1) School of Mathematics, University of Minnesota, Minneapolis, MN, USA,
<EMAIL_ADDRESS>
2) Faculty of Mathematics, Bielefeld University, Germany, chistyak@math-
uni.bielefeld.de.

2) Faculty of Mathematics, Bielefeld University, Germany, goetze@math-
uni.bielefeld.de.

4) Research supported by the NSF grant DMS-2154001 and the GRF – SFB 1283/2
2021 – 317210226

# Richter’s local limit theorem,
its refinement, and related results

S. G. Bobkov1,4 missing ,  G. P. Chistyakov2,4 missing and  F. Götze2,4
missingmissing In memoriam Gennadiy P. Chistyakov *May 1, 1945
$\dagger$December 30, 2022

###### Abstract.

We give a detailed exposition of the proof of Richter’s local limit theorem in
a refined form, and establish the stability of the remainder term in this
theorem under small perturbations of the underlying distribution (including
smoothing). We also discuss related quantitative bounds for characteristic
functions and Laplace transforms.

###### Key words and phrases:

Central limit theorem, local limit theorem, characteristic functions

###### 2010 Mathematics Subject Classification:

Primary 60E, 60F

##  1\. Introduction and Formulation of the Results

Let $(X_{n})_{n\geq 1}$ be independent copies of a random variable $X$ with
mean ${\mathbb{E}}X=0$ and variance ${\rm Var}(X)=1$. Throughout, we assume
without mentioning that the normalized sum

$Z_{n}=\frac{X_{1}+\dots+X_{n}}{\sqrt{n}}$

has a bounded density $p_{n_{0}}$ for some $n=n_{0}$, that is,
$p_{n_{0}}(x)\leq M$ for all $x\in{\mathbb{R}}$ with some constant $M$. Then
all $Z_{n}$ with $n\geq 2n_{0}$ have continuous bounded densities $p_{n}(x)$.
An asymptotic behavior of these densities describing their closeness to the
normal density

$\varphi(x)=\frac{1}{\sqrt{2\pi}}\,e^{-x^{2}/2},\quad x\in{\mathbb{R}},$

is governed by several local limit theorems. First of all, there is a uniform
local limit theorem due to Gnedenko

$\sup_{x}|p_{n}(x)-\varphi(x)|\rightarrow 0\quad{\rm as}\ n\rightarrow\infty.$

Under higher order moment assumptions, say if ${\mathbb{E}}\,|X|^{m}<\infty$
for an integer $m\geq 3$, this statement may be considerably sharpened in the
form of a non-uniform local limit theorem

$\sup_{x}\,(1+|x|^{m})\,|p_{n}(x)-\varphi_{m}(x)|=o\big{(}n^{-\frac{m-2}{2}}\big{)},$
(1.1)

where $\varphi_{m}$ denotes the Edgeworth correction of $\varphi$ of order $m$
(cf. [11], [16], [17]). In various applications, this relation is typically
effective in the range $|x|\leq\sqrt{c\log n}$, since then the ratio
$p_{n}(x)/\varphi(x)$ remains close to 1 (for a suitable $c$). For example,
(1.1) is crucial in the study of rates in the entropic central limit theorem,
rates for Rényi divergences of finite orders and for the relative Fisher
information ([5]-[7]).

As for larger regions, the asymptotic behavior of $p_{n}(x)$ is governed by
the following remarkable theorem due to Richter [18], assuming the finiteness
of an exponential moment for the random variable $X$.

Theorem 1.1. Suppose that, for some $b>0$,

${\mathbb{E}}\,e^{b|X|}<\infty.$ (1.2)

Then, for $x=o(\sqrt{n})$, the densities of $Z_{n}$ admit the representation

$\frac{p_{n}(x)}{\varphi(x)}=\exp\Big{\\{}\frac{x^{3}}{\sqrt{n}}\,\lambda\Big{(}\frac{x}{\sqrt{n}}\Big{)}\Big{\\}}\,\Big{(}1+O\Big{(}\frac{1+|x|}{\sqrt{n}}\Big{)}\Big{)}.$
(1.3)

Here, $\lambda(\tau)$ represents an analytic function in some neighborhood of
zero.

It was shown by Amosova [1] that the condition (1.2) is necessary for the
existence of a representation like (1.3) in the region $|x|=o(\sqrt{n})$ with
some analytic function $\lambda$.

The function $\lambda$ in (1.3) is representable as a power series, called the
Cramér series,

$\lambda(\tau)=\sum_{k=0}^{\infty}\lambda_{k}\tau^{k},$ (1.4)

which is absolutely convergent in some disc $|\tau|<\tau_{0}$ of the complex
plane. It has appeared in the work by Cramér [9] in a similar representation
for the ratio of the tails of distribution functions of $Z_{n}$ and the
standard normal law (cf. also [10], [14], [15]).

Let us also mention that (1.3) is stated in Richter’s work in a slightly
different form with $O(\frac{|x|}{\sqrt{n}})$ in the last brackets and for
$|x|>1$. A similar result is proved in the book by Ibragimov and Linnik [13]
under the assumption that $X$ has a bounded continuous density.

As a consequence of (1.3), one immediately obtains, for example, that

$\frac{p_{n}(x)}{\varphi(x)}\rightarrow 1\quad{\rm as}\ n\rightarrow\infty$
(1.5)

uniformly in the region $|x|=o(n^{1/6})$. In the region
$c_{0}n^{1/6}\leq|x|\leq c_{1}n^{1/2}$, the behavior may be quite different,
and in order to describe it, the appearance of the term
$O(\frac{1+|x|}{\sqrt{n}})$ in (1.3) is non-desirable. The purpose of this
paper is to give a detailed exposition of the proof of Theorem 1.1, clarifying
the meaning of the leading coefficient in (1.4) and replacing this term with
an $n$-depending quantity. We basically employ the tools of [13] and derive
the following refinement.

Theorem 1.2. Let the conditions of Theorem $1.1$ be fulfilled, and $n\geq
2n_{0}$. There is a constant $\tau_{0}>0$ with the following property. With
$\tau=x/\sqrt{n}$, we have for $|\tau|\leq\tau_{0}$

$\frac{p_{n}(x)}{\varphi(x)}=e^{n\tau^{3}\lambda(\tau)-\mu(\tau)}\,\big{(}1+O(n^{-1}(\log
n)^{3})\big{)},$ (1.6)

where $\mu(\tau)$ is an analytic function in $|\tau|\leq\tau_{0}$ such that
$\mu(0)=0$.

As we will see in Section 5,

$\displaystyle\lambda(\tau)$ $\displaystyle=$
$\displaystyle\frac{1}{m!}\,\gamma_{m}\tau^{m-3}+O(|\tau|^{m-2}),$
$\displaystyle\mu(\tau)$ $\displaystyle=$
$\displaystyle\frac{1}{2(m-2)!}\,\gamma_{m}\tau^{m-2}+O(|\tau|^{m-1}),$

where $\gamma_{m}$ ($m\geq 3$) is the first non-zero cumulant of the random
variable $X$ (assuming that it is not normal). Equivalently, $m$ is the
smallest positive integer such that ${\mathbb{E}}X^{m}\neq{\mathbb{E}}Z^{m}$,
where $Z$ is a standard normal random variable, in which case

$\gamma_{m}={\mathbb{E}}X^{m}-{\mathbb{E}}Z^{m}.$

With this refinement, it should be clear that the relation (1.5) holds true
uniformly over all $x$ in the potentially larger region

$|x|\leq\varepsilon_{n}\,n^{\frac{1}{2}-\frac{1}{m}}\quad(\varepsilon_{n}\rightarrow
0).$

For example, if the distribution of $X$ is symmetric about the origin, then
$\gamma_{3}=0$, so that necessarily $m\geq 4$.

Another consequence of (1.6), which cannot be obtained on the basis of (1.3),
is needed in the study of the central limit theorem (CLT) with respect to the
Rényi divergence of infinite order (including the rate of convergence). Let us
recall that the Rényi divergence of a finite order $\kappa>0$ from the
distribution of $Z_{n}$ to the standard normal law is defined by

$D_{\kappa}(p_{n}||\varphi)=\frac{1}{\kappa-1}\log\int_{-\infty}^{\infty}\Big{(}\frac{p_{n}(x)}{\varphi(x)}\Big{)}^{\kappa}\,\varphi(x)\,dx.$

As a function of $\kappa$, it is non-decreasing, representing a strong
distance-like quantity. In the range $0<\kappa<1$, it is metrically equivalent
to the total variation, that is, $L^{1}$-distance between $p_{n}$ and
$\varphi$. The case $\kappa=1$ corresponds to the relative entropy (Kullback-
Leibler’s distance)

$D(p_{n}||\varphi)=\lim_{\kappa\rightarrow
1}D_{\kappa}(p_{n}||\varphi)=\int_{-\infty}^{\infty}p_{n}(x)\,\log\frac{p_{n}(x)}{\varphi(x)}\,dx,$

and another important case $\kappa=2$ leads to the function of the
$\chi^{2}$-Pearson distance. So far, information-theoretic CLT’s of the form
$D_{\kappa}(p_{n}||\varphi)\rightarrow 0$ as $n\rightarrow\infty$ have been
completely characterized in terms of the distribution of $X$ (that is, in the
i.i.d. situation). However, such a statement remains fully open for the limit
distance

$D_{\infty}(p_{n}||\varphi)=\lim_{\kappa\rightarrow\infty}D_{\kappa}(p_{n}||\varphi)=\sup_{x}\,\log\,\frac{p_{n}(x)}{\varphi(x)}.$

Equivalently, the problem is to find conditions under which the related
quantity (the Tsallis distance of infinite order)

$T_{\infty}(p_{n}||\varphi)=\sup_{x}\,\frac{p_{n}(x)-\varphi(x)}{\varphi(x)}$

tends to zero for growing $n$ (note that one may not put the absolute values
sign, since all $p_{n}$ may be compactly supported).

As a first natural step towards this variant of the CLT, we consider the
problem of the convergence for the restricted Tsallis distance with the above
suprema taken over growing intervals $|x|=O(\sqrt{n})$. With this in mind,
Theorem 1.2 allows one to prove the following:

Corollary 1.3. Under the conditions of Theorem $1.1$, suppose that $m$ is
even, $m\geq 4$, and $\gamma_{m}<0$. There exist constants $\tau_{0}>0$ and
$c>0$ with the following property. If $|\tau|\leq\tau_{0}$, $\tau=x/\sqrt{n}$,
then

$\frac{p_{n}(x)-\varphi(x)}{\varphi(x)}\leq\frac{c(\log n)^{3}}{n}.$ (1.7)

Here, the condition about cumulants is fulfilled, for example, when the random
variable $X$ is strongly subgaussian in the sense that

${\mathbb{E}}\,e^{tX}\leq e^{t^{2}/2}\quad{\rm for\ all}\ t\in{\mathbb{R}}$
(1.8)

(recall that that ${\mathbb{E}}X^{2}=1$, while the condition ${\mathbb{E}}X=0$
is necessary). This interesting class of probability distributions is rather
rich, and we refer the reader to [8] for discussions and various examples. Our
main motivation stemmed from the fact that the strong subgaussianity is
necessary for the convergence $D_{\infty}(p_{n}||\varphi)\rightarrow 0$. As we
have recently learned, (1.8) had previously appeared under the name “sharp
subgaussianity” in the work by Guionnet and Husson [12] for a completely
different reason as a condition to have LDPs for the largest eigenvalue of
Wigner matrices with the same rate function as in the case of Gaussian
entries.

One important issue which is not addressed in the formulation of Theorem 1.2
is how one can control the involved constant in the $O$-remainder term in
(1.6). To better quantify this asymptotic representation, we actually prove
the following statement using the same analytic functions $\lambda(\tau)$ and
$\mu(\tau)$.

Theorem 1.4. Assume that ${\mathbb{E}}\,e^{\alpha|X|}\leq 2$ $(\alpha>0)$.
There exist absolute positive constants $C$ and $c$ such that, whenever $n\geq
n_{1}$ and $\tau=x/\sqrt{n}$, $|\tau|\leq\tau_{0}$, we have

$\frac{p_{n}(x)}{\varphi(x)}=e^{n\tau^{3}\lambda(\tau)-\mu(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)},$ (1.9)

where $|B|\leq C$ and

$n_{1}=CM^{4}n_{0}^{2}\,\alpha^{-12},\quad\tau_{0}=\frac{c\alpha^{3}}{M^{2}n_{0}}.$

This statement should be useful in applications to smoothed distributions in
order to guarantee that the constant in the remainder term may be chosen to be
common for all distributions under consideration.

In order to make the proofs/arguments more transparent and self-contained, we
include a short review of various related results – partly technical, but
often interesting in themselves – about maxima of densities, analytic
characteristic functions and log-Laplace transforms. The rest of the paper is
organized as follows. In Sections 2 we recall basic properties of the maximum
of convolved densities and then develop their applications to bounding
restricted integrals of powers of characteristic functions (Section 3). In
section 4, we discuss behavior of analytic characteristic functions near the
origin. Section 5 is devoted to the so-called saddle points and associated
Taylor expansions for the log-Laplace transforms. Here we also analyze the
functions $\lambda(\tau)$ and $\mu(\tau)$. Sections 6 deals with contour
integration needed to establish preliminary representations for $p_{n}(x)$.
Final steps in the proof of Theorem 1.4 are made in Section 7. The proof of
Corollary 1.3 is postponed to Section 8.

##  2\. Maximum of Convolved Densities

Convolved densities are known to have improved smoothing properties. First,
let us emphasize the following general fact (which explains the condition
$n\geq 2n_{0}$ mentioned before Theorem 1.1).

Proposition 2.1. If independent random variables $\xi_{1},\dots,\xi_{m}$
$(m\geq 2)$ have bounded densities, then the sum $S_{m}=\xi_{1}+\dots+\xi_{m}$
has a bounded uniformly continuous density vanishing at infinity.

Proof. Denote by $q_{k}$ the densities of $\xi_{k}$ and assume that
$q_{k}(x)\leq M_{k}$ for all $x\in{\mathbb{R}}$ with some constants $M_{k}$
($k\leq m$). By the Plancherel theorem, for the characteristic functions
$g_{k}(t)={\mathbb{E}}\,e^{it\xi_{k}}$, we have

$\displaystyle\int_{-\infty}^{\infty}|g_{k}(t)|^{m}\,dt$ $\displaystyle\leq$
$\displaystyle\int_{-\infty}^{\infty}|g_{k}(t)|^{2}\,dt\,=\,2\pi\int_{-\infty}^{\infty}q_{k}(x)^{2}\,dx$
$\displaystyle\leq$ $\displaystyle
2\pi\int_{-\infty}^{\infty}M_{k}q_{k}(x)\,dx\,=\,2\pi M_{k},$

where we used the property $|g_{k}(t)|\leq 1$, $t\in{\mathbb{R}}$. Hence, by
Hölder’s inequality, the characteristic function $g(t)=g_{1}(t)\dots g_{m}(t)$
of $S_{m}$ is integrable and has $L^{1}$-norm

$\displaystyle\int_{-\infty}^{\infty}|g(t)|\,dt$ $\displaystyle\leq$
$\displaystyle\Big{(}\int_{-\infty}^{\infty}|g_{1}(t)|^{m}\,dt\Big{)}^{1/m}\dots\Big{(}\int_{-\infty}^{\infty}|g_{m}(t)|^{m}\,dt\Big{)}^{1/m}$
(2.1) $\displaystyle\leq$ $\displaystyle 2\pi\,(M_{1}\dots
M_{m})^{1/m}\,<\,\infty.$

One may conclude that the random variable $S_{m}$ has a bounded, uniformly
continuous density expressed by the inversion Fourier formula

$q(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-itx}g(t)\,dt,\quad
x\in{\mathbb{R}}.$ (2.2)

Since $g$ is integrable, it also follows that $q(x)\rightarrow 0$ as
$|x|\rightarrow\infty$ (by the Riemann-Lebesgue lemma). ∎

Consider the functional

$M(\xi)={\rm ess\,sup}_{x}\ q(x),$

where $\xi$ is a random variable with density $q$ (one may put $M(\xi)=\infty$
in all other cases). Since, by (2.2),

$q(x)\leq\frac{1}{2\pi}\int_{-\infty}^{\infty}|g(t)|\,dt$

for all $x\in{\mathbb{R}}$, the inequality (2.1) also implies that

$M(S_{m})\leq(M(\xi_{1})\dots M(\xi_{m}))^{1/m}.$ (2.3)

This shows in particular that $M(\xi)$ may not increase by adding to $\xi$ an
independent random variable. However, the relation (2.3) does not correctly
reflect the behavior of $M(S_{m})$ with respect to the growing parameter $m$,
especially in the i.i.d. situation. A more precise statement is described in
the following relation, where the geometric mean of maxima is replaced with
the harmonic mean.

Proposition 2.2. Given independent random variables $\xi_{k}$, $1\leq k\leq
m$, one has

$\frac{1}{M(S_{m})^{2}}\geq\frac{1}{2}\sum_{k=1}^{m}\frac{1}{M(\xi_{k})^{2}}.$
(2.4)

This bound may be viewed as a counterpart of the entropy power inequality in
Information Theory. It may be obtained by combining Rogozin’s maximum-of-
density theorem with Ball’s bound on the volume of slices of the cube. Namely,
it was shown in [19] that, if the values $M_{k}=M(\xi_{k})$ are fixed,
$M(S_{m})$ is maximized for $\xi_{k}$ uniformly distributed in the intervals
of length $1/M_{k}$. Of course, in this case $M(S_{m})$ has a rather
complicated structure as a function in variables $M_{1},\dots,M_{m}$.

On the other hand, if $T_{m}=a_{1}\eta_{1}+\dots+a_{m}\eta_{m}$, where
$\eta_{k}$ are independent and uniformly distributed in $(0,1)$, and the
coefficients satisfy $a_{1}^{2}+\dots+a_{m}^{2}=1$, then

$1\leq M(T_{m})\leq\sqrt{2},$ (2.5)

cf. [2]. In geometric language, this is the same as saying that $1\leq|Q\cap
H|\leq\sqrt{2}$, where $Q=(0,1)^{m}$ is the unit cube, $H$ is an arbitrary
hyperplane in ${\mathbb{R}}^{m}$ passing through the center of the cube, and
$|\,\cdot\,|$ stands for the $(m-1)$-dimensional volume. To obtain (2.4), put

$a_{k}=\frac{1}{aM_{k}},\quad a^{2}=\sum_{k=1}^{m}\frac{1}{M_{k}^{2}},$

so that, by the upper bound in (2.5),

$M\Big{(}\sum_{k=1}^{m}\frac{1}{M_{k}}\,\eta_{k}\Big{)}=M\Big{(}a\sum_{k=1}^{m}a_{k}\,\eta_{k}\Big{)}=\frac{1}{a}M(T_{m})\leq\frac{1}{a}\sqrt{2}.$
(2.6)

Since, by [19], $M(S_{m})$ does not exceed the first term in (2.6), we get
$M(S_{m})\leq\frac{1}{a}\sqrt{2}$, that is, (2.4).

With this argument, this relation is mentioned in [3], where its
multidimensional analog is derived by applying the Hausdorff-Young inequality
with best constants (due to Beckner and Lieb).

Remark 2.3. Modulo a universal constant, the left inequality in (2.5) may be
extended to the more general setting. Namely, if a random variable $\xi$ has a
density $q$ with a finite standard deviation $\sigma$, then

$M(\xi)\geq\frac{1}{12\,\sigma}.$ (2.7)

Here equality is attained for the uniform distribution on arbitrary bounded
intervals of the real line. This relation is well-known; as an early reference
one can mention Statuljavičus [20], p. 651, where (2.7) is stated without
proof. Since it is used below, let us include a short argument. For
normalization, one may assume that $M(\xi)=1$ and ${\mathbb{E}}\xi=0$. In this
case, the tail function $H(x)={\mathbb{P}}\\{|\xi|\geq x\\}$ has a Lipschitz
semi-norm at most 2, implying that $H(x)\geq 1-2x$ for all $x\geq 0$. This
gives

$\sigma^{2}=\int_{-\infty}^{\infty}x^{2}\,q(x)\,dx=2\int_{0}^{\infty}xH(x)\,dx\geq
2\int_{0}^{1/2}x(1-2x)\,dx=\frac{1}{12}.$

##  3\. $L^{p}$-Norms of Characteristic Functions and Orlicz Norms

One useful consequence of (2.4) is the next bound on $L^{2m}$-norms of
characteristic functions.

Proposition 3.1. If $g(t)$ is the characteristic function of a random variable
$\xi$, then for any integer $m\geq 1$,

$\frac{1}{2\pi}\int_{-\infty}^{\infty}|g(t)|^{2m}\,dt\leq\frac{1}{\sqrt{m}}\,M(\xi).$
(3.1)

Proof. We apply Proposition 2.2 to $2m$ summands
$\xi_{1},-\xi_{1}^{\prime},\dots,\xi_{m},-\xi_{m}^{\prime}$, assuming that
$\xi_{k}$, $\xi_{k}^{\prime}$ are independent copies of $\xi$. Introduce the
symmetrized random variable $\tilde{S}_{m}=S_{m}-S_{m}^{\prime}$, where
$S_{m}^{\prime}$ is an independent copy of $S_{m}$. By (2.4), we then get

$M(\tilde{S}_{m})\leq\frac{1}{\sqrt{m}}\,M(\xi).$

In addition, $\tilde{S}_{m}$ has characteristic function $|g(t)|^{2m}$. If
$M(\xi)$ is finite, one may apply Proposition 2.1 and conclude that
$\tilde{S}_{m}$ has a bounded continuous density $q_{m}(x)$ which is vanishing
at infinity. Moreover, $q_{m}(x)$ is maximized at $x=0$, and its value at this
point is described by the inversion formula (2.2) which gives

$M(\tilde{S}_{m})=q_{m}(0)=\frac{1}{2\pi}\int_{-\infty}^{\infty}|g(t)|^{2m}\,dt.$

∎

Using (2.3), one can obtain a similar relation, but without the factor
$1/\sqrt{m}$ in (3.1).

When $M(\xi)$ is finite and $m$ is large, this bound may be considerably
sharpened asymptotically with respect to $m$ when restricting the integration
to the regions $|t|\geq\varepsilon>0$. Before making this precise, first let
us note that, since the random variable $\xi$ has a density, we have

$\delta_{g}(\varepsilon)=\max_{|t|\geq\varepsilon}|g(t)|<1$ (3.2)

for all $\varepsilon>0$. This holds by continuity of $g$, and since $|g(t)|<1$
for all $t\neq 0$ (which is true for any non-lattice distribution), while
$g(t)$ tends to zero as $t\rightarrow\infty$, by the Riemann-Lebesgue lemma.
By the way, this property remains to hold in the more general situation, where
the $m$-fold convolution of the distribution of $\xi$ with itself has a
density (while the distribution of $\xi$ might be not absolutely continuous).
Indeed, in that case, (3.2) may be applied to $g^{m}$, and it remains to
notice that this relation does not depend on $m$.

The property (3.2) may be quantified using, for example, the following
observation due to Statuljavičus [20].

Proposition 3.2. If a random variable $\xi$ has a bounded density with
$M=M(\xi)$ and finite variance $\sigma^{2}={\rm Var}(\xi)$, $\sigma>0$, then
its characteristic function $g$ satisfies, for all $\varepsilon>0$,

$\delta_{g}(\varepsilon)\leq\exp\Big{\\{}-\frac{\varepsilon^{2}}{96\,M^{2}\,(2\sigma\varepsilon+\pi)^{2}}\Big{\\}}.$
(3.3)

This relation may be extended to non-bounded densities $q$, in which case the
parameter $M$ should be replaced with quantiles of the random variable
$q(\xi)$. The moment condition may also be removed, and instead it is
sufficient to deal with quantiles of $|\xi-\xi^{\prime}|$, where
$\xi^{\prime}$ is an independent copy of $\xi$; cf. [4] for details.

Returning to (3.1) and applying (3.3) with $\varepsilon\leq 1$, we then have

$\int_{|t|\geq\varepsilon}|g(t)|^{4m}\,dt\,\leq\,\delta_{g}(\varepsilon)^{2m}\int_{-\infty}^{\infty}|g(t)|^{2m}\,dt\,\leq\,\frac{2\pi
M}{\sqrt{m}}\,\exp\Big{\\{}-\frac{m\varepsilon^{2}}{CM^{2}}\Big{\\}}$

with some absolute constant $C$. Thus, the resulting bound decays
asymptotically fast in $m$.

Let us derive a similar bound in the scheme of independent copies
$(X_{n})_{n\geq 1}$ of the random variable $X$ with ${\rm Var}(X)=1$, assuming
that the normalized sum $Z_{n}$ has a bounded density for $n=n_{0}$ with
$M=M(Z_{n_{0}})$. Consider the characteristic function
$f(t)={\mathbb{E}}\,e^{itX}$. We apply Propositions 3.1-3.2 with
$\xi=X_{1}+\dots+X_{n_{0}}$ in which case $g(t)=f(t)^{n_{0}}$ and
$M(\xi)=\frac{1}{\sqrt{n_{0}}}\,M$. Then, for any $1\leq m\leq n/2n_{0}$, by
(3.1),

$\displaystyle\int_{|t|\geq\varepsilon}|f(t)|^{n}\,dt$ $\displaystyle=$
$\displaystyle\int_{|t|\geq\varepsilon}|f(t)|^{n-2mn_{0}}\,|g(t)|^{2m}\,dt$
$\displaystyle\leq$
$\displaystyle\delta_{f}(\varepsilon)^{n-2mn_{0}}\int_{-\infty}^{\infty}|g(t)|^{2m}\,dt\,\leq\,\frac{2\pi
M}{\sqrt{mn_{0}}}\,\delta_{f}(\varepsilon)^{n-2mn_{0}}.$

If $n\geq 4n_{0}$, let us choose $m=[\frac{n}{4n_{0}}]$. Then
$n-2mn_{0}\geq\frac{n}{2}$, while $m\geq\frac{n}{8n_{0}}$, and we arrive at

$\int_{|t|\geq\varepsilon}|f(t)|^{n}\,dt\leq\frac{4\pi
M}{\sqrt{2n}}\,\delta_{f}(\varepsilon)^{n/2}.$

By (3.3) with $\varepsilon\leq 1$, we also have

$(\delta_{f}(\varepsilon))^{n_{0}}\leq\exp\Big{\\{}-\frac{\varepsilon^{2}}{96\,M^{2}\,(2\sqrt{n_{0}}+\pi)^{2}}\Big{\\}},$

which may be simplified to

$\delta_{f}(\varepsilon)\leq\exp\Big{\\{}-\frac{\varepsilon^{2}}{96\,(2+\pi)^{2}\,n_{0}M^{2}}\Big{\\}}.$

Combining the two bounds, one may summarize.

Corollary 3.3. Let ${\rm Var}(X)=1$, and suppose that $Z_{n}$ has a density
for $n=n_{0}$ bounded by $M$. Then, for all $0<\varepsilon\leq 1$ and $n\geq
4n_{0}$, the characteristic function $f$ of $X$ satisfies

$\int_{|t|\geq\varepsilon}|f(t)|^{n}\,dt\leq\frac{4\pi
M}{\sqrt{2n}}\,\exp\Big{\\{}-\frac{n\varepsilon^{2}}{Cn_{0}M^{2}}\Big{\\}},\quad
C=5200.$ (3.4)

##  4\. Behavior of Characteristic Functions near Zero

While the boundedness of the density is important to control integrability
properties of powers of the characteristic function of a random variable $X$,
the condition (1.2) on the finiteness of an exponential moment of $X$
guarantees that the characteristic function

$f(z)={\mathbb{E}}\,e^{izX},\quad z=t+iy,\quad t,y\in{\mathbb{R}},$

is well-defined and analytic in the strip $|y|=|{\rm Re}(z)|<b$ of the complex
plane. Equivalently, we will assume throughout that, for some $\alpha>0$,

${\mathbb{E}}\,e^{\alpha|X|}\leq 2.$ (4.1)

This parameter is more convenient to quantify the behavior of $f(z)$ near
zero.

For example, using $xe^{-x}\leq e^{-1}$ ($x\geq 0$) and assuming that
$|y|\leq\frac{\alpha}{2}$, we then have

$\displaystyle|f^{\prime}(z)|$ $\displaystyle=$
$\displaystyle|\,{\mathbb{E}}\,Xe^{izX}|\,\leq\,{\mathbb{E}}\,|X|\,e^{|yX|}\,\leq\,{\mathbb{E}}\,|X|\,e^{\alpha|X|/2}$
$\displaystyle=$
$\displaystyle{\mathbb{E}}\,|X|\,e^{-\alpha|X|/2}\,e^{\alpha|X|}\,\leq\,\frac{4}{\alpha
e}.$

Hence $|f(z)-1|\leq\frac{4}{\alpha e}\,|z|$ (since $f(0)=1$). Thus, we obtain:

Lemma 4.1. For all complex numbers $z$ in the disc $|z|\leq\frac{\alpha}{2}$,

$|f^{\prime}(z)|\leq\frac{4}{\alpha e},\quad|f(z)-1|\leq\frac{2}{e}.$

This allows one to consider the log-Laplace transform

$K(z)=\log{\mathbb{E}}\,e^{zX}=\log f(-iz)$

as an analytic function in the disc $|z|\leq\frac{\alpha}{2}$. Since it has
derivative $K^{\prime}(z)=-i\,\frac{f^{\prime}(-iz)}{f(-iz)},$ from Lemma 4.1
we get that in this disc

$|K^{\prime}(z)|\leq\frac{6}{\alpha},\quad|K(z)|\leq 3.$ (4.2)

One may also bound the derivatives of all orders.

Lemma 4.2. For all complex numbers $z$ in the disc $|z|\leq\frac{\alpha}{4}$,

$|K^{(k)}(z)|\leq 3k!\,\Big{(}\frac{4}{\alpha}\Big{)}^{k},\quad k=1,2,\dots$
(4.3)

Moreover, if ${\mathbb{E}}X=0$, ${\mathbb{E}}X^{2}=1$, then

$|K^{\prime\prime\prime}(z)|\leq\frac{8}{\alpha^{3}},\quad|z|\leq\frac{\alpha}{16}.$
(4.4)

As a consequence,

$|K^{\prime\prime}(z)-1|\leq\frac{1}{2},\quad|z|\leq\frac{\alpha^{3}}{16}.$
(4.5)

Thus, these derivatives have at most a factorial growth in absolute value with
respect to the growing parameter $k$. For the particular orders $k=2$ and
$k=3$, and under our moment assumptions, the bound (4.3) may be refined in a
smaller disc according to (4.4)-(4.5).

Proof. To obtain (4.3), one may apply Cauchy’s formula

$K^{(k)}(z)=\frac{k!}{2\pi i}\int_{|w-z|=r}\frac{K(w)}{w^{k+1}}\,dw$

with $r=\frac{\alpha}{4}$ together with the second bound in (4.2).

Turning to the refined bounds, note that in terms of the Laplace transform
$L(z)={\mathbb{E}}\,e^{zX}$, we have $K^{\prime}=L^{\prime}/L$ and

$K^{\prime\prime\prime}=\frac{L^{\prime\prime\prime}}{L}-3\,\frac{L^{\prime\prime}L^{\prime}}{L^{2}}+2\,\frac{L^{\prime
3}}{L^{3}}.$ (4.6)

For $x\geq 0$ and $p=1,2,3$, we use the elementary inequality
$x^{p}e^{-x}\leq(p/e)^{p}$. Suppose that $|z|\leq(1-c)\,\alpha$ with
$\frac{1}{2}\leq c<1$. Since $L^{(p)}(z)={\mathbb{E}}\,X^{p}\,e^{zX}$, we then
have

$|L^{(p)}(z)|\leq{\mathbb{E}}\,|X|^{p}\,e^{(1-c)\,\alpha|X|}.$

Hence, by (4.1),

$|L^{(p)}(z)|\leq\,{\mathbb{E}}\,|X|^{p}\,e^{-c\alpha|X|}\,e^{\alpha|X|}\,\leq\,2\,\Big{(}\frac{p}{c\alpha
e}\Big{)}^{p}.$

In particular,

$|L^{\prime}(z)|\leq\frac{2}{c\alpha
e},\quad|L^{\prime\prime}(z)|\leq\frac{8}{(c\alpha
e)^{2}},\quad|L^{\prime\prime\prime}(z)|\leq\frac{54}{(c\alpha e)^{3}},$ (4.7)

so

$|L(z)-1|\leq\frac{2}{c\alpha e}\,|z|\leq\frac{2(1-c)}{ce},\qquad|L(z)|\geq
1-\frac{2(1-c)}{ce}.$

Putting $q^{-1}=1-\frac{2(1-c)}{ce}$, from (4.6) it follows that

$|K^{\prime\prime\prime}(z)|\leq(c\alpha
e)^{-3}\,(54\,q+48\,q^{2}+16\,q^{3}).$

Choosing $c=15/16$, we have $q=(1-\frac{2}{15\,e})^{-1}<1.06$, and the last
expression becomes smaller than $8\alpha^{-3}$. Hence

$|K^{\prime\prime\prime}(z)|\leq\frac{8}{\alpha^{3}},\quad|K^{\prime\prime}(z)-1|\leq\frac{8}{\alpha^{3}}\,|z|$

for $|z|\leq\frac{\alpha}{16}$, where we used $K^{\prime\prime}(0)=1$. The
last inequality readily implies (4.5). ∎

We shall now show that $|f(z)|$ is bounded away from 1 in a certain region
near zero.

Lemma 4.3. Let ${\mathbb{E}}X=0$, ${\mathbb{E}}X^{2}=1$. For all complex
numbers $z=t+iy$ with $|t|\leq\frac{\alpha^{3}}{8}$,
$|y|\leq\frac{1}{2}\,|t|$, we have

$|f(z)|\leq e^{-t^{2}/5}.$ (4.8)

Proof. Using $f^{\prime}(0)=0$, $f^{\prime\prime}(0)=-1$, one may start with
an integral Taylor formula

$f(z)=1-\frac{1}{2}\,z^{2}+\frac{1}{2}\,z^{3}\int_{0}^{1}f^{\prime\prime\prime}(sz)(1-s)^{2}\,ds.$

This equality is needed in the disc $|z|\leq r$ of radius
$r=\frac{1}{8}\sqrt{\frac{5}{4}}\,\alpha^{3}$. By the triangle inequality, we
then have

$|f(z)|\leq\Big{|}1-\frac{1}{2}\,z^{2}\Big{|}+\frac{A}{6}\,|z|^{3},\quad
A=\max_{|z|\leq r}|f^{\prime\prime\prime}(z)|.$ (4.9)

First, let us check that

$\Big{|}1-\frac{1}{2}\,z^{2}\Big{|}\leq 1-\frac{1}{3}\,t^{2},$ (4.10)

which actually holds in the larger region $|y|\leq\frac{1}{2}\,|t|$,
$|t|\leq\frac{1}{4}$. In this case, $t^{2}-y^{2}\geq\frac{3}{4}\,t^{2}$ and
$|ty|\leq\frac{1}{2}\,t^{2}$, implying

$\displaystyle\Big{|}1-\frac{1}{2}\,z^{2}\Big{|}^{2}$ $\displaystyle=$
$\displaystyle 1-(t^{2}-y^{2})+\frac{1}{4}\,(t^{2}-y^{2})^{2}+(ty)^{2}$
$\displaystyle\leq$ $\displaystyle
1-\frac{3}{4}\,t^{2}+\frac{1}{2}\,t^{4}\,\leq\,\Big{(}1-\frac{1}{3}\,t^{2}\Big{)}^{2},$

where we used $|t|\leq\frac{1}{4}$ on the last step. Thus, (4.10) follows.

Turning to the maximum in (4.9), one may apply the last bound in (4.7) valid
for $|z|\leq(1-c)\,\alpha$. Hence, we have the constraint $(1-c)\alpha\geq r$,
which is fulfilled for the choice $c=1-\frac{1}{8}\sqrt{\frac{5}{4}}$ (due to
$\alpha<1$, cf. Lemma 4.4 below). In this case, we get

$|f^{\prime\prime\prime}(z)|\leq\frac{54}{(c\alpha
e)^{3}}<\frac{4.3}{\alpha^{3}}.$

For $z=t+iy$, $|y|\leq\frac{1}{2}\,|t|$, we have
$|z|^{3}\leq(\frac{5}{4})^{3/2}\,|t|^{3}$, and (4.9)-(4.10) therefore give

$\displaystyle|f(z)|$ $\displaystyle\leq$ $\displaystyle
1-\frac{1}{3}\,t^{2}+\frac{4.3}{6\alpha^{3}}\,\Big{(}\frac{5}{4}\Big{)}^{3/2}\,|t|^{3}$
$\displaystyle\leq$ $\displaystyle
1-\frac{1}{3}\,t^{2}+\frac{4.3}{48}\,\Big{(}\frac{5}{4}\Big{)}^{3/2}\,t^{2}\,\leq\,1-\frac{1}{5}\,t^{2}.$

∎

Finally, let us make a few remarks about the relationship between the
conditions (1.2) and (4.1). When the random variable $X$ has a finite
exponential moment, and $\alpha$ is optimal, then (4.1) becomes an equality.
In this case, the quantity $\frac{1}{\alpha}$ represents the Orlicz norm of
$X$ generated by the Young function $\psi(x)=e^{|x|}-1$, $x\in{\mathbb{R}}$:

$\|X\|_{\psi}=\inf\big{\\{}\lambda>0:{\mathbb{E}}\,\psi(X/\lambda)\leq
1\big{\\}}.$

If ${\mathbb{E}}X^{2}=1$, the parameter $\alpha$ may not be large, since the
$L^{2}$-norm is dominated by the $L^{\psi}$-norm. More precisely, using
$x^{2}e^{-x}\leq 4e^{-2}$ ($x\geq 0$), we have

$\alpha^{2}={\mathbb{E}}\,(\alpha X)^{2}\leq
4e^{-2}\,{\mathbb{E}}\,e^{\alpha|X|}=8e^{-2},$

implying $\alpha\leq 2e^{-1}\sqrt{2}<1.05$. In fact, this bound may be
sharpened.

Lemma 4.4. If ${\mathbb{E}}X^{2}=1$, and $(4.1)$ holds, then $\alpha<1$.

Proof. We may assume that $X\geq 0$ and then we need to show that
${\mathbb{E}}\,e^{X}>2$. It is easy to check that $x+\frac{1}{6}\,x^{3}\geq
ax^{2}$ for all $x\geq 0$ with the optimal constant $a=\frac{2}{\sqrt{6}}$.
Since ${\mathbb{E}}X^{k}\geq({\mathbb{E}}X^{2})^{k/2}=1$ for $k\geq 2$, we get

$\displaystyle{\mathbb{E}}\,e^{X}$ $\displaystyle=$ $\displaystyle
1+\frac{1}{2}\,{\mathbb{E}}X^{2}+{\mathbb{E}}\,\Big{(}X+\frac{1}{6}\,X^{3}\Big{)}+\sum_{k=4}^{\infty}\frac{1}{k!}\,{\mathbb{E}}X^{k}$
$\displaystyle\geq$
$\displaystyle\frac{3}{2}+a+\sum_{k=4}^{\infty}\frac{1}{k!}\,=\,e-\frac{7}{6}+\frac{2}{\sqrt{6}}\,>\,2.36.$

∎

Note that if we start with a more general condition
$B={\mathbb{E}}e^{b|X|}<\infty$ as in Theorem 1.1, (4.1) is fulfilled for a
certain constant $\alpha>0$. Indeed, if $B\leq 2$, then one may take
$\alpha=b$. Otherwise,

${\mathbb{E}}\,e^{\varepsilon
b|X|}\leq({\mathbb{E}}\,e^{b|X|})^{\varepsilon}\leq B^{\varepsilon}=2$

for $\varepsilon=\frac{1}{\log_{2}(B)}$. Hence $\alpha=\varepsilon
b=\frac{b}{\log_{2}(B)}$ works as well. The two cases may be united by taking

$\alpha=\frac{b}{\log_{2}(\max(B,2))}.$ (4.11)

##  5\. Saddle Point and Taylor Expansions

Assume that ${\mathbb{E}}X=0$, ${\mathbb{E}}X^{2}=1$, and
${\mathbb{E}}\,e^{\alpha|X|}\leq 2$ ($\alpha>0$). Since the log-Laplace
transform $K(z)=\log{\mathbb{E}}\,e^{zX}$ was defined as an analytic function
in the disc $|z|\leq\frac{\alpha}{2}$ of the complex plane, it may be expanded
as an absolutely convergent power series

$K(z)=\frac{1}{2}\,z^{2}+\sum_{k=3}^{\infty}\frac{\gamma_{k}}{k!}\,z^{k}.$

Here, the coefficients $\gamma_{k}=K^{k)}(0)$ are called cumulants of $X$.
Every $\gamma_{k}$ represents a certain polynomial in moments of $X$ up to
order $k$. In particular, $\gamma_{3}={\mathbb{E}}X^{3}$ and
$\gamma_{4}={\mathbb{E}}X^{4}-3$.

Similarly,

$K^{\prime}(z)=z+\sum_{k=2}^{\infty}\frac{\gamma_{k+1}}{k!}\,z^{k}.$

The next object is important for contour integration.

Definition 5.1. Given $\tau\in{\mathbb{C}}$, a saddle point is a solution
$z_{0}=z_{0}(\tau)$ of the equation

$K^{\prime}(z)=\tau.$ (5.1)

Thus, a saddle point is the solution of

$\tau=z+\sum_{k=2}^{\infty}\frac{\gamma_{k+1}}{k!}\,z^{k}.$ (5.2)

Proposition 5.2. In the disc $|\tau|\leq\frac{\alpha^{3}}{32}$, the equation
$(5.1)$ has a unique solution $z_{0}(\tau)$. Moreover, it represents an
injective analytic function satisfying $z_{0}^{\prime}(0)=1$ and

$|z_{0}(\tau)|\leq
2\tau\leq\frac{\alpha^{3}}{16},\quad|\tau|\leq\frac{\alpha^{3}}{32}.$ (5.3)

Proof. Let us use (5.2) as the definition of the analytic function
$\tau=K^{\prime}(z)$. If $\tau$ is sufficiently small, say
$|\tau|\leq\tau_{0}$, this equality may be inverted as a power series in
$\tau$,

$z=z_{0}(\tau)=\tau-\frac{\gamma_{3}}{2}\,\tau^{2}+\frac{3\gamma_{3}^{2}-\gamma_{4}}{6}\,\tau^{3}+\dots$

Let us indicate an explicit expression for $\tau_{0}$ in the form of a
positive function of $\alpha$.

By Lemma 4.2, cf. (4.4),

$|\tau^{\prime}(z)-1|\leq\frac{8}{\alpha^{3}}\,|z|,\quad|z|\leq\frac{\alpha}{16}.$
(5.4)

One may use this relation for $|z|\leq\frac{\alpha^{3}}{16}$, since
$\alpha<1$, by Lemma 4.4. Given two points $z_{1}$ and $z_{2}$ in the disc
$|z|\leq\frac{\alpha^{3}}{16}$, define the path $z_{t}=(1-t)z_{1}+tz_{2}$
connecting these points. We have

$\tau(z_{2})-\tau(z_{1})=(z_{2}-z_{1})\Big{(}1+\int_{0}^{1}(\tau^{\prime}(z_{t})-1)\,dt\Big{)},$
(5.5)

implying

$|\tau(z_{2})-\tau(z_{1})|\geq|z_{2}-z_{1}|\Big{(}1-\int_{0}^{1}|\tau^{\prime}(z_{t})-1|\,dt\Big{)}.$

Since $|z_{t}|\leq\frac{\alpha^{3}}{16}$, it follows from (5.4) that

$|\tau(z_{2})-\tau(z_{1})|\geq\frac{1}{2}\,|z_{2}-z_{1}|.$

As a consequence, the map $z\rightarrow\tau(z)$ is injective in the disc
$|z|\leq\frac{\alpha^{3}}{16}$. In addition, since $\tau(0)=0$, we have

$|\tau(z)|\geq\frac{1}{2}\,|z|.$ (5.6)

Therefore, the image of the circle $|z|=\frac{\alpha^{3}}{16}$ under this map
represents a closed curve on the complex plane outside the circle
$|\tau|=\frac{\alpha^{3}}{32}$. Since the image of the disc
$|z|\leq\frac{\alpha^{3}}{16}$ under $\tau$ is a connected set, while
$\tau(0)=0$, this set must contain the disc $|\tau|\leq\frac{\alpha^{3}}{32}$.
Thus, the inverse map $z_{0}(\tau)=\tau^{-1}$ is well-defined and represents a
holomorphic injective function in $|\tau|\leq\frac{\alpha^{3}}{32}$ satisfying
(5.3), by (5.6), and $z_{0}^{\prime}(0)=1$, by (5.4). Hence, one may take
$\tau_{0}=\frac{\alpha^{3}}{32}$.

In addition, $z_{0}(\tau)$ takes real values for real $\tau$. Indeed, since
all cumulants are real numbers, $\tau(z)$ is real for real $z$, so is the
inverse function $z_{0}$. Also, by (5.5),

$\tau=z_{0}(\tau)\Big{(}1+\int_{0}^{1}(\tau^{\prime}(tz_{0}(\tau))-1)\,dt\Big{)},$

which shows that $z_{0}(\tau)>0$ as long as $0<\tau\leq\frac{\alpha^{3}}{32}$
(since the expression under the integral sign is a real-valued function whose
absolute value does not exceed $1/2$). ∎

It is natural to determine the leading term in the Taylor expansion for
$z_{0}(\tau)$ when expanding this function as a power series in $\tau$.
Assuming that $X$ is not normal, let $\gamma_{m}$ ($m\geq 3$) be the first
non-zero cumulant of $X$. Then, as $|z|\rightarrow 0$,

$K(z)=\frac{1}{2}\,z^{2}+\frac{\gamma_{m}}{m!}\,z^{m}+O(|z|^{m+1}),$

so that

$K^{\prime}(z)=z+\frac{\gamma_{m}}{(m-1)!}\,z^{m-1}+O(|z|^{m})$ (5.7)

and

$K^{\prime\prime}(z)=1+\frac{\gamma_{m}}{(m-2)!}\,z^{m-2}+O(|z|^{m-1}).$ (5.8)

Since, $z_{0}(\tau)=\tau+O(|\tau|^{2})$ as $\tau\rightarrow 0$, cf.
Proposition 5.2, we get from (5.7)

$\displaystyle\tau$ $\displaystyle=$ $\displaystyle
K^{\prime}(z_{0}(\tau))\,=\,z_{0}(\tau)+\frac{\gamma_{m}}{(m-1)!}\,z_{0}(\tau)^{m-1}+O(|z_{0}(\tau)|^{m})$
$\displaystyle=$ $\displaystyle
z_{0}(\tau)+\frac{\gamma_{m}}{(m-1)!}\,\tau^{m-1}+O(|\tau|^{m}).$

Therefore,

$z_{0}(\tau)=\tau-\frac{\gamma_{m}}{(m-1)!}\,\tau^{m-1}+O(|\tau|^{m}).$ (5.9)

Next, let us write down the Taylor expansion around the point
$z_{0}=z_{0}(\tau)$:

$K(z)-\tau z=K(z_{0})-\tau
z_{0}+\sum_{k=2}^{\infty}\frac{\rho_{k}}{k!}\,(z-z_{0})^{k},\quad\rho_{k}=K^{(k)}(z_{0}).$
(5.10)

Here, we used the property that the function $K(z)-\tau z$ has derivative
$K^{\prime}(z_{0})-\tau=0$ at the saddle point $z=z_{0}$. Thus, the linear
term in (5.10) corresponding to $k=1$ is vanishing. As for the free term
corresponding to $k=0$, we have

$K(z_{0})=\sum_{k=2}^{\infty}\frac{\gamma_{k}}{k!}\,z_{0}^{k}$

and

$\tau
z_{0}=z_{0}K^{\prime}(z_{0})=\sum_{k=2}^{\infty}\frac{\gamma_{k}}{(k-1)!}\,z_{0}^{k}.$

Hence

$K(z_{0})-\tau
z_{0}=-\sum_{k=2}^{\infty}\frac{k-1}{k!}\,\gamma_{k}z_{0}^{k}=-\frac{1}{2}\,z_{0}^{2}-\frac{1}{3}\,\gamma_{3}z_{0}^{3}+\dots$
(5.11)

Using (5.9) and (5.11), we actually have

$\displaystyle K(z_{0})-\tau z_{0}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\,z_{0}^{2}-\frac{m-1}{m!}\,\gamma_{m}z_{0}^{m}+\dots$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\,\Big{(}\tau-\frac{\gamma_{m}}{(m-1)!}\,\tau^{m-1}+O(|\tau|^{m})\Big{)}^{2}$
$\displaystyle-\
\frac{m-1}{m!}\,\gamma_{m}\Big{(}\tau-\frac{\gamma_{m}}{(m-1)!}\,\tau^{m-1}+O(|\tau|^{m})\Big{)}^{m}+\dots$

which is simplified to

$\displaystyle K(z_{0})-\tau z_{0}$ $\displaystyle=$
$\displaystyle-\frac{1}{2}\,\tau^{2}+\frac{1}{m!}\,\gamma_{m}\tau^{m}+O(|\tau|^{m+1})$
(5.12) $\displaystyle=$
$\displaystyle-\frac{1}{2}\,\tau^{2}+\tau^{3}\lambda(\tau).$

Thus, applying Proposition 5.2 and recalling that $K(z)$ is analytic in
$|z|\leq\frac{\alpha}{2}$ (Lemma 4.1), we obtain:

Proposition 5.3. The function

$\lambda(\tau)=\frac{1}{\tau^{3}}\,\Big{(}K(z_{0}(\tau))-\tau
z_{0}(\tau)+\frac{1}{2}\,\tau^{2}\Big{)}$

is well-defined and analytic in the disc $|\tau|\leq\frac{\alpha^{3}}{32}$.
Moreover, as $\tau\rightarrow 0$,

$\lambda(\tau)=\frac{1}{m!}\,\gamma_{m}\tau^{m-3}+O(|\tau|^{m-2}).$ (5.13)

Definition 5.4. Being an analytic function, $\lambda(\tau)$ is represented as
a power series in the disc $|\tau|\leq\frac{\alpha^{3}}{32}$. It is called
Cramér’s series.

It follows that $\lambda(\tau)$ is bounded for small $\tau$, but we will need
to quantify this property in terms of the parameter $\alpha$. Recall that
${\mathbb{E}}X=0$, ${\mathbb{E}}X^{2}=1$ and ${\mathbb{E}}\,e^{\alpha|X|}\leq
2$ ($\alpha>0$).

Proposition 5.5. We have

$|\lambda(\tau)|\leq 700\,\alpha^{-3},\quad|\tau|\leq\frac{\alpha^{3}}{64}.$
(5.14)

Proof. Proposition 5.2 allows us to apply Cauchy’s formula, which yields

$z_{0}^{\prime\prime}(\tau)=\frac{4}{2\pi
i}\int_{|\xi-\tau|=r}\frac{z_{0}(\xi)}{(\xi-\tau)^{3}}\,d\xi$

with $r=\frac{\alpha^{3}}{64}$. Moreover, the latter implies, by (5.3),

$|z_{0}^{\prime\prime}(\tau)|\,\leq\,\frac{4}{r^{2}}\,\max_{|\xi-\tau|=r}|z_{0}(\xi)|\,\leq\,\frac{4}{r^{2}}\cdot\frac{\alpha^{3}}{16}\,=\,2^{12}\,\alpha^{-3}.$
(5.15)

Next, we note that, by Definition 5.1 of the saddle point, cf. (5.1), the
function

$\psi(\tau)=K(z_{0}(\tau))-\tau z_{0}(\tau)+\frac{1}{2}\,\tau^{2}$

has the first three derivatives

$\displaystyle\psi^{\prime}(\tau)$ $\displaystyle=$ $\displaystyle
K^{\prime}(z_{0}(\tau))z_{0}^{\prime}(\tau)-z_{0}(\tau)-\tau
z_{0}^{\prime}(\tau)+\tau\,=\,\tau-z_{0}(\tau),$
$\displaystyle\psi^{\prime\prime}(\tau)$ $\displaystyle=$ $\displaystyle
1-z_{0}^{\prime}(\tau),$ $\displaystyle\psi^{\prime\prime\prime}(\tau)$
$\displaystyle=$ $\displaystyle-z_{0}^{\prime\prime}(\tau).$

Since $\psi(0)=\psi^{\prime}(0)=\psi^{\prime\prime}(0)=0$, we may apply the
Taylor integral formula together with (5.15) to conclude that

$\displaystyle|\psi(\tau)|$ $\displaystyle\leq$
$\displaystyle\frac{|\tau|^{3}}{6}\,\max_{|\xi|\leq|\tau|}|\psi^{\prime\prime\prime}(\xi)|$
$\displaystyle\leq$
$\displaystyle\frac{|\tau|^{3}}{6}\,\max_{|\xi|\leq|\tau|}|z_{0}^{\prime\prime}(\xi)|\,\leq\,\frac{1}{6}\cdot
2^{12}\,\alpha^{-3}.$

As $\psi(\tau)=\tau^{3}\lambda(\tau)$, the relation (5.14) follows. ∎

Let us now introduce another analytic function which appears in the
representation (1.6) of Theorem 1.2.

Proposition 5.6. The function

$\mu(\tau)=\frac{1}{2}\,\log K^{\prime\prime}(z_{0}(\tau))$

is well-defined and analytic in the disc $|\tau|\leq\frac{\alpha^{3}}{32}$.
Moreover, as $\tau\rightarrow 0$,

$\mu(\tau)=\frac{1}{2(m-2)!}\,\gamma_{m}\tau^{m-2}+O(|\tau|^{m-1}).$ (5.16)

Proof. By (5.3), $|z_{0}(\tau)|\leq\frac{\alpha^{3}}{16}$. Hence, by Lemma
4.2, $K^{\prime\prime}(z_{0}(\tau))$ takes values in the disc with center at 1
of radius $1/2$. Thus, the principal value of $\log
K^{\prime\prime}(z_{0}(\tau))$ is well-defined and represents an analytic
function in $|\tau|\leq\frac{\alpha^{3}}{32}$. Moreover, by (5.8)-(5.9),

$\displaystyle K^{\prime\prime}(z_{0}(\tau))$ $\displaystyle=$ $\displaystyle
1+\frac{\gamma_{m}}{(m-2)!}\,z_{0}(\tau)^{m-2}+O(|z_{0}(\tau)|^{m-1})$
$\displaystyle=$ $\displaystyle
1+\frac{\gamma_{m}}{(m-2)!}\,\tau^{m-2}+O(|\tau|^{m-1}).$

Taking the logarithm of this expression, we arrive at (5.16). ∎

Let us also mention that the function $K(z)$ is convex and has a positive
second derivative on the real line, more precisely – on the interval where it
is finite. Hence $\mu(\tau)$ is real-valued for real $\tau$.

##  6\. Contour Integration

Let $(X_{n})_{n\geq 1}$ be independent copies of a random variable $X$ with
${\mathbb{E}}X=0$, ${\rm Var}(X)=1$, and characteristic function
$f(t)={\mathbb{E}}\,e^{itX}$. We now consider the normalized sum

$Z_{n}=\frac{X_{1}+\dots+X_{n}}{\sqrt{n}},$

assuming that $M=M(Z_{n_{0}})$ is finite. As already discussed in Section 2,
in this case all $Z_{n}$ with $n\geq 2n_{0}$ have continuous bounded densities
expressed by the inversion formula

$p_{n}(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-itx}f_{n}(t)\,dt,\quad
x\in{\mathbb{R}},$

where

$f_{n}(t)=f\Big{(}\frac{t}{\sqrt{n}}\Big{)}^{n}$

denotes the characteristic functions of $Z_{n}$. Equivalently,

$p_{n}(x)=\frac{\sqrt{n}}{2\pi}\int_{-\infty}^{\infty}e^{-itx\sqrt{n}}\,f(t)^{n}\,dt.$
(6.1)

Using contour integration, one can cast this formula in a different form
involving the log-Laplace transform $K(z)=\log{\mathbb{E}}\,e^{zX}$ and the
saddle point $z_{0}=z_{0}(\tau)$ for the real value $\tau=x/\sqrt{n}$. This is
a preliminary step towards Theorems 1.2 and 1.4.

As before, let ${\mathbb{E}}\,e^{\alpha|X|}\leq 2$ with a parameter
$\alpha>0$.

Lemma 6.1 Let $n\geq 4n_{0}$. If $0<\varepsilon\leq\frac{\alpha^{3}}{16}$ and
$|\tau|\leq\frac{\varepsilon}{2}$, then

$p_{n}(x)=\frac{\sqrt{n}}{2\pi}\int_{-\varepsilon}^{\varepsilon}\exp\Big{\\{}n(K(z_{0}+it)-\tau(z_{0}+it))\Big{\\}}\,dt+\theta
R_{n}$ (6.2)

with $|\theta|\leq 1$ and

$R_{n}=5M\exp\Big{\\{}-\frac{n\varepsilon^{2}}{Cn_{0}M^{2}}\Big{\\}},\quad
C=5200.$ (6.3)

Proof. Applying Corollary 3.3, we get from (6.1) that, for any
$\varepsilon\in(0,1]$,

$\Big{|}\,p_{n}(x)-\frac{\sqrt{n}}{2\pi}\int_{-\varepsilon}^{\varepsilon}e^{-itx\sqrt{n}}\,f(t)^{n}\,dt\Big{|}\leq
4M\exp\Big{\\{}-\frac{n\varepsilon^{2}}{Cn_{0}M^{2}}\Big{\\}}$ (6.4)

with $C=5200$.

Due to the assumption on $\varepsilon$, we may apply Lemma 4.3 which gives

$|f(\pm\varepsilon+iy)|\leq e^{-\varepsilon^{2}/5}\ \ {\rm whenever}\
|y|\leq\frac{\varepsilon}{2}.$ (6.5)

Assuming for definiteness that $x\geq 0$, we take the rectangle contour

$L=L_{1}+L_{2}+L_{3}+L_{4}$

with segment parts

$L_{1}=[-\varepsilon,\varepsilon]$, $L_{2}=[\varepsilon,\varepsilon-ih]$,

$L_{3}=[\varepsilon-ih,-\varepsilon-ih]$, $L_{4}=[-\varepsilon-
ih,-\varepsilon]$,

where $h>0$ is chosen to satisfy $h\leq\frac{\varepsilon}{2}$. With this
choice the complex numbers $z=t+iy$ with $|t|\leq\varepsilon$, $|y|\leq h$ lie
in the domain of the definition of $K(z)$. Then, by Cauchy’s theorem,

$\displaystyle\int_{L_{1}}e^{-izx\sqrt{n}}\,f(z)^{n}\,dz+\int_{L_{2}}e^{-izx\sqrt{n}}\,f(z)^{n}\,dz$
$\displaystyle\hskip-142.26378pt+\
\int_{L_{3}}e^{-izx\sqrt{n}}\,f(z)^{n}\,dz+\int_{L_{4}}e^{-izx\sqrt{n}}\,f(z)^{n}\,dz\,=\,0.$

Note that in the lower half-plane $z=t-iy$, $0\leq y\leq h$, we have
$|e^{-izx\sqrt{n}}|=e^{-yx\sqrt{n}}\leq 1$. Moreover, $|f(z)|$ is bounded away
from 1 on $L_{2}$ and $L_{4}$ according to (6.5) which gives

$\Big{|}\int_{L_{2}}\Big{|}+\Big{|}\int_{L_{4}}\Big{|}\leq\varepsilon\,e^{-n\varepsilon^{2}/5}\leq\frac{1}{16}\,e^{-n\varepsilon^{2}/5}.$

To simplify, note that

$4M\exp\Big{\\{}-\frac{n\varepsilon^{2}}{Cn_{0}M^{2}}\Big{\\}}+\frac{1}{16}\,e^{-n\varepsilon^{2}/5}\leq
R_{n},$

where we used $M\geq\frac{1}{12}$ (cf. Remark 2.3). Combining this bound with
(6.4), we arrive at

$\displaystyle p_{n}(x)$ $\displaystyle=$
$\displaystyle\frac{\sqrt{n}}{2\pi}\int_{L_{3}}e^{-izx\sqrt{n}}\,f(z)^{n}\,dz+\theta
R_{n}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{n}}{2\pi}\int_{-\varepsilon}^{\varepsilon}e^{-i(t-ih)\,x\sqrt{n}}\,f(t-ih)^{n}\,dt+\theta
R_{n}.$

Using the log-Laplace transform, let us rewrite the above as a contour
integral

$p_{n}(x)=\frac{\sqrt{n}}{2\pi
i}\int_{h-i\varepsilon}^{h+i\varepsilon}\exp\\{n(K(z)-\tau z)\\}\,dz+\theta
R_{n}$

with $\tau=x/\sqrt{n}$ and apply it with $h=z_{0}=z_{0}(\tau)$. Due to the
requirement $0\leq\tau\leq\frac{\varepsilon}{2}$, we have
$0\leq\tau\leq\frac{\alpha^{3}}{32}$ and $0\leq
z_{0}\leq\frac{\alpha^{3}}{16}$, according to (5.3), so that Proposition 5.2
is applicable. After the change of variable, we thus obtain (6.2)-(6.3). ∎

As a next step, let us show that, at the expense of a small error, the
integration in (6.2) may be restricted to the interval $|t|\leq t_{n}$ with

$t_{n}=n^{-1/2}\,\sqrt{8\log n},\quad n\geq 4n_{0}.$

This can be achieved under stronger conditions such as

$|\tau|\leq\frac{\varepsilon}{80\,M^{2}n_{0}},\quad
0\leq\varepsilon\leq\frac{\alpha^{3}}{80}.$ (6.6)

Indeed, using (5.10) and (5.12) in the representation (6.2), one may rewrite
(6.2) as

$p_{n}(x)=\frac{\sqrt{n}}{2\pi}\int_{-\varepsilon}^{\varepsilon}\exp\bigg{\\{}n\,\Big{(}\sum_{k=2}^{\infty}\rho_{k}\frac{(it)^{k}}{k!}\Big{)}\bigg{\\}}\,dt\
e^{n\,(-\frac{1}{2}\,\tau^{2}+\tau^{3}\lambda(\tau))}+\theta R_{n},$

where $\tau=x\sqrt{n}$ and $\rho_{k}=K^{(k)}(z_{0})$. Equivalently

$\frac{p_{n}(x)}{\varphi(x)}=\frac{\sqrt{n}}{\sqrt{2\pi}}\,e^{n\tau^{3}\lambda(\tau)}\int_{-\varepsilon}^{\varepsilon}\exp\bigg{\\{}n\,\Big{(}\sum_{k=2}^{\infty}\rho_{k}\frac{(it)^{k}}{k!}\Big{)}\bigg{\\}}\,dt+\theta
R_{n}e^{x^{2}/2}.$ (6.7)

Here, the new remainder term

$R_{n}e^{x^{2}/2}=5M\exp\Big{\\{}-\frac{\varepsilon^{2}n}{CM^{2}n_{0}}+\tau^{2}n\Big{\\}},\quad
C=5200,$

is still exponentially small with respect to $n$ due to the first condition in
(6.6), which strengthens the assumption $|\tau|\leq\frac{\varepsilon}{2}$ in
Lemma 6.1 (recall that $M\geq\frac{1}{12}$). In this case, the expression in
the exponent will be of order $-\frac{cn\varepsilon^{2}}{n_{0}M^{2}}$ up to an
absolute constant $c>0$. Hence, (6.7) yields

$\frac{p_{n}(x)}{\varphi(x)}=\frac{\sqrt{n}}{\sqrt{2\pi}}\,e^{n\tau^{3}\lambda(\tau)}\int_{-\varepsilon}^{\varepsilon}\exp\bigg{\\{}n\,\Big{(}\sum_{k=2}^{\infty}\rho_{k}\frac{(it)^{k}}{k!}\Big{)}\bigg{\\}}\,dt+\theta
R_{n},$ (6.8)

where

$R_{n}=5M\exp\Big{\\{}-\frac{c\varepsilon^{2}n}{M^{2}n_{0}}\Big{\\}}.$ (6.9)

Now, by Lemma 4.2,

$\frac{1}{2}\leq\rho_{2}\leq\frac{3}{2},\qquad|\rho_{k}|\leq
3k!\,\Big{(}\frac{4}{\alpha}\Big{)}^{k}\quad(k\geq 3).$ (6.10)

It follows that

$\displaystyle{\rm
Re}\bigg{(}\sum_{k=2}^{\infty}\rho_{k}\,\frac{(it)^{k}}{k!}\bigg{)}$
$\displaystyle=$
$\displaystyle-\rho_{2}\,\frac{t^{2}}{2}+\sum_{k=2}^{\infty}(-1)^{k}\rho_{2k}\,\frac{t^{2k}}{(2k)!}$
$\displaystyle\leq$
$\displaystyle-\frac{1}{4}\,t^{2}+3\sum_{k=2}^{\infty}\Big{(}\frac{4t}{\alpha}\Big{)}^{2k}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}\,t^{2}+3\,(16t^{2})\sum_{k=2}^{\infty}\frac{(4t)^{2k-2}}{\alpha^{2k}}\,\leq\,-\frac{1}{8}\,t^{2},$

where we used $\alpha<1$ and $|t|\leq\varepsilon\leq\frac{1}{80}\,\alpha^{3}$
so as to bound the last sum, according to the second assumption in (6.6).
Hence, when restricted to $|t|\geq t_{n}$, the absolute value of the integral
in (6.8) does not exceed

$2\int_{t_{n}}^{\infty}e^{-nt^{2}/8}\,dt=\frac{4}{\sqrt{n}}\int_{\frac{t_{n}}{2}\sqrt{n}}^{\infty}e^{-s^{2}/2}\,ds<\frac{2\sqrt{2\pi}}{\sqrt{n}}\,e^{-nt_{n}^{2}/8}=\frac{2\sqrt{2\pi}}{n^{3/2}}.$

As a result, assuming the conditions (6.6),

$\displaystyle\frac{p_{n}(x)}{\varphi(x)}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{n}}{\sqrt{2\pi}}\,e^{n\tau^{3}\lambda(\tau)}\int_{|t|\leq
t_{n}^{\prime}}\exp\bigg{\\{}n\,\Big{(}\sum_{k=2}^{\infty}\rho_{k}\frac{(it)^{k}}{k!}\Big{)}\bigg{\\}}\,dt$
(6.11) $\displaystyle+\
2\theta_{1}n^{-1}e^{n\tau^{3}\lambda(\tau)}+\theta_{2}R_{n},$

where $t_{n}^{\prime}=\min(t_{n},\varepsilon)$, $|\theta_{j}|\leq 1$, and
where $R_{n}$ is now defined in (6.9).

##  7\. Proof of Theorem 1.4

As a final step, we need to explore an asymptotic behavior of the integral in
(6.11), where we recall that $\rho_{k}=K^{(k)}(z_{0})$, $z_{0}=z_{0}(\tau)$
being the saddle point for $\tau=x/\sqrt{n}$. In view of the conditions in
(6.6), we choose

$\varepsilon=\frac{\alpha^{3}}{80},\quad\tau_{0}=\frac{\varepsilon}{80\,M^{2}n_{0}}=\frac{c_{0}\alpha^{3}}{M^{2}\,n_{0}}$

with $c_{0}=1/6400$. Note that with this choice the definition (6.9) becomes

$R_{n}=5M\exp\Big{\\{}-\frac{c_{1}\alpha^{6}n}{M^{2}n_{0}}\Big{\\}},$ (7.1)

where $c_{1}>0$ is an absolute constant. Suppose that $|\tau|\leq c\tau_{0}$
with a constant $0<c\leq 1$ to be chosen later on.

The integrand in (6.11) may be written as

$u_{n}(t)=\exp\Big{\\{}-n\rho_{2}\,\frac{t^{2}}{2}+n\rho_{3}\,\frac{(it)^{3}}{6}+nv(t)\Big{\\}}$

with

$v(t)=\sum_{k=4}^{\infty}\rho_{k}\,\frac{(it)^{k}}{k!}.$

First assume that $n\geq n_{1}=\max(4n_{0},\varepsilon^{-4})$ which insures
that $t_{n}^{\prime}=t_{n}$ (since $\varepsilon<\frac{1}{80}$). As $|t|\leq
t_{n}$, from (6.10) it follows that

$nv(t)=O(nt^{4})=B\alpha^{-4}\,\frac{(\log n)^{2}}{n}.$ (7.2)

Here and below $B$ denotes a quantity, perhaps different in different places,
bounded by an absolute constant. With this convention, since
$n\geq(\frac{80}{\alpha^{3}})^{4}$, we also have $nv(t)=B$, and by (6.10) with
$k=3$,

$n\rho_{3}\,t^{3}=B\alpha^{-3}\,\frac{(\log n)^{3/2}}{\sqrt{n}}=B.$ (7.3)

So, one may use the Taylor expansion $e^{x}=1+x+O(x^{2})$ in a bounded
interval $|x|\leq B$ with

$x=n\rho_{3}\,\frac{(it)^{3}}{6}+nv(t).$

From (7.2)-(7.3) and using again $n\geq B\alpha^{-12}$ together with
$\alpha<1$, we have

$(nv(t))^{2}=B\alpha^{-8}\,\frac{1}{n}\,\frac{(\log n)^{4}}{n}=\frac{B}{n}$

and

$(n\rho_{3}t^{3})\cdot(nv(t))=B\alpha^{-7}\,\frac{1}{n}\,\frac{(\log
n)^{7/2}}{\sqrt{n}}=\frac{B\alpha^{-2}}{n}.$

Since

$(n\rho_{3}t^{3})^{2}=B\alpha^{-6}\,\frac{(\log n)^{3}}{n},$

which dominates (7.2) and the previous two expressions, we obtain that

$\displaystyle u_{n}(t)$ $\displaystyle=$ $\displaystyle
e^{-n\rho_{2}t^{2}/2+x}$ $\displaystyle=$ $\displaystyle
e^{-n\rho_{2}t^{2}/2}\,\big{(}1+x+Bx^{2}\big{)}$ $\displaystyle=$
$\displaystyle
e^{-n\rho_{2}t^{2}/2}\,\Big{(}1+n\rho_{3}\,\frac{(it)^{3}}{6}+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)}.$

Hence

$\int_{|t|\leq t_{n}}u_{n}(t)\,dt=\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)}\int_{|t|\leq t_{n}}e^{-n\rho_{2}t^{2}/2}\,\,dt,$

and (6.11) is simplified to

$\displaystyle\frac{p_{n}(x)}{\varphi(x)}$ $\displaystyle=$
$\displaystyle\frac{\sqrt{n}}{\sqrt{2\pi}}\,e^{n\tau^{3}\lambda(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)}\int_{|t|\leq t_{n}}e^{-n\rho_{2}t^{2}/2}\,\,dt$ (7.4)
$\displaystyle+\ 2\theta_{1}n^{-1}e^{n\tau^{3}\lambda(\tau)}+\theta_{2}R_{n}.$

Next, one may extend the integration in (7.4) to the whole real line at the
expense of an error not exceeding

$\displaystyle\int_{|t|>t_{n}}e^{-n\rho_{2}t^{2}/2}\,\,dt$ $\displaystyle=$
$\displaystyle\frac{2}{\sqrt{\rho_{2}n}}\int_{t_{n}\sqrt{\rho_{2}n}}^{\infty}e^{-s^{2}/2}\,ds$
$\displaystyle<$
$\displaystyle\frac{\sqrt{2\pi}}{\sqrt{\rho_{2}n}}\,e^{-n\rho_{2}t_{n}^{2}/2}\,\leq\,\frac{2\sqrt{\pi}}{\sqrt{n}}\,e^{-nt_{n}^{2}/4}\,=\,\frac{2\sqrt{\pi}}{\sqrt{n}}\,n^{-2},$

where we used $\rho_{2}\geq\frac{1}{2}$. The latter bound is dominated by
$B\alpha^{-6}\,\frac{(\log n)^{3}}{n}$, and since the integral over the whole
real line is equal to $\frac{\sqrt{2\pi}}{\sqrt{\rho_{2}n}}$, we obtain from
(7.4) a simpler representation

$\frac{p_{n}(x)}{\varphi(x)}=\frac{1}{\sqrt{\rho_{2}}}\,e^{n\tau^{3}\lambda(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)}+Bn^{-1}e^{n\tau^{3}\lambda(\tau)}+BR_{n}.$

Here, the first remainder term may be absorbed in the brackets, so that this
formula is further simplified to

$\frac{p_{n}(x)}{\varphi(x)}=\frac{1}{\sqrt{\rho_{2}}}\,e^{n\tau^{3}\lambda(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}+B\,e^{-n\tau^{3}\lambda(\tau)}R_{n}\Big{)}.$ (7.5)

Moreover, one may eliminate the factor $e^{-n\tau^{3}\lambda(\tau)}$ in front
of $R_{n}$ by choosing a smaller value of $c_{1}$ in (7.1) for a proper
absolute constant $c$ appearing in the assumption $|\tau|\leq c\tau_{0}$. To
this end, it will be sufficient to require that

$|\psi(\tau)|\leq\frac{c_{1}\alpha^{6}}{2M^{2}n_{0}},\quad\psi(\tau)=\tau^{3}\lambda(\tau).$

Recall that, by Proposition 5.5, $|\psi(\tau)|\leq
700\,\alpha^{-3}\,|\tau|^{3}$ in the interval
$|\tau|\leq\frac{1}{64}\,\alpha^{3}$ (which is larger than
$|\tau|\leq\tau_{0}$). Hence, the above bound holds true, as long as

$|\tau|\leq\frac{c_{1}^{1/3}}{(1400\,M^{2}n_{0})^{1/3}}\,\alpha^{3}.$ (7.6)

Since $M\geq\frac{1}{12}$, we have $(M^{2}n_{0})^{1/3}\leq
12^{4/3}M^{2}n_{0}$. Hence (7.6) may be strengthened to $|\tau|\leq c\tau_{0}$
with a suitable constant $c>0$. Under this condition, from (7.5) we thus get

$\frac{p_{n}(x)}{\varphi(x)}=\frac{1}{\sqrt{\rho_{2}}}\,e^{n\tau^{3}\lambda(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}+BR_{n}\Big{)},$ (7.7)

where $R_{n}$ is still defined as in (7.1) with a new constant $c_{1}$.

It is now useful to note that the last error term in this representation is
dominated by the second last one for sufficiently large $n$. Indeed, using
$e^{-y}\leq 2/y^{2}$ ($y>0$), we have

$R_{n}\leq
2\,\Big{(}\frac{M^{2}n_{0}}{c_{1}\alpha^{6}n}\Big{)}^{2}\leq\alpha^{-6}\,\frac{1}{n},$

where the last inequality holds true for $n\geq CM^{4}n_{0}^{2}\,\alpha^{-12}$
with an absolute constant $C>0$. This condition is slightly stronger than
$n\geq n_{1}$ which was assumed before. As a result, (7.7) then yields

$\frac{p_{n}(x)}{\varphi(x)}=\frac{1}{\sqrt{\rho_{2}}}\,e^{n\tau^{3}\lambda(\tau)}\Big{(}1+B\alpha^{-6}\,\frac{(\log
n)^{3}}{n}\Big{)}.$

It remains to recall Proposition 5.6 according to which

$\rho_{2}^{-1/2}=K^{\prime\prime}(z_{0}(\tau))^{-1/2}=e^{-\mu(\tau)},$

and then we arrive at the assertion in Theorem 1.4 (which is a refinement of
Theorem 1.2).

Let us also note that the case $2n_{0}\leq n<n_{1}$ is not interesting, since
then $|x|\leq\tau_{0}n_{1}$, and (1.6) holds true by choosing a suitable
constant in $O$ in (1.6).

##  8\. Proof of Corollary 1.3

Starting from (5.13) and (5.16), we have

$\displaystyle n\tau^{3}\lambda(\tau)-\mu(\tau)$ $\displaystyle=$
$\displaystyle\frac{n}{m!}\,\gamma_{m}\tau^{m}+O(|\tau|^{m+1})-\frac{1}{2(m-2)!}\,\gamma_{m}\tau^{m-2}+O(|\tau|^{m-1})$
$\displaystyle=$
$\displaystyle\frac{\gamma_{m}}{m!}\,\tau^{m-2}\,\Lambda(\tau).$

Here

$\Lambda(\tau)\,=\,n\tau^{2}-\frac{m(m-1)}{2}+nO(\tau^{3})+O(\tau)\,\geq\,\frac{1}{2}\,\Big{[}n\tau^{2}-\frac{m(m-1)}{2}\Big{]},$

which is bounded away from zero, if $|\tau|\leq\tau_{1}$ for some constant
$\tau_{1}>0$ and $n\tau^{2}=x^{2}\geq m^{2}$. In this case, (1.6) immediately
yields the desired relation (1.7).

In the remaining bounded interval $|x|\leq m$, this argument does not work,
and it is better to employ the Chebyshev-Edgeworth expansion for the
correction $\varphi_{m}(x)$ in (1.1) (which depends on $n$ as well). In terms
of the first non-zero cumulant, (1.1) may be written more accurately as

$p_{n}(x)=\varphi(x)+\frac{\gamma_{m}}{m!}\,H_{m}(x)\varphi(x)\,n^{-\frac{m-2}{2}}+\frac{1}{1+|x|^{m}}\,o\big{(}n^{-\frac{m-2}{2}}\big{)},$

where $H_{m}(x)$ denotes the Chebyshev-Hermite polynomial of degree $m$. As a
consequence, for any constant $x_{0}>0$,

$\sup_{|x|\leq
x_{0}}\frac{|p_{n}(x)-\varphi(x)|}{\varphi(x)}=O\big{(}n^{-\frac{m-2}{2}}\big{)},$

which is stronger than (1.7), since $m$ is even ($m\geq 4$).

Acknowledgment. We thank the referees for careful reading and valuable
remarks.

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