# Triviality of mean-field $\varphi^{4}_{4}$-theories using the flow equations

Pierre WANG<EMAIL_ADDRESS>Centre de Physique Théorique (CPHT),
CNRS, UMR 7644 Institut Polytechnique de Paris, 91128 Palaiseau, France
Christoph KOPPER<EMAIL_ADDRESS>Centre de Physique
Théorique (CPHT), CNRS, UMR 7644 Institut Polytechnique de Paris, 91128
Palaiseau, France

(August, 2024)

# Triviality proof for mean-field $\varphi^{4}_{4}$-theories

Pierre WANG<EMAIL_ADDRESS>Centre de Physique Théorique (CPHT),
CNRS, UMR 7644 Institut Polytechnique de Paris, 91128 Palaiseau, France
Christoph KOPPER<EMAIL_ADDRESS>Centre de Physique
Théorique (CPHT), CNRS, UMR 7644 Institut Polytechnique de Paris, 91128
Palaiseau, France

(August, 2024)

###### Abstract

The differential equations of the Wilson renormalization group are a powerful
tool to study the Schwinger functions of Euclidean quantum field theory. In
particular renormalization theory can be based entirely on inductively
bounding their perturbatively expanded solutions. Recently the solutions of
these equations for scalar field theory have been analysed rigorously without
recourse to perturbation theory, at the cost of restricting to the mean-field
approximation [1]. In particular it was shown there that one-component
$\varphi^{4}_{4}$-theory is trivial if the bare coupling constant of the UV
regularized theory is not large. This paper presents progress w.r.t. [1]:
1\. The upper bound on the bare coupling is sent to infinity and the proof is
extended to $O(N)$ vector models.
2\. The unphysical infrared cutoff used in [1] for technical simplicity is
replaced by a physical mass.

##  1 Introduction

#####

Quantum field theory is the fundamental framework of theoretical physics. It
comprises both quantum mechanics and special relativity and acts as a powerful
tool to study systems with a large or infinite number of degrees of freedom.
Euclidean field theory is used in statistical mechanics in order to study
critical behavior. Relativistic field theory is related to the Euclidean
theory via analytic continuation. In perturbative quantum field theory, one
typically expands the correlation functions which appear in transition
amplitudes, in a power series w.r.t. a coupling constant $\lambda$ which
represents the strength of the interaction. Feynman graph amplitudes are
contributions to these correlation functions, they are typically of order
$\lambda^{V}$ if the number of vertices is V.

Such expansions may lead to contributions which are ill-defined. E.g. in the
$\varphi^{4}_{4}$-theory the first order correction to the two-point function
is UV-divergent. A standard procedure is then to first regularize the theory,
and to renormalize it afterwards through the introduction of counterterms. In
$\varphi^{4}_{4}$-theory in four dimensions, one adds counterterms in the
lagrangian for the mass, the coupling constant and the wavefunction so that
the previously divergent graphs eventually become finite.

The differential flow equations of the Wilson renormalization group [2]-[3]
were introduced for the first time by Wegner and Houghton in 1972 [4].
Polchinski in his seminal paper [5] proved the perturbative renormalizability
of the scalar $\varphi^{4}_{4}$-theory using these flow equations. Instead of
dealing with the combinatorial complexity of Feynman diagrams, he analyzed the
Schwinger functions as a whole. They are regularized by an UV-cutoff and an
infrared cutoff, also called flow parameter. The flow equations are
differential equations whose solutions are the regularized connected amputated
Schwinger functions. The solutions of the perturbative flow equations can be
bounded using an inductive scheme. These bounds are sufficiently strong to
prove renormalizability [6]. Later on, the proof of perturbative
renormalizability was extended to the massless $\varphi^{4}_{4}$-theory [7],
to the non-linear $\sigma$-model [8]. The flow equations were also used to
prove rigourously renormalizability of spontaneously broken $SU(2)$ Yang-Mills
theory [9]-[10], perturbative renormalizability in Minkowski space [11]. Other
results in mathematical physics which have been established using the flow
equations include the convergence of the operator product expansion in
perturbation theory [12] and local Borel summability of the perturbation
expansion for Euclidean massive $\varphi^{4}_{4}$-theory [13],[14].

Our paper is concerned with the so-called triviality of the $\varphi^{4}$
theory in four dimensions. In the standard model, this theory appears as the
pure Higgs sector after spontaneous symmetry breaking when ignoring the
coupling of the Higgs field to the gauge fields and to the fermionic fields.
Aizenman and Fröhlich in [15],[16] and [17] proved the triviality of the
continuum limit of lattice regularized Euclidean one-component
$\varphi^{4}$-theory in $d>4$ dimensions, in the sense that the truncated
four-point function of the theory vanishes in this limit. Recently Aizenman
and Duminil-Copin extended the proof to $d=4$ using multi-scale analysis in
[18]. The question whether the Standard model is trivial or not remains open,
in particular because the aforementioned proofs did not consider scalar fields
coupled to other fields such as gauge fields or fermionic fields. It is worth
to note that it has been shown that the continuum limit of lattice QED in
dimensions greater than four is trivial [19] while the question in four
dimensions remains open.

In this paper we take up and extend the work from [1]. We again restrict to
the mean field approximation. The paper is organized as follows. In Sect.2.1
we recall the flow equations for $O(N)$ vector models. The mean-field
approximation of these flow equations is presented in Sect.2.2. In Sect.2.3 we
comment on the analyticity of the solutions of the flow equations w.r.t. the
flow parameter called $\alpha$ and its consequences on the uniqueness of the
solutions of the flow equations for fixed boundary conditions. In Sect.3 we
prove the triviality of the mean-field $O(N)$ vector models, which include in
particular pure $\varphi_{4}^{4}$-theory, for any value of the bare coupling
using the technical IR cutoff from [1]. In Sect.3.1, we prove the existence of
a trivial solution of the mean-field flow equations then in Sect.3.2 we prove
the uniqueness of the trivial solution for fixed mean-field boundary
conditions. We end this section commenting on the large $N$ limit in Sect.3.3.
In Sect.4 we replace the technical IR cutoff from [1] by the physical one of a
massive theory. In Sect.4.1 we derive the flow equations in this case, in
Sect.4.2 we again prove triviality of $\varphi^{4}_{4}$-theory in the new
setting.

##  2 Flow equations in the mean-field approximation

###  2.1 The flow equations for the $O(N)$-model

#####

We want to analyse self-interacting $N$-component vector scalar field theories
on four-dimensional Euclidean space. We assume the theory has
${\mathbb{Z}}_{2}$ symmetry under $\varphi\mapsto-\varphi$. An $N$-component
vector model with $O(N)$ symmetry was first introduced by Stanley in [20] to
generalize the Ising model ($N=1$), the XY model ($N=2$) and the Heisenberg
model $(N=3)$. In the continuum description the scalar field $\varphi$ has now
$N$ real components
$\varphi(x)=\left[\varphi_{1}(x),\cdots,\varphi_{N}(x)\right]^{T}\,$, and the
lagrangian has a global $O(N)$ symmetry. The behaviour of the solutions of the
$O(N)$ model in the large $N$ limit has been studied in detail in [21] and by
Moshe and Zinn-Justin in [22], carrying out an expansion in powers of
$\frac{1}{N}\,$. The result is a non-trivial theory which turns out to be the
exact solution of the spherical model [23]. We will follow the steps from [1]
to derive the flow equations for the $O(N)$-model and perform the mean-field
approximation afterwards. Using $O(N)$ symmetry, the form of the mean-field
flow equations generalizes those of the single-component theory.

To derive the flow equations, we base ourselves on [24], [1]. We adopt the
following convention and shorthand notation for the Fourier transform

$f(x)=\int_{p}e^{ipx}\hat{f}(p)\
,\quad\int_{p}:=\int\frac{d^{4}p}{(2\pi)^{4}}\ .$

This implies for the functional derivatives $\frac{\delta}{\delta\varphi(x)}$

$\dfrac{\delta}{\delta\varphi(x)}=(2\pi)^{4}\int_{p}e^{-ipx}\dfrac{\delta}{\delta\hat{\varphi}(p)}\
.$

We introduce the following regularized propagator

$\hat{C}^{\alpha_{0},\alpha}(p,m):=\dfrac{1}{p^{2}+m^{2}}\Big{(}\exp(-\alpha_{0}(p^{2}+m^{2}))-\exp(-\alpha(p^{2}+m^{2}))\Big{)}\geq
0\;,$ (1)

where $m$ is the mass of the field. Here $\alpha_{0}>0$ acts as an ultraviolet
cutoff and $\alpha\in[\alpha_{0},+\infty)$ is the flow parameter. By taking
the limits $\alpha_{0}\rightarrow 0$ and $\alpha\rightarrow+\infty$ we recover
the usual Euclidean propagator in momentum space, namely

$\lim\limits_{\alpha\rightarrow+\infty}\,\lim\limits_{\alpha_{0}\rightarrow
0}\hat{C}^{\alpha_{0},\alpha}(p,m)=\frac{1}{p^{2}+m^{2}}\ .$ (2)

With the chosen convention of the Fourier transform, the regularized
propagator in position space $C^{\alpha_{0},\alpha}(x-y,m)$ (also called
covariance) reads

$C^{\alpha_{0},\alpha}(x-y,m)=\int_{p}e^{ip(x-y)}\hat{C}^{\alpha_{0},\alpha}(p,m)\
.$ (3)

The regularized propagator of the $N$-component massive vector scalar field
theory in momentum space, is then given by a diagonal $N\times N$ matrix to be
called $C^{\alpha,\alpha_{0}}_{N}(x-y,m)$. Its elements read

${\hat{C}}_{N;\;ij}^{\alpha_{0},\alpha}(p,m)={\hat{C}}^{\alpha_{0},\alpha}(p,m)\delta_{ij}\;.$
(4)

This regularized propagator is positive and satisfies

  * •

${\hat{C}}_{N;\;ij}^{\alpha_{0},\alpha}(p,m)$ is analytic w.r.t. $\alpha$.

  * •

${\hat{C}}_{N;\;ij}^{\alpha_{0},\alpha_{0}}(p,m)=0$ 111The corresponding
Gaussian measure corresponds in this case to a $\delta$-type measure on
function space..

  * •

At $\alpha$ and $i$ fixed, ${\hat{C}}_{N;\;ii}^{\alpha_{0},\alpha}(p,m)$ falls
off more rapidly than any power of $|p|$.

We will consider bare interactions of the form

$L^{\mathcal{V}}_{0,N}(\varphi)=\int_{\mathcal{V}}d^{4}x\Big{[}b_{0}(\alpha_{0})\sum_{1\leq
i\leq N}(\partial\varphi_{i}(x))^{2}+\sum_{n\in
2{\mathbb{N}}}c_{0,n}(\alpha_{0})\varphi(x)^{n}\Big{]}\;,$ (5)

where $\varphi^{2n}(x):=(\varphi^{2}(x))^{n}$ for $n\geq 1$,
$\varphi^{2}(x):=\sum_{i=1}^{N}\varphi_{i}^{2}(x)\,$, and $\mathcal{V}$ is a
finite volume in ${\mathbb{R}}^{4}$ This bare interaction lagrangian is $O(N)$
invariant. The constants $b_{0}(\alpha_{0})$, $c_{0,n}(\alpha_{0})$ are called
the bare couplings. The quantities in the sum for $n\geq 6$ are called
irrelevant terms, while $b_{0}(\alpha_{0}),c_{0,2}(\alpha_{0})$ and
$c_{0,4}(\alpha_{0})$ are relevant terms. Generally the relevant terms are
required in order that the renormalized physical quantities such as the
renormalized mass or the renormalized coupling constant are finite upon
removing the UV cutoff. In the mean-field approximation to be considered soon
the constant $b_{0}(\alpha_{0})$ vanishes, because in this case the field
variable $\varphi$ becomes a constant.

The functional integral with the bare lagrangian
$L^{\mathcal{V}}_{0,N}(\varphi)$ is well-defined if for some constant
$C^{\mathcal{V}}_{N}\in{\mathbb{R}}$, depending on $\mathcal{V}$ and $N$

$-\infty<C^{\mathcal{V}}<L_{0,N}^{\mathcal{V}}(\varphi)\;,\quad\varphi\in\mbox{supp}(\mu^{\alpha_{0},\alpha}_{N})\;,$
(6)

where $\mu_{N}^{\alpha_{0},\alpha}$ is the normalized Gaussian measure
associated to the propagator $C^{\alpha_{0},\alpha}_{N}$. Some properties of
Gaussian measures taylored for our purposes, can be found in [24], more
information can be found in [25]. We collected a few items in Appendix A.1.
The field $\varphi$ is supposed to belong to the support of the Gaussian
measure $\mu^{\alpha_{0},\alpha}_{N}$. Since the regularized propagator
$\hat{C}^{\alpha_{0},\alpha}(p,m)$ falls off more rapidly than any power of
$|p|$ in momentum space, its support is contained on smooth functions in
position space, see e.g. [26], so that the quantities in $L_{0}^{\mathcal{V}}$
i.e. $\varphi^{2}(x),\varphi^{4}(x),\cdots$ are well-defined. Here we will not
discuss the infinite volume limit explicitly, for more details see [24]. It
can be taken once we have passed to the connected amputated Schwinger
functions (see below). We will thus drop the subscript $\mathcal{V}$.

The correlation or Schwinger functions are defined as

$\langle\varphi_{i_{1}}(x_{1})\varphi_{i_{2}}(x_{2})\cdots\varphi_{i_{n}}(x_{n})\rangle^{\alpha_{0},\alpha}:=\dfrac{1}{Z^{\alpha_{0},\alpha}_{N}}\int
d\mu^{\alpha_{0},\alpha}_{N}(\varphi)e^{-L_{0,N}(\varphi)}\varphi_{i_{1}}(x_{1})\varphi_{i_{2}}(x_{2})\cdots\varphi_{i_{n}}(x_{n})\;,$
(7)

where $d\mu^{\alpha_{0},\alpha}_{N}$ is the Gaussian measure associated with
the regularized propagator (4). The generating functional of the regularized
connected amputated Schwinger functions (CAS)
$e^{-L^{\alpha_{0},\alpha}_{N}(\varphi)}$ at scale $\alpha$ satisfies

$e^{-L^{\alpha_{0},\alpha}_{N}(\varphi)}=\dfrac{1}{Z^{\alpha_{0},\alpha}_{N}}\int
d\mu^{\alpha_{0},\alpha}_{N}(\phi)\ e^{-L_{0,N}(\phi+\varphi)}\ .$ (8)

Expanding $L^{\alpha_{0},\alpha}_{N}(\varphi)$ in a formal power series in
$\hat{\varphi}_{i}$ gives

$L^{\alpha_{0},\alpha}_{N}(\varphi)=\sum_{n\in 2{\mathbb{N}}}\sum_{1\leq
i_{1},\cdots,i_{n}\leq
N}\int_{p_{1},p_{2},\cdots,p_{n}}\bar{\mathcal{L}}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})\hat{\varphi}_{i_{1}}(p_{1})\cdots\hat{\varphi}_{i_{n}}(p_{n})\
.$ (9)

The CAS distributions
$\bar{\mathcal{L}}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})$ or moments of $L^{\alpha_{0},\alpha}_{N}$ can be
factorized due to translation invariance as

$\bar{\mathcal{L}}^{\alpha_{0},\alpha}_{n,i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})=\delta^{4}\Big{(}\sum_{i=1}^{n}p_{i}\Big{)}\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n}),\quad p_{n}=-p_{1}-\cdots-p_{n-1}\ .$ (10)

The CAS functions to be called
$\,\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})$ are obtained through functional derivation

$\dfrac{(2\pi)^{4n}}{n!}\dfrac{\delta}{\delta\hat{\varphi}_{i_{1}}(p_{1})}\cdots\dfrac{\delta}{\delta\hat{\varphi}_{i_{n}}(p_{n})}L^{\alpha_{0},\alpha}_{N}(\varphi)|_{\varphi\equiv
0}=\delta^{4}\Big{(}\sum_{i=1}^{n}p_{i}\Big{)}\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})\ .$ (11)

The flow equations are obtained on deriving (8) w.r.t. $\alpha\,$ using the
change of covariance formula (175)

$\partial_{\alpha}L^{\alpha_{0},\alpha}_{N}=\dfrac{1}{2}\sum_{i=1}^{N}\Big{\langle}\dfrac{\delta}{\delta\varphi_{i}},\dot{C}^{\alpha}\dfrac{\delta}{\delta\varphi_{i}}\Big{\rangle}L^{\alpha_{0},\alpha}_{N}-\dfrac{1}{2}\sum_{i=1}^{N}\Big{\langle}\dfrac{\delta}{\delta\varphi_{i}}L^{\alpha_{0},\alpha}_{N},\dot{C}^{\alpha}\dfrac{\delta}{\delta\varphi_{i}}L^{\alpha_{0},\alpha}_{N}\Big{\rangle}\;,$
(12)

where $\dot{C}^{\alpha}=\partial_{\alpha}C^{\alpha_{0},\alpha}$. Using
(9),(10) in (12), the flow equations for the moments
$\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots i_{n}}$ can be written as

$\begin{split}\partial_{\alpha}\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(p_{1},\cdots,p_{n})&=\binom{n+2}{2}\sum_{j=1}^{N}\int_{k}\dot{C}^{\alpha}(k,m)\mathcal{L}^{\alpha_{0},\alpha}_{n+2;i_{1}i_{2}\cdots
i_{n}jj}(p_{1},\cdots,p_{n},k,-k)\\\
&-\dfrac{1}{2}\sum_{n_{1}+n_{2}=n+2}\sum_{j=1}^{N}n_{1}n_{2}\mathbb{S}\Big{[}\mathcal{L}^{\alpha_{0},\alpha}_{n_{1};i_{1}i_{2}\cdots
i_{n_{1}-1}j}(p_{1},\cdots,p_{n_{1}-1},q)\dot{C}^{\alpha}(q,m)\\\
&\mathcal{L}^{\alpha_{0},\alpha}_{n_{2};ji_{n_{1}}i_{n_{1}+1}\cdots
i_{n}}(-q,p_{n_{1}},\cdots,p_{n})\Big{]}\;,\end{split}$ (13)

with $q:=-p_{1}-p_{2}-\cdots-p_{n_{1}-1}$. Here $\mathbb{S}$ is a
symmetrisation operator which permutes the pairs $(i_{j},p_{j})$. It averages
over the permutations $\sigma\in S_{n}\,$ such that
$\sigma(1)<\sigma(2)<\cdots<\sigma(n_{1}-1)$ and
$\sigma(n_{1})<\sigma(n_{1}+1)<\cdots<\sigma(n)\,$. Since we considered a
theory with a $\mathbb{Z}_{2}$-symmetry, only even moments (in $n,n_{1}$ and
$n_{2}$) are nonvanishing.

The FEs are an infinite system of non-linear differential equations, the
solutions of which are the CAS functions. On imposing renormalization
conditions, one can prove the perturbative renormalizability of the
regularized theory through an inductive scheme, see [24] and references given
there.

###  2.2 The mean field approximation for the $O(N)$-model

#####

In the mean field approximation, the $n$-point functions are assumed to be
momentum independent. We set

$A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}:=\mathcal{L}^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}(0,\cdots,0)\ .$ (14)

In statistical physics the critical behaviour of Ising type systems in $d>4$
dimensions is exactly obtained [16], [17] in the mean-field approximation. We
now derive the mean-field flow equations following [1]. The mean field
effective action $L_{mf}^{\alpha_{0},\alpha}(\phi)$ is expanded as a formal
power series in the constant field $\varphi\in{\mathbb{R}}$

$L_{mf}^{\alpha_{0},\alpha}(\varphi)=\sum_{n\in
2{\mathbb{N}}}A_{n}^{\alpha_{0},\alpha}\varphi^{n}\ .$ (15)

We first make the technical simplification from [1] and set $m=0$ in
$\hat{C}^{\alpha_{0},\alpha}(p,m)$. We then the analyse the theory in the
interval $\alpha\in[\alpha_{0},\frac{1}{m^{2}}]$ so that the upper limit on
$\alpha$ takes the role of the infrared cutoff, thus replacing the mass. The
existence of the UV-limit means that the mean-field solutions have a finite
limit at $\alpha=1$, when the UV-cutoff $\frac{1}{\alpha_{0}}$ is sent to
infinity.

The regularized propagator then reads

$\dfrac{e^{-\alpha_{0}p^{2}}-e^{-\frac{1}{m^{2}}p^{2}}}{p^{2}}\underset{p^{2}\ll
1}{=}\frac{1}{m^{2}}-\alpha_{0}+O(p^{2})\ .$ (16)

Then the mean-field flow equations read

$\begin{split}\partial_{\alpha}A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}&=\binom{n+2}{2}\frac{c}{\alpha^{2}}\sum_{j=1}^{N}A^{\alpha_{0},\alpha}_{n+2;i_{1}i_{2}\cdots
i_{n}jj}\\\
&-\dfrac{1}{2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}\sum_{j=1}^{N}\mathbb{S}\Big{[}A^{\alpha_{0},\alpha}_{n_{1};i_{1}i_{2}\cdots
i_{n_{1}-1}j}A^{\alpha_{0},\alpha}_{n_{2};ji_{n_{1}}i_{n_{1}+1}\cdots
i_{n}}\Big{]},\quad c:=\frac{1}{16\pi^{2}}\ .\end{split}$ (17)

Without any further input, the flow equations (17) do not allow to construct
inductively the CAS functions because we can only compute the contraction of
$A^{\alpha_{0},\alpha}_{n+2,i_{1}i_{2}\cdots i_{n+2}}$ w.r.t. its last two
indices from the CAS functions
$A^{\alpha_{0},\alpha}_{n^{\prime};i_{1}i_{2}\cdots i_{n^{\prime}}}$,
$n^{\prime}\leq n$. The mean-field solutions for the $\,O(N)\,$ model of the
flow equations satisfy by assumption the following properties:

  * •

(P1): $A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots i_{n}}=0$ if $n$ is odd.

  * •

(P2): $A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots i_{n}}$ is symmetric under
any permutation of its indices $i_{1},i_{2},\cdots,i_{n}$ .

  * •

(P3): $A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots i_{n}}$ is $O(N)$-invariant
in the following sense: let $O$ be an orthogonal matrix i.e.
$O^{T}O=O^{T}O=I$, then

$O_{i_{1}j_{1}}O_{i_{2}j_{2}}\cdots
O_{i_{n}j_{n}}A^{\alpha_{0},\alpha}_{n;j_{1}j_{2}\cdots
j_{n}}=A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots i_{n}}\ .$ (18)

Properties (P1), (P2) and (P3) require knowledge on the $O(N)$-invariant
symmetric tensors. Some facts are collected in Appendix A.2. We recall the
symmetric part of a rank $n$-tensor T by

$T_{(i_{1}i_{2}\cdots i_{n})}:=\frac{1}{n!}\sum_{\sigma\in
S_{n}}T_{i_{\sigma(1)}i_{\sigma(2)}i_{\sigma(n-1)}i_{\sigma(n)}}\ .$ (19)

and we define

$F_{i_{1}i_{2}\cdots
i_{n}}:=\delta_{(i_{1}i_{2}}\delta_{i_{3}i_{4}}\cdots\delta_{i_{n-1}i_{n})}=\frac{1}{n!}\sum_{\sigma\in
S_{n}}\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}\;.$
(20)

From Proposition A.1, we set

$A^{\alpha_{0},\alpha}_{n;i_{1}i_{2}\cdots
i_{n}}=A^{\alpha_{0},\alpha}_{n}F_{i_{1}i_{2}\cdots i_{n}}$ (21)

with $A^{\alpha_{0},\alpha}_{n}$ smooth.

The mean field flow equations (17) can now be rewritten as

$\begin{split}\partial_{\alpha}A^{\alpha_{0},\alpha}_{n}F_{i_{1}i_{2}\cdots
i_{n}}&=\binom{n+2}{2}\frac{c}{\alpha^{2}}A^{\alpha_{0},\alpha}_{n+2}\sum_{j=1}^{N}F_{i_{1}i_{2}\cdots
i_{n}jj}\\\
&-\dfrac{1}{2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}A^{\alpha_{0},\alpha}_{n_{1}}A^{\alpha_{0},\alpha}_{n_{2}}\sum_{j=1}^{N}\mathbb{S}\Big{[}F_{i_{1}i_{2}\cdots
i_{n_{1}-1}j}F_{ji_{n_{1}}i_{n_{1}+1}\cdots i_{n}}\Big{]}\ .\end{split}$ (22)

#####

From Proposition A.2, the flow equations (22) reduce to a much simpler form

$\begin{split}\partial_{\alpha}A^{\alpha_{0},\alpha}_{n}&=\binom{n+2}{2}\dfrac{N+n}{n+1}\frac{c}{\alpha^{2}}A^{\alpha_{0},\alpha}_{n+2}-\dfrac{1}{2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}A^{\alpha_{0},\alpha}_{n_{1}}A^{\alpha_{0},\alpha}_{n_{2}}\
.\end{split}$ (23)

Because of the factor $\binom{n+2}{2}$, iterated integration w.r.t. $\alpha$
of approximate solutions of the infinite dynamical system for the functions
$\,A_{n}^{\alpha_{0},\alpha}\,$ would produce at each step factors of order
$n^{2}\,$. Therefore this procedure is unstable w.r.t. $n$ from a
combinatorial point of view. As a consequence, we will follow the strategy
from [1]:

  * •

Start from a smooth two-point function $A_{2}^{\alpha_{0},\alpha}$ 222We will
specify the properties of two-point function $A_{2}^{\alpha_{0},\alpha}$ later
on. and fix boundary conditions at the bare scale $\alpha=\alpha_{0}\,$.

  * •

Smooth solutions $A_{n}^{\alpha_{0},\alpha}$ can then be constructed
inductively using (23). Their properties depend on
$A_{2}^{\alpha_{0},\alpha}\,$.

Adopting the change of function and variable

$f_{n}(\mu):=\frac{1}{n}\alpha^{2-\frac{n}{2}}c^{\frac{n}{2}-1}A^{\alpha_{0},\alpha}_{n}\;,\quad\mu:=\ln\left(\dfrac{\alpha}{\alpha_{0}}\right)\;,$
(24)

the mean-field flow equations read

$f_{n+2}(\mu)=\dfrac{1}{n+N}\sum_{n_{1}+n_{2}=n+2}f_{n_{1}}(\mu)f_{n_{2}}(\mu)+\dfrac{n-4}{n(n+N)}f_{n}(\mu)+\dfrac{2}{n(n+N)}\partial_{\mu}f_{n}(\mu)\;,\quad
n\geq 2\;,$ (25)

for $\mu\in[0,\mu_{\max}]\,$ where
$\mu_{\max}:=\ln\left(\frac{1}{\alpha_{0}}\right)$.

#####

In [1] locally analytic smooth solutions $f_{n}(\mu)$, uniformly bounded
w.r.t. $\mu$ with bare mean-field action locally analytic w.r.t. $\varphi$,
were shown to exist. A subclass of these solutions are smooth solutions which
vanish at $\mu=0$ upon removing the UV-cutoff so that they are asymptotically
free in the ultraviolet.

######  Remark.

  * •

The statements in [1] on the local analyticity w.r.t. $\mu$ of uniformly
bounded smooth solutions $f_{n}(\mu)$ and the analyticity of the bare-mean
field action w.r.t. $\varphi$ remain valid for $N>1$.

###  2.3 Local analyticity w.r.t. $\alpha$ of the mean-field CAS functions

#####

We recall that the regularized propagator in momentum space
$\hat{C}^{\alpha_{0},\alpha}(p,m)$ introduced in (1) is analytic w.r.t.
$\alpha$. If we construct the solutions $A_{n+2}^{\alpha_{0},\alpha}$ of the
FEs (17) as indicated, we have the following analyticity and uniqueness
statements

######  Proposition 2.1.

  * •

Let $A_{n}^{\alpha_{0},\alpha}$ be mean-field smooth solutions of (17). The
boundary conditions are assumed to be induced by a two-point function
$A_{2}^{\alpha_{0},\alpha}$ and its derivatives at $\alpha=\alpha_{0}$ which
is locally analytic w.r.t. $\alpha$. Then $A_{n}^{\alpha_{0},\alpha}$ is
locally analytic w.r.t. $\alpha\,$.

  * •

Let $A_{2}^{\alpha_{0},\alpha}$ and $\tilde{A}_{2}^{\alpha_{0},\alpha}$ be
locally analytic w.r.t. $\alpha$. If $A_{n}^{\alpha_{0},\alpha}$ and
$\tilde{A}_{n}^{\alpha_{0},\alpha}$ are constructed from
$A_{2}^{\alpha_{0},\alpha}$ and $\tilde{A}_{2}^{\alpha_{0},\alpha}$
respectively, using the flow equations (17) and if

$\partial_{\alpha}^{k}\tilde{A}_{2}^{\alpha_{0},\alpha}|_{\alpha=\alpha_{0}}=\partial_{\alpha}^{k}A_{2}^{\alpha_{0},\alpha}|_{\alpha=\alpha_{0}}\
,\quad k\geq 0\;,$ (26)

then we have for arbitrary $\alpha$

$\partial_{\alpha}^{k}\tilde{A}_{n}^{\alpha_{0},\alpha}=\partial_{\alpha}^{k}A_{n}^{\alpha_{0},\alpha},\quad
k\geq 0\,,\quad n\geq 2\ .$ (27)

###### Proof.

The proof of the first statement proceeds by induction in $n$. It obviously
holds for $n=2$. From (23), we have

$A_{n+2}^{\alpha_{0},\alpha}=\dfrac{2}{c(n+N)(n+2)}\alpha^{2}\partial_{\alpha}A_{n}^{\alpha_{0},\alpha}+\dfrac{\alpha^{2}}{c(n+N)(n+2)}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}A_{n_{1}}^{\alpha_{0},\alpha}A_{n_{2}}^{\alpha_{0},\alpha}\
.$ (28)

which implies the statement using the induction hypothesis.

The second statement is proven by induction in $N=n+2k$, going up in $n$ for
fixed $N\,$. It then follows directly from the fact that locally analytic
functions are uniquely defined by their Taylor expansions within their radius
of convergence, and from the fact that sums of products of locally analytic
functions are again locally analytic. ∎

Due to Proposition 2.1 locally analytic mean-field solutions
$A_{n}^{\alpha_{0},\alpha}$ are unique for fixed boundary conditions at the
bare scale, if we start from a locally analytic two-point function
$A_{2}^{\alpha_{0},\alpha}$.

##  3 Triviality in the mean field approximation in the presence of an IR
cutoff

#####

Let us recall the notion of triviality in perturbative quantum field theory.
From the point of view of perturbative quantum field theory, the effective
coupling constant $g(\lambda)$ is a function of the energy scale $\lambda$.
Its behaviour is described by the beta function defined by

$\beta(g(\lambda)):=\lambda\dfrac{dg}{d\lambda}(\lambda)\;.$ (29)

Note that in practice $\beta(g(\lambda))$ can only be calculated to a finite
order in the perturbative expansion. In asymptotically free theories, $\beta$
is negative so that the coupling constant vanishes at high energies,
$g(\lambda)\longrightarrow 0$ for $\lambda\longrightarrow+\infty$. For non-
asymptotically free QFTs, such as QED or $\varphi^{4}_{4}$-theory, $\beta$ is
positive. Thus the effective coupling constant grows logarithmically with
$\lambda$. We define $g(0)$ the renormalized coupling constant and
$g(\Lambda)$ the bare coupling where $\Lambda$ is the UV-cutoff. Keeping the
value of $g(\Lambda)$ fixed and removing the UV-cutoff, if one obtains
$g(0)=0$, the theory is said to be trivial or Gaussian. Another manifestation
of triviality stems from the so-called Landau pole. The effective coupling
constant $g(\lambda)$ grows if $\beta(g)$ is positive. It diverges at a finite
$\lambda_{L}$ called the Landau pole. Of course perturbation theory is no more
reliable at this point. The singularity disappears for $g(0)\longrightarrow 0$
thus implying triviality. In our context, we use the logarithmic energy scale
$\mu$ and we say that the theory is trivial if

$\lim\limits_{\mu_{\max}\rightarrow+\infty}f_{4}(\mu_{\max})=0\;,$ (30)

while keeping the bare value $f_{4}(0)$ fixed. Note that the limit
$\mu_{\max}\rightarrow+\infty$, i.e. $\alpha_{0}\rightarrow 0$ corresponds to
removing the UV-cutoff.

#####

Now we turn to the triviality of the pure mean-field $O(N)$
$\varphi^{4}_{4}$-theory. First we prove the existence of solutions of (25)
which vanish in the UV-limit for fixed mean-field boundary conditions. Then we
will prove the uniqueness of the mean-field trivial solution.

###  3.1 Existence of smooth trivial solutions of the mean-field FE

#####

We consider the following bare lagrangian without irrelevant terms i.e.
$c_{0,n}=0$, $n\geq 6$

$L_{0,N}^{\mathcal{V}}(\varphi)=\int_{\mathcal{V}}d^{4}x\Big{(}c_{0,2}\varphi^{2}(x)+c_{0,4}\varphi^{4}(x)\Big{)}$
(31)

and the following (fixed) mean-field boundary conditions following from (14),
(15), (24) and (31):

$f_{2}(0)=2(2\pi)^{4}\alpha_{0}c_{0,2},\quad f_{4}(0)=4\pi^{2}c_{0,4},\quad
f_{n}(0)=0,\quad n\geq 6\;.$ (32)

#####

A direct consequence of (32) is

######  Lemma 3.1.

For smooth solutions $f_{n}(\mu)$ of (25) with boundary conditions (32), we
have

$\partial_{\mu}^{l}f_{n}(0)=0\;,\quad n\geq 6,\;0\leq l\leq\frac{n}{2}-3\;.$
(33)

###### Proof.

See [1]. ∎

From Lemma 3.1, we can set

$f_{n}(\mu)=\mu^{\frac{n}{2}-2}g_{n}(\mu)\;,\quad n\geq 4\;,$ (34)

where $g_{n}(\mu)$ are smooth. We can then rewrite the dynamical system (25)
as

$\displaystyle\begin{split}\mu^{2}g_{n+2}&=\dfrac{1}{n+N}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}g_{n_{1}}g_{n_{2}}+\mu\dfrac{1}{n+N}g_{n}\left(2f_{2}+1-\dfrac{4}{n}\right)\\\
&+\dfrac{n-4}{n(n+N)}g_{n}+\dfrac{2}{n(n+N)}\mu\partial_{\mu}g_{n}\;.\end{split}$
(35)

Expanding $f_{2}$ and $g_{n}$ as formal Taylor series around $\mu=0$

$f_{2}(\mu)=\sum_{k\geq 0}f_{2,k}\mu^{k},\quad g_{n}(\mu)=\sum_{k\geq
0}g_{n,k}\mu^{k}\;,$ (36)

we get

$\displaystyle f_{2,k+1}$
$\displaystyle=\dfrac{1}{k+1}\left((N+2)g_{4,k}+f_{2,k}-\sum_{\nu=0}^{k}f_{2,\nu}f_{2,k-\nu}\right),$
(37)
$\displaystyle\begin{split}g_{n,k+2}&=-\dfrac{n-4}{n+2k}g_{n,k+1}-\dfrac{2n}{n+2k}\sum_{\nu=0}^{k+1}g_{n,\nu}f_{2,k+1-\nu}-\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\sum_{\nu=0}^{k+2}g_{n_{1},\nu}g_{n_{2},k+2-\nu}\\\
&+\dfrac{n(n+N)}{n+2k}g_{n+2,k}\;.\end{split}$ (38)

The first line of (37) corresponds to (35) at $n=2$ while the second and third
lines of (37) correspond to (35) for $n\geq 4$. Regularity at $\mu=0$ implies
for $n\geq 4$

$\displaystyle\dfrac{n-4}{n}g_{n,0}$
$\displaystyle+\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}g_{n_{1},0}g_{n_{2},0}=0\;,$ (39)
$\displaystyle\dfrac{n-2}{n}g_{n,1}$
$\displaystyle+2\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}g_{n_{1},0}g_{n_{2},1}+g_{n,0}\left(2f_{2,0}+1-\dfrac{4}{n}\right)=0\;.$
(40)

#####

In [1] it was proven for $N=1$

######  Theorem 3.1 (Triviality in pure weakly-coupled mean-field
$\varphi^{4}$-theory).

For boundary conditions (32) such that

$0\leq
c_{0,4}\leq\dfrac{\varepsilon}{2^{7}\pi^{2}},\quad|c_{0,2}|\leq\Lambda_{0}^{2}\dfrac{\varepsilon}{2^{7}\pi^{4}},\quad
0\leq\varepsilon\leq 10^{-2},\;\Lambda_{0}^{-2}=\alpha_{0}\;.$ (41)

there exist smooth solutions of (25) $f_{n}\in C^{\infty}([0,\mu_{\max}])$
which vanish in the UV-limit, i.e. in the limit
$\mu_{\max}\longrightarrow+\infty$.

###### Proof.

See [1]. ∎

#####

The key point of the proof is the construction of a two-point function
$f_{2}(\mu)$ such that the mean-field smooth solutions $f_{n}(\mu)$ turn out
to be trivial. In [1], the ansatz for $f_{2}(\mu)$ is defined by

$f_{2}(\mu)=\sum_{n\geq 1}b_{n}\dfrac{x_{n}^{n-1}}{1+x_{n}^{n}},\quad\forall
n\geq 1,\;x_{n}:=n\mu\;.$ (42)

######  Remark.

The ansatz proposed in (42) is not analytic at $\mu=0$.

By expanding $f_{2}(\mu)$ as in (36), its Taylor coefficients can be rewritten
as

$f_{2,k}=(k+1)^{k}\sum_{\rho=1}^{k+1}b_{\\{\frac{k+1}{\rho}\\}}(-1)^{\rho-1}\dfrac{1}{\rho^{k}}\;,$
(43)

where by convention $b_{0}=0$ and

$\left\\{\frac{m}{n}\right\\}:=\left\\{\begin{array}[]{ll}\frac{m}{n}&\mbox{if
}\frac{m}{n}\in{\mathbb{N}}\\\ 0&\mbox{otherwise.}\end{array}\right.$ (44)

From (42)-(43), $f_{2,0}=b_{1}$ and $f_{2,1}=2b_{2}-b_{1}$ where
$f_{2,1}=3f_{4,0}-f_{2,0}(f_{2,0}-1)$, therefore the values of $b_{1}$ and
$b_{2}$ are fixed by the free choice of $f_{2,0}$ and $f_{4,0}$. The
$b_{n}$’s, $n\geq 3$ are then uniquely determined by (37)-(39) because of the
boundary conditions (32) and the smoothness condition. From (43) we also have
for $n\geq 1$

$b_{n+1}=\frac{f_{2,n}}{(n+1)^{n}}-\sum_{\rho=2}^{n+1}b_{\\{\frac{n+1}{\rho}\\}}(-1)^{\rho-1}\dfrac{1}{\rho^{n}}\;.$
(45)

#####

The fact that $f_{2}(\mu)$ is well defined on $[0,\mu_{\max}]$ follows from

######  Proposition 3.1.

Under the assumptions of Theorem 3.1 and choosing the two-point function
$f_{2}(\mu)$ as in (42), the following bounds hold:

$|b_{n}|\leq 4\left(\dfrac{3}{4}\right)^{n}\varepsilon\;,\quad n\geq 1\;.$
(46)

###### Proof.

See [1]. ∎

#####

Note that this result implies
$\lim\limits_{\mu_{\max}\rightarrow+\infty}f_{2}(\mu_{\max})=0$. The
triviality follows then from these bounds with the aid of the flow equations.
This triviality result in [1] is weaker than the triviality statements
[16]-[18] as they do not require any upper bound on the value of the bare
coupling constant. The aim of this section is to extend Theorem 3.1 to
arbitrarily large values of the mean field couplings. We will follow the steps
in [1]: choose the two-point function as in (42), derive bounds on $g_{n,k}$
and $f_{2,k}$ for given on $g_{4,0}$ and $f_{2,0}$ and derive bounds on
$b_{n}$ which imply that $f_{2}(\mu)$ is well-defined on $[0,\mu_{\max}]$. We
start proving

######  Lemma 3.2.

Let $f_{n}$ be solutions of (25) which respect the boundary conditions (32).
For given $f_{2,0},f_{4,0}$ we choose $K>1$ sufficiently large such that

$|f_{2,0}|\leq\dfrac{\sqrt{K}}{4},\quad|f_{4,0}|=|g_{4,0}|\leq\dfrac{\sqrt{K}}{32}\;.$
(47)

Then

$|f_{2,1}|\leq\dfrac{KN}{2},\quad|g_{4,1}|\leq\dfrac{K}{32}\;,$ (48)

and for $n\geq 6$

$|g_{n,0}|\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\;,\quad|g_{n,1}|\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{n^{2}}\left(1+\dfrac{nK}{2}\right)\;.$
(49)

###### Proof.

From (37) we have

$|f_{2,1}|=|(N+2)g_{4,0}+f_{2,0}-f_{2,0}^{2}|\leq\dfrac{KN}{2}\;.$ (50)

and

$|g_{4,1}|=4|g_{4,0}f_{2,0}|\leq\dfrac{K}{32}\;.$

We proceed by induction in $n$. For $n\geq 6$, we find from (39) and for $K$
large enough

$|g_{n,0}|\leq\dfrac{n}{n-4}\dfrac{1}{4}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+1-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})^{2}}\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\;.$

From (40), for $n\geq 6$ and choosing $K>4$

$\displaystyle|g_{n,1}|$
$\displaystyle\leq\dfrac{2n}{n-2}\dfrac{1}{2}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+1-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})^{2}}\left(1+\dfrac{n_{2}K}{2}\right)+\dfrac{n}{n-2}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\left(\dfrac{\sqrt{K}}{2}+1-\dfrac{4}{n}\right)$
$\displaystyle\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{n^{2}}\left(1+\dfrac{nK}{2}\right)\;.$

The previous bounds for the sums over $n_{1}$ can be checked for $n\leq 10$.
For $n\geq 12$, we use Lemma A.1 in Appendix A.3. ∎

#####

We define for $n_{1},n_{2}\in 2{\mathbb{N}}\backslash\\{2\\}$ and
$k,\nu\in{\mathbb{N}}_{0}$

$g(n_{1},n_{2},k,\nu):=\dfrac{\left|\frac{n_{1}}{4}+\nu-3\right|!\left|\frac{n_{2}}{4}+k-\nu-1\right|!\;}{(k+2-\nu)!\;\nu!}\;,$
(51)

where for $n\in{\mathbb{C}}\backslash{\mathbb{Z}}_{0}^{-}$ we define
$n!\;:=\Gamma(n+1)$ with $\Gamma$ the Gamma function. Furthermore we also
extend the definition of the binomial coefficient $\binom{n}{k}$ to
$n,k\in{\mathbb{C}}\backslash{\mathbb{Z}}_{0}^{-}$ such that
$n-k\in{\mathbb{C}}\backslash{\mathbb{Z}}_{0}^{-}$ by

$\binom{n}{k}=\dfrac{n!}{k!\;(n-k)!}\;.$

We also define for $l\in(0,1)$, $n_{1},n_{2}\in
2{\mathbb{N}}\backslash\\{2\\}$ and $k,a,b\in{\mathbb{N}}_{0}$

$F(n_{1},n_{2},k,a,b,l):=\sum_{\nu=a}^{k+2-b}\dfrac{\left|\frac{n_{1}}{4}+\nu-3\right|!\left|\frac{n_{2}}{4}+k-\nu-1\right|!\;}{[(k+2-\nu)!\;\nu!]^{l}}\;.$
(52)

#####

These two quantities will appear in the proofs of the bounds on $g_{n,k}$.
Before establishing bounds on $g_{n,k}$ we establish useful lemmas.

######  Lemma 3.3.

For $n_{1},n_{2},k,a,b\in{\mathbb{N}}_{0}$ such that

  * •

$a+b\leq k+2$

  * •

$3-\frac{n_{1}}{4}\leq a$ and $3-\frac{n_{2}}{4}\leq b$

we have

$\mathcal{S}(n_{1},n_{2},k,a,b):=\sum_{\nu=a}^{k+2-b}g(n_{1},n_{2},k,\nu)\leq\dfrac{1}{\frac{n_{1}+n_{2}}{4}+a+b-5}\;\dfrac{\left(\frac{n_{1}+n_{2}}{4}+k-3\right)!}{(k+2-(a+b))!}\;.$
(53)

###### Proof.

We will use the following equality, found in Sect.1.10 of [27]

$\sum_{\nu=0}^{m}\binom{a+\nu}{\nu}\binom{r+m-\nu}{m-\nu}=\binom{a+r+m+1}{m},\quad
a,r\geq 0,\;m\in\mathbb{N}\;.$ (54)

Using $(\ref{combinatorial})$ we get

$\begin{split}\mathcal{S}(n_{1},n_{2},k,a,b)&=\sum_{\nu=0}^{k+2-(a+b)}\dfrac{\left(\frac{n_{1}}{4}-3+a+\nu\right)!\;\left(\frac{n_{2}}{4}-3+b+k+2-(a+b)-\nu\right)!}{(\nu+a)!\;(k+2-a-\nu)\;!}\\\
&\leq\sum_{\nu=0}^{k+2-(a+b)}\dfrac{\left(\frac{n_{1}}{4}-3+a+\nu\right)!\;\left(\frac{n_{2}}{4}-3+b+k+2-(a+b)-\nu\right)!}{\nu!\;(k+2-(a+b)-\nu)\;!}\\\
&\leq\sum_{\nu=0}^{k+2-(a+b)}\left(\frac{n_{1}}{4}-3+a\right)!\;\left(\frac{n_{2}}{4}-3+b\right)!\;\binom{\frac{n_{1}}{4}+a-3+\nu}{\nu}\\\
&\binom{\frac{n_{2}}{4}+b-3+k+2-(a+b)-\nu}{k+2-(a+b)-\nu}\\\
&\leq\left(\frac{n_{1}}{4}-3+a\right)!\;\left(\frac{n_{2}}{4}-3+b\right)!\;\binom{\frac{n_{1}+n_{2}}{4}+k-3}{k+2-(a+b)}\\\
&\leq\dfrac{1}{\frac{n_{1}+n_{2}}{4}+a+b-5}\;\dfrac{\left(\frac{n_{1}+n_{2}}{4}+k-3\right)!}{(k+2-(a+b))!}\;,\end{split}$
(55)

where we used

$a!\;b!\leq(a+b)!\;,\quad a,b\geq 0\;.$

∎

A consequence of Lemma 3.3 is

######  Lemma 3.4.

Let $n_{1},n_{2},k,a$ and $l\in(0,1)$ such that

  * •

$2a\leq k+2$

  * •

$3-\frac{n_{1}}{4}\leq a$ and $3-\frac{n_{2}}{4}\leq a$

then:

$F(n_{1},n_{2},k,a,a,l)\leq[a!(k+2-a)!]^{1-l}\dfrac{1}{\frac{n_{1}+n_{2}}{4}+2a-5}\;\dfrac{\left(\frac{n_{1}+n_{2}}{4}+k-3\right)!}{(k+2-2a)!}\;.$
(56)

###### Proof.

We use the following identity

$\nu!\;(k+2-\nu)!\leq a!\;(k+2-a)!\;,\quad a\leq\nu\leq k+2-a\;.$ (57)

Then from Lemma 3.3, it follows that

$\begin{split}F(n_{1},n_{2},k,a,a,l)&\leq[a!\;(k+2-a)!]^{1-l}\mathcal{S}(n_{1},n_{2},k,a,a)\\\
&\leq[a!(k+2-a)!]^{1-l}\dfrac{1}{\frac{n_{1}+n_{2}}{4}+2a-5}\;\dfrac{\left(\frac{n_{1}+n_{2}}{4}+k-3\right)!}{(k+2-2a)!}\;.\end{split}$
(58)

∎

The following proposition shows that the coefficients $g_{n,k}$ and $f_{2,k}$
grow at most as $[k!]^{\frac{3}{4}}$. This will allow us later to show that
the function $f_{2}(\mu)$ and then also the $f_{n}(\mu)$ are well-defined for
$\mu\in[0,\mu_{\max}]$, see (86) and Propositions 3.3-3.5.

######  Proposition 3.2.

Under the same assumptions as in Lemma 3.2, we have for $n\geq 4,\;k\geq
2,\;N\geq 1$,

$|g_{n,k}|\leq
N^{\frac{n}{2}+k-2}K^{\frac{n}{2}+k-\frac{3}{2}}\left|\dfrac{n}{4}+k-3\right|!\;\dfrac{1}{(k!)^{\frac{1}{4}}}\;,\quad|f_{2,k}|\leq
N^{k+1}K^{k+\frac{1}{2}}\dfrac{|k-3|!}{(|k-1|!)^{\frac{1}{4}}}\;.$ (59)

###### Proof.

We proceed by induction going up in $M=n+k$. For a fixed value of $M=n+k$ we
go up in $k$. To initialize the induction, the bounds for $k\leq 1$ for
$g_{n,k}$ and $f_{2,k}$ follow from Lemma 3.2.

We will first bound $g_{n,k+2}$. Using (38) we can then bound $g_{n,k+2}$,
knowing the bounds on the terms appearing on the r.h.s. We proceed term by
term.

  1. 1.

We see that for the cases where $\frac{n}{4}+k-3<0$, namely $(n=4,k=0,1)$,
$(n=6,k=0,1)$, $(n=8,k=0)$ and $(n=10,k=0)$, the bounds follow from Lemma 3.2.

  2. 2.

We look at the different terms on the r.h.s of (38).

     * •

First term: this term vanishes for $n=4$. For $n\geq 6$ and $k=0$ we use the
bounds in Lemma 3.2 to obtain

$\begin{split}\dfrac{n-4}{n+2}|g_{n,1}|&\leq
N^{\frac{n}{2}-1}\dfrac{n-4}{n+2}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{n^{2}}\left(1+\dfrac{nK}{2}\right)\\\
&\leq N^{\frac{n}{2}-1}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+1}}{n}\\\ &\leq
N^{\frac{n}{2}}K^{\frac{n}{2}+2-\frac{3}{2}}\left(\frac{n}{4}-1\right)!\;\frac{1}{K}\;.\end{split}$
(60)

Then for $n\geq 6$ and $k\geq 1$, using the induction hypothesis we obtain

$\begin{split}&\dfrac{n-4}{n+2k}|g_{n,k+1}|\leq
N^{\frac{n}{2}+k-1}\dfrac{n-4}{n+2k}K^{\frac{n}{2}+k-\frac{1}{2}}\left|\dfrac{n}{4}+k-2\right|!\;\dfrac{1}{[(k+1)!]^{\frac{1}{4}}}\\\
&\leq
N^{\frac{n}{2}+k-1}K^{\frac{n}{2}+k-\frac{1}{2}}\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{(k+2)^{\frac{1}{4}}}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{n-4}{n+2k}\dfrac{1}{(n+4k-4)}\\\
&\leq
N^{\frac{n}{2}+k-1}K^{\frac{n}{2}+k+2-\frac{3}{2}}\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{1}{K}\;.\end{split}$
(61)

     * •

Second term: We can rewrite the second term as follows

$\begin{split}\dfrac{2n}{n+2k}\sum_{\nu=0}^{k+1}|g_{n,\nu}f_{2,k+1-\nu}|&=\dfrac{2n}{n+2k}\Big{(}|g_{n,0}f_{2,k+1}|+|g_{n,1}f_{2,k}|+|g_{n,k+1}f_{2,0}|\\\
&+|g_{n,k}f_{2,1}|+|g_{n,k-1}f_{2,2}|+\sum_{\nu=2}^{k-2}|g_{n,\nu}f_{2,k+1-\nu}|\Big{)}\;.\end{split}$
(62)

For the terms with $\nu\leq 1$, we use the bounds in Lemma 3.2 to get:

       * –

$\nu=0$:

$\begin{split}\dfrac{2n}{n+2k}|g_{n,0}f_{2,k+1}|&\leq\dfrac{2n}{n+2k}N^{k+2}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}K^{k+\frac{3}{2}}\dfrac{|k-2|!}{(k!)^{\frac{1}{4}}}\\\
&\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{1}{\sqrt{K}}\;,\end{split}$
(63)

where we used

$\dfrac{((k+1)(k+2))^{\frac{1}{4}}|k-2|!}{2n^{2}}\leq\left(\frac{n}{4}+k-1\right)!\;,\quad
n\geq 4,k\geq 0\;.$ (64)

       * –

$\nu=1$:

$\begin{split}\dfrac{2n}{n+2k}|g_{n,1}f_{2,k}|&\leq\dfrac{2n}{n+2k}N^{k+1}\dfrac{3K^{\frac{n}{2}-\frac{1}{2}}}{4n}K^{k+\frac{1}{2}}\dfrac{|k-3|!}{(|k-1|!)^{\frac{1}{4}}}\\\
&\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{1}{\sqrt{K}}\;,\end{split}$
(65)

where we used the following inequalities, valid for $n\geq 4,\;k\geq 0$,

$\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{n^{2}}\left(1+\dfrac{nK}{2}\right)\leq\dfrac{3K^{\frac{n}{2}-\frac{1}{2}}}{4n},\quad\dfrac{|k-3|!}{16n(|k-1|!)^{\frac{1}{4}}}\leq\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;.$
(66)

From Lemma 3.2 and (37), we have

$|f_{2,2}|\leq\dfrac{1}{2}\left((N+2)|g_{4,1}|+|f_{2,1}|+2|f_{2,0}f_{2,1}|\right)\leq
NK\sqrt{K}\;.$ (67)

       * –

$k-1\leq\nu\leq k+1$: it is clear that $|g_{n,\nu}f_{2,k+1-\nu}|$ are bounded
by

$N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{C}{\sqrt{K}}\;,$
(68)

where $C$ is a constant which does not depend on $n,k$.

       * –

$2\leq\nu\leq k-2$ We now bound the remaining sum in (62). Note that this sum
is non-zero only if $k\geq 4$, so we assume $k\geq 4$ from now on. The
remaining sum is bounded by

$N^{\frac{n}{2}+k}K^{\frac{n}{4}+k+\frac{1}{2}}\dfrac{n}{2\sqrt{K}(n+2k)}F\left(n,4,k-2,2,2,\frac{1}{4}\right)\;.$
(69)

Then from Lemma 3.4 we have

$\begin{split}\dfrac{n}{n+2k}F\left(n,4,k-2,2,2,\frac{1}{4}\right)&\leq\dfrac{n}{n+2k}[(k-2)!]^{\frac{3}{4}}\dfrac{4}{n}\dfrac{\left(\frac{n}{4}+k-4\right)!}{(k-4)!}\\\
&\leq
4\;\dfrac{\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{4}}}\dfrac{\Big{(}k(k+1)(k+2)\Big{)}^{\frac{1}{4}}}{n+2k}\\\
&\leq
4\;\dfrac{\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{4}}}\;.\end{split}$
(70)

Summing (63), (65), (68) and (70) we find

$\dfrac{2n}{n+2k}\sum_{\nu=0}^{k+1}|g_{n,\nu}f_{2,k+1-\nu}|\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{C_{2}}{\sqrt{K}}\;,$
(71)

where $C_{2}>0$ is a suitable constant.

     * •

Third term is bounded by

$\begin{split}&\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\sum_{\nu=0}^{k+2}|g_{n_{1},\nu}g_{n_{2},k+2-\nu}|\leq\dfrac{2n}{n+2k}I+N^{\frac{n}{2}+k-1}\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}n}{\sqrt{K}(n+2k)}\times\\\
&\Bigg{[}\sum_{\begin{subarray}{c}4\leq n_{1}\leq 10\\\ n_{1}\leq n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}F\left(n_{1},n_{2},k,2,2,\frac{1}{4}\right)+\sum_{\begin{subarray}{c}12\leq
n_{1}\leq n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}F\left(n_{1},n_{2},k,0,0,\frac{1}{4}\right)\Bigg{]}\;,\end{split}$
(72)

where we define

$I:=\sum_{\begin{subarray}{c}4\leq n_{1}\leq 10\\\ n_{1}\leq n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}|g_{n_{1},0}g_{n_{2},k+2}|+|g_{n_{1},1}g_{n_{2},k+1}|+|g_{n_{1},k+2}g_{n_{2},0}|+|g_{n_{1},k+1}g_{n_{2},1}|\;.$
(73)

The first and the last term in the r.h.s. of (72) contains a finite number of
terms which does not depend on $n$. Using the bounds from Lemma 3.2 and the
induction hypothesis, it is not too hard to prove that

$I\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\dfrac{C^{\prime}}{\sqrt{K}}\;,$
(74)

where $C^{\prime}$ is a constant which does not depend on $n,k$. Then we use
Lemma 3.4 to obtain respectively

$\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}4\leq n_{1}\leq 10\\\ n_{1}\leq
n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}F\left(n_{1},n_{2},k,2,2,\frac{1}{4}\right)\leq\dfrac{4n}{n+2k}\dfrac{[2k!]^{\frac{3}{4}}}{\frac{n}{4}-\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-\frac{5}{2}\right)!}{(k-2)!}$
(75)

and333$12\leq n_{1}\leq n_{2}$ implies $n\geq 22$, so that the denominators
are non-zero.

$\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}12\leq n_{1}\leq n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}F\left(n_{1},n_{2},k,0,0,\frac{1}{4}\right)\leq\frac{n}{2}\dfrac{2}{\frac{n}{4}-\frac{9}{2}}\dfrac{\left(\frac{n}{4}+k-\frac{5}{2}\right)!}{[(k+2)!]^{\frac{1}{4}}}\;.$
(76)

Using the inequality

$\left(m+\frac{1}{2}\right)!\leq 2m!\;\sqrt{m+\frac{1}{2}},\quad
m\in{\mathbb{N}}\;,$

we obtain finally

$\begin{split}\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}4\leq n_{1}\leq 10\\\
n_{1}\leq n_{2}\\\
n_{1}+n_{2}=n+2\end{subarray}}F\left(n_{1},n_{2},k,2,2,\frac{1}{4}\right)&\leq\dfrac{8n}{n+2k}\dfrac{2^{\frac{3}{4}}}{\frac{n}{4}-\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;\sqrt{\frac{n}{4}+k-\frac{5}{2}}}{[k!]^{\frac{1}{4}}}\\\
&\leq
C^{\prime\prime}\dfrac{\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{4}}}\;,\end{split}$
(77)

where $C^{\prime\prime}$ does not depend on $n,k$.

#####

Summing over all contributions, we get the bound

$\dfrac{n}{n+2k}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}\sum_{\nu=0}^{k+2}|g_{n_{1},\nu}g_{n_{2},k+2-\nu}|\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\dfrac{C_{3}}{\sqrt{K}}\;,$
(78)

for some finite positive constant $C_{3}$.

     * •

Fourth term: First, for $k\leq 1$, we use the bounds in Lemma 3.2 to obtain

       * –

$k=0$

$\dfrac{n(n+N)}{n}|g_{n+2,0}|\leq
N^{\frac{n}{2}}K^{\frac{n}{2}+\frac{1}{2}}\left(\frac{n}{4}-1\right)!\;\dfrac{1}{K}\;.$
(79)

       * –

$k=1$

$\dfrac{n(n+N)}{n+2}|g_{n+2,1}|\leq
N^{\frac{n}{2}+1}K^{\frac{n}{2}+\frac{3}{2}}\left(\frac{n}{4}\right)!\;\dfrac{1}{K}\;.$
(80)

For $k\geq 2$ we get

$\displaystyle\begin{split}&\dfrac{n(n+N)}{n+2k}|g_{n+2,k}|\leq\dfrac{n(n+N)}{n+2k}N^{\frac{n}{2}+k-1}K^{\frac{n}{2}+k-\frac{1}{2}}\left(\dfrac{n}{4}+k+\dfrac{1}{2}-3\right)!\;\dfrac{1}{(k!)^{\frac{1}{4}}}\\\
&\leq\dfrac{N^{\frac{n}{2}+k}\;K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{2n(n+1)\sqrt{\frac{n}{4}+k-\frac{5}{2}}[(k+1)(k+2)]^{\frac{1}{4}}}{(n+2k)(n+4k-4)(n+4k-8)K}\\\
&\leq
N^{\frac{n}{2}+k}K^{\frac{n}{2}+k+\frac{1}{2}}\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{4}}}\;\dfrac{1}{K}\;.\end{split}$
(81)

Summing (61), (71), (78) and (81), we obtain

$\begin{split}|g_{n,k+2}|&\leq\left[\frac{1}{K}+\frac{C_{2}+C_{3}}{\sqrt{K}}+\frac{1}{K}\right]N^{\frac{n}{2}+k}\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{[(k+2)!]^{\frac{1}{4}}}\left(\dfrac{n}{4}+k-1\right)!\\\
&\leq
N^{\frac{n}{2}+k}\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{[(k+2)!]^{\frac{1}{4}}}\left(\dfrac{n}{4}+k-1\right)!\;.\end{split}$
(82)

#####

We will now bound the $f_{2,k}$. One can check term by term that the bound
(59) is true for $k\leq 6$. For $K$ large enough, and for $k>6$ we obtain from
(37) by induction

$\begin{split}|f_{2,k+1}|&\leq\dfrac{1}{k+1}\Bigg{(}(N+2)N^{k}\dfrac{K^{k+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}+N^{k}K^{k+\frac{1}{2}}\dfrac{(k-3)!}{[(k-1)!]^{\frac{1}{4}}}\\\
&+N^{k+1}K^{k+1}\sum_{\nu=0}^{k}\dfrac{|\nu-3|!|k-\nu-3|!}{[|\nu-1|!\;|k-\nu-1|!]^{\frac{1}{4}}}\Bigg{)}\;.\end{split}$
(83)

We will bound each term in the r.h.s. of (83). We will again proceed term by
term.

     * •

First term: we trivially have

$\displaystyle\dfrac{1}{(k+1)}\dfrac{K^{k+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}$
$\displaystyle\leq\dfrac{1}{K(k+1)}\dfrac{K^{k+1+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}\;.$

     * •

Second term: obviously

$\displaystyle\dfrac{K^{k+\frac{1}{2}}}{k+1}\dfrac{(k-3)!}{((k-1)!)^{\frac{1}{4}}}\leq\dfrac{k^{\frac{1}{4}}}{K(k+1)(k-2)}\dfrac{K^{k+1+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}\;.$

     * •

Third term: we note that
$[(\nu-1)!\;(k-\nu-1)!]^{\frac{3}{4}}\leq[(k-2)!]^{\frac{3}{4}}\leq\left(\frac{k!}{k(k-1)}\right)^{\frac{3}{4}}$
for $3\leq\nu\leq k-3$. Then the terms summed over $3\leq\nu\leq k-3$ are
bounded by

$\begin{split}\dfrac{(k!)^{\frac{3}{4}}}{(k+1)(k(k-1))^{\frac{3}{4}}}\sum_{\nu=3}^{k-3}\dfrac{(\nu-3)!\;(k-\nu-3)!}{(\nu-1)!\;(k-\nu-1)!}&\leq\frac{(k!)^{\frac{3}{4}}}{(k-5)(k+1)(k(k-1))^{\frac{3}{4}}}\\\
&\leq\dfrac{(k-2)!}{(k!)^{\frac{1}{4}}}\;.\end{split}$ (84)

The remaining terms are symmetric under $\nu^{\prime}\mapsto k-\nu^{\prime}$
so that we restrict ourselves to $0\leq\nu\leq 2$. For $k>6$, they are bounded
by

       1. (a)

$\nu=0$:

$\dfrac{6(k-3)!}{((k-1)!)^{\frac{1}{4}}}\leq
6\dfrac{(k-2)!}{(k!)^{\frac{1}{4}}}\;,$

       2. (b)

$\nu=1$:

$\dfrac{4(k-4)!}{((k-2)!)^{\frac{1}{4}}}\leq
4\dfrac{(k-2)!}{(k!)^{\frac{1}{4}}}\;,$

       3. (c)

$\nu=2$:

$\dfrac{(k-5)!}{((k-3)!)^{\frac{1}{4}}}\leq\dfrac{(k-2)!}{(k!)^{\frac{1}{4}}}\;,$

Then by summing we get

$\dfrac{1}{k+1}\sum_{\nu=0}^{k}\dfrac{|\nu-3|!|k-\nu-3|!}{(|\nu-1|!\;|k-\nu-1|!)^{\frac{1}{4}}}\leq\left(1+\frac{25}{112}\right)\dfrac{(k-2)!}{(k!)^{\frac{1}{4}}}\;.$

Altogether we find

$|f_{2,k+1}|\leq\left(\frac{1}{K}+\frac{1}{K}+\frac{2}{\sqrt{K}}\right)N^{k+2}\dfrac{K^{k+1+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}\leq\dfrac{K^{k+1+\frac{1}{2}}(k-2)!}{(k!)^{\frac{1}{4}}}\;.$
(85)

This ends the proof.

∎

#####

Using the bound (59) in (45) we have for $n\geq 1$

$|b_{n+1}|\leq
c_{n,N}+\sum_{\rho=2}^{n+1}|b_{\\{\frac{n+1}{\rho}\\}}|\dfrac{1}{\rho^{n}}\;,\quad
c_{n,N}:=N^{n+1}K^{n+\frac{1}{2}}\dfrac{|n-3|!}{(|n-1|!)^{\frac{1}{4}}(n+1)^{n}}\;.$
(86)

Note that this bound is sharper than the one obtained in (46) and in [1] due
to the factor $(\frac{1}{|n-1|!})^{\frac{1}{4}}$ .We set $C_{N}:=\sum_{n\geq
0}c_{n,N}<+\infty$. Now we establish bounds on the $b_{n}$’s. We prove the
following

######  Proposition 3.3.

There exists $C(N,K)>0$ such that

$|b_{n}|\leq C(N,K)\frac{n^{2}}{2^{n}},\quad n\geq 1\;.$ (87)

###### Proof.

The proof is done by induction in $n\in{\mathbb{N}}$. For $n=1$,
$b_{1}=f_{2,0}$ and the bound is obtained choosing any constant
$C(N,K)\geq\frac{N\sqrt{K}}{2}$. For $n\in{\mathbb{N}}$, we use (86) to obtain

$\begin{split}|b_{n+1}|&\leq
c_{n,N}+\sum_{\rho=2}^{n+1}C(N,K)\left(\frac{n+1}{\rho}\right)^{2}\dfrac{1}{2^{\frac{n+1}{\rho}}\rho^{n}}\\\
&\leq
c_{n,N}+C(N,K)\dfrac{(n+1)^{2}}{2^{n+3}}+(n+1)^{2}C(N,K)\sum_{\rho=3}^{n+1}\dfrac{1}{2^{\frac{n+1}{\rho}}\rho^{n+2}}\;.\end{split}$
(88)

There exists a constant $\tilde{C}(N,K)>0$ such that

$c_{m,N}\leq\tilde{C}(N,K)\frac{(m+1)^{2}}{2^{m+3}},\quad m\geq 1\;.$ (89)

Moreover we have

$\begin{split}\sum_{\rho=3}^{n+1}\dfrac{1}{2^{\frac{n+1}{\rho}}\rho^{n+2}}\leq\sum_{\rho=3}^{n+1}\dfrac{1}{\rho^{n+2}}\leq\int_{2}^{n+1}\dfrac{dx}{x^{n+2}}=\dfrac{1}{2^{n+1}(n+1)}-\dfrac{1}{(n+1)^{n+2}}\leq\dfrac{1}{2^{n+2}}\;.\end{split}$
(90)

From (89) and (90), we get

$|b_{n+1}|\leq\left[\frac{\tilde{C}(N,K)}{4}+\frac{C(N,K)}{4}+\frac{C(N,K)}{2}\right]\frac{(n+1)^{2}}{2^{n+1}}\leq
C(N,K)\frac{(n+1)^{2}}{2^{n+1}}\;,$ (91)

if we choose $C(N,K)\geq\max(\tilde{C}(N,K),\frac{\sqrt{K}}{2})$. ∎

We also prove the following

######  Proposition 3.4.

For $l\geq 0$, $n>l+1$, we have

$\left|\partial_{\mu}^{l}\frac{x_{n}^{n-1}}{1+x_{n}^{n}}\right|\leq\frac{n^{n+l-1}\mu^{n-l-1}}{1+x_{n}^{n}}\mathcal{C}_{l}\;,\quad\mu\in[0,\mu_{\max}]\;,$
(92)

where the integers $\mathcal{C}_{l}$ are defined by $\mathcal{C}_{0}=1$ and

$\mathcal{C}_{l+1}=1+\sum_{j=0}^{l}\binom{l+1}{j}\mathcal{C}_{j}\leq
4^{l+1}(l+1)!\;.$ (93)

###### Proof.

We prove the proposition by induction in $l\geq 0$.

  * •

The case $l=0$ is obvious.

  * •

We use the following formula

$\left(\dfrac{f}{g}\right)^{(l)}=\frac{1}{g}\left[f^{(l)}-l!\;\sum_{j=1}^{l}\frac{g^{(l+1-j)}}{(l+1-j)!}\frac{1}{(j-1)!}\left(\dfrac{f}{g}\right)^{(j-1)}\right]\;,$
(94)

for $f,g$ smooth functions with $g>0$. See Appendix A.4 for a proof. We have
for $n>l+2$ and $\mu\geq 0$

$\displaystyle\left|\partial_{\mu}^{l+1}\frac{x_{n}^{n-1}}{1+x_{n}^{n}}\right|$
$\displaystyle\leq\dfrac{1}{1+x_{n}^{n}}\Bigg{[}n^{n-1}\prod_{m=0}^{l}(n-1-m)\;\mu^{n-2-l}$
$\displaystyle+(l+1)!\sum_{j=1}^{l+1}\dfrac{1}{(l+2-j)!\;(j-1)!}n^{n}\prod_{m=0}^{l+1-j}(n-m)\;\mu^{n-l-2+j}\frac{n^{n+j-2}\mu^{n-j}}{1+x_{n}^{n}}\mathcal{C}_{j-1}\Bigg{]}$
$\displaystyle\leq\frac{1}{1+x_{n}^{n}}\left[n^{n-1}n^{l+1}\mu^{n-l-2}+\mu^{n-l-2}\sum_{j=1}^{l+1}\binom{l+1}{j-1}n^{l+2-j}n^{n+j-2}C_{j-1}\right]$
$\displaystyle\leq\dfrac{n^{n+l}\mu^{n-l-2}}{1+x_{n}^{n}}\left[1+\sum_{j=0}^{l}\binom{l+1}{j}\mathcal{C}_{j}\right]\leq\dfrac{n^{n+l}\mu^{n-l-2}}{1+x_{n}^{n}}\mathcal{C}_{l+1}\;.$

The bound on $C_{l}$ can be straightforwardly proven by induction in $l$. ∎

#####

Now we can prove the last result concerning the behaviour of the mean-field
smooth solutions $f_{n}(\mu)$ in the UV-limit.

######  Proposition 3.5.

  * •

$f_{2}(\mu)$ is well defined on $[0,\mu_{\max}]$ and

$\quad\lim\limits_{\mu_{\max}\longrightarrow+\infty}\partial_{\mu}^{l}f_{2}(\mu_{\max})=0,\quad
l\geq 0\;.$ (95)

  * •

The functions $\partial_{\mu}^{l}f_{n}(\mu)$, $l\geq 0,n\geq 4$ are well
defined on $[0,\mu_{\max}]$ and

$\lim\limits_{\mu_{\max}\longrightarrow+\infty}\partial_{\mu}^{l}f_{n}(\mu_{\max})=0,\quad
n\geq 4,\;l\geq 0\;.$ (96)

###### Proof.

As a consequence of Proposition 3.4, for $l\geq 0$, $n>l+1$ and $\mu\geq 0$

$\left|\partial_{\mu}^{l}\frac{x_{n}^{n-1}}{1+x_{n}^{n}}\right|\leq
n^{2l}\dfrac{x_{n}^{n-l-1}}{1+x_{n}^{n}}C_{l}\;.$ (97)

For $0\leq l<n-1$, the function $g(\mu)=\dfrac{(n\mu)^{n-l-1}}{1+(n\mu)^{n}}$
defined on ${\mathbb{R}}_{+}$ reaches its maximum at
$\tilde{\mu}=\frac{1}{n}\left(\frac{n}{l+1}-1\right)^{\frac{1}{n}}$ and

$\|g\|_{\infty}=g(\tilde{\mu})=\frac{l+1}{n}\left(\frac{n}{l+1}-1\right)^{1-\frac{l+1}{n}}\leq
1\;.$ (98)

Proposition 3.4 and the bounds (87) imply the uniform convergence of $f_{2}$
and its derivatives on $[0,\mu_{\max}]$. As a result we have

$\forall l\geq 0,\;\forall m\geq
1,\quad\lim\limits_{\mu\longrightarrow+\infty}\partial_{\mu}^{l}\dfrac{x_{m}^{m-1}}{1+x_{m}^{m}}=0\quad\Longrightarrow\quad\forall
l\geq
0,\quad\lim\limits_{\mu_{\max}\longrightarrow+\infty}\partial_{\mu}^{l}f_{2}(\mu_{\max})=0\;.$
(99)

#####

The proof of the second statement is done by induction in $n+l$, going up in
$n$. We can then easily check the case $n=4$ using (25) and (95). For $n\geq
4$, we obtain by differentiating (25) $l$ times w.r.t. $\mu$

$\partial_{\mu}^{l}f_{n+2}=\dfrac{2}{n(n+N)}\partial_{\mu}^{l+1}f_{n}+\dfrac{n-4}{n(n+N)}\partial_{\mu}^{l}f_{n}+\dfrac{1}{n+N}\sum_{n_{1}+n_{2}=n+2}\quad\sum_{l_{1}+l_{2}=l}\binom{l}{l_{1}}\partial_{\mu}^{l_{1}}f_{n_{1}}\partial_{\mu}^{l_{2}}f_{n_{2}}\;.$
(100)

Using the induction hypothesis, $\partial_{\mu}^{l}f_{n+2}(\mu)$ are well
defined on $[0,\mu_{\max}]$ and they vanish when
$\mu_{\max}\longrightarrow+\infty$. ∎

#####

Collecting our findings, we can now state our existence result.

######  Theorem 3.2.

Consider a $O(N)$ vector model $\varphi^{4}_{4}$-theory of bare interaction
lagrangian (31). Let $f_{n}(\mu)$ be smooth solutions of the mean-field flow
equations (25) and we consider the mean-field boundary conditions (32) with

$0<c_{0,4}<+\infty,\quad|c_{0,2}|<+\infty\;.$ (101)

There exist smooth solutions of (25) $f_{n}(\mu)\in
C^{\infty}([0,\mu_{\max}])$ such that they vanish in the UV-limit, i.e.

$\lim\limits_{\mu_{\max}\longrightarrow+\infty}f_{n}(\mu_{\max})=0,\quad n\geq
2\;.$ (102)

###### Proof.

For $c_{0,4}<+\infty$ and $|c_{0,2}|<+\infty$ fixed, there exists $K$ such
that

$0\leq
c_{0,4}\leq\dfrac{\sqrt{K}}{2^{7}\pi^{2}},\quad|c_{0,2}|\leq\Lambda_{0}^{2}\dfrac{\sqrt{K}}{2^{7}\pi^{4}},\quad\Lambda_{0}^{-2}=\alpha_{0}\;,$
(103)

Using Lemma 3.2 and Proposition 3.3, we choose $K$ such that the bounds (59)
hold. Then we choose a smooth two-point function $f_{2}(\mu)$ as in (42). From
Proposition 3.3, Proposition 3.4 and Proposition 3.5, the obtained smooth
solutions $f_{n}(\mu)$ vanish in the UV-limit.

∎

#####

We finish this section with a few remarks

######  Remarks.

  * •

The limit $\mu_{\max}\rightarrow+\infty$ or $\alpha_{0}\rightarrow 0$ is
equivalent to removing the UV-cutoff. In statistical mechanics we fix a
lattice with a fixed spacing $h$ which corresponds to a fixed UV-cutoff. We
can then interpret $\mu_{\max}\rightarrow+\infty$ as the limit
$\alpha\rightarrow+\infty$ at $\alpha_{0}$ fixed.

  * •

The bounds derived in Proposition 3.2 could be sharpened. For the triviality
statement, they are sufficient.

###  3.2 Uniqueness of the mean-field trivial solution

#####

So far, we proved that for the mean-field $O(N)$-model
$\varphi^{4}_{4}$-theory has a trivial solution for fixed mean-field
condition. Here we prove the uniqueness of the trivial solution we
constructed. We restrict for simplicity of notation to the case $N=1$. The
more general case $N>1$ can be treated analogously.

The mean-field FE (25) for can be obtained again following Felder’s steps [28]
by considering the continuum limit of the hierarchical model, introduced by
Dyson [29]. The effective action at the scale $L^{-1}\lambda$, where $L>1$, is
related to the effective action at the scale $\lambda$ by

$e^{-u(L^{-1}\lambda,x)}=\int
d\mu_{L}(y)e^{-L^{4}u(\lambda,L^{-1}x+y)}\;,\quad\lambda\in(0,\Lambda_{0}]\;,$
(104)

where $\mu_{L}$ is the one-dimensional Gaussian measure defined by

$d\mu_{L}(y):=\dfrac{1}{\sqrt{2\pi(L-1)}}e^{-\frac{y^{2}}{2(L-1)}}dy\;.$ (105)

Both the r.h.s and the l.h.s of (104) have a limit when $L\longrightarrow 1$,
since the Gaussian measure $\mu_{L}$ becomes a Dirac measure in the limit
$L\longrightarrow 1$. Then taking the $L$-derivative of (104) and evaluating
at $L=1$ yields the partial differential equation

$-\lambda\partial_{\lambda}u=\frac{1}{2}\partial_{xx}u-\frac{1}{2}\left(\partial_{x}u\right)^{2}+4u-x\partial_{x}u\;.$
(106)

If we expand $u(\lambda,x)$ as a power series in $x$

$u(\lambda,x)=\sum_{n\in
2{\mathbb{N}}}\dfrac{(2)^{\frac{n}{2}}f_{n}(\lambda)}{n}x^{n}\;,$ (107)

then the moments $f_{n}(\lambda)$ satisfy the dynamical system

$-\frac{1}{n(n+1)}\lambda\partial_{\lambda}f_{n}=f_{n+2}-\frac{1}{n+1}\sum_{n_{1}+n_{2}=n+2}f_{n_{1}}f_{n_{2}}+\frac{4}{n(n+1)}f_{n}-\frac{1}{n+1}f_{n}\;.$
(108)

Setting $\lambda=\Lambda_{0}e^{-\frac{\mu}{2}}$, we obtain again the FE (25).
In [28], Felder derived (106) in two ways:

  * •

Simplyfing Wilson’s renormalization group equations [2],[3] using the local
potential approximation [30].

  * •

Considering the continuum limit of the recursion relation of the hierarchical
model introduced by Gallavotti [31].

He analyzed rigourously the global solutions of (106) in great generality and
concluded that in $d=4-\varepsilon$, the non-trivial fixed point solution
$u_{4}$ in $3<d<4$ dimensions vanishes. Nevertheless his analysis does not
exclude the existence of fixed points other than those he found. The momentum
expansion (107) may not be valid for arbitrarily large $x\in{\mathbb{R}}$. We
prove that for the mean-field moments $f_{n}(\mu)$ constructed in Sect.3.1,
$u(\lambda,x)$ is locally analytic w.r.t. $x$. We proceed as follows: first we
bound $\partial_{\mu}^{l}f_{2}(\mu)$, then we bound inductively
$\partial_{\mu}^{l}f_{n}(\mu)$ using the mean-field FE (25). Finally we obtain
bounds for $f_{n}(\mu)$. We introduce the new variable $X:=n\mu$ and we define

$p_{n}(X)=\dfrac{X^{n-1}}{1+X^{n}}\;.$ (109)

######  Proposition 3.6.

We have for $l\in{\mathbb{N}}_{0}$ and $n\in{\mathbb{N}}$

$|\partial_{X}^{l}p_{n}(X)|\leq\left\\{\begin{array}[]{ll}\dfrac{(l+2)!\;3^{l+1}e^{\frac{3l}{n}}}{n^{l}\mu^{2l+1}}&X\in(0,3)\\\
\dfrac{3^{l+1}l!}{\mu^{l}n^{l}}&X\geq 3\;.\end{array}\right.$ (110)

###### Proof.

For $X\in(0,3)$, the proof is done by induction in $l\in{\mathbb{N}}_{0}$. The
case $l=0$ is obvious. For $l>0$, we use (94). Inserting the induction
hypothesis in the r.h.s of (94) gives

$\begin{split}|\partial_{\mu}^{l}p_{n}(\mu)|&\leq\dfrac{1}{1+X^{n}}\Bigg{[}\prod_{i=0}^{l-1}(n-1-i)X^{n-1-l}+l!\sum_{j=1}^{l}\dfrac{\prod_{i=0}^{l-j}(n-i)X^{n-l-1+j}}{(l+1-j)!\;(j-1)!}\frac{(j+1)!\;3^{j}e^{\frac{3(j-1)}{n}}}{n^{j-1}\mu^{2j-1}}\Bigg{]}\\\
&\leq\dfrac{(l+2)!}{n^{l}}\Bigg{[}\dfrac{3^{l}}{n\mu^{2l+1}(l+2)!}+3^{l}e^{\frac{3(l-1)}{n}}\sum_{j=1}^{l}\dfrac{1}{(l+1-j)!}\frac{1}{\mu^{l+j}}\Bigg{]}\\\
&\leq\dfrac{(l+2)!\;3^{l+1}e^{\frac{3l}{n}}}{n^{l}\mu^{2l+1}}.\end{split}$
(111)

For $X\geq 3$, we expand $p_{n}(X)$ as a power series in $\frac{1}{X}$, then
we have

$\begin{split}|\partial_{X}^{l}p_{n}(X)|&\leq\sum_{k=0}^{\infty}\dfrac{(nk+l)!}{(nk)!}\dfrac{1}{X^{nk+l+1}}\\\
&\leq\dfrac{2^{l}l!}{X^{l+1}}\sum_{k=0}^{\infty}\dfrac{2^{nk}}{X^{nk}}\leq\dfrac{3^{l+1}l!}{\mu^{l}n^{l}}\;.\end{split}$
(112)

∎

Now we prove bounds for the derivatives of $f_{2}(\mu)$

######  Proposition 3.7.

We have for a constant $K_{1}(1,K)$

$\left|\partial_{\mu}^{l}f_{2}(\mu)\right|\leq\frac{K_{1}(1,K)^{l+1}(l+2)!\;}{M_{l}(\mu)}\;,\quad
l\geq 0,\quad\mu\in(0,\mu_{\max}]\;,$ (113)

where we defined

$M_{l}(\mu):=\min\\{\mu^{2l+1},\mu^{l}\\}\;.$ (114)

###### Proof.

From Proposition (3.6) and Proposition (3.3)

$\begin{split}|\partial_{\mu}^{l}f_{2}(\mu)|&\leq
C(1,K)\Bigg{[}\sum_{n<\frac{3}{\mu}}\dfrac{n^{2}}{2^{n}}\dfrac{(l+2)!\;3^{l+1}e^{3l}}{\mu^{2l+1}}+\sum_{n\geq\frac{3}{\mu}}\dfrac{n^{2}}{2^{n}}\dfrac{l!\;3^{l+1}}{\mu^{l}}\Bigg{]}\\\
&\leq\frac{K_{1}(1,K)^{l+1}(l+2)!\;}{M_{l}(\mu)}\;,\end{split}$ (115)

if we choose $K_{1}(1,K)>24C(1,K)$.

∎

#####

Now we prove bounds for the derivatives of $f_{n}(\mu)$. It is convenient to
distinguish $\mu<1$ and $\mu\geq 1$. We have

######  Proposition 3.8.

Let $f_{n}(\mu)$ be smooth mean-field solutions of the FE (25) and we assume
that the derivatives of the two-point function $\partial_{\mu}^{l}f_{2}(\mu)$
satisfies the bounds (113). Then we have for a constant
$K_{2}(1,K)>K_{1}(1,K)$

$|\partial_{\mu}^{l}f_{n}(\mu)|\leq\dfrac{K_{2}(1,K)^{n+l-1}}{(l+1)^{2}}\dfrac{(n+l)!\;}{n!}\dfrac{1}{\mu^{2l+n-1}},\quad
n\geq 2,\;,l\geq 0\;,\mu<1\;,$ (116)

and

$|\partial_{\mu}^{l}f_{n}(\mu)|\leq\dfrac{K_{2}(1,K)^{n+l-1}}{(l+1)^{2}}\dfrac{(n+l)!\;}{n!},\quad
n\geq 2,\;,l\geq 0\;,\mu\geq 1\;.$ (117)

###### Proof.

The proof is done by induction in $n+l$, going up in $n$. We will prove our
statement for $0<\mu<1$. For $n=2$, bounds follow from (113) since
$M_{l}(\mu)=\mu^{2l+1}$. For $n>2$ we differentiate (25) $l$ times w.r.t.
$\mu$ and we insert the induction hypothesis. We get

$\begin{split}|\partial_{\mu}^{l}f_{n+2}(\mu)|&\leq\frac{2}{n(n+1)}\dfrac{K_{2}(1,K)^{n+l}}{(l+2)^{2}}\dfrac{(n+l+1)!\;}{n!}\frac{1}{\mu^{2l+2+n-1}}\\\
&+\frac{1}{n+1}\dfrac{K_{2}(1,K)^{n+l-1}}{(l+1)^{2}}\dfrac{(n+l)!\;}{n!}\frac{1}{\mu^{2l+n-1}}\\\
&+\dfrac{K_{2}(1,K)^{n+l}}{n+1}\sum_{n_{1}+n_{2}=n+2}\sum_{l_{1}+l_{2}=l}\binom{l}{l_{1}}\dfrac{(n_{1}+l_{1})!\;(n_{2}+l_{2})!}{(l_{1}+1)^{2}(l_{2}+1)^{2}}\dfrac{1}{n_{1}!\;n_{2}!}\dfrac{1}{\mu^{2l+n}}\;.\end{split}$
(118)

We will proceed term by term.

  * •

First term: it is bounded by

$\dfrac{K_{2}(1,K)^{n+l+1}}{(l+1)^{2}}\dfrac{(n+l+2)!\;}{(n+2)!}\dfrac{1}{\mu^{2l+n+1}}\dfrac{2}{K_{2}(1,K)}\;.$
(119)

  * •

Second term: since $\mu<1$, it is bounded by

$\dfrac{K_{2}(1,K)^{n+l+1}}{(l+1)^{2}}\dfrac{(n+l+2)!\;}{(n+2)!}\dfrac{1}{\mu^{2l+n+1}}\dfrac{1}{4K_{2}(1,K)^{2}}\;.$
(120)

  * •

Third term: We use the Vandermonde inequality

$\binom{l}{l_{1}}\binom{n+2}{n_{1}}\leq\binom{n+l+2}{n_{1}+l_{1}}\;.$ (121)

Then we obtain

$\begin{split}\binom{l}{l_{1}}\dfrac{(n_{1}+l_{1})!\;(n_{2}+l_{2})!}{n_{1}!\;n_{2}!}&=\binom{l}{l_{1}}\binom{n+2}{n_{1}}\dfrac{(n_{1}+l_{1})!\;(n_{2}+l_{2})!}{(n+2)!}\\\
&\leq\dfrac{(n+l+2)!}{(n+2)!}\;.\end{split}$ (122)

Since

$\sum_{l_{1}+l_{2}=l}\dfrac{1}{(l_{1}+1)^{2}(l_{2}+1)^{2}}\leq\dfrac{1}{(l+2)^{2}}\left[2\sum_{l_{1}=0}^{l}\frac{1}{(l_{1}+1)^{2}}+\dfrac{2}{l+2}\sum_{l_{1}=0}^{l}\dfrac{1}{l_{1}+1}\right]\;,$
(123)

then we obtain the following bound

$\dfrac{K_{2}(1,K)^{n+l+1}}{(l+1)^{2}}\dfrac{(n+l+2)!\;}{(n+2)!}\dfrac{1}{\mu^{2l+n+1}}\dfrac{6}{K_{2}(1,K)}\;,$
(124)

using again the fact that $0<\mu<1$.

Summing the three bounds (119), (120) and (124) yields

$\begin{split}|\partial_{\mu}^{l}f_{n}(\mu)|&\leq\left[\dfrac{8}{K_{2}(1,K)}+\dfrac{1}{4K_{2}(1,K)^{2}}\right]\dfrac{K_{2}(1,K)^{n+l+1}}{(l+1)^{2}}\dfrac{(n+l+2)!\;}{(n+2)!}\dfrac{1}{\mu^{2l+n+1}}\\\
&\leq\dfrac{K_{2}(1,K)^{n+l+1}}{(l+1)^{2}}\dfrac{(n+l+2)!\;}{(n+2)!}\dfrac{1}{\mu^{2l+n+1}}\;.\end{split}$
(125)

For $\mu\geq 1$, we can bound the negative power of $\mu$ in (113) by $1$ and
then we proceed as above. ∎

#####

From Proposition (3.8), the mean-field effective action $u(\mu,x)$ is locally
analytic w.r.t. $x$ for $\mu>0$. Its radius of convergence $R(\mu)$ is such
that

$R(\mu)\geq\dfrac{\min\\{\mu,1\\}}{\sqrt{2}K_{2}(1,K)},\quad\mu\in(0,\mu_{\max}]\;.$
(126)

For $\mu=0$, the boundary conditions (32) imply that $u(0,x)$ is polynomial in
$x$, the solution $u(\mu,x)$ is well-defined for $\mu\in[0,\mu_{\max}]$. Then
we claim that for fixed mean-field boundary conditions, the solution
$u(\mu,x)$ is unique.

######  Theorem 3.3 (Uniqueness of the mean-field trivial solution for the
pure $\varphi_{4}^{4}$-theory).

We consider the bare interaction lagrangian of a pure $\varphi^{4}_{4}$-theory
(31). For fixed mean-field boundary conditions (32), consider smooth mean-
field solutions $f_{n}(\mu)$ of the mean-field FE (25) which satisfy (32) such
that the corresponding mean-field effective action $u(\lambda,x)$ is locally
analytic w.r.t. $x\in{\mathbb{R}}$. Then they are unique.

###### Proof.

For fixed mean-field boundary conditions (32), let $f_{n}(\mu)$ be the mean-
field solutions of (25) constructed in Sect.3.1 which satisfy (32). Let
$\tilde{f}_{n}(\mu)$ be solutions of the mean-field FE (25) which satisfy
$f_{n}(0)=\tilde{f}_{n}(0)$. We assume that the corresponding mean-field
effective action $\tilde{u}(\lambda,x)$ is locally analytic w.r.t. $x$ for
$\lambda<\Lambda_{0}$. Then $u(\Lambda_{0},z)=\tilde{u}(\Lambda_{0},z)$ for
any arbitrary $z\in{\mathbb{R}}$ since they are polynomial in $z$. It follows
from (104) that for a fixed $\lambda<\Lambda_{0}$,
$u(\lambda,x)=\tilde{u}(\lambda,x)$ for $0<|x|\leq\varepsilon_{\lambda}$,
$\varepsilon_{\lambda}>0$. Thus it implies that
$f_{n}(\mu)=\tilde{f}_{n}(\mu)$. ∎

#####

The extension to the more general case $N>1$ is done by considering the mean-
field effective action $u(\lambda,\bm{x})$, $\bm{x}\in{\mathbb{R}}^{N}$. We
recall the Euclidean scalar product in ${\mathbb{R}}^{N}$

$(\bm{x},\bm{y}):=\sum_{i=1}^{N}x_{i}y_{i},\quad|\bm{y}|^{2}:=(\bm{y},\bm{y})=\sum_{i=1}^{N}y_{i}^{2}\;.$
(127)

The relation (104) is generalized as follows

$e^{-u(L^{-1}\lambda,\bm{x})}=\int
d\mu_{N,L}(\bm{y})e^{-L^{4}u(\lambda,L^{-1}\bm{x}+\bm{y})}\;,\quad\lambda\in(0,\Lambda_{0}]\;,$
(128)

where $\mu_{N,L}$ is the $N$-dimensional Gaussian measure defined by

$d\mu_{N,L}(\bm{y}):=\dfrac{1}{\sqrt{2\pi(L-1)}}e^{-\frac{|\bm{y}|^{2}}{2(L-1)}}\prod_{i=1}^{N}dy_{i}\;.$
(129)

The generalization of (106) is

$-\lambda\partial_{\lambda}u=\frac{1}{2}\Delta u-\frac{1}{2}|\nabla
u|^{2}+4u-(\bm{x},\nabla u)\;.$ (130)

Expanding $u(\lambda,\bm{x})$ as a power series in $|\bm{x}|$

$u(\lambda,\bm{x})=\sum_{n\in
2{\mathbb{N}}}\dfrac{(2)^{\frac{n}{2}}f_{n}(\lambda)}{n}|\bm{x}|^{n}\;,$ (131)

we find (25) upon setting $\lambda=\Lambda_{0}e^{-\frac{\mu}{2}}$. Extensions
of Propositions 3.6-3.8 and Theorem 3.3 to $N>1$ are immediate.

###  3.3 The $1/N$-expansion

In cases where $N$ may be considered to be large, the large $N$-expansion in
powers of $\frac{1}{N}$ is complementary to the perturbative expansion in the
coupling constant. This expansion is based on rescaling the coupling constant
as $g\to g/N\,$. Using this expansion the universal properties of critical
systems obtained in an expansion in $\varepsilon=4-d$ can be obtained at fixed
dimension but in an expansion w.r.t. $\frac{1}{N}$ instead [22]. Here we want
to show as a cross-check that we can recover the behaviour in $\frac{1}{N}$ in
our framework. We choose the following bare interaction lagrangian

$L_{0}(\varphi)=\int
d^{4}x\Big{(}c_{0,2}\varphi^{2}(x)+\dfrac{c_{0,4}}{N}\varphi^{4}(x)\Big{)}\ .$
(132)

######  Lemma 3.5.

Let $f_{n}$ be solutions of (25) which respect the boundary conditions (32).
We assume that for some $K$ sufficiently large:

$|f_{2,0}|\leq\dfrac{\sqrt{K}}{4},\quad|f_{4,0}|=|g_{4,0}|\leq\dfrac{\sqrt{K}}{32N}\;.$
(133)

Then

$|f_{2,1}|\leq\dfrac{K}{2},\quad|g_{4,1}|\leq\dfrac{K}{32N}\;,$ (134)
$|g_{n,0}|\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2N^{\frac{n}{2}-1}n^{2}},\quad|g_{n,1}|\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{N^{\frac{n}{2}-1}n^{2}}\left(1+\dfrac{nK}{2}\right)\;,\quad
n\geq 6\;,$ (135)

Furthermore

$|g_{n,k}|\leq\dfrac{1}{N^{\frac{n}{2}-1}}K^{\frac{n}{2}+k-\frac{3}{2}}\left|\dfrac{n}{4}+k-3\right|!\;\dfrac{1}{(k!)^{\frac{1}{4}}}\;,\quad|f_{2,k}|\leq
K^{k+\frac{1}{2}}\dfrac{|k-3|!}{(|k-1|!)^{\frac{1}{4}}}\;,\quad n\geq 4,k\geq
0\ .$ (136)

###### Proof.

We have

$|f_{2,1}|\leq\dfrac{N+2}{N}\dfrac{\sqrt{K}}{32}+\dfrac{\sqrt{K}}{4}+\dfrac{K}{16}\leq\frac{K}{2}\;,\quad|g_{4,1}|\leq
4\dfrac{\sqrt{K}}{4}\dfrac{\sqrt{K}}{32N}=\dfrac{K}{32N}\;,$ (137)

since $N\geq 1$. We can then proceed by induction in $n\,$, and the r.h.s of
(39),(40) give the correct bound w.r.t. $N\,$ on $g_{n,0}\,,\ g_{n,1}\,$. The
proof of (136) is identical to the proof of Proposition (3.2) up to the
following changes: in the r.h.s. of (37), the factor $(n+N)g_{n+2,k}$ produces
a term $\frac{n+N}{N}$ which is obviously bounded by $n+1$. For the bound on
$f_{2,k+1}$, the term $(N+2)g_{4,k}$ is bounded by $\frac{N+2}{N}\leq 3$. It
is then easy to check that the claimed behaviour w.r.t. $N\,$ is true. ∎

Due to these bounds, the bounds (87) still hold and using the results in
Sect.3 we construct the trivial solution as before. The behavior in $N$ of the
bounds derived in Proposition 3.5 is sharp, and we see that the two-point
function $f_{2}$ does not blow up w.r.t. $N$ while the four-point function
behaves as $\frac{1}{N}$ in the large $N$ limit, in agreement with the results
obtained from partial resummed perturbation theory [21].

##  4 The case of the theory with a physical IR cutoff

#####

In the previous section we constructed the trivial solution using a technical
IR cutoff supposed to take the role of the mass and set the mass equal to
zero. Here we will work with the true propagator of the massive theory. We
choose a regularized flowing propagator which preserves the analyticity
properties w.r.t. $\alpha\,$. We will again prove triviality of the mean-field
massive $\varphi^{4}_{4}$-theory. Here we restrict to the case $N=1\,$ in
order not to overload the proof with technicalities. But it also goes through
for $N>1\,$.

We will adapt the flow equations to the new scheme. Then we will construct
uniformly bounded mean-field solutions. The bounds we will derive prove the
analyticity of the mean-field action $L_{mf}^{\alpha_{0},\alpha}$. The trivial
solution will be constructed using again the ansatz introduced in (42). We
follow the steps as from Sect.3.1.

###  4.1 The flow equations for the massive theory

We assume $\alpha_{0}<\frac{1}{2m^{2}}$. We choose the following regularized
propagator

$\tilde{C}^{\alpha_{0},\alpha}(p,m)=\dfrac{e^{-\alpha_{0}(p^{2}+m^{2})}-e^{-\alpha(p^{2}+m^{2})}(\frac{1}{m^{2}}+\alpha_{0}-\alpha)m^{2}}{p^{2}+m^{2}}\;,\quad\alpha\in\left[\alpha_{0},\frac{1}{m^{2}}+\alpha_{0}\right]\
.$ (138)

We also verify the required properties

$\tilde{C}^{\alpha_{0},\alpha_{0}}(p,m)=0\
,\quad\lim\limits_{\alpha_{0}\rightarrow
0}\lim\limits_{\alpha\rightarrow\frac{1}{m^{2}}+\alpha_{0}}\tilde{C}^{\alpha_{0},\alpha}(p,m)=\frac{1}{p^{2}+m^{2}}\
,\quad\tilde{C}^{\alpha_{0},\alpha}(p,m)\geq 0\ .$ (139)

At fixed $\alpha$, $\tilde{C}^{\alpha_{0},\alpha}(p,m)$ falls off more rapidly
than any power of $|p|$ and is smooth and locally analytic w.r.t. $\alpha$. We
set $\beta_{0}:=\alpha_{0}m^{2}$ and $\beta:=\alpha m^{2}$. Then we have

$\tilde{C}^{\beta_{0},\beta}(k,m)=\dfrac{e^{-\frac{\beta_{0}}{m^{2}}(p^{2}+m^{2})}-e^{-\frac{\beta}{m^{2}}(p^{2}+m^{2})}(1+\beta_{0}-\beta)}{p^{2}+m^{2}}\;,\quad\beta\in\left[\beta_{0},1+\beta_{0}\right]\
.$ (140)

Proceeding as before, see Sect.2.1, we obtain the flow equations for the CAS
in expanded form as

$\begin{split}\partial_{\beta}\mathcal{L}^{\beta_{0},\beta}_{n}(p_{1},\cdots,p_{n})&=\binom{n+2}{2}\int_{k}\dot{\tilde{C}}^{\beta}(k,m)\mathcal{L}^{\beta_{0},\beta}_{n+2}(k,-k,p_{1},\cdots,p_{n})\\\
&-\dfrac{1}{2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}\mathbb{S}\Bigg{(}\mathcal{L}^{\beta_{0},\beta}_{n_{1}}(p_{1},\cdots,p_{n_{1}-1},q)\dot{\tilde{C}}^{\beta}(q,m)\mathcal{L}^{\beta_{0},\beta}_{n_{2}}(-q,p_{n_{1}},\cdots,p_{n})\Bigg{)}\;,\end{split}$
(141)

where
$\dot{\tilde{C}}^{\beta}(k,m)=\partial_{\beta}\tilde{C}^{\beta_{0},\beta}(k,m)\,$.
In the mean field approximation, we substitute
$\mathcal{L}^{\beta_{0},\beta}_{n}(p_{1},\cdots,p_{n})$ with
$A_{n}^{\beta_{0},\beta}:=\mathcal{L}^{\beta_{0},\beta}_{n}(0,\cdots,0)$. We
obtain from (141)

$\begin{split}\partial_{\beta}A_{n}^{\beta_{0},\beta}&=\binom{n+2}{2}I(\beta)A_{n+2}^{\beta_{0},\beta}\\\
&-\frac{1}{2m^{2}}e^{-\beta}(2+\beta_{0}-\beta)\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}A_{n_{1}}^{\beta_{0},\beta}A_{n_{2}}^{\beta_{0},\beta}\;,\quad\beta\in[\beta_{0},1+\beta_{0}]\;,\end{split}$
(142)

where
$I(\beta):=m^{2}c(1+\beta_{0}-\beta)\frac{e^{-\beta}}{\beta^{2}}+m^{2}\int_{k}\frac{e^{-\beta(k^{2}+1)}}{k^{2}+1}$
with $c:=\frac{1}{16\pi^{2}}\,$.

######  Remark.

For the regularized propagator (1) we would have to substitute $I(\beta)$ with
$\,c\,\frac{e^{-\beta}}{\beta^{2}}\,$.
Therefore bounding $A^{\beta_{0},\beta}_{n+2}$ from
$A^{\beta_{0},\beta}_{n^{\prime}}$, $n^{\prime}\leq n$ and from
$\partial_{\beta}A^{\beta_{0},\beta}_{n}$ would produce bounds, on dividing by
$I(\beta)$, which blow up for $\beta\longrightarrow+\infty$. For the choice
(138) $\beta$ is limited by $1+\beta_{0}$.

We again factor out the scaling factor $\beta^{\frac{n}{2}-2}$ and the
combinatorial factor setting

$A_{n}^{\beta_{0},\beta}:=\beta^{\frac{n}{2}-2}e^{\frac{\beta
n}{2}}\frac{1}{m^{n-4}}\frac{1}{n}a_{n}(\beta)\;,\quad\beta\in[\beta_{0},1+\beta_{0}]\;,$
(143)

where we removed the upper index $\beta_{0}$ on the r.h.s for shortness. Then
the mean-field system (142) reads

$\begin{split}G(\beta)\
a_{n+2}(\beta)&=\dfrac{2}{n(n+1)}\,\beta\partial_{\beta}a_{n}(\beta)+\dfrac{n-4}{n(n+1)}a_{n}(\beta)+\frac{1}{n+1}\,\beta\,a_{n}(\beta)\\\
&+\dfrac{1}{n+1}(2+\beta_{0}-\beta)\sum_{n_{1}+n_{2}=n+2}a_{n_{1}}(\beta)\,a_{n_{2}}(\beta)\;,\end{split}$
(144)

where $G(\beta):=c\,(1+\beta_{0}-\beta)\,+\,\beta^{2}\
e^{\beta}\int_{k}\frac{e^{-\beta(k^{2}+1)}}{k^{2}+1}\ $. The integral
appearing in the expression of $G(\beta)$ can be rewritten as

$\beta^{2}e^{\beta}\int_{k}\frac{e^{-\beta(k^{2}+1)}}{k^{2}+1}=\,2\,c\,\beta\int_{0}^{+\infty}\dfrac{u^{3}e^{-u^{2}}}{u^{2}+\beta}\;du\;,$

so that the limit $\beta\rightarrow 0$ is finite. Moreover it is easy to see
that $G(\beta_{0})=c+\mathcal{O}(\beta_{0})$ when $\beta_{0}\rightarrow 0$,
meaning that $G(\beta)$ takes a role analogous to $c$ in the theory at
$m=0\,$, when we compare (144) to (23) for $N=1$.

We perform a change of variable defining
$\mu:=\ln\left(\dfrac{\beta}{\beta_{0}}\right)$ so that
$\beta\partial_{\beta}=\partial_{\mu}$. Setting $f_{n}(\mu)=a_{n}(\beta)\,$ we
get

$\begin{split}H(\mu)f_{n+2}(\mu)&=\dfrac{2}{n(n+1)}\partial_{\mu}f_{n}(\mu)+\dfrac{n-4}{n(n+1)}f_{n}(\mu)+\frac{1}{n+1}\,\beta_{0}\
e^{\mu}\,f_{n}(\mu)\\\
&+\dfrac{1}{n+1}(2+\beta_{0}\,(1-e^{\mu}))\sum_{n_{1}+n_{2}=n+2}f_{n_{1}}(\mu)f_{n_{2}}(\mu)\;,\quad\mu\in\left[0,\tilde{\mu}_{\max}\right]\;,\end{split}$
(145)

with $\,\tilde{\mu}_{\max}:=\ln\left(1+\dfrac{1}{\beta_{0}}\right)$ and

$\,H(\mu):=G(\beta_{0}e^{\mu})=c(1+\beta_{0})-c\,\beta_{0}\,e^{\mu}+h(\mu)\;,$
(146)

where we set

$h(\mu)=\beta_{0}^{2}\,e^{2\mu}\int_{k}\dfrac{e^{-\beta_{0}e^{\mu}k^{2}}}{k^{2}+1}\
.$ (147)

From Lemma B.1, we can absorb the factor $H(\mu)$ in a new non-singular change
of function

$\tilde{f}_{n}(\mu)=(H(\mu))^{\frac{n}{2}-1}f_{n}(\mu)\ .$ (148)

Then, the dynamical system (145) reads

$\begin{split}\tilde{f}_{n+2}(\mu)&=\dfrac{2}{n(n+1)}\partial_{\mu}\tilde{f}_{n}(\mu)+\dfrac{n-4}{n(n+1)}\tilde{f}_{n}(\mu)-\dfrac{n-2}{n(n+1)}\partial_{\mu}\log(H(\mu))\tilde{f}_{n}(\mu)\\\
&+\frac{1}{n+1}\beta_{0}e^{\mu}\tilde{f}_{n}(\mu)+\dfrac{1}{n+1}(2+\beta_{0}(1-e^{\mu}))\sum_{n_{1}+n_{2}=n+2}\tilde{f}_{n_{1}}(\mu)\tilde{f}_{n_{2}}(\mu)\;,\end{split}$
(149)

which can also be written as

$\tilde{f}_{4}(\mu)=\dfrac{1}{3}\Big{(}\partial_{\mu}\tilde{f}_{2}(\mu)-\tilde{f}_{2}(\mu)+\beta_{0}\
e^{\mu}\,\tilde{f}_{2}(\mu)\,+\,(2+\beta_{0}\,(1-e^{\mu}))\tilde{f}_{2}^{2}(\mu)\Big{)}$
(150)
$\begin{split}\tilde{f}_{n+2}(\mu)&=\dfrac{2}{n(n+1)}\partial_{\mu}\tilde{f}_{n}(\mu)+\dfrac{n-4}{n(n+1)}\tilde{f}_{n}(\mu)-\dfrac{n-2}{n(n+1)}\partial_{\mu}\log(H(\mu))\tilde{f}_{n}(\mu)\\\
&+\frac{1}{n+1}\,\beta_{0}\
e^{\mu}\,\tilde{f}_{n}(\mu)+\dfrac{2}{n+1}\,(2+\beta_{0}\,(1-e^{\mu}))\,\tilde{f}_{2}(\mu)\tilde{f}_{n}(\mu)\\\
&+\dfrac{1}{n+1}\,(2+\beta_{0}\,(1-e^{\mu}))\\!\\!\\!\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\tilde{f}_{n_{1}}(\mu)\tilde{f}_{n_{2}}(\mu),\quad
n\geq 4\ .\end{split}$ (151)

The flow equations (150)-(151) include additional terms which are
$\mu$-dependent as compared to the flow equations (25), but they retain a
similar form.

###  4.2 Triviality of massive $\varphi_{4}^{4}$ theory

For the triviality proof we proceed in close analogy with Sect.3.1. Most
technical proofs are deferred to the Appendix. Due to Lemma B.4,
$\partial_{\mu}\log(H(\mu))$ is locally analytic around $\mu=0\,$. For $|\mu|$
sufficiently small

$\\!\\!\\!\partial_{\mu}\log(H(\mu))=\sum_{k\geq 0}h_{k}\mu^{k}\;,\ \
h_{k}=\dfrac{\partial_{\mu}^{k+1}\log(H(\mu))|_{\mu=0}}{k!}\ ,\ \ |h_{k}|\leq
c\,C^{k+1}\,(5e)^{k+2}2^{k+1}\ .$ (152)

The bare mean-field boundary conditions for $\tilde{f}_{n}(\mu)$ are

$\tilde{f}_{2}(0)=2(2\pi)^{4}\alpha_{0}e^{-\beta_{0}}c_{0,2},\quad\tilde{f}_{4}(0)=(2\pi)^{4}e^{-2\beta_{0}}H(0)c_{0,4},\quad\tilde{f}_{n}(0)=0,\quad
n\geq 6\ .$ (153)

First we will factor out a power of $\mu$ in $\tilde{f}_{n}(\mu)$ for $n\geq
4$.

######  Lemma 4.1.

For smooth solutions $\tilde{f}_{n}(\mu)$ of (151) with boundary conditions
(32), we have

$\partial_{\mu}^{l}\tilde{f}_{n}(0)=0\;,\quad n\geq 6,\;0\leq
l\leq\frac{n}{2}-3\ .$ (154)

###### Proof.

See Appendix B.2. ∎

From Lemma 4.1, we can write

$\tilde{g}_{n}(\mu)=\mu^{2-\frac{n}{2}}\tilde{f}_{n}(\mu),\quad n\geq 4\;,$
(155)

where $\tilde{g}_{n}(\mu)$ is smooth w.r.t. $\mu$. The FEs (151) can be
written in terms of $\tilde{g}_{n}(\mu)$

$\begin{split}\mu^{2}\tilde{g}_{n+2}&=\dfrac{n-4}{n(n+1)}\tilde{g}_{n}+\dfrac{2}{n(n+1)}\mu\partial_{\mu}\tilde{g}_{n}+\dfrac{1}{n+1}(2+\beta_{0}-\beta_{0}e^{\mu})\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\tilde{g}_{n_{1}}\tilde{g}_{n_{2}}\\\
&+\mu\dfrac{1}{n+1}\tilde{g}_{n}\left(2(2+\beta_{0}-\beta_{0}\,e^{\mu})\tilde{f}_{2}+\beta_{0}e^{\mu}+1-\dfrac{4}{n}-(1-\frac{2}{n})\partial_{\mu}\log(H(\mu))\right)\
.\end{split}$ (156)

We write the formal Taylor expansion of $\tilde{f}_{2}$ and $\tilde{g}_{n}$
around $\mu=0$

$\tilde{f}_{2}(\mu)=\sum_{k\geq
0}\tilde{f}_{2,k}\mu^{k}\;,\quad\tilde{g}_{n}(\mu)=\sum_{k\geq
0}\tilde{g}_{n,k}\mu^{k}\ .$ (157)

From (157), (152) and the FEs (150), (151), we deduce the relations between
the coefficients of the formal Taylor expansion of $\tilde{f}_{2}$ and
$\tilde{g}_{n}$

$\begin{split}\tilde{f}_{2,k+1}&=\frac{1}{k+1}\Big{(}3\tilde{g}_{4,k}+\tilde{f}_{2,k}-(2+\beta_{0})\sum_{\nu=0}^{k}\tilde{f}_{2,\nu}\tilde{f}_{2,k-\nu}-\beta_{0}\sum_{\nu=0}^{k}\frac{1}{(k-\nu)!}\tilde{f}_{2,\nu}\\\
&+\beta_{0}\sum_{\nu=0}^{k}\frac{1}{\nu!}\sum_{\nu^{\prime}=0}^{k-\nu}\tilde{f}_{2,\nu^{\prime}}\tilde{f}_{2,k-\nu-\nu^{\prime}}\Big{)}\
.\end{split}$ (158)

$\begin{split}\tilde{g}_{n,k+2}&=\dfrac{n(n+1)}{n+2k}\tilde{g}_{n+2,k}-\dfrac{n-4}{n+2k}\tilde{g}_{n,k+1}+\dfrac{n-2}{n+2k}\sum_{\nu=0}^{k+1}\tilde{g}_{n,\nu}h_{k+1-\nu}\\\
&-\dfrac{\beta_{0}n}{n+2k}\sum_{\nu=0}^{k+1}\frac{1}{(k+1-\nu)!}\tilde{g}_{n,\nu}-\dfrac{(2+\beta_{0})n}{n+2k}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\sum_{\nu=0}^{k+2}\tilde{g}_{n_{1},\nu}\tilde{g}_{n_{2},k+2-\nu}\\\
&+\dfrac{\beta_{0}n}{n+2k}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\sum_{\nu=0}^{k+2}\frac{1}{\nu!}\sum_{\nu^{\prime}=0}^{k+2-\nu-\nu^{\prime}}\tilde{g}_{n_{1},\nu^{\prime}}\tilde{g}_{n_{2},k+2-\nu-\nu^{\prime}}\\\
&-2\dfrac{(2+\beta_{0})n}{n+2k}\sum_{\nu=0}^{k+1}\tilde{g}_{n,\nu}\tilde{f}_{2,k+1-\nu}+\dfrac{2\beta_{0}n}{n+2k}\sum_{\nu=0}^{k+1}\frac{1}{\nu!}\sum_{\nu^{\prime}=0}^{k+1-\nu}\tilde{g}_{n,\nu^{\prime}}\tilde{f}_{2,k+1-\nu-\nu^{\prime}}\
.\end{split}$ (159)

Regularity at $\mu=0$ implies that

$0=\dfrac{n-4}{n}\tilde{g}_{n,0}+2\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},0}\;.$ (160)

and

$\begin{split}0&=\dfrac{n-2}{n}\tilde{g}_{n,1}+\dfrac{n-4}{n}\tilde{g}_{n,0}+\beta_{0}\tilde{g}_{n,0}-\dfrac{n-2}{n}h_{0}\tilde{g}_{n,0}+4\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},1}\\\
&-\beta_{0}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},0}+4\tilde{g}_{n,0}\tilde{f}_{2,0}\
.\end{split}$ (161)

Using (160) we can rewrite (161) as

$\dfrac{n-2}{n}\tilde{g}_{n,1}+4\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},1}+\tilde{g}_{n,0}\left(4\tilde{f}_{2,0}+(1-\frac{4}{n})(1+\frac{\beta_{0}}{2})+\beta_{0}-(1-\frac{2}{n})h_{0}\right)=0\
.$ (162)

Now we derive bounds on $\tilde{g}_{n,k}$ and $\tilde{f}_{2,k}$.

######  Lemma 4.2.

Let $\tilde{f}_{n}$ be smooth solutions of (150),(151). For given
$\tilde{f}_{2,0},\tilde{f}_{4,0}$ we choose $K$ sufficiently large such that

$|\tilde{f}_{2,0}|\leq\dfrac{\sqrt{K}}{16},\quad|\tilde{f}_{4,0}|=|\tilde{g}_{4,0}|\leq\dfrac{\sqrt{K}}{32}\
.$ (163)

Then

$|\tilde{f}_{2,1}|\leq\dfrac{K}{2},\quad|\tilde{g}_{4,1}|\leq\dfrac{K}{32}\;,$
(164)

and for $n\geq 4$

$|\tilde{g}_{n,0}|\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}},\quad|\tilde{g}_{n,1}|\leq\dfrac{K^{\frac{n}{2}-\frac{1}{2}}}{n}\
.$ (165)

###### Proof.

See Appendix B.2. ∎

######  Proposition 4.1.

Under the same assumptions as in Lemma 4.2, choosing $K$ large enough we have

$|\tilde{g}_{n,k}|\leq
K^{\frac{n}{2}+k-\frac{3}{2}}\left|\dfrac{n}{4}+k-3\right|!\;\dfrac{1}{(k!)^{\frac{1}{8}}}\;,\quad|\tilde{f}_{2,k}|\leq
K^{k+\frac{1}{2}}\dfrac{|k-3|!}{(|k-1|!)^{\frac{1}{8}}}\;,\quad n\geq 4,\
k\geq 0\ .$ (166)

###### Proof.

See Appendix B.2. ∎

#####

Choosing a smooth two-point function of the form (42), the sequence
$(b_{n})_{n\geq 1}$ satisfies bounds of the same type as (87) ($N=1$ in our
setting). Since we chose the same two-point function as in Sect.3, Lemma 3.4
remains valid and so does Proposition 3.5. Then, the solutions
$\tilde{f}(\mu)$ are well-defined on $[0,\tilde{\mu}^{\max}]$ and vanish in
the UV-limit. Therefore the extension of Theorem 3.2 to the massive theory is
straightforward.

######  Theorem 4.1 (Triviality of pure mean-field $\varphi^{4}$-theory for
the theory with a physical IR cutoff).

Consider the $\varphi^{4}_{4}$-theory of bare interaction lagrangian (31) for
$N=1$. Let $\tilde{f}_{n}(\mu)$ be smooth solutions of the mean-field flow
equations (149) and the corresponding mean-field boundary conditions (153)
with

$0<c_{0,4}<+\infty,\quad|c_{0,2}|<+\infty\;.$ (167)

There exist smooth solutions of (149) $\tilde{f}_{n}(\mu)\in
C^{\infty}([0,\tilde{\mu}_{\max}])$ such that they vanish in the UV-limit,
i.e.

$\lim\limits_{\tilde{\mu}_{\max}\rightarrow+\infty}\tilde{f}_{n}(\tilde{\mu}_{\max})=0,\quad
n\geq 2\;.$ (168)

###### Proof.

The proof is the same as for Theorem 3.2. ∎

The uniqueness of the solutions can be proven following the reasoning in
Sect.3.2. The differences remain purely technical as the coefficients in the
r.h.s of (149) are $\mu$-dependent analytic functions in our new setting.

##  Appendix A Appendix A

###  A.1 Properties of Gaussian measures

We consider a Gaussian probability measure $d\mu$ on the space of continuous
real-valued functions $C(\Omega)$, where $\Omega$ is a finite (simply
connected compact) volume in ${\mathbb{R}}^{d}$, $d\geq 1\,$.

####  A.1.1 Covariance of a Gaussian measure

We recall here the definition of the covariance of a Gaussian measure, details
can be found in [25].

A Gaussian measure of mean zero is uniquely characterized by its covariance
$C(x,y)$

$\int d\mu_{C}(\phi)\,\phi(x)\phi(y)=\tilde{C}(x,y)=\tilde{C}(y,x)\;.$ (169)

$\tilde{C}\,$ is a positive non-degenerate bilinear form defined on
$\mathcal{C}^{\infty}(\Omega)\times\mathcal{C}^{\infty}(\Omega)\,$. We assume
that $\tilde{C}(x,y)$ is translation invariant, then $C(z):=\tilde{C}(x,y)\,,\
z=x-y\,$, is well defined. Using the notations

$\langle\phi,J\rangle=\int_{\Omega}d^{d}x\,\phi(x)J(x)\;,\quad\langle
J,CJ\rangle=\int_{\Omega}d^{d}xd^{d}y\,J(x)C(x-y)J(y)$ (170)

with $J\in\mathcal{C}^{\infty}(\Omega)$, the generating functional of the
correlation functions is

$\int d\mu_{C}(\phi)e^{\langle\phi,J\rangle}=e^{\frac{1}{2}\langle
J,CJ\rangle}\;.$ (171)

The generating functional is also called the characteristic functional of the
Gaussian measure $\mu_{C}$. For $C=(-\Delta+I)^{-1}$, where $\Delta$ denotes
the Laplacian operator in ${\mathbb{R}}^{d}$, the corresponding Gaussian
measure $\mu_{C}$ is supported on distributions with
$1-\frac{d}{2}-\varepsilon$ continuous derivatives, $\varepsilon>0$. For a
regularized propagator, the Fourier transform of which falls off rapidly in
momentum space, the Gaussian measure is supported on smooth functions.

####  A.1.2 Properties of Gaussian measures

We list here some properties of Gaussian measures. Proofs can be found in
[25].

  * •

Integration by parts: Let $A(\phi)$ be a polynomial in $\phi(x)$ and its
derivatives $\partial_{\mu}\phi(x)$.

$\int d\mu_{C}(\phi)\phi(x)A(\phi)=\int
d\mu_{C}(\phi)\int_{\Omega}dy\;C(x-y)\dfrac{\delta}{\delta\phi(y)}A(\phi)\;.$
(172)

  * •

Translation of a Gaussian measure: Let $C$ be a covariance. Under a change of
variable $\phi=\varphi¨+\psi$ for $\varphi\in\mbox{supp}(\mu_{C})$ and $\psi$
such that its Fourier transform $\hat{\psi}(p)$ is compactly supported.

$d\mu_{C}(\phi)=e^{-\frac{1}{2}\langle\psi,C^{-1}\psi\rangle}e^{-\langle
C^{-1}\psi,\varphi\rangle}d\mu_{C}(\varphi)\;.$ (173)

  * •

Decomposition of the covariance: Assume that

$C=C_{1}+C_{2}\;,\quad C_{i}>0\;.$

Then for $A(\phi)$ as in (172)

$\int d\mu_{C}(\phi)A(\phi)=\int d\mu_{C_{1}}(\phi_{1})\int
d\mu_{C_{2}}(\phi_{2})A(\phi_{1}+\phi_{2})\;.$ (174)

  * •

Infinitesimal change of covariance: We assume the covariance depends on a
parameter $t$, and is differentiable w.r.t. $t$

$C(x-y)\equiv C_{t}(x-y)\;,\quad\dot{C}_{t}(x-y):=\dfrac{d}{dt}C_{t}(x-y)\;.$

Let $F(\phi)$ be a smooth functional, integrable w.r.t. $\mu_{C_{t}}$ $\forall
t\,$. We have

$\dfrac{d}{dt}\int d\mu_{C_{t}}(\phi)F(\phi)=\dfrac{1}{2}\int
d\mu_{C_{t}}(\phi)\left\langle\dfrac{\delta}{\delta\phi},\dot{C}_{t}\dfrac{\delta}{\delta\phi}\right\rangle
F(\phi)\;.$ (175)

###  A.2 Isotropic Cartesian tensors

####  A.2.1 Isotropic Cartesian tensors

######  Definition A.1 (Cartesian tensors).

Let $X$ be an Euclidean space of finite dimension $N\geq 1$. We identify $X$
with its dual space $X^{\ast}$. A rank $n$-tensor T is an element of
$\bigotimes_{i=1}^{n}X$. Assuming that we work with an orthonormal basis, we
do not need to distinguish the contravariant and the covariant components of a
tensor. Then, T $\in\bigotimes_{i=1}^{n}X$ is called a Cartesian rank $n$
tensor. Its components are denoted by $T_{i_{1}i_{2}\cdots i_{n}}$.

######  Definition A.2 (Isotropic Cartesian tensors).

A Cartesian rank $n$ tensor T is said to be isotropic if for any matrix $M\in
SO(N)$

$M_{i_{1}j_{1}}M_{i_{2}j_{2}}\cdots M_{i_{n}j_{n}}T_{j_{1}j_{2}\cdots
j_{n}}=T_{i_{1}i_{2}\cdots i_{n}}\;.$ (176)

######  Proposition A.1.

Let T be a real rank $n$-tensor, $n\in 2{\mathbb{N}}\,$. If T is symmetric and
$O(N)$-invariant, then it is of the form

$T_{i_{1}i_{2}\cdots i_{n}}=A\sum_{\sigma\in
S_{n}}\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}\;,\quad
A\in{\mathbb{R}}\;,$ (177)

where $S_{n}$ denotes the set of permutations in $\,\\{1,\cdots,n\\}\,$.

###### Proof.

The most general forms of real isotropic Cartesian rank $n$ tensors are

  * •

$n<N$:

$T_{i_{1}i_{2}\cdots i_{n}}=\sum_{\sigma\in S_{n}}\lambda_{\sigma}\
\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}\;,\quad\lambda_{\sigma}\in{\mathbb{R}}\
.$ (178)

  * •

$n=N$:

$T_{i_{1}i_{2}\cdots i_{n}}=\sum_{\sigma\in S_{n}}\lambda_{\sigma}\
\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}+\mu\
\varepsilon_{i_{1}i_{2}\cdots
i_{n}}\;,\quad\lambda_{\sigma},\mu\in{\mathbb{R}}\ .$ (179)

  * •

$n>N$ and $N$ even:

$\\!\\!\\!\\!T_{i_{1}i_{2}\cdots i_{n}}=\sum_{\sigma\in
S_{n}}\lambda_{\sigma}\
\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}+\sum_{\sigma\in
S_{n}}\mu_{\sigma}\ \varepsilon_{i_{\sigma(1)}\cdots
i_{\sigma(N)}}\delta_{i_{\sigma(N+1)}i_{\sigma(N+2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}\;,$
(180)

where $\lambda_{\sigma},\mu_{\sigma}\in{\mathbb{R}}\,$.

Here $\varepsilon_{i_{1}i_{2}\cdots i_{n}}$ is the Levi-Civita tensor defined
by

$\varepsilon_{i_{1}i_{2}\cdots
i_{n}}=\left\\{\begin{array}[]{lll}1&\mbox{if}\;(i_{1}i_{2}\cdots i_{n})\mbox{
is an even permutation of}\;(1,2,\cdots,n)\\\ -1&\mbox{if}\;(i_{1}i_{2}\cdots
i_{n})\mbox{ is an odd permutation of}\;(1,2,\cdots,n)\\\
0&\mbox{otherwise.}\end{array}\right.$ (181)

For a proof see [32].

If T is $O(N)$-invariant, it is an isotropic Cartesian tensor. It then takes
the form (178), (179),(180) depending on $n\,$. We consider the reflection $R$
in the hyperplane through the origin, orthogonal to $\bm{e_{k}}$, $1\leq k\leq
N$, where $\bm{e_{k}}$ denotes a canonical basis vector of
${\mathbb{R}}^{N}\,$. The matrix expression of $R$ is given by

$R_{ij}=\delta_{ij}-2\delta_{ik}\delta_{jk}\ .$ (182)

Then we have

$R_{i_{1}j_{1}}\cdots R_{i_{N}j_{N}}\varepsilon_{j_{1}\cdots
j_{N}}=\mbox{det}(R)\varepsilon_{i_{1}\cdots i_{N}}=-\varepsilon_{i_{1}\cdots
i_{N}}\ .$ (183)

Then from (183) and (178), (179), (180), symmetric and $O(N)$-invariant
tensors take the form (177). ∎

####  A.2.2 Contraction of isotropic Cartesian tensors

We recall the definition of $F_{i_{1}i_{2}\cdots i_{n}}$

$F_{i_{1}i_{2}\cdots
i_{n}}:=\delta_{(i_{1}i_{2}}\delta_{i_{3}i_{4}}\cdots\delta_{i_{n-1}i_{n})}:=\frac{1}{n!}\sum_{\sigma\in
S_{n}}\delta_{i_{\sigma(1)}i_{\sigma(2)}}\cdots\delta_{i_{\sigma(n-1)}i_{\sigma(n)}}\;.$
(184)

######  Proposition A.2.

We have the following identities:

$\sum_{j=1}^{N}F_{i_{1}i_{2}\cdots
i_{n}jj}=\dfrac{N+n}{n+1}F_{i_{1}i_{2}\cdots
i_{n}}\;,\quad\sum_{j=1}^{N}\mathbb{S}\Big{[}F_{i_{1}i_{2}\cdots
i_{n_{1}-1}j}F_{i_{n_{1}}i_{n_{1}+1}\cdots i_{n}j}\Big{]}=F_{i_{1}i_{2}\cdots
i_{n}}\ .$ (185)

###### Proof.

Let $F(\bm{x})$ be the generating series of $F_{i_{1}i_{2}\cdots i_{n}}$
defined by

$F(\bm{x}):=\sum_{n\in
2{\mathbb{N}}}\sum_{i_{1},i_{2},\cdots,i_{n}}x_{i_{1}}\cdots
x_{i_{n}}F_{i_{1}i_{2}\cdots
i_{n}}=\sum_{n=1}^{\infty}|\bm{x}|^{2n}\;,\quad|\bm{x}|^{2}:=\sum_{i=1}^{N}x_{i}^{2},\quad\bm{x}\in{\mathbb{R}}^{N}\
.$

The result of the action of the Laplacian on $F(\bm{x})$ is

$\Delta F(\bm{x})=\sum_{j=1}^{N}\partial^{2}_{j}F(\bm{x})=2N+\sum_{n\in
2{\mathbb{N}}}\sum_{i_{1},i_{2},\cdots,i_{n}}x_{i_{1}}\cdots
x_{i_{n}}(n+2)(n+1)\sum_{j=1}^{N}F_{i_{1}i_{2}\cdots i_{n}jj}\ .$ (186)

On the other hand we have for $n\in{\mathbb{N}}$

$\sum_{j=1}^{N}\partial^{2}_{j}|\bm{x}|^{2n}=2n(N+2n-2)|\bm{x}|^{2n-2}$ (187)

and therefore

$\begin{split}\Delta F(\bm{x})&=2N+\sum_{n\in
2{\mathbb{N}},n>2}n(N+n-2)|\bm{x}|^{n-2}\\\ &=2N+\sum_{n\in
2{\mathbb{N}}}(n+2)(N+n)|\bm{x}|^{n}\\\ &=2N+\sum_{n\in
2{\mathbb{N}}}\sum_{i_{1},\cdots,i_{n}}x_{i_{1}}\cdots
x_{i_{n}}(n+2)(N+n)F_{i_{1}\cdots i_{n}}\ .\end{split}$ (188)

Now (186) and (188) imply

$\sum_{j=1}^{N}F_{i_{1}i_{2}\cdots
i_{n}jj}=\dfrac{N+n}{n+1}F_{i_{1}i_{2}\cdots i_{n}}\;.$ (189)

For the second identity in (185), we compute

$\begin{split}\|\nabla
F(x)\|^{2}:=\sum_{j=1}^{N}(\partial_{j}F(\bm{x}))^{2}&=4|\bm{x}|^{2}\sum_{n_{1},n_{2}\geq
1}n_{1}n_{2}|\bm{x}|^{2(n_{1}+n_{2}-2)}=|\bm{x}|^{2}\sum_{n_{1}\in
2{\mathbb{N}},n_{2}\in 2{\mathbb{N}}}n_{1}n_{2}|\bm{x}|^{n_{1}+n_{2}-2}\\\
&=4|\bm{x}|^{2}+\sum_{n\in
2{\mathbb{N}},n>2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}\sum_{i_{1},\cdots,i_{n}}x_{i_{1}}\cdots
x_{i_{n}}F_{i_{1}\cdots i_{n}}\ .\end{split}$ (190)

And on the other hand, we have

$\|\nabla F(x)\|^{2}=4|\bm{x}|^{2}+\sum_{n\in
2{\mathbb{N}},n>2}\sum_{n_{1}+n_{2}=n+2}n_{1}n_{2}\sum_{i_{1},\cdots,i_{n}}x_{i_{1}}\cdots
x_{i_{n}}\sum_{j=1}^{N}F_{(i_{1}\cdots i_{n_{1}-1}j}F_{ji_{n_{1}}\cdots
i_{n})}\;,$ (191)

leading to

$\sum_{j=1}^{N}F_{(i_{1}\cdots i_{n_{1}-1}j}F_{ji_{n_{1}}\cdots
i_{n})}=F_{i_{1}\cdots i_{n}}\;.$ (192)

From the definition of a symmetric part of a tensor T in (19) and the fact
that $\mathbb{S}$ is an average operator, we obtain

$\sum_{j=1}^{N}\mathbb{S}\Big{[}F_{i_{1}i_{2}\cdots
i_{n_{1}-1}j}F_{ji_{n_{1}}i_{n_{1}+1}\cdots i_{n}}\Big{]}=F_{i_{1}\cdots
i_{n}}\ .$ (193)

∎

###  A.3 Bound on a sum

######  Lemma A.1.

For $n\geq 12$

$\frac{n}{n-2}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq 4,n_{i}\in
2{\mathbb{N}}\end{subarray}}\dfrac{1}{n_{1}^{2}(n+2-n_{1})^{2}}\leq\frac{1}{n^{2}}\;.$
(194)

###### Proof.

First we have for $n\geq 12$

$\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq 4,n_{i}\in
2{\mathbb{N}}\end{subarray}}\dfrac{1}{n_{1}^{2}(n+2-n_{1})^{2}}\leq\dfrac{1}{16}\sum_{\begin{subarray}{c}n_{1}+n_{2}=\frac{n}{2}+1\\\
n_{i}\geq
2,n_{i}\in{\mathbb{N}}\end{subarray}}\dfrac{1}{n_{1}^{2}(\frac{n}{2}+1-n_{1})^{2}}\;.$

We use the decomposition

$\dfrac{1}{X^{2}(X-A)^{2}}=\dfrac{1}{A^{2}}\left(\dfrac{1}{X^{2}}+\dfrac{1}{(X-A)^{2}}+\dfrac{2}{AX}-\dfrac{2}{A(X-A)}\right),\quad
A>0\;.$

We get

$\begin{split}&\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4,n_{i}\in 2{\mathbb{N}}\end{subarray}}\dfrac{1}{n_{1}^{2}(n+2-n_{1})^{2}}\\\
&\leq\dfrac{1}{4(n+2)^{2}}\sum_{2\leq
n_{1}\leq\frac{n}{2}-1}\Bigg{(}\dfrac{1}{n_{1}^{2}}+\frac{1}{(\frac{n}{2}+1-n_{1})^{2}}+\dfrac{2}{(\frac{n}{2}+1)n_{1}}+\dfrac{2}{(\frac{n}{2}+1)(\frac{n}{2}+1-n_{1})}\Bigg{)}\\\
&\leq\dfrac{1}{2(n+2)^{2}}\left(\zeta(2)-1+\frac{n-4}{n+2}\right)\leq\frac{5}{6(n+2)^{2}}\
,\end{split}$

where we used the fact that $\sum_{2\leq
n_{1}\leq\frac{n}{2}-1}\frac{1}{n_{1}}\leq\frac{n-4}{4}\,$. Therefore we have
for $n\geq 12$

$\displaystyle\dfrac{n}{n-2}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq 4\end{subarray}}\dfrac{1}{n_{1}^{2}(n+2-n_{1})^{2}}$
$\displaystyle\leq\dfrac{5}{6(n+2)^{2}}\frac{n}{n-2}\leq\frac{5}{6n^{2}}\frac{n^{2}}{(n+2)^{2}}\frac{n}{n-2}\leq\frac{1}{n^{2}}\;.$

∎

###  A.4 Derivatives of $\frac{f}{g}$

We prove

######  Proposition A.3.

For $f,g$ smooth with $g>0$,

$\left(\dfrac{f}{g}\right)^{(l)}=\frac{1}{g}\left[f^{(l)}-l!\;\sum_{j=1}^{l}\frac{g^{(l+1-j)}}{(l+1-j)!}\frac{1}{(j-1)!}\left(\dfrac{f}{g}\right)^{(j-1)}\right]\;.$
(195)

###### Proof.

The proof is done by induction in $l\in{\mathbb{N}}$. For $l=1$, the statement
is easily verified. Then differentiating (195) and using the induction
hypothesis, we obtain

$\begin{split}\left(\dfrac{f}{g}\right)^{(l+1)}&=\dfrac{f^{(l+1)}}{g}-\dfrac{g^{\prime}f^{(l)}}{g^{2}}+\frac{g^{\prime}}{g^{2}}\sum_{j=1}^{l}\binom{l}{j-1}g^{(l+1-j)}\left(\dfrac{f}{g}\right)^{(j-1)}\\\
&-\frac{1}{g}\sum_{j=1}^{l}\binom{l}{j-1}\Big{(}g^{(l+2-j)}\left(\dfrac{f}{g}\right)^{(j-1)}+g^{(l+1-j)}\left(\dfrac{f}{g}\right)^{(j)}\Big{)}\\\
&=\dfrac{f^{(l+1)}}{g}-\frac{g^{\prime}}{g}\left(\dfrac{f}{g}\right)^{(l)}-\frac{g^{(l+1)}}{g}\dfrac{f}{g}-l\frac{g^{\prime}}{g}\left(\dfrac{f}{g}\right)^{(l)}\\\
&-\frac{1}{g}\sum_{j=2}^{l}\Bigg{[}\binom{l}{j-1}+\binom{l}{j-2}\Bigg{]}g^{(l+2-j)}\left(\dfrac{f}{g}\right)^{(j-1)}\\\
&=\frac{1}{g}\left[f^{(l+1)}-(l+1)!\;\sum_{j=1}^{l+1}\frac{g^{(l+2-j)}}{(l+2-j)!}\frac{1}{(j-1)!}\left(\dfrac{f}{g}\right)^{(j-1)}\right]\;,\end{split}$
(196)

where we used

$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k},\quad n\in{\mathbb{N}}_{0}\,,\
k\in{\mathbb{N}}\;.$ (197)

∎

##  Appendix B Appendix B

In this appendix, we prove the different lemmas and propositions stated in
Sect.4. The bounds we obtain are expressed in terms of positive constants
$C\,,\ C_{i}\,$, $i=1,\ldots,11\,$ chosen sufficiently large and then for $K$
sufficiently large, depending on these constants.

###  B.1 Bounds on the functions $H(\mu)\,,\ h(\mu)$

Here we prove bounds on the functions $H(\mu)\,,\ h(\mu)$ introduced in
(146)-(147), and on their derivatives.

######  Lemma B.1.

$0<\dfrac{1}{H(\mu)}\leq C\ ,\quad\mu\in[0,{\mu}_{\max}]\ .$ (198)

###### Proof.

We recall that we can choose $\alpha_{0}\leq\frac{1}{2m^{2}}$ so that
$\beta_{0}\leq\frac{1}{2}$. Obviously $H(\mu)>0$ for $\mu\in[0,{\mu}_{\max}]$.
For $\mu\in[0,-\ln(2\beta_{0})]$, we have $H(\mu)\geq
c(1+\beta_{0}-\frac{1}{2})\geq c>0\,$. For
$\,\mu\in[-\ln(2\beta_{0}),{\mu}_{\max}]\,$ we have

$H(\mu)\geq
h(\mu)\geq\dfrac{1}{4}\int_{k}\dfrac{e^{-k^{2}}}{k^{2}+1}\geq\dfrac{1}{4e}\int_{k,|k|\leq
1}\dfrac{1}{k^{2}+1}\geq\dfrac{1}{8e}\int_{k,|k|\leq 1}1=\dfrac{c}{16e}\ .$
(199)

Choosing $\,C:=256\,e\,\pi^{2}\,$, the bound (198) is satisfied. ∎

######  Lemma B.2.

For $l\geq 0$ and $\mu\in[0,{\mu}_{\max}]$

$|\partial_{\mu}^{l}h(\mu)|\leq c\,(5\,e)^{l}|l-1|!\ .$ (200)

###### Proof.

The proof is by induction in $l\geq 0\,$. First note $|h(\mu)|\leq c\,$ for
$\,\mu\in[0,{\mu}_{\max}]\,$. For $l=1$

$|h^{\prime}(\mu)|=|(2+\beta_{0}e^{\mu})h(\mu)-c\beta_{0}e^{\mu}|\leq(4+2\beta_{0})c\leq
c\,5\,e\ ,$

since $\beta_{0}\leq\frac{1}{2}\,$. Using Leibniz’ s rule, we obtain for
$l\geq 1$

$\partial_{\mu}^{l}h(\mu)=\sum_{0\leq l_{1}\leq
l-1}\binom{l-1}{l_{1}}(2\delta_{l_{1},0}+\beta_{0}e^{\mu})\partial_{\mu}^{l-1-l_{1}}h(\mu)-c\,\beta_{0}\,e^{\mu}\
.$ (201)

Inserting the induction hypothesis in the r.h.s of (201) we get

$\begin{split}|\partial_{\mu}^{l}h(\mu)|&\leq\sum_{0\leq l_{1}\leq
l-1}\binom{l-1}{l_{1}}(3+\beta_{0})(5e)^{l-1-l_{1}}(l-1-l_{1})!\;c+c(1+\beta_{0})\\\
&\leq(3+\beta_{0})(5e)^{l-1}\sum_{0\leq l_{1}\leq
l-1}\binom{l-1}{l_{1}}(l-1-l_{1})!\;c+c\,(1+\beta_{0})\\\
&\leq(3+\beta_{0})(5e)^{l-1}(l-1)!\;e\,c+c(1+\beta_{0})\\\ &\leq
c\,(5e)^{l}(l-1)!\left(\frac{7}{10}+\frac{3}{2\cdot 8^{l}}\right)\leq
c\,(5e)^{l}(l-1)!\ .\end{split}$ (202)

∎

######  Lemma B.3.

For $l\geq 0$, $\mu\in[0,{\mu}_{\max}]\,$,

$|\partial_{\mu}^{l}H(\mu)|\leq 3\,c\,(5e)^{l}|l-1|!\ .$ (203)

###### Proof.

From Lemma B.2

$|\partial_{\mu}^{l}H(\mu)|\,\leq\,|c\,(1+\beta_{0})\,\delta_{l,0}\,-\,c\,\beta_{0}\
e^{\mu}|+|\partial_{\mu}^{l}h(\mu)|\,\leq\,3\,c+\,c\,(5e)^{l}|l-1|!\,\leq\,3\,c\,(5e)^{l}|l-1|!\
.$ (204)

∎

######  Lemma B.4.

For $l\geq 1$ and $\mu\in[0,{\mu}_{\max}]\,$,

$|\partial_{\mu}^{l}\log(H(\mu))|\,\leq\,c\,C^{l}\,(5e)^{l+1}\,2^{l}\,(l-1)!\
.$ (205)

###### Proof.

For $l=1$ the bounds derived for $h,h^{\prime}$ in Lemmas B.1 and B.2 give :

$|\partial_{\mu}\log H(\mu)|\,\leq\,c\,C\,(5+\frac{3}{2})\,\leq\,c\,C\,50\
e^{2}\ .$ (206)

For $l\geq 1$ we have using (94)

$\partial_{\mu}^{l+1}\log
H(\mu)=\partial_{\mu}^{l}\left(\frac{H^{\prime}(\mu)}{H(\mu)}\right)=\dfrac{1}{H(\mu)}\left[\partial_{\mu}^{l+1}H(\mu)-l!\sum_{j=0}^{l-1}\dfrac{\partial_{\mu}^{l-j}H(\mu)}{(l-j)!\;j!}\partial_{\mu}^{j}\left(\dfrac{H^{\prime}(\mu)}{H(\mu)}\right)\right]\
.$ (207)

Since
$\partial_{\mu}^{j-1}\left(\dfrac{H^{\prime}(\mu)}{H(\mu)}\right)=\partial_{\mu}^{j}\log
H(\mu)$, we can proceed inductively on the r.h.s of (207). Using Lemma B.3, we
get

$\begin{split}|\partial_{\mu}^{l+1}\log H(\mu)|&\leq
C\Bigl{[}c\,(5e)^{l+1}l!+l!\sum_{j=1}^{l}\dfrac{4c(5e)^{l+1-j}(l-j)!}{(l+1-j)!\;(j-1)!}cC^{j}5^{j+1}2^{j}(j-1)!\Bigr{]}\\\
&\leq
c\,C^{l+1}(5e)^{l+2}\Bigl{[}\dfrac{l!}{5eC^{l}}+4cl!\;\sum_{j=1}^{l}2^{j}\Bigr{]}\\\
&\leq
c\,C^{k+1}(5e)^{l+2}2^{l+1}l!\Bigl{[}\frac{1}{10eC}+\frac{1}{4\pi^{2}}\Bigr{]}\leq
c\ C^{l+1}\ (5e)^{l+2}\ 2^{l+1}l!\ .\end{split}$ (208)

∎

###  B.2 Bounds on the coefficients $\,\tilde{f}_{n,k}\,\ \tilde{g}_{n,k}\,$

See 4.1

###### Proof.

The proof is done by induction in $N=n+2l$, going up in $l$. We start at $N=6$
and we have from the boundary conditions (153)

$\tilde{f}_{6}(0)=0\ .$

For $0\leq l<\frac{n}{2}-3$, we use (149) and we solve it for
$\partial_{\mu}^{l+1}\tilde{f}_{n}(0)$. Using the induction hypothesis, we
obtain $\partial_{\mu}^{l+1}\tilde{f}_{n}(0)=0$ since in the products

$\partial_{\mu}^{l_{1}}\tilde{f}_{n_{1}}(0)\partial_{\mu}^{l_{2}}\tilde{f}_{n_{2}}(0)\;,$

the constraints $n_{1}+n_{2}=n+2$ and $l_{1}+l_{2}\leq l$ and
$l<\frac{n}{2}-3$ imply that either $l_{1}\leq\frac{n_{1}}{2}-3\,$ or
$\,l_{2}\leq\frac{n_{2}}{2}-3\,$. ∎

See 4.2

###### Proof.

From (158) we have

$|\tilde{f}_{2,1}|=|3\tilde{g}_{4,0}+(1-\beta_{0})\tilde{f}_{2,0}-2\tilde{f}_{2,0}^{2}|\leq\dfrac{K}{2}\;,$

and from (162)

$|\tilde{g}_{4,1}|=2|\tilde{g}_{4,0}|\left|4\tilde{f}_{2,0}+\beta_{0}-\frac{1}{2}h_{0}\right|\leq\dfrac{K}{32}$

choosing $\sqrt{K}>7\,c\,C>4\,$. We proceed by induction in $n$. For $n\geq
6$, we find from (160) and from (162) for $K$ large enough

$|\tilde{g}_{n,0}|\leq\dfrac{n}{n-4}\frac{1}{2}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+1-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})^{2}}\leq\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\
,$
$|\tilde{g}_{n,1}|\leq\dfrac{2n}{n-2}(n-2)\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+2-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})^{2}}+\dfrac{n}{n-2}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\left(\dfrac{\sqrt{K}}{4}+2+\frac{1}{2}7cC\right)\leq\dfrac{K^{\frac{n}{2}-\frac{1}{2}}}{n}\
.$

For $n\leq 10$ the previous bounds for the sum over $n_{1}$ can be checked
explicitly, for $n\geq 12$ we use Lemma A.1 in Appendix A.3. ∎

See 4.1

###### Proof.

The proof proceeds by induction in $N=n+k$ going up in $k$, as in the proof of
Proposition 3.2. For $k\leq 1$, the bounds follow from Lemma 4.2. In the r.h.s
of (159), the first, second, fifth and seventh terms can be treated as in the
proof of Proposition 3.2. So we focus on the remaining terms.

  * •

Third term: we separate the terms summed over $2\leq\nu\leq k-2$ and the
remaining terms. Using Lemma 4.2 we have, choosing $K>\,10\,e\,C$

    * –

$\nu=0$:

$\begin{split}\dfrac{n-2}{n+2k}|\tilde{g}_{n,0}h_{k+1}|&\leq\dfrac{n-2}{n+2k}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}c\,C^{k+2}(5e)^{k+3}2^{k+2}\\\
&\leq
K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{8}}}5\,e\,c\
.\end{split}$ (209)

    * –

$\nu=1$:

$\begin{split}\dfrac{n-2}{n+2k}|\tilde{g}_{n,1}h_{k}|&\leq\dfrac{n-2}{n+2k}\dfrac{K^{\frac{n}{2}-\frac{1}{2}}}{n}c\,C^{k+1}(5e)^{k+2}2^{k+1}\\\
&\leq
K^{\frac{n}{2}+k+\frac{1}{2}}\left(\frac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{8}}}5\,e\,c\
.\end{split}$ (210)

For $k-1\leq\nu\leq k+1$, we have the following bounds

$|\tilde{g}_{n,\nu}|\leq
K^{\frac{n}{2}+\nu-\frac{3}{2}}\left(\frac{n}{4}+k-1\right)!\;\frac{1}{[(k+2)!]^{\frac{1}{8}}}\;.$
(211)

Using (152) we obtain

$\dfrac{n-2}{n+2k}\sum_{\nu=k-1}^{k+1}|\tilde{g}_{n,\nu}h_{k+1-\nu}|\leq\dfrac{3}{4}5ecK^{\frac{n}{2}+k+\frac{1}{2}}\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{8}}}\
.$

The remaining sum can be bounded for $k\geq 4$ using Lemma 3.4 and (152)

$\begin{split}\dfrac{n-2}{n+2k}\sum_{\nu=2}^{k-2}|\tilde{g}_{n,\nu}h_{k+1-\nu}|&\leq\dfrac{n-2}{n+2k}K^{\frac{n}{4}+k+\frac{1}{2}}\frac{5ec}{4}\sum_{\nu=2}^{k-2}\frac{(\frac{n}{4}+\nu-3)!}{[\nu!]^{\frac{1}{8}}}\\\
&\leq\dfrac{n-2}{n+2k}K^{\frac{n}{4}+k+\frac{1}{2}}\frac{5ec}{4}\mathcal{F}\Bigl{(}n,4,k-2,2,2,\frac{1}{8}\Bigr{)}\\\
&\leq\dfrac{n-2}{n+2k}K^{\frac{n}{4}+k+\frac{1}{2}}5ec\dfrac{2^{\frac{7}{8}}[(k-2)!]^{\frac{7}{8}}}{n}\dfrac{\left(\frac{n}{4}+k-3\right)!}{(k-4)!}\\\
&\leq
K^{\frac{n}{4}+k+\frac{1}{2}}\,10\,e\,c\left(\dfrac{n}{4}+k-1\right)!\;\dfrac{1}{[(k+2)!]^{\frac{1}{8}}}\
.\end{split}$ (212)

Then the third term is bounded by

$K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\frac{1}{3}\
.$ (213)

  * •

Fourth term: we proceed similarly as for the third term, then the fourth term
is bounded by

$K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\frac{{C}_{2}}{\sqrt{K}}\
.$ (214)

  * •

Sixth term: Looking at the terms corresponding to $\nu\geq k+1$ and using the
bounds from Lemma 4.2 we get

    * –

$\nu=k+1$

$\dfrac{2}{(k+1)!}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}|\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},1}|\leq\dfrac{1}{(k+1)!}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+2-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})}$
$\leq\dfrac{1}{\sqrt{K}}\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{(k+1)!}\dfrac{1}{n}\leq\dfrac{1}{\sqrt{K}}K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\
.$

    * –

$\nu=k+2$

$\dfrac{1}{(k+2)!}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\ n_{i}\geq
4\end{subarray}}|\tilde{g}_{n_{1},0}\tilde{g}_{n_{2},0}|\leq\dfrac{1}{4(k+2)!}\sum_{\begin{subarray}{c}n_{1}+n_{2}=n+2\\\
n_{i}\geq
4\end{subarray}}\dfrac{K^{\frac{n}{2}-\frac{3}{2}+1-\frac{3}{2}}}{n_{1}^{2}(n+2-n_{1})^{2}}$
$\leq\dfrac{1}{4\sqrt{K}}\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{(k+2)!}\dfrac{1}{n^{2}}\leq\dfrac{1}{4\sqrt{K}}K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\
.$

For the remaining part of the sum, we substitute
$\mathcal{F}(n_{1},n_{2},k,c,c,\frac{1}{4})$, $c\in\\{0,2\\}$ from the
analysis of the third term in the proof of Proposition 3.2 by
$\mathcal{F}(n_{1},n_{2},k-\nu,c,c,\frac{1}{8})$, $c\in\\{0,2\\}$. Then, the
sixth term is bounded by

$\frac{1}{2\sqrt{n+2k}}\dfrac{C_{3}}{\sqrt{K}}K^{\frac{n}{2}+k+\frac{1}{2}}\sum_{\nu=0}^{k}\frac{1}{\nu!}\dfrac{\left(\frac{n}{4}+k-1-\nu\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}\
.$ (215)

One can then extract the terms in the sum corresponding to $\nu\leq 1$ and
bound them by
$\frac{\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{8}}}\,$. The residual
sum is non-zero for $k\geq 2$, it is bounded by

$\sum_{\nu=2}^{k}\frac{1}{\nu!}\dfrac{\left(\frac{n}{4}+k-1-\nu\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}=\sum_{\nu=0}^{k-2}\frac{1}{(k-\nu)!}\dfrac{\left(\frac{n}{4}-1+\nu\right)!}{[(\nu+2)!]^{\frac{1}{8}}}\leq\dfrac{\left(\frac{n}{4}+k-3\right)!}{[(k-2)!]^{\frac{1}{8}}}e\
.$

Then we obtain

$\frac{1}{2\sqrt{n+2k}}\sum_{\nu=2}^{k}\frac{1}{\nu!}\dfrac{\left(\frac{n}{4}+k-1-\nu\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}\leq\frac{e}{2}\dfrac{1}{\sqrt{n+2k}}\dfrac{\left(\frac{n}{4}+k-3\right)!}{[(k-2)!]^{\frac{1}{8}}}\leq\frac{e}{2}\dfrac{\left(\frac{n}{4}+k-1\right)!}{[(k+2)!]^{\frac{1}{8}}}\
.$

Finally the sixth term is bounded by

$K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\frac{C_{4}}{\sqrt{K}}\;.$
(216)

  * •

Eighth term: first the term in the sum corresponding to $\nu=k+1$ is

$\dfrac{n}{n+2k}\dfrac{1}{(k+1)!}|\tilde{g}_{n,0}\tilde{f}_{2,0}|\leq\dfrac{\sqrt{K}}{16}\dfrac{K^{\frac{n}{2}-\frac{3}{2}}}{2n^{2}}\leq
K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\frac{1}{32nK}\
.$

We follow the steps from the proof of Proposition 3.2 for the second term but
substituting $\mathcal{F}(n,4,k-2,2,2,\frac{1}{4})$ by
$\mathcal{F}(n,4,k-2-\nu,2,2,\frac{1}{8})$. Thus we have

$\dfrac{n}{n+2k}\mathcal{F}\left(n,4,k-2-\nu,2,2,\frac{1}{8}\right)\leq\dfrac{n}{n+2k}[(k-2)!]^{\frac{7}{8}}\dfrac{4}{n}\dfrac{\left(\frac{n}{4}+k-\nu-4\right)!}{(k-\nu-4)!}$
$\leq
4\;\dfrac{\left(\frac{n}{4}+k-\nu-1\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}\dfrac{\Big{(}k(k+1)(k+2)\Big{)}^{\frac{1}{8}}}{n+2k}\leq\frac{4}{\sqrt{n+2k}}\;\dfrac{\left(\frac{n}{4}+k-1-\nu\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}\
.$

Therefore we can bound the eighth term by

$\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}\,{C}_{5}}{\sqrt{K}\sqrt{n+2k}}\sum_{\nu=0}^{k}\dfrac{1}{\nu!}\dfrac{\left(\frac{n}{4}+k-1-\nu\right)!}{[(k+2-\nu)!]^{\frac{1}{8}}}\
.$ (217)

From the analysis of the sixth term, we deduce that the eighth term is finally
bounded by

$K^{\frac{n}{2}+k+\frac{1}{2}}\dfrac{\left(\frac{n}{4}+k-1\right)!\;}{[(k+2)!]^{\frac{1}{8}}}\frac{C_{6}}{\sqrt{K}}\;.$
(218)

Summing (213), (214), (216) and (218) with the already treated terms in the
proof of Proposition 3.2, we obtain

$|\tilde{g}_{n,k+2}|\leq\left[\frac{1}{3}+\frac{C_{7}}{K}+\frac{C_{8}}{\sqrt{K}}\right]\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{[(k+2)!]^{\frac{1}{8}}}\left(\dfrac{n}{4}+k-1\right)!\leq\dfrac{K^{\frac{n}{2}+k+\frac{1}{2}}}{[(k+2)!]^{\frac{1}{8}}}\left(\dfrac{n}{4}+k-1\right)!\
.$ (219)

#####

The bounds for $\tilde{f}_{2,k}$, for $k\leq 1$, follow from Lemma 4.2. The
bounds for $k\leq 6$ can be checked by hand noting that we can always factor
out $\dfrac{1}{\sqrt{K}}$ in the r.h.s of (158) using the induction hypothesis
so that the bound holds choosing $K$ sufficiently large. For $k>6$ we will
focus on the last two terms in the r.h.s. of (158) since the other terms are
treated as in the proof of Proposition 3.2.

  * •

Fourth term: it is bounded by

$\dfrac{\beta_{0}}{(k+1)}\dfrac{K^{k+\frac{1}{2}+1}}{K}\sum_{\nu=0}^{k}\dfrac{|\nu-3|!}{(k-\nu)!\;[|\nu-1|!]^{\frac{1}{8}}}\
.$ (220)

The sum can be bounded as follows

$\begin{split}\sum_{\nu=0}^{k}\dfrac{|\nu-3|!}{(k-\nu)!\;[|\nu-1|!]^{\frac{1}{8}}}&=\sum_{\nu=3}^{k}\dfrac{(\nu-3)!}{(k-\nu)!\;[(\nu-1)!]^{\frac{1}{8}}}+\dfrac{6}{k!}+\dfrac{2}{(k-1)!}+\dfrac{1}{(k-2)!}\\\
&\leq[(k-1)!]^{\frac{7}{8}}\sum_{\nu=3}^{k}\dfrac{(\nu-3)!}{(k-\nu)!\;(\nu-1)!}\\\
&\leq[(k-1)!]^{\frac{7}{8}}\sum_{\nu=3}^{k}\dfrac{5}{(k+1-\nu)^{2}(\nu-1)^{2}}\\\
&\leq[(k-1)!]^{\frac{7}{8}}\dfrac{15}{k}\leq\dfrac{15(k-2)!}{[(k-1)!]^{\frac{1}{8}}}\
.\end{split}$ (221)

Taking into account the factor $\frac{1}{k+1}$, the fourth term is bounded by

$K^{k+1+\frac{1}{2}}\dfrac{(k-2)!}{[k!]^{\frac{1}{8}}}\dfrac{C_{9}}{K}\ .$
(222)

  * •

Fifth term: we separate the sum as follows:

$\dfrac{1}{2(k+1)}\sum_{\nu=0}^{k-2}\frac{1}{\nu!}\sum_{\nu^{\prime}=0}^{k-\nu}|\tilde{f}_{2,\nu^{\prime}}\tilde{f}_{k-\nu-\nu^{\prime}}|+\dfrac{1}{(k+1)(k-1)!}|\tilde{f}_{2,0}\tilde{f}_{2,1}|+\dfrac{1}{2(k+1)!}\tilde{f}_{2,0}^{2}\
.$

The bounds for the last two terms follow from Lemma 4.2. The sum is bounded as
in the proof of Proposition 3.2 by

$\dfrac{K^{k+1+\frac{1}{2}}}{\sqrt{K}}\sum_{\nu=0}^{k-2}\dfrac{1}{\nu!}\dfrac{(k-2-\nu)!}{[(k-\nu)!]^{\frac{1}{8}}}\;.$
(223)

This sum is then treated as in the fourth term. Therefore we have the final
bound

$K^{k+1+\frac{1}{2}}\dfrac{(k-2)!}{[k!]^{\frac{1}{8}}}\,\dfrac{C_{10}}{\sqrt{K}}\
.$ (224)

Then, we obtain

$|\tilde{f}_{2,k+1}|\leq
K^{k+1+\frac{1}{2}}\dfrac{(k-2)!}{[k!]^{\frac{1}{8}}}\dfrac{C_{11}}{\sqrt{K}}\leq
K^{k+1+\frac{1}{2}}\dfrac{(k-2)!}{[k!]^{\frac{1}{8}}}\ .$ (225)

∎

## Acknowledgement

#####

The authors are much indebted to Riccardo Guida for his remarks and
suggestions.

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