(\nu)+s^{\aleph},\quad&\text{if
}\lambda\geq\nu,\\\ s^{\aleph},\quad&\text{if }0\leq\lambda<\nu,\end{cases}$
(3.1)

where $G_{\lambda}$ is defined by (2.4) and $s^{\aleph}$ is defined by (2.9).
Some typical graphs of the functions $A_{\nu}^{\aleph}(\lambda)$ are shown in
Figure 3.

The crucial step in the proof of Theorem 2.3 comes from the following bounds
on the number of zeros of Bessel functions and their derivatives below a given
number.

######  Proposition 3.1.

Let $\nu\geq 0$ and $\lambda\geq 0$. Then

$\\#\left\\{k\in\mathbb{N}:j_{\nu,k}\leq\lambda\right\\}\leq\left\lfloor
A_{\nu}^{\mathrm{D}}(\lambda)\right\rfloor$ (3.2)

and

$\\#\left\\{k\in\mathbb{N}:j^{\prime}_{\nu,k}\leq\lambda\right\\}\geq\left\lfloor
A_{\nu}^{\mathrm{N}}(\lambda)\right\rfloor.$ (3.3)

######  Remark 3.2.

For $\lambda\in[0,\nu]$, the inequalities (3.2) and (3.3) become the trivial
identities $0=\left\lfloor s^{\aleph}\right\rfloor=0$. ∎

######  Remark 3.3.

For $\nu=0$, the inequality (3.2) is equivalent to
$j_{0,k}\geq\pi\left(k-\frac{1}{4}\right)$, which was proved in [Het70]. ∎

###### Proof of Proposition 3.1.

We recall the representations of the Bessel functions of the first and second
kind, $J_{\nu}$ and $Y_{\nu}$, and their derivatives in terms of the so-called
_modulus functions_ $M_{\nu}$ and $N_{\nu}$ and the _phase functions_
$\theta_{\nu}$ and $\phi_{\nu}$,

$\displaystyle J_{\nu}(x)$ $\displaystyle=M_{\nu}(x)\cos\theta_{\nu}(x),$
$\displaystyle\quad Y_{\nu}(x)$ $\displaystyle=M_{\nu}(x)\sin\theta_{\nu}(x),$
$\displaystyle J^{\prime}_{\nu}(x)$
$\displaystyle=N_{\nu}(x)\cos\phi_{\nu}(x),$ $\displaystyle\quad
Y^{\prime}_{\nu}(x)$ $\displaystyle=N_{\nu}(x)\sin\phi_{\nu}(x)$

(see [DLMF, Eqs. 10.18.4–5]). We will be using various properties of the phase
functions below; for a review of these properties see [Hor17]. We will be only
considering the cases $\nu\geq 0$ and $x\geq 0$ for which the moduli
$M_{\nu}(x)$ and $N_{\nu}(x)$ are both positive.

Let us concentrate first on (3.2). We have $J_{\nu}(x_{0})=0$ if and only if
$\cos\theta_{\nu}(x_{0})=0$, and so if and only if

$\frac{1}{\pi}\theta_{\nu}(x_{0})+\frac{1}{2}\in\mathbb{Z}.$ (3.4)

Note that the phase function $\theta_{\nu}(x)$ satisfies
$\theta_{\nu}(x)\to-\frac{\pi}{2}$ as $x\to+0$ [DLMF, Eq. 10.18.3], and that
it is monotone increasing for $x\in(0,+\infty)$ [Hor17, Theorem 1], therefore
(3.4) can be replaced by

$B^{\mathrm{D}}_{\nu}(x_{0})\in\mathbb{N},$

where

$B^{\mathrm{D}}_{\nu}(x):=\frac{1}{\pi}\theta_{\nu}(x)+\frac{1}{2},$ (3.5)

and therefore

$\\#\left\\{k\in\mathbb{N}:j_{\nu,k}\leq\lambda\right\\}=\left\lfloor
B^{\mathrm{D}}_{\nu}(\lambda)\right\rfloor.$ (3.6)

We have

$G_{\lambda}(\nu)=\frac{\lambda}{\pi}-\frac{\nu}{2}+O\left(\frac{1}{\lambda}\right)\qquad\text{as
}\lambda\to+\infty.$ (3.7)

Using the asymptotics101010See also [HBRV15] for all the coefficients of the
full asymptotic expansion and some useful remarks. [Hor17, Eq. (21)], [DLMF,
Eq. 10.18.18],

$\theta_{\nu}(\lambda)=\lambda-\frac{\pi}{2}\left(\nu+\frac{1}{2}\right)+O\left(\frac{1}{\lambda}\right)\qquad\text{as
}\lambda\to+\infty,$

and (3.5), (3.7), we obtain

$B^{\mathrm{D}}_{\nu}(\lambda)-A^{\mathrm{D}}_{\nu}(\lambda)\to
0\qquad\text{as }\lambda\to+\infty.$

Further,

$\frac{\mathrm{d}B^{\mathrm{D}}_{\nu}(\lambda)}{\mathrm{d}\lambda}=\frac{1}{\pi}\theta^{\prime}_{\nu}(\lambda)>\frac{\sqrt{\lambda^{2}-\nu^{2}}}{\pi\lambda}=\frac{\mathrm{d}A^{\mathrm{D}}_{\nu}(\lambda)}{\mathrm{d}\lambda}$

for $\lambda\geq\nu$ by [Hor17, Eq. (56)]. As we additionally have

$\frac{\mathrm{d}B^{\mathrm{D}}_{\nu}(\lambda)}{\mathrm{d}\lambda}>0=\frac{\mathrm{d}A^{\mathrm{D}}_{\nu}(\lambda)}{\mathrm{d}\lambda}$

for $\lambda\in(0,\nu]$, the function
$B^{\mathrm{D}}_{\nu}(\lambda)-A^{\mathrm{D}}_{\nu}(\lambda)$ is monotone
increasing on $(0,+\infty)$ and tends to zero at infinity. Thus, we have
proved that

$B^{\mathrm{D}}_{\nu}(\lambda)<A^{\mathrm{D}}_{\nu}(\lambda)\qquad\text{for
}\lambda\in[0,+\infty).$ (3.8)

Combining (3.6) and (3.8) proves (3.2).

We now prove (3.3). In the same manner we have $J^{\prime}_{\nu}(x_{0})=0$ if
and only if

$\frac{1}{\pi}\phi_{\nu}(x_{0})+\frac{1}{2}\in\mathbb{Z}.$ (3.9)

We note that the phase function $\phi_{\nu}$ satisfies
$\phi_{\nu}(x)\to\frac{\pi}{2}$ as $x\to+0$ [DLMF, Eq. 10.18.3]. Also,
$\phi_{\nu}(x)$ is monotone increasing for $x\in(\nu,+\infty)$ and monotone
decreasing for $x\in(0,\nu)$ [Hor17, Theorem 1], with
$\phi_{\nu}(\nu)>-\frac{\pi}{2}$ [Hor17, formula (60)]. Thus, the condition
(3.9) can be replaced by

$B^{\mathrm{N}}_{\nu}(x_{0})\in\mathbb{N},$

where

$B^{\mathrm{N}}_{\nu}(x):=\frac{1}{\pi}\phi_{\nu}(x)+\frac{1}{2},$ (3.10)

and therefore

$\\#\left\\{k\in\mathbb{N}:j^{\prime}_{\nu,k}\leq\lambda\right\\}=\left\lfloor
B^{\mathrm{N}}_{\nu}(\lambda)\right\rfloor.$ (3.11)

Using the asymptotics [Hor17, Eq. (22)], [DLMF, Eq. 10.18.21],

$\phi_{\nu}(\lambda)=\lambda-\frac{\pi}{2}\left(\nu-\frac{1}{2}\right)+O\left(\frac{1}{\lambda}\right)\qquad\text{as
}\lambda\to+\infty,$

and (3.10), (3.7), we get

$B^{\mathrm{N}}_{\nu}(\lambda)-A^{\mathrm{N}}_{\nu}(\lambda)\to
0\qquad\text{as }\lambda\to+\infty.$

Also,

$\frac{\mathrm{d}B^{\mathrm{N}}_{\nu}(\lambda)}{\mathrm{d}\lambda}=\frac{1}{\pi}\phi^{\prime}_{\nu}(\lambda)<\frac{\sqrt{\lambda^{2}-\nu^{2}}}{\pi\lambda}=\frac{\mathrm{d}A^{\mathrm{N}}_{\nu}(\lambda)}{\mathrm{d}\lambda}$

for $\lambda\geq\nu$ by [Hor17, formula following Eq. (58)]. As we
additionally have

$\frac{\mathrm{d}B^{\mathrm{N}}_{\nu}(\lambda)}{\mathrm{d}\lambda}<0=\frac{\mathrm{d}A^{\mathrm{N}}_{\nu}(\lambda)}{\mathrm{d}\lambda}$

for $\lambda\in(0,\nu]$, the function
$B^{\mathrm{N}}_{\nu}(\lambda)-A^{\mathrm{N}}_{\nu}(\lambda)$ is monotone
decreasing on $(0,+\infty)$ and tends to zero at infinity. Thus, we have
proved that

$B^{\mathrm{N}}_{\nu}(\lambda)>A^{\mathrm{N}}_{\nu}(\lambda)\qquad\text{for
}\lambda\in[0,+\infty).$ (3.12)

Combining (3.11) and (3.12) proves (3.3). ∎

We illustrate inequalities (3.8) and (3.12) in Figure 3.

Figure 3: An illustration of inequalities (3.8) and (3.12). The plots of
$B^{\mathrm{D}}_{\nu}(\lambda)$ and $B_{\nu}^{\mathrm{N}}(\lambda)$ are drawn
using the recipe from [Hor17]. We remark that $B^{\mathrm{N}}_{\nu}(\lambda)$
has a minimum at $\lambda=\nu$.

The bound (3.2) can be also proved without relying on the properties of the
Bessel phase function. Instead, one uses the known asymptotics of the Bessel
zeros and the Sturm comparison theorem. We present this alternative argument
below for an interested reader. The bound (3.3) can be proved in the same
manner; we omit the details.

Let $\nu\geq 0$, and let us consider the function

$\sin\left(\pi A^{\mathrm{D}}_{\nu}(x)\right).$

Let $a_{\nu,k}$ be its $k$th zero in $[\nu,+\infty)$, ordered increasingly.
Obviously,

$\\#\left\\{k\in\mathbb{N}:a_{\nu,k}\leq\lambda\right\\}=\left\lfloor
A^{\mathrm{D}}_{\nu}(\lambda)\right\rfloor.$ (3.13)

We will prove

######  Lemma 3.4.

$a_{\nu,k}\leq j_{\nu,k}$ for all $k\in\mathbb{N}$.

Together with (3.13), Lemma 3.4 immediately implies (3.2).

###### Proof of Lemma 3.4.

The function $U_{\nu}(x):=\sqrt{x}J_{\nu}(x)$ satisfies the differential
equation

$U^{\prime\prime}_{\nu}(x)+\left(1-\frac{\nu^{2}-1/4}{x^{2}}\right)U_{\nu}(x)=0.$
(3.14)

Consider the function

$V_{\nu}(x):=\frac{\sqrt{x}}{(x^{2}-\nu^{2})^{1/4}}\sin\left(\pi
A^{\mathrm{D}}_{\nu}(x)+b\right)$

with some $b\in\left[0,\frac{\pi}{4}\right)$. Then

$\displaystyle V^{\prime}_{\nu}(x)$
$\displaystyle=-\frac{\nu^{2}}{2\sqrt{x}\left(x^{2}-\nu^{2}\right)^{5/4}}\sin\left(\pi
A^{\mathrm{D}}_{\nu}(x)+b\right)+\frac{\left(x^{2}-\nu^{2}\right)^{1/4}}{\sqrt{x}}\cos\left(\pi
A^{\mathrm{D}}_{\nu}(x)+b\right),$ $\displaystyle V^{\prime\prime}_{\nu}(x)$
$\displaystyle=\left(\frac{6\nu^{2}x^{2}-\nu^{4}}{4x^{3/2}(x^{2}-\nu^{2})^{9/4}}-\frac{(x^{2}-\nu^{2})^{3/4}}{x^{3/2}}\right)\sin\left(\pi
A^{\mathrm{D}}_{\nu}(x)+b\right).$

Therefore,

$V^{\prime\prime}_{\nu}(x)+\left(1-\frac{\nu^{2}}{x^{2}}-\frac{\nu^{2}(6x^{2}-\nu^{2})}{4x^{2}(x^{2}-\nu^{2})^{2}}\right)V_{\nu}(x)=0\qquad\text{for
}x\in(\nu,+\infty).$ (3.15)

Denote by $v_{\nu,k}=v_{\nu,k}(b)$ the $k$th zero of the function $V_{\nu}$.
By the definitions of $V_{\nu}$ and $A_{\nu}^{\mathrm{D}}$,

$\sqrt{v_{\nu,k}^{2}-\nu^{2}}-\nu\arccos\frac{\nu}{v_{\nu,k}}+\frac{\pi}{4}+b=\pi
k.$

As

$\sqrt{v_{\nu,k}^{2}-\nu^{2}}=v_{\nu,k}+O\left(v_{\nu,k}^{-1}\right)\qquad\text{and}\qquad\arccos\frac{\nu}{v_{\nu,k}}=\frac{\pi}{2}+O\left(v_{\nu,k}^{-1}\right)\qquad\text{as
}k\to\infty,$

we have

$v_{\nu,k}(b)=\pi\left(k+\frac{\nu}{2}-\frac{1}{4}\right)-b+O\left(k^{-1}\right)\qquad\text{as
}k\to\infty.$

On the other hand the asymptotics

$j_{\nu,k}=\pi\left(k+\frac{\nu}{2}-\frac{1}{4}\right)+O\left(k^{-1}\right)\qquad\text{as
}k\to\infty.$

is well known, see for example [DLMF, Eq. 10.21.19].

Suppose that $b>0$. Then there exists $K\in\mathbb{N}$ such that

$j_{\nu,k}>v_{\nu,k}(b)\qquad\text{for }k\geq K.$ (3.16)

The coefficient in front of $U_{\nu}$ in (3.14) is greater than the
coefficient in front of $V_{\nu}$ in (3.15):

$1-\frac{\nu^{2}-1/4}{x^{2}}>1-\frac{\nu^{2}}{x^{2}}>1-\frac{\nu^{2}}{x^{2}}-\frac{\nu^{2}(6x^{2}-\nu^{2})}{4x^{2}(x^{2}-\nu^{2})^{2}}\qquad\text{for
all }x\in[\nu,+\infty).$

By the Sturm comparison theorem there is a zero of $U_{\nu}$ between
$v_{\nu,k}(b)$ and $v_{\nu,k+1}(b)$. So, if $j_{\nu,k_{0}}\leq
v_{\nu,k_{0}}(b)$ for some number $k_{0}$, then $j_{\nu,k_{0}+1}\leq
v_{\nu,k_{0}+1}(b)$, and by induction $j_{\nu,k}\leq v_{\nu,k}(b)$ for all
$k\geq k_{0}$ which contradicts (3.16). Therefore, (3.16) holds for all
natural $k$.

Finally, each function $v_{\nu,k}(b)$ is continuous in $b$. Thus,

$j_{\nu,k}\geq v_{\nu,k}(0)=a_{\nu,k}.$

∎

Returning now to the proof of Theorem 2.3, we rewrite (2.8) as

$\mathcal{P}_{d}^{\aleph}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda-d/2+1\right\rfloor}\kappa_{d,m}\left\lfloor
A_{m+d/2-1}^{\aleph}(\lambda)\right\rfloor.$ (3.17)

Theorem 2.3 now immediately follows from Proposition 3.1 with account of
(3.17), (2.2), and (2.3).

## §4. From the weighted lattice point count towards Pólya’s conjecture

By Theorem 2.3, the Dirichlet Pólya’s conjecture for $\mathbb{B}^{d}$ would
follow immediately if we can prove that

$\mathcal{P}_{d}^{\mathrm{D}}(\lambda)<W_{d}(\lambda)$ (4.1)

for all $\lambda\in(0,+\infty)$, where $W_{d}(\lambda)$ is defined by (1.8).

Similarly, the Neumann Pólya’s conjecture for $\mathbb{D}$ would follow
immediately if we can prove that

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)>W_{2}(\lambda)$ (4.2)

for all $\lambda\in(\Lambda_{0},+\infty)$; we note that we have already dealt
with $\lambda\leq\Lambda_{0}$ by Lemma 1.3.

We establish (4.1) in the following cases, which will be dealt with
separately.

######  Theorem 4.1.

The inequality

$\mathcal{P}_{2}^{\mathrm{D}}(\lambda)<W_{2}(\lambda)=\frac{\lambda^{2}}{4}$
(4.3)

holds for all $\lambda>0$.

Theorem 4.1 will be proved in §5. Together with Theorem 2.3, it implies
Theorem 1.2 in the planar case.

######  Theorem 4.2.

The inequalities (4.1) hold for all $d\geq 3$ and $\lambda>0$.

Theorem 4.2 will be proved in §7. Together with Theorem 2.3, it implies
Theorem 1.2 for higher-dimensional balls.

In the Neumann case, the situation is more delicate, as we cannot expect (4.2)
to hold for all values of $\lambda\in(0,+\infty)$ since
$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)$ is identically zero for
$\lambda<\frac{\pi}{4}$, see Figure 4.

Figure 4: A numerical experiment: the computed
$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)/W_{2}(\lambda)-1$ as a function of
$\lambda$.

We prove the following results.

######  Theorem 4.3.

The inequality

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)>W_{2}(\lambda)=\frac{\lambda^{2}}{4}$

holds for all $\lambda\geq\Lambda_{1}$, where $\Lambda_{1}$ is given by (1.9).

Theorem 4.3 will be proved in §6. Together with Theorem 2.3, it implies
Theorem 1.4.

We are further able to eliminate the remaining gap in the Neumann case.

######  Theorem 4.4.

The inequality (4.2) holds for any
$\lambda\in\left(\Lambda_{0},\Lambda_{1}\right)$.

The proof of this result, presented in §8, is computer-assisted. Theorem 4.4
implies Theorem 1.5.

In all cases, we deal with estimating a (weighted) count of (shifted) lattice
points under the graph of a particular function $G_{\lambda}$. Such problems
have been extensively studied in number theory, going back to the Gauss circle
problem. Important contributions in the general case can be traced through the
works of van der Corput [vdC23] and Krätzel [Krä00] to some very recent
results of Laugesen and Liu [Liu17, LauLiu18a]. In particular, [LauLiu18b,
Proposition 15] is directly applicable (with account of the fact that Laugesen
and Liu do not count the points on the vertical axis and do not double-count
the points inside) to our shifted lattice point count
$\mathcal{P}_{2}^{\mathrm{D}}(\lambda)$, yielding the bound

$\mathcal{P}_{2}^{\mathrm{D}}(\lambda)\leq\frac{\lambda^{2}}{4}+\left(\frac{2}{3}+\frac{1}{\pi}-\frac{\sqrt{3}}{2\pi}\right)\lambda\approx\frac{\lambda^{2}}{4}+0.7093\lambda.$

Unfortunately, since the coefficient in front of $\lambda$ in this formula is
positive, this bound is weaker than our required bound (4.3). We need
therefore to obtain sharper lattice point count bounds than those available
generally, and to do so we additionally use some properties of the
_derivative_ of the function $G_{\lambda}$ in addition to the properties of
the function itself, see Theorems 5.1 and 6.1, and also Remarks 5.2 and 6.3
for an informal explanation.

For future use, we summarise below some elementary properties of the function
$G_{\lambda}$.

The first lemma is checked by a direct calculation.

######  Lemma 4.5.

The function $G_{\lambda}:[0,\lambda]\to\left[0,\frac{\lambda}{\pi}\right]$
defined by (2.4) is a strictly monotone decreasing convex $C^{1}$ function
with

$\displaystyle\qquad G_{\lambda}(0)$ $\displaystyle=\frac{\lambda}{\pi},$
$\displaystyle\qquad G_{\lambda}(\lambda)$ $\displaystyle=0,$ $\displaystyle
G^{\prime}_{\lambda}(z)$
$\displaystyle=-\frac{1}{\pi}\arccos\frac{z}{\lambda},$ $\displaystyle\qquad
G^{\prime}_{\lambda}(0)$ $\displaystyle=-\frac{1}{2},$ $\displaystyle\qquad
G^{\prime}_{\lambda}(\lambda)$ $\displaystyle=0,$ $\displaystyle
G^{\prime\prime}_{\lambda}(z)$
$\displaystyle=\frac{1}{\pi\sqrt{\lambda^{2}-z^{2}}}.$

We can therefore define the inverse function
$G^{-1}_{\lambda}:\left[0,\frac{\lambda}{\pi}\right]\to[0,\lambda]$ which is
also monotone decreasing and convex. Sometimes, it will be also convenient for
us to consider $G_{\lambda}$ on the interval
$\left[0,\left\lceil\lambda\right\rceil\right]$ by extending it by zero to
$\left(\lambda,\left\lceil\lambda\right\rceil\right]$: the resulting function,
which we for simplicity denote by the same symbol, remains monotone
decreasing, convex, and $C^{1}$.

######  Lemma 4.6.

Let $\beta\geq 0$. Then

$\int_{0}^{\lambda}z^{\beta}G_{\lambda}(z)\,\mathrm{d}z=\frac{\Gamma\left(\frac{\beta+1}{2}\right)\lambda^{\beta+2}}{4\sqrt{\pi}\,(\beta+2)\,\Gamma\left(\frac{\beta+4}{2}\right)}.$

In particular,

$\int_{0}^{\lambda}G_{\lambda}(z)\,\mathrm{d}z=\frac{\lambda^{2}}{8}.$ (4.4)

###### Proof.

In fact, the identity can be checked using computer algebra software, but we
include a proof for the sake of completeness. After a change of variables
$z=\lambda\cos\tau$, we obtain

$\int_{0}^{\lambda}z^{\beta}G_{\lambda}(z)\,\mathrm{d}z=\frac{\lambda^{\beta+2}}{\pi}\int_{0}^{\pi/2}(\cos\tau)^{\beta}(\sin\tau-\tau\cos\tau)\sin\tau\,d\tau.$
(4.5)

By [DLMF, Eqs. 5.12.1–2],

$\int_{0}^{\pi/2}(\cos\tau)^{\rho}(\sin\tau)^{\sigma}\mathrm{d}\tau=\frac{\Gamma\left(\frac{\rho+1}{2}\right)\Gamma\left(\frac{\sigma+1}{2}\right)}{2\Gamma\left(\frac{\rho+\sigma+2}{2}\right)}$

for any $\rho,\sigma\geq 0$. Therefore,

$\int_{0}^{\pi/2}(\cos\tau)^{\beta}(\sin\tau)^{2}\mathrm{d}\tau=\frac{\sqrt{\pi}\Gamma\left(\frac{\beta+1}{2}\right)}{4\Gamma\left(\frac{\beta+4}{2}\right)},$
$\int_{0}^{\pi/2}\tau(\cos\tau)^{\beta+1}\sin\tau
d\tau=-\left.\frac{\tau(\cos\tau)^{\beta+2}}{\beta+2}\right|_{0}^{\pi/2}+\frac{1}{\beta+2}\int_{0}^{\pi/2}(\cos\tau)^{\beta+2}d\tau=\frac{\sqrt{\pi}\,\Gamma\left(\frac{\beta+3}{2}\right)}{2\,(\beta+2)\,\Gamma\left(\frac{\beta+4}{2}\right)},$

and so

$\int_{0}^{\pi/2}(\cos\tau)^{\beta}(\sin\tau-\tau\cos\tau)\sin\tau\,\mathrm{d}\tau=\frac{\sqrt{\pi}\Gamma\left(\frac{\beta+1}{2}\right)}{4(\beta+2)\Gamma\left(\frac{\beta+4}{2}\right)}.$

Substituting this into (4.5) we get the result. ∎

######  Corollary 4.7.

Let $d\in\mathbb{N}$, $d\geq 2$. Then

$\frac{2}{(d-2)!}\int_{0}^{\lambda}z^{d-2}G_{\lambda}(z)\,\mathrm{d}z=W_{d}(\lambda)=w_{d}\lambda^{d}.$

###### Proof.

Applying Lemma 4.6 with $\beta=d-2$ we get

$\frac{2}{(d-2)!}\int_{0}^{\lambda}z^{d-2}G_{\lambda}(z)\,\mathrm{d}z=\frac{\Gamma\left(\frac{d-1}{2}\right)\lambda^{d}}{2\,d\,\sqrt{\pi}\,(d-2)!\,\Gamma\left(\frac{d+2}{2}\right)}.$

The duplication formula [DLMF, Eq. 5.5.5]

$\sqrt{\pi}(d-2)!=\sqrt{\pi}\Gamma(d-1)=2^{d-2}\Gamma\left(\frac{d-1}{2}\right)\Gamma\left(\frac{d}{2}\right)$

implies

$\frac{\Gamma\left(\frac{d-1}{2}\right)\lambda^{d}}{2d\sqrt{\pi}(d-2)!\,\Gamma\left(\frac{d+2}{2}\right)}=\frac{\lambda^{d}}{2^{d-1}\,d\,\Gamma\left(\frac{d}{2}\right)\Gamma\left(\frac{d+2}{2}\right)}=\frac{\lambda^{d}}{2^{d}\left(\Gamma\left(\frac{d+2}{2}\right)\right)^{2}}=W_{d}(\lambda).$

∎

An important role in our study in the Neumann case will be played by the
inverse function value $G_{\lambda}^{-1}\left(\frac{1}{4}\right)$ (defined for
all $\lambda\geq\frac{\pi}{4}$). We will use the following bounds.

######  Lemma 4.8.

We have

$G_{\lambda}^{-1}\left(\frac{1}{4}\right)<\lambda-1$ (4.6)

for all $\lambda\geq 2$. Additionally, for any
$\sigma\in\left(0,\frac{\pi}{2}\right]$ we have

$G_{\lambda}^{-1}\left(\frac{1}{4}\right)\geq\lambda\cos\sigma$ (4.7)

whenever

$\lambda\geq r_{1}(\sigma):=\frac{\pi}{4(\sin\sigma-\sigma\cos\sigma)}.$ (4.8)

###### Proof.

Since $G_{\lambda}$ is monotone decreasing, the claim (4.6) is equivalent to

$G_{\lambda}(\lambda-1)<\frac{1}{4}.$ (4.9)

We have $G_{2}(1)-\frac{1}{4}=\frac{\sqrt{3}}{\pi}-\frac{7}{12}<0$, and
additionally

$\frac{\mathrm{d}}{\mathrm{d}\lambda}\left(G_{\lambda}(\lambda-1)\right)=\frac{1}{\pi}\left(\frac{1}{\lambda}\sqrt{2\lambda-1}-\arccos\left(1-\frac{1}{\lambda}\right)\right)<0$

as

$\cos\left(\frac{1}{\lambda}\sqrt{2\lambda-1}\right)>1-\frac{2\lambda-1}{2\lambda^{2}}>1-\frac{1}{\lambda},$

thus implying (4.9) for $\lambda\geq 2$.

Similarly, the claim (4.7) is equivalent to

$G_{\lambda}(\lambda\cos\sigma)=\frac{\lambda(\sin\sigma-\sigma\cos\sigma)}{\pi}\geq\frac{1}{4},$

given (4.8). ∎

## §5. Proof of Theorem 4.1

We first state the following

######  Theorem 5.1.

Let $b>0$, and let $g$ be a non-negative decreasing convex function on $[0,b]$
such that $g(b)=0$ and

$\left|g(z)-g(w)\right|\leq\frac{1}{2}|z-w|$ (5.1)

for all $z,w\in[0,b]$. Then

$\left\lfloor g(0)+\frac{1}{4}\right\rfloor+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{1}{4}\right\rfloor\leq
2\int_{0}^{b}g(z)\,\mathrm{d}z.$ (5.2)

The equality is possible only if $g$ is identically zero on $[0,b]$.

######  Remark 5.2.

We explain here, very informally, the ideas behind the proof of Theorem 5.1.
The area under the graph of the function $g$ on the interval $[m,m+1]$ is
approximately equal to the area under the straight line passing through the
points $(m,g(m))$ and $(m+1,g(m+1))$, so

$\int_{m}^{m+1}g(z)\mathrm{d}z\approx\frac{1}{2}\left(g(m)+g(m+1)\right).$

Summing up these equalities over $m$ we obtain

$2\int_{0}^{b}g(z)\mathrm{d}z\approx g(0)+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}g(m).$

If a number $x$ is chosen randomly then $\left\lfloor x\right\rfloor\approx
x-\frac{1}{2}$ on average. Thus, $\left\lfloor
g(m)+\frac{1}{4}\right\rfloor\approx g(m)-\frac{1}{4}$, and these extra
contributions of $-\frac{1}{4}$ ensure the sign of the inequality in (5.2). In
order to prove Theorem 5.1 rigorously we divide the graph by horizontal lines
$y=n$, with $n=0,1,\dots,\left\lfloor g(0)\right\rfloor$, and we consider what
happens in the intervals where $n+1\geq g(z)\geq n$. The values of
$\left\lfloor g(m)+1/4\right\rfloor$ there are either $n$ or $n+1$. The point
$m$ is “bad” if $g(m)\geq n+\frac{3}{4}$ and thus $\left\lfloor
g(m)+1/4\right\rfloor=n+1$: these “bad” points contribute more to the sum than
we expect “on average”. The convexity of the function $g$ and condition (5.1)
ensure that the number of such “bad” points is not greater than half of the
total number of integer points in an interval, and this yields the required
estimate in the interval we are considering. ∎

In order to prove Theorem 5.1 we require the following

######  Lemma 5.3.

Let $i,j\in\mathbb{Z}$, $i<j$. Let $g$ be a decreasing convex function on
$[i,j+1]$ satisfying (5.1) for all $z,w\in[i,j+1]$. Assume additionally that

$n+1\geq g(i+1)\geq\dots\geq g(j)\geq n\geq g(j+1)$ (5.3)

for some $n\in\mathbb{Z}$. Then

$\frac{1}{2}\left\lfloor
g(i)+\frac{1}{4}\right\rfloor+\sum_{m=i+1}^{j-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(j)+\frac{1}{4}\right\rfloor\leq\int_{i}^{j}g(z)\,\mathrm{d}z.$ (5.4)

###### Proof of Lemma 5.3.

The validity of the claim does not change if we add a constant integer number
to the function $g$. So, without loss of generality we can assume $n=0$, so
that (5.3) becomes

$1\geq g(i+1)\geq\dots\geq g(j)\geq 0\geq g(j+1).$ (5.5)

Additionally, (5.1) implies that $g(i)\leq g(i+1)+\frac{1}{2}\leq\frac{3}{2}$.

Set

$K=\\#\left\\{m\in\\{i,\dots,j\\}:\frac{3}{4}\leq g(m)\right\\},$

and consider four cases.

Case $K=0$.

The left-hand side of (5.4) is zero, and the right-hand side is non-negative
by (5.5).

Case $K=1$.

Here

$\frac{5}{4}>g(i)\geq\frac{3}{4}>g(i+1)\geq\dots\geq g(j)\geq 0,$

and the left-hand side of (5.4) is equal to $\frac{1}{2}$. The assumption
(5.1) yields $g\left(i+\frac{1}{2}\right)\geq\frac{1}{2}$, therefore by non-
negativity and convexity of $g$ on $[i,j]$,

$\int_{i}^{j}g(z)\,\mathrm{d}z\geq\int_{i}^{i+1}g(z)\,\mathrm{d}z\geq
g\left(i+\frac{1}{2}\right)\geq\frac{1}{2}.$

Case $K=2$.

Here

$\frac{3}{2}\geq g(i)\geq g(i+1)\geq\frac{3}{4}>g(i+2)\geq\dots\geq g(j)\geq
0\geq g(j+1),$

and the left-hand side of (5.4) equals $\frac{3}{2}$. By (5.1) we have $j\geq
i+2$, and therefore by non-negativity and convexity of $g$,

$\int_{i}^{j}g(z)\,\mathrm{d}z\geq\int_{i}^{i+2}g(z)\,\mathrm{d}z\geq
2g(i+1)\geq\frac{3}{2}.$

Case $K\geq 3$.

Here

$\frac{3}{2}\geq g(i)\geq g(i+1)\geq\dots\geq
g(i+K-1)\geq\frac{3}{4}>g(i+K)\geq\dots\geq g(j)\geq 0\geq g(j+1).$

The left-hand side of (5.4) is equal to $K-\frac{1}{2}$. By convexity of $g$,

$(K-1)g(i+1)+(K-2)g(i+2K-2)\geq(2K-3)g(i+K-1),$

and therefore

$g(i+2K-2)\geq\frac{2K-5}{4(K-2)}>0.$

Thus, $j\geq i+2K-2$. Next,

$\int_{i}^{j}g(z)\,\mathrm{d}z\geq\int_{i}^{i+2K-2}g(z)\,\mathrm{d}z\geq(2K-2)g(i+K-1)\geq\frac{3K-3}{2}\geq
K-\frac{1}{2}$

as $K\geq 3$.

∎

######  Remark 5.4.

One can easily see from the proof that the equality in (5.4) is attained in
the following three cases only:

  * •

$g(z)\equiv n$ on $[i,j]$ (if $K=0$);

  * •

$j=i+1$ and $g(z)=n+\frac{i-z}{2}+\frac{3}{4}$ on $[i,i+1]$ (if $K=1$);

  * •

$j=i+2$ and $g(z)=n+s(i+1-z)+\frac{3}{4}$ on $[i,i+2]$ with
$s\in\left[\frac{3}{8},\frac{1}{2}\right]$ (if $K=2$).

∎

We can now proceed to the proof of Theorem 5.1 proper.

###### Proof of Theorem 5.1.

Let

$N=\left\lfloor g(0)\right\rfloor.$

If $N=0$, then applying Lemma 5.3 with $i=0$, $j=\left\lfloor b\right\rfloor$,
and $g(z)$ extended by zero for $z\in\left(b,\left\lfloor
b\right\rfloor+1\right]$, gives

$\left\lfloor g(0)+\frac{1}{4}\right\rfloor+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{1}{4}\right\rfloor\leq
2\int_{0}^{b}g(z)\,\mathrm{d}z,$

and therefore (5.2).

Assume now $N\geq 1$. For $k=0,1,\dots,N$, let

$L_{k}:=\max\left\\{m\in\\{0,\dots,\left\lfloor b\right\rfloor\\}:g(m)\geq
k\right\\},$ (5.6)

see Figure 5. Therefore, we have

$0\leq L_{N}<L_{N-1}<\dots<L_{0}=\left\lfloor b\right\rfloor,$

where the strict inequalities $L_{k}<L_{k-1}$, $k=1,\dots,N$, follow from
(5.1).

Figure 5: The numbers $L_{k}$, see (5.6) and also Remark 5.5.

We will consider two cases depending on whether $L_{N}=0$.

Case $L_{N}=0$.

We write

$\begin{split}&\qquad\qquad\left\lfloor
g(0)+\frac{1}{4}\right\rfloor+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{1}{4}\right\rfloor\\\
&=2\sum_{k=0}^{N-1}\left(\frac{1}{2}\left\lfloor
g(L_{k+1})+\frac{1}{4}\right\rfloor+\sum_{m=L_{k+1}+1}^{L_{k}-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(L_{k})+\frac{1}{4}\right\rfloor\right).\end{split}$ (5.7)

For each $k\in\\{0,\dots,N-1\\}$ we have

$k+1>g\left(L_{k+1}+1\right)\dots\geq g\left(L_{k}\right)\geq k\geq
g\left(L_{k}+1\right),$

and applying Lemma 5.3 with $i=L_{k+1}$, $j=L_{k}$, and $n=k$ yields

$\frac{1}{2}\left\lfloor
g(L_{k+1})+\frac{1}{4}\right\rfloor+\sum_{m=L_{k+1}+1}^{L_{k}-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(L_{k})+\frac{1}{4}\right\rfloor\leq\int_{L_{k+1}}^{L_{k}}g(z)\,\mathrm{d}z.$
(5.8)

Substituting (5.8) into (5.7) gives

$\left\lfloor g(0)\right\rfloor+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{1}{4}\right\rfloor\leq
2\int_{0}^{\left\lfloor b\right\rfloor}g(z)\,\mathrm{d}z\leq
2\int_{0}^{b}g(z)\,\mathrm{d}z,$ (5.9)

as required, where in the last inequality we used non-negativity of $g$.

Case $L_{N}>0$.

We write

$\begin{split}&\qquad\qquad\left\lfloor
g(0)+\frac{1}{4}\right\rfloor+2\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{1}{4}\right\rfloor\\\
&=2\left(\frac{1}{2}\left\lfloor
g(0)+\frac{1}{4}\right\rfloor+\sum_{m=1}^{L_{N}-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(L_{N})+\frac{1}{4}\right\rfloor\right)\\\
&+2\sum_{k=0}^{N-1}\left(\frac{1}{2}\left\lfloor
g(L_{k+1})+\frac{1}{4}\right\rfloor+\sum_{m=L_{k+1}+1}^{L_{k}-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(L_{k})+\frac{1}{4}\right\rfloor\right).\end{split}$ (5.10)

We have

$N+1>g(0)\geq\dots\geq g\left(L_{N}-1\right)\geq g\left(L_{N}\right)\geq
N>g\left(L_{N}+1\right),$

therefore applying Lemma 5.3 with $i=0$, $j=L_{N}$, and $n=N$, we get

$2\left(\frac{1}{2}\left\lfloor
g(0)+\frac{1}{4}\right\rfloor+\sum_{m=1}^{L_{N}-1}\left\lfloor
g(m)+\frac{1}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(L_{N})+\frac{1}{4}\right\rfloor\right)\leq
2\int_{0}^{L_{N}}g(z)\,\mathrm{d}z.$ (5.11)

Substituting (5.11) and (5.8) into (5.10) gives (5.9).

Finally, assume that we have the equality in (5.2). Due to Remark 5.4, the
function $g$ is linear on each interval $[L_{k+1},L_{k}]$, and either
$\operatorname{dist}(g(L_{k}),\mathbb{Z})\geq\frac{1}{4}$ or $g\equiv 0$ on
$[L_{k+1},L_{k}]$. The situation when
$\operatorname{dist}(g(L_{k}),\mathbb{Z})\geq\frac{1}{4}$ and $g\equiv 0$ on
$[L_{k},L_{k-1}]$ is impossible due to continuity of $g$. If
$\operatorname{dist}(g(L_{k}),\mathbb{Z})\geq\frac{1}{4}$ for all $k$, then in
particular $g(L_{0})=g(\left\lfloor B\right\rfloor)\geq\frac{1}{4}$, and the
last inequality in (5.9) is strict. Therefore, the equality in (5.2) requires
$g\equiv 0$ on the whole interval $[0,b]$. ∎

######  Remark 5.5.

If $g$ is strictly monotone on $[0,b]$, then the inverse function $g^{-1}$ is
well-defined on $[0,g(0)]$, and the definitions (5.6) may be equivalently
rewritten as

$L_{k}=\left\lfloor g^{-1}(k)\right\rfloor,\qquad k=0,\dots,\left\lfloor
g(0)\right\rfloor.$

∎

We finally use Theorem 5.1 to prove Theorem 4.1.

###### Proof of Theorem 4.1.

We apply (5.2) with $b=\lambda$ and $g=G_{\lambda}$ (which we can do since
Lemma 4.5 ensures that (5.1) holds in this case), and use (4.4), giving the
bound (4.3) and therefore confirming the validity of the Dirichlet Pólya’s
conjecture for the disk. ∎

## §6. Proof of Theorem 4.3

We start by stating

######  Theorem 6.1.

Let $b>0$, and let $g$ be a non-negative decreasing convex function on $[0,b]$
such that $g(0)\geq\frac{1}{4}$, $g(b)=0$, and (5.1) holds for all
$z,w\in[0,b]$. Let

$M_{0}=M_{g,0}:=1+\max\left\\{m\in\\{0,\dots,\left\lfloor
b\right\rfloor\\}:g(m)\geq\frac{1}{4}\right\\},$

and assume that $M_{0}\leq b$. Then

$\sum_{m=0}^{\left\lfloor b\right\rfloor}\left\lfloor
g(m)+\frac{3}{4}\right\rfloor\geq\int_{0}^{b}g(z)\,\mathrm{d}z-\frac{b-3M_{0}}{8}.$
(6.1)

######  Remark 6.2.

If $g$ is strictly monotone on $[0,b]$, then the inverse function $g^{-1}$ is
well-defined on $[0,g(0)]$, and

$M_{0}=\left\lfloor g^{-1}\left(\frac{1}{4}\right)\right\rfloor+1,$ (6.2)

cf. Remark 5.5. ∎

######  Remark 6.3.

Once more, we first outline a very informal plan of proving Theorem 6.1. As we
have argued in Remark 5.2, we have $\left\lfloor
g(m)+\frac{3}{4}\right\rfloor\approx g(m)+\frac{1}{4}$, which should in
principle ensure the correct inequality sign in (6.1). “Bad” points are now
the points with $n\leq g(m)\leq n+\frac{1}{4}$. So, we divide the graph of $g$
by the horizontal lines at $y=n+\frac{1}{4}$, where $n=0,1,\dots,\left\lfloor
g(0)+\frac{3}{4}\right\rfloor$. Again, this guarantees that the number of
“bad” points in each resulting interval is less than half the total number of
points there. This still leaves an unresolved issue of points $m$ lying under
the tail of the graph of $g$, where $0\leq g(m)<\frac{1}{4}$. Such points make
no contribution to the left-hand side of (6.1), but the tail does contribute
to the integral: consider, for example, a toy case of a function
$g(z)=\frac{10-z}{80}$ on the interval $[0,10]$. To account for that, we
subtract an additional term in the right-hand side of (6.1). ∎

Before proceeding to the proof of Theorem 6.1, we require

######  Lemma 6.4.

Let $i,j\in\mathbb{Z}$, $i<j$. Let $g$ be a decreasing convex function on
$[i,j]$. Assume that

$n+\frac{1}{4}>g(i)\geq\dots\geq g(j-1)\geq n-\frac{3}{4}>g(j)$

for some $n\in\mathbb{Z}$. Then

$\frac{1}{2}\left\lfloor
g(i)+\frac{3}{4}\right\rfloor+\sum_{m=i+1}^{j-1}\left\lfloor
g(m)+\frac{3}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g(j)+\frac{3}{4}\right\rfloor\geq\int_{i}^{j}g(z)\,\mathrm{d}z+\frac{j-i}{4}-\frac{1}{2}.$
(6.3)

###### Proof of Lemma 6.4.

The left-hand side of (6.3) is equal to $(j-i)n-\frac{1}{2}$. By convexity of
$g$,

$\int_{i}^{j}g(z)\,\mathrm{d}z\leq\frac{j-i}{2}\left(g(j)+g(i)\right)\leq(j-i)\left(n-\frac{1}{4}\right).$

∎

###### Proof of Theorem 6.1.

As $g(0)\geq\frac{1}{4}$, we have

$N:=\left\lfloor g(0)+\frac{3}{4}\right\rfloor\geq 1.$ (6.4)

For $k=1,\dots,N-1$, denote

$M_{k}=M_{g,k}:=1+\max\left\\{m\in\left\\{0,\dots,\left\lfloor
b\right\rfloor\right\\}:g(m)\geq k+\frac{1}{4}\right\\}.$

We also denote $M_{N}=M_{g,N}:=0$. The assumption (5.1) yields $M_{k}>M_{k+1}$
for all $k=0,\dots,N-1$. Therefore, by Lemma 6.4,

$\begin{split}&\qquad\frac{1}{2}\left\lfloor
g(0)+\frac{3}{4}\right\rfloor+\sum_{m=1}^{\left\lfloor
b\right\rfloor}\left\lfloor g(m)+\frac{3}{4}\right\rfloor\\\
&=\sum_{n=0}^{N-1}\left(\frac{1}{2}\left\lfloor
g\left(M_{n+1}\right)+\frac{3}{4}\right\rfloor+\sum_{m=M_{n+1}+1}^{M_{n}-1}\left\lfloor
g(m)+\frac{3}{4}\right\rfloor+\frac{1}{2}\left\lfloor
g\left(M_{n}\right)+\frac{3}{4}\right\rfloor\right)\\\
&\geq\sum_{n=0}^{N-1}\left(\int_{M_{n+1}}^{M_{n}}g(z)\,\mathrm{d}z+\frac{M_{n}-M_{n+1}}{4}-\frac{1}{2}\right)=\int_{0}^{M_{0}}g(z)\,\mathrm{d}z+\frac{M_{0}}{4}-\frac{N}{2}.\end{split}$

Due to (6.4) we get

$\sum_{m=0}^{\left\lfloor b\right\rfloor}\left\lfloor
g(m)+\frac{3}{4}\right\rfloor\geq\int_{0}^{M_{0}}g(z)\,\mathrm{d}z+\frac{M_{0}}{4}.$

By convexity of $g$,

$\int_{M_{0}}^{b}g(z)\,\mathrm{d}z\leq\frac{1}{2}g(M_{0})(b-M_{0})\leq\frac{b-M_{0}}{8},$

and we get (6.1). ∎

We can now proceed to the proof of Theorem 4.3. We start with

######  Proposition 6.5.

Let $\lambda\geq 2$. Then

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)>\frac{\lambda^{2}}{4}+\frac{1}{4}R_{2}(\lambda),$
(6.5)

where

$R_{2}(\lambda):=3G_{\lambda}^{-1}\left(\frac{1}{4}\right)-\lambda\left(1+\frac{4}{\pi}\right)-3.$

###### Proof.

We have, with account of $G_{\lambda}(0)=\frac{\lambda}{\pi}$,

$\begin{split}\mathcal{P}_{2}^{\mathrm{N}}(\lambda)&=\left\lfloor
G_{\lambda}(0)+\frac{3}{4}\right\rfloor+2\sum_{m=1}^{\left\lfloor\lambda\right\rfloor}\left\lfloor
G_{\lambda}(m)+\frac{3}{4}\right\rfloor\\\
&=-\left\lfloor\frac{\lambda}{\pi}+\frac{3}{4}\right\rfloor+2\sum_{m=0}^{\left\lfloor\lambda\right\rfloor}\left\lfloor
G_{\lambda}(m)+\frac{3}{4}\right\rfloor.\end{split}$

We apply Theorem 6.1 to the sum in the right-hand side, with $g=G_{\lambda}$,
$b=\lambda$, and

$M_{G_{\lambda},0}=\left\lfloor
G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right\rfloor+1$

(see Remark 6.2). Given that $\lambda\geq 2$, we take into account the bound
(4.6) (which ensures that $M_{G_{\lambda},0}\leq\lambda$), and the value of
the integral from Lemma 4.6, leading to

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)-\frac{\lambda^{2}}{4}\geq\frac{3\left\lfloor
G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right\rfloor+3-\lambda-4\left\lfloor\frac{\lambda}{\pi}+\frac{3}{4}\right\rfloor}{4}.$

Finally, we use

$x-1<\left\lfloor x\right\rfloor\leq x$

to obtain (6.5). ∎

We now have

######  Proposition 6.6.

Let $\lambda\geq\Lambda_{1}$, where $\Lambda_{1}$ is given by (1.9). Then
$R_{2}(\lambda)\geq 0$.

###### Proof of Proposition 6.6.

Let $\sigma=\arccos\frac{5}{6}\in\left(0,\frac{\pi}{2}\right]$, and let

$\lambda\geq r_{1}(\sigma)=\frac{3\pi/2}{\sqrt{11}-5\arccos\frac{5}{6}},$

see (4.8). Then by Lemma 4.8, the bound (4.7) holds. Therefore,

$R_{2}(\lambda)\geq\lambda\left(3\cos\sigma-1-\frac{4}{\pi}\right)-3=\lambda\frac{3\pi-8}{2\pi}-3,$

and is non-negative if

$\lambda\geq\frac{6\pi}{3\pi-8}=\Lambda_{1}>0.$

As $\Lambda_{1}>r_{1}(\sigma)$ (with the inequality easy to prove rigorously
using the method of verified rational approximations discussed in §8), the
result follows. ∎

Theorem 4.3 now immediately follows from Propositions 6.6 and 6.5.

## §7. Proof of Theorem 4.2

.

We have the following “dimension reduction” formula.

######  Theorem 7.1.

Let $d\geq 3$. Then

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda)=\sum_{n=0}^{\left\lfloor\lambda-\frac{d}{2}+1\right\rfloor}\binom{n+d-3}{d-3}\tilde{\mathcal{P}}^{\mathrm{D}}_{n+\frac{d}{2}-1}\left(\lambda\right),$
(7.1)

where for $r\in[0,\lambda]$ we denote by

$\tilde{\mathcal{P}}^{\mathrm{D}}_{r}(\lambda):=\left\lfloor\tilde{G}_{\lambda,r}(0)+\frac{1}{4}\right\rfloor+2\sum_{j=1}^{\left\lfloor\lambda-r\right\rfloor}\left\lfloor\tilde{G}_{\lambda,r}(j)+\frac{1}{4}\right\rfloor=\left\lfloor
G_{\lambda}(r)+\frac{1}{4}\right\rfloor+2\sum_{j=1}^{\left\lfloor\lambda-r\right\rfloor}\left\lfloor
G_{\lambda}(j+r)+\frac{1}{4}\right\rfloor$ (7.2)

the “standard” two-dimensional weighted shifted lattice point count under the
graph of the function $\tilde{G}_{\lambda,r}(t):=G_{\lambda}(t+r)$,
$t\in[0,\lambda-r]$.

We remark that in comparison to our original definition

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda-\frac{d}{2}+1\right\rfloor}\kappa_{d,m}\left\lfloor
G_{\lambda}\left(m+\frac{d}{2}-1\right)+\frac{1}{4}\right\rfloor,$

see (2.8), where the weights $\kappa_{d,m}$ are attached at each individual
abscissa $m$, the formula in the right-hand side of (7.1) attaches weights
$\binom{n+d-3}{d-3}$ to the whole counts
$\tilde{\mathcal{P}}^{\mathrm{D}}_{n+\frac{d}{2}-1}\left(\lambda\right)$,
which we will later estimate using the previously proven Theorem 5.1. Note
also that for $\lambda<\frac{d}{2}-1$, the equation (7.1) becomes the trivial
identity $0=0$ by our notational convention for sums, see Remark 2.2.

###### Proof of Theorem 7.1.

With account of (7.2), the right-hand side of (7.1) reads

$\sum_{n=0}^{\left\lfloor\lambda-\frac{d}{2}+1\right\rfloor}\binom{n+d-3}{d-3}\left(\left\lfloor
G_{\lambda}\left(n+\frac{d}{2}-1\right)+\frac{1}{4}\right\rfloor+2\sum_{j=1}^{\left\lfloor\lambda-n-\frac{d}{2}+1\right\rfloor}\left\lfloor
G_{\lambda}\left(j+n+\frac{d}{2}-1\right)+\frac{1}{4}\right\rfloor\right).$

For a fixed $m\in\\{0\\}\cup\mathbb{N}$, the contribution of $\left\lfloor
G_{\lambda}\left(m+\frac{d}{2}-1\right)+\frac{1}{4}\right\rfloor$ in this
expression appears with the factor

$\begin{split}\binom{m+d-3}{d-3}+2\sum_{j=0}^{m-1}\binom{j+d-3}{d-3}&=\sum_{j=0}^{m}\binom{j+d-3}{d-3}+\sum_{j=0}^{m-1}\binom{j+d-3}{d-3}\\\
&=\sum_{i=d-3}^{m+d-3}\binom{i}{d-3}+\sum_{i=d-3}^{m-d-4}\binom{i}{d-3}=\binom{m+d-2}{d-2}+\binom{m+d-3}{d-2}\\\
&=\binom{m+d-1}{d-1}-\binom{m+d-3}{d-1}=\kappa_{d,m},\end{split}$

where we have used the standard identity [DLMF, Eq. 26.3.7]

$\sum_{i=l}^{r}\binom{i}{l}=\binom{r+1}{l+1},\qquad r\geq l,$ (7.3)

and another identity [DLMF, Eq. 26.3.5],

$\binom{k}{l}=\binom{k+1}{l+1}-\binom{k}{l+1},\qquad k\geq l.$

Thus, the contributions of $\left\lfloor
G_{\lambda}\left(m+\frac{d}{2}-1\right)+\frac{1}{4}\right\rfloor$ in both
sides of (7.1) coincide. ∎

Before proceeding to the proof of Theorem 4.2 we will introduce some
additional notation and state some auxiliary facts which will be required
later. Let, for $x\geq 0$,

$\Pi_{n}(x):=\prod_{j=1}^{n}(x+j)=(x+1)\cdots(x+n)\quad\text{for
}n\in\mathbb{N},\qquad\Pi_{0}(x):=1.$

The function $\Pi_{n}(x)$ is closely related to _Pochhammer’s symbol_ , or the
_rising factorial_ $(x)_{n}:=x\cdots(x+n-1)$ (for which numerous other
notation is also used) in the sense that $\Pi_{n}(x)=(x+1)_{n}$. We also have

$\binom{i+j}{i}=\frac{(i+1)\cdots(i+j)}{j!}=\frac{\Pi_{j}(i)}{j!}=\frac{\Pi_{i}(j)}{i!}.$

Let us introduce, for $d\geq 3$, a piecewise-constant function

$f_{d}(t):=\begin{cases}0,&\qquad\text{if }t<\frac{d}{2}-1,\\\
\frac{\Pi_{d-2}(m)}{(d-2)!}=\binom{m+d-2}{d-2},&\qquad\text{if }\left\lfloor
t-\frac{d}{2}+1\right\rfloor=m\geq 0,\end{cases}$ (7.4)

and let

$F_{d}(z):=\int_{0}^{z}f_{d}(t)\,\mathrm{d}t.$

In what follows we will require an upper polynomial bound on $F_{d}(z)$. If
$z<\frac{d}{2}-1$, then $F_{d}(z)=0$. Let $z\geq\frac{d}{2}-1$ and
$m=\left\lfloor z-\frac{d}{2}+1\right\rfloor\geq 0$. Then

$\begin{split}F_{d}(z)=\int_{0}^{z}f_{d}(t)\,\mathrm{d}t&=\left(\sum_{k=0}^{m-1}\int_{k+d/2-1}^{k+d/2}f_{d}(t)\,\mathrm{d}t\right)+\int_{m+d/2-1}^{z}f_{d}(t)\,\mathrm{d}t\\\
&=\left(\sum_{k=0}^{m-1}\binom{k+d-2}{d-2}\right)+\left(z-m-\frac{d}{2}+1\right)\binom{m+d-2}{d-2}\\\
&=\binom{m+d-2}{d-1}+\left(z-m-\frac{d}{2}+1\right)\binom{m+d-2}{d-2}\\\
&=\frac{\Pi_{d-2}(m)}{(d-1)!}\left((d-1)z-(d-2)m-\frac{(d-1)(d-2)}{2}\right),\end{split}$
(7.5)

where we used (7.3) to evaluate the sum of binomial coefficients.

To establish a bound on $F_{d}(z)$, we apply the AM-GM inequality

$l\left(\beta_{1}\cdots\beta_{\ell}\right)^{1/l}\leq\beta_{1}+\dots+\beta_{l},\qquad
l\in\mathbb{N},\quad\beta_{1},\dots,\beta_{l}\geq 0,$

with $l=d-1$ and

$\beta_{j}=\begin{cases}(m+j)\Pi_{d-1}(m)&\qquad\text{if }j\leq d-2,\\\
z^{d-1}&\qquad\text{if }j=d-1,\end{cases}$

yielding

$\begin{split}(d-1)z\Pi_{d-2}(m)&\leq\Pi_{d-2}(m)\left((m+1)+\dots+(m+d-2)\right)+z^{d-1}\\\
&=\left((d-2)m+\frac{(d-2)(d-1)}{2}\right)\Pi_{d-2}(m)+z^{d-1}.\end{split}$

Collecting together the terms with $\Pi_{d-2}(m)$ and substituting the
resulting bound into the right-hand side of (7.5), we deduce that

$F_{d}(z)\leq\tilde{F}_{d}(z):=\frac{z^{d-1}}{(d-1)!}\qquad\text{for all
}z\geq 0.$ (7.6)

We will require an auxiliary

######  Lemma 7.2.

Let $f$ be a locally integrable function on $[0,+\infty)$, and let
$\tilde{F}\in C^{1}[0,+\infty)$ with $\tilde{F}(0)=0$ be such that

$F(z):=\int_{0}^{z}f(t)\,\mathrm{d}t\leq\tilde{F}(z)\qquad\text{for all }z\geq
0.$

Let $b>0$, and let $g\in C^{1}[0,b]$ be a decreasing function such that
$g(b)=0$. Then

$\int_{0}^{b}f(z)g(z)\,\mathrm{d}z\leq\int_{0}^{b}\tilde{F}^{\prime}(z)g(z)\,\mathrm{d}z.$
(7.7)

###### Proof.

Integrating by parts, we get

$\begin{split}\int_{0}^{b}\tilde{F}^{\prime}(z)g(z)\,\mathrm{d}z-\int_{0}^{b}f(z)g(z)\,\mathrm{d}z&=\int_{0}^{b}\left(\tilde{F}^{\prime}(z)-F^{\prime}(z)\right)g(z)\,\mathrm{d}z\\\
&=\left.\left(\tilde{F}(z)-F(z)\right)g(z)\right|_{0}^{b}-\int_{0}^{b}\left(\tilde{F}(z)-F(z)\right)g^{\prime}(z)\,\mathrm{d}z\geq
0,\end{split}$

since $\tilde{F}(0)=F(0)=g(b)=0$ and
$\left(\tilde{F}(z)-F(z)\right)g^{\prime}(z)\leq 0$, thus proving (7.7). ∎

We now proceed to the proof of Theorem 4.2 proper.

###### Proof of Theorem 4.2.

First of all, note that for $\lambda<\frac{d}{2}-1$ there is nothing to prove
as in this case $\mathcal{P}^{\mathrm{D}}_{d}(\lambda)=0$. For
$\lambda\geq\frac{d}{2}-1$, we apply Theorem 5.1 with
$g=\tilde{G}_{\lambda,r}$, $r=n+\frac{d}{2}-1$, and
$b=\lambda-r=\lambda-n-\frac{d}{2}+1$ to the right-hand side of (7.2), giving

$\tilde{\mathcal{P}}^{\mathrm{D}}_{n+\frac{d}{2}-1}(\lambda)<2\int_{0}^{\lambda-n-\frac{d}{2}+1}G_{\lambda}\left(t+n+\frac{d}{2}-1\right)\,\mathrm{d}t=2\int_{n+\frac{d}{2}-1}^{\lambda}G_{\lambda}(z)\,\mathrm{d}z$

for $n=0,\dots,\left\lfloor\lambda-\frac{d}{2}+1\right\rfloor$. We now
substitute the results into (7.1), yielding, with account of (7.3), the bound

$\begin{split}\mathcal{P}^{\mathrm{D}}_{d}(\lambda)&<2\sum_{n=0}^{\left\lfloor\lambda-\frac{d}{2}+1\right\rfloor}\binom{n+d-3}{d-3}\int_{n+\frac{d}{2}-1}^{\lambda}G_{\lambda}(z)\,\mathrm{d}z\\\
&=2\int_{\frac{d}{2}-1}^{\lambda}\left(\sum_{n=0}^{\left\lfloor
z-\frac{d}{2}+1\right\rfloor}\binom{n+d-3}{d-3}\right)G_{\lambda}(z)\,\mathrm{d}z\\\
&=2\int_{\frac{d}{2}-1}^{\lambda}\binom{\left\lfloor
z-\frac{d}{2}+1\right\rfloor+d-2}{d-2}G_{\lambda}(z)\,\mathrm{d}z.\end{split}$
(7.8)

Using notation (7.4), we rewrite (7.8) as

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda)<2\int_{0}^{\lambda}f_{d}(z)G_{\lambda}(z)\,\mathrm{d}z.$
(7.9)

We now apply Lemma 7.2 to the right-hand side of (7.9), taking $f=f_{d}$,
$g=G_{\lambda}$, and $\tilde{F}(z)=\tilde{F}_{d}(z)=\frac{z^{d-1}}{(d-1)!}$ by
(7.6), which gives

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda)<2\int_{0}^{\lambda}\tilde{F}^{\prime}_{d}(z)G_{\lambda}(z)\,\mathrm{d}z=\frac{2}{(d-2)!}\int_{0}^{\lambda}z^{d-2}G_{\lambda}(z)\,\mathrm{d}z.$

Finally, applying Corollary 4.7 gives

$\mathcal{P}^{\mathrm{D}}_{d}(\lambda)<W_{d}(\lambda)$

as required. ∎

## §8. Closing the gap: the proof of Theorem 4.4

We describe the algorithm (based on the two Principles stated in §1) of
verifying the statement

$\mathcal{P}_{2}^{\mathrm{N}}(\lambda)>\frac{\lambda^{2}}{4}\qquad\text{for
all }\lambda\in[3,14],$ (8.1)

with the lattice point counts given explicitly by

$\mathcal{P}^{\mathrm{N}}_{2}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda\right\rfloor}\kappa_{2,m}\left\lfloor
G_{\lambda}\left(m\right)+\frac{3}{4}\right\rfloor.$

The first Principle is realised with the help of the following simple lemma,
which ensures that our algorithm described below requires only a finite number
of steps.

######  Lemma 8.1.

If the inequality (8.1) holds for a particular $\lambda_{0}$, that is, we have

$e(\lambda_{0}):=\mathcal{P}_{2}^{\mathrm{N}}(\lambda_{0})-\frac{\lambda_{0}^{2}}{4}>0,$

this inequality also holds for all
$\lambda\in\left(\lambda_{0},\lambda_{0}+\delta(\lambda_{0})\right)=\left(\lambda_{0},\sqrt{\lambda_{0}^{2}+4e(\lambda_{0})}\right)$,
where

$\delta(\lambda_{0}):=\sqrt{\lambda_{0}^{2}+4e(\lambda_{0})}-\lambda_{0}.$

###### Proof.

The result immediately follows from the facts that
$\mathcal{P}^{\mathrm{N}}_{2}(\lambda)$ is non-decreasing in $\lambda$ and
that $\Lambda=\lambda_{0}+\delta(\lambda_{0})$ is the positive root of the
equation $\frac{\Lambda^{2}}{4}=\frac{\lambda_{0}^{2}}{4}+e(\lambda_{0})$. ∎

To implement the second Principle, we work with rational numbers only. Let,
for $x\in\mathbb{R}$, $\underline{x}\leq x\leq\overline{x}$, where
$\underline{x},\overline{x}\in\mathbb{Q}$ are some lower and upper rational
approximations of $x$. The function $G_{\lambda}(z)$ does not as a rule take
rational values even for rational $\lambda$ and $z$, so to overcome this we
work instead with

$\underline{\mathcal{P}}^{\mathrm{N}}_{2}(\lambda)=\sum_{m=0}^{\left\lfloor\lambda\right\rfloor}\kappa_{2,m}\left\lfloor\underline{G}_{\lambda}\left(m\right)+\frac{3}{4}\right\rfloor,\qquad\lambda\in\mathbb{Q},$

where

$\underline{G}_{\lambda}(z)=\frac{1}{\overline{\pi}}\left(\underline{\sqrt{\lambda^{2}-z^{2}}}-z\operatorname{\overline{arccos}}\frac{z}{\lambda}\right),\qquad
z\in\mathbb{Q}\cap[0,\lambda].$

Of course, $\underline{G}_{\lambda}(z)\leq G_{\lambda}(z)$ and therefore
$\underline{\mathcal{P}}^{\mathrm{N}}_{2}(\lambda)\leq\mathcal{P}^{\mathrm{N}}_{2}(\lambda)$,
for $\lambda,z\in\mathbb{Q}$. Obviously, taking integer parts of rational
numbers, as well as other arithmetic operations on them is exact and does not
introduce any numerical errors. We now describe how to construct _verified
rational approximations_ of the square roots $\underline{\sqrt{\cdot}}$ and
arccosines $\operatorname{\overline{arccos}}(\cdot)$. For the former, any
guess (say, obtained from numerics) can be directly verified by taking squares
and comparing rationals, which is rigorous. To verify our approximations of
arccosines (taken at rational points) one may proceed as follows. Define the
functions
$\operatorname{\underline{cos}},\operatorname{\overline{cos}}:\mathbb{Q}\cap\left(0,\frac{\pi}{2}\right]\to\mathbb{Q}$
as

$\operatorname{\overline{cos}}x:=T_{12}[\cos](x),\qquad\operatorname{\underline{cos}}x:=T_{14}[\cos](x),$

where $T_{t}[\cos](x)$ is the Taylor polynomial of $\cos x$ at $x=0$ of degree
$t$, so that $\operatorname{\underline{cos}}x<\cos
x<\operatorname{\overline{cos}}x$, see, for example, [Dör65, Problem
15].111111We thank the referee for pointing out this reference. Then

$\operatorname{\overline{cos}}\left(\overline{\beta}\right)<x<\operatorname{\underline{cos}}\left(\underline{\beta}\right)\text{\qquad
implies\qquad}\underline{\beta}:=\operatorname{\underline{arccos}}x<\beta=\arccos
x<\operatorname{\overline{arccos}}x=:\overline{\beta},$

and the verification is again reduced to elementary operations on rationals.
In the same manner,

$\underline{\pi}=3\operatorname{\underline{arccos}}\frac{1}{2}\quad\text{and}\quad\overline{\pi}=3\operatorname{\overline{arccos}}\frac{1}{2}$

provide verified rational approximations for $\pi$.

To finish describing our process, we need also to rationalise the square root
appearing in the definition of $\delta(\lambda)$: we effectively replace
$e(\lambda)$ by a smaller number

$\underline{e}(\lambda):=\underline{\mathcal{P}}^{\mathrm{N}}_{2}(\lambda)-\frac{\lambda^{2}}{4},$
(8.2)

and also replace $\delta(\lambda)$ by a smaller number

$\underline{\delta}(\lambda):=\underline{\sqrt{\lambda^{2}+4\underline{e}(\lambda)}}-\lambda,$
(8.3)

where a verification is again by taking squares.

######  Remark 8.2.

In practice, we use the following process to find lower and upper rational
approximations of a number $x\in\mathbb{R}$ (which may be a square root, or an
arccosine). Throughout, we fix a relatively small number $\varepsilon$ (say,
$\varepsilon=10^{-3}$) as an accuracy parameter. We find numerically some
approximation $x_{0}$ of $x$ (which may be above or below $x$) with some
better accuracy. Then, we define $\underline{x}$ as the rational number in the
interval $[x_{0}-3\varepsilon,x_{0}-\varepsilon]$ with the smallest possible
denominator, and $\overline{x}$ as the rational number in the interval
$[x_{0}+\varepsilon,x_{0}+3\varepsilon]$ with the smallest possible
denominator, using a modification of a fast algorithm for traversing the
Stern–Brocot tree [For07]. As we always verify the resulting approximations
using the procedures described above, we do not in fact depend on the quality
of an original numerical “guess” $x_{0}$ as long as $|x_{0}-x|<\varepsilon$. ∎

Thus, our main algorithms work as follows. In order to prove (8.1) for
$\lambda\in[\Lambda_{0},\Lambda_{1}]$, we move upwards: set
$\lambda=\underline{\Lambda_{0}}$, compute the margin $\underline{e}(\lambda)$
from (8.2), set
$\lambda_{\text{new}}=\lambda+\underline{\delta}\left(\lambda\right)$ using
(8.3), and continue the process. If the margins are positive on each step, the
process will stop successfully if after a finite number of steps we reach
$\lambda>\overline{\Lambda_{1}}$, see Figure 6.

$\lambda\leftarrow\underline{\Lambda_{0}}$

$\text{stepnumber}\leftarrow 0$

while $\lambda\leq\overline{\Lambda_{1}}$ do

$\text{stepnumber}\leftarrow\text{stepnumber}+1$

$p\leftarrow\underline{\mathcal{P}}^{\mathrm{N}}_{2}(\lambda)$

$e\leftarrow p-\frac{\lambda^{2}}{4}$

if $e>0$ then

$\lambda\leftarrow\underline{\sqrt{\lambda^{2}+4e}}$

else

print “Proof failed ☹”; exit

end if

end while

print “Success in ”, stepnumber, “steps ☺”

The algorithm for proving (8.1) Figure 6: The basic algorithm.

The algorithm works extremely fast (when implemented in Mathematica, see the
footnote on the title page), thus proving Theorem 4.4: in principle, with
enough patience the whole implementation can be done by hand. We summarise its
outcomes in Table LABEL:table:2.

Table 1: Detailed output of the computer-assisted algorithm. Step | $\lambda$ | $\underline{e}(\lambda)$ | $\underline{\delta}(\lambda)$
---|---|---|---
$1$ | $3<\Lambda_{0}$ | $\frac{3}{4}$ | $\frac{6}{13}$
$2$ | $\frac{45}{13}$ | $\frac{1355}{676}$ | $\frac{223}{221}$
$3$ | $\frac{76}{17}$ | $\frac{868}{289}$ | $\frac{584}{493}$
$4$ | $\frac{164}{29}$ | $\frac{3368}{841}$ | $\frac{995}{783}$
$5$ | $\frac{187}{27}$ | $\frac{11687}{2916}$ | $\frac{29}{27}$
$6$ | $8$ | $3$ | $\frac{43}{60}$
$7$ | $\frac{523}{60}$ | $\frac{57671}{14400}$ | $\frac{227}{260}$
$8$ | $\frac{374}{39}$ | $\frac{6098}{1521}$ | $\frac{719}{897}$
$9$ | $\frac{239}{23}$ | $\frac{10591}{2116}$ | $\frac{339}{368}$
$10$ | $\frac{181}{16}$ | $\frac{4103}{1024}$ | $\frac{11}{16}$
$11$ | $12$ | $6$ | $\frac{24}{25}$
$12$ | $\frac{324}{25}$ | $\frac{2506}{625}$ | $\frac{241}{400}$
$13$ | $\frac{217}{16}$ | $\frac{7183}{1024}$ | $\frac{271}{272}$
$14$ | $\frac{495}{34}>\Lambda_{1}$ |  |
|  |  |
|  |  |

## §9. Proof of Theorem 1.12

Let $S_{\alpha}$ be a circular sector of aperture $0<\alpha\leq 2\pi$. The
eigenvalues of the Dirichlet and Neumann Laplacians on $S_{\alpha}$ are easily
found by separation of variables. They are all simple, and are given by

$\lambda_{m,k}=\left(j_{\frac{m\pi}{\alpha},k}\right)^{2}\qquad
m\in\mathbb{N},\quad k\in\mathbb{N},$

and

$\mu_{m,k}=\left(j^{\prime}_{\frac{m\pi}{\alpha},k}\right)^{2},\qquad
m\in\mathbb{N}\cup\\{0\\},\quad k\in\mathbb{N},$

respectively. Therefore, the corresponding eigenvalue counting functions are

$\mathcal{N}^{\mathrm{D}}_{S_{\alpha}}(\lambda)=\sum_{m=1}^{\left\lfloor\frac{\alpha\lambda}{\pi}\right\rfloor}\\#\left\\{k\in\mathbb{N}:j_{\frac{m\pi}{\alpha},k}\leq\lambda\right\\}\qquad\text{and}\qquad\mathcal{N}^{\mathrm{N}}_{S_{\alpha}}(\lambda)=\sum_{m=0}^{\left\lfloor\frac{\alpha\lambda}{\pi}\right\rfloor}\\#\left\\{k\in\mathbb{N}:j^{\prime}_{\frac{m\pi}{\alpha},k}\leq\lambda\right\\},$

where the both sums are finite since $j_{\nu,1}>j^{\prime}_{\nu,1}\geq\nu$.

Assume for the moment that the sector $S_{\alpha}$ contains a half-disk, that
is $\alpha\in[\pi,2\pi]$. By Proposition 3.1 with account of (3.1), (2.4) and
(2.9), we have

$\mathcal{N}^{\mathrm{D}}_{S_{\alpha}}(\lambda)\leq\sum_{m=1}^{\left\lfloor\frac{\alpha\lambda}{\pi}\right\rfloor}\left\lfloor
G_{\lambda}\left(\frac{m\pi}{\alpha}\right)+\frac{1}{4}\right\rfloor.$ (9.1)

Set

$g_{\lambda,\alpha}(t):=G_{\lambda}\left(\frac{\pi t}{\alpha}\right),\qquad
t\in\left[0,b\right],$

where

$b:=\frac{\alpha\lambda}{\pi}.$

Then $g_{\lambda,\alpha}$ is a monotone decreasing convex function on $[0,b]$
with $g(b)=0$; moreover,

$g^{\prime}_{\lambda,\alpha}(t)=\frac{\pi}{\alpha}G^{\prime}_{\lambda}\left(\frac{\pi
t}{\alpha}\right)\in\left[-\frac{\pi}{2\alpha},0\right]\subset\left[-\frac{1}{2},0\right]$

due to our assumption $\alpha\geq\pi$. With this notation, the right-hand side
of (9.1) becomes

$\sum_{m=1}^{\left\lfloor b\right\rfloor}\left\lfloor
g_{\lambda,\alpha}(m)+\frac{1}{4}\right\rfloor,$

and we can estimate it from above directly by Theorem 5.1 with
$g=g_{\lambda,\alpha}$, giving

$\sum_{m=1}^{\left\lfloor b\right\rfloor}\left\lfloor
g_{\lambda,\alpha}(m)+\frac{1}{4}\right\rfloor<\int_{0}^{\alpha\lambda/\pi}g_{\lambda,\alpha}(t)\,\mathrm{d}t=\frac{\alpha}{\pi}\int_{0}^{\lambda}G_{\lambda}(z)\,\mathrm{d}z=\frac{\alpha\lambda^{2}}{8\pi}.$

Substituting this into (9.1) proves the Dirichlet Pólya’s conjecture for
sectors containing a half-disk.

We now turn to the Neumann problem in $S_{\alpha}$, still assuming that
$\alpha\in[\pi,2\pi]$. Following the same argument as in Lemma 1.3, we
conclude that the Neumann Pólya’s conjecture for $S_{\alpha}$ holds for
$\lambda\leq 2\sqrt{\frac{6\pi}{\alpha}}$. We therefore may assume that

$\lambda>2\sqrt{\frac{6\pi}{\alpha}}\geq 2\sqrt{3}$ (9.2)

from now on.

Using once more Proposition 3.1 yields

$\mathcal{N}^{\mathrm{N}}_{S_{\alpha}}(\lambda)\geq\sum_{m=0}^{\left\lfloor\frac{\alpha\lambda}{\pi}\right\rfloor}\left\lfloor
G_{\lambda}\left(\frac{m\pi}{\alpha}\right)+\frac{3}{4}\right\rfloor=\sum_{m=0}^{\left\lfloor
b\right\rfloor}\left\lfloor g_{\lambda,\alpha}(m)+\frac{3}{4}\right\rfloor,$
(9.3)

with the same function $g_{\lambda,\alpha}$ and parameter $b$ as above.

We are now going to use Theorem 6.1 with $g=g_{\lambda,\alpha}$ to estimate
the right-hand side of (9.3) from below. Note that
$g_{\lambda,\alpha}(0)=\frac{\lambda}{\pi}>\frac{1}{4}$ by (9.2). Also, by
(6.2)

$M_{0}=M_{g_{\lambda,\alpha},0}=\left\lfloor
g_{\lambda,\alpha}^{-1}\left(\frac{1}{4}\right)\right\rfloor+1=\left\lfloor\frac{\alpha}{\pi}G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right\rfloor+1,$

and in order to apply Theorem 6.1 we need to ensure that $M_{0}\leq
b=\frac{\alpha\lambda}{\pi}$. But this is true since, with account of
$\alpha\geq\pi$,

$M_{0}-b\leq\frac{\alpha}{\pi}G_{\lambda}^{-1}\left(\frac{1}{4}\right)+1-\frac{\alpha\lambda}{\pi}\leq\frac{\alpha}{\pi}\left(G_{\lambda}^{-1}\left(\frac{1}{4}\right)+1-\lambda\right)<0$

by (4.6). Then, (6.1) implies

$\sum_{m=0}^{\left\lfloor b\right\rfloor}\left\lfloor
g_{\lambda,\alpha}(m)+\frac{3}{4}\right\rfloor\geq\frac{\alpha\lambda^{2}}{8\pi}-\frac{b-3M_{0}}{8}.$
(9.4)

We now reason as in the proof of Theorem 4.3: if we can show that $b-3M_{0}<0$
for all $\lambda\geq 2\sqrt{3}$, this would prove, via the combination of
(9.3) and (9.4), that
$\mathcal{N}^{\mathrm{N}}_{S_{\alpha}}(\lambda)>\frac{\alpha\lambda^{2}}{8\pi}$.
We have

$b-3M_{0}=\frac{\alpha\lambda}{\pi}-3\left\lfloor\frac{\alpha}{\pi}G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right\rfloor-3\leq\frac{\alpha}{\pi}\left(\lambda-3G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right)<\frac{\alpha}{\pi}\left(\lambda-2G_{\lambda}^{-1}\left(\frac{1}{4}\right)\right).$

We now apply the second statement of Lemma 4.8 with $\sigma=\frac{\pi}{3}$
which guarantees that $b-3M_{0}<0$ for

$\lambda>r_{1}\left(\frac{\pi}{3}\right)=\frac{3\pi}{6\sqrt{3}-2\pi}.$

As $\frac{3\pi}{6\sqrt{3}-2\pi}<2\sqrt{3}$, this finishes the proof of Theorem
1.12 for sectors $S_{\alpha}$ of aperture $\alpha\in[\pi,2\pi]$.

To complete the proof of Theorem 1.12 we now need to consider the case
$\alpha\in(0,\pi)$. Set

$\ell:=\left\lfloor\frac{2\pi}{\alpha}\right\rfloor\geq 2.$

Then $\tilde{\alpha}=\ell\alpha\in[\pi,2\pi]$, Pólya’s conjecture holds for
$S_{\tilde{\alpha}}$, and $S_{\alpha}$ tiles $S_{\tilde{\alpha}}$. By Theorem
1.8, Pólya’s conjecture holds for $S_{\alpha}$.

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