# Old and new on the 3-convex functions

Dan-Ştefan Marinescu  National College ”Iancu de Hunedoara”, Hunedoara,
Romania<EMAIL_ADDRESS>and  Constantin P. Niculescu  Department of
Mathematics, University of Craiova, Craiova 200585, Romania
<EMAIL_ADDRESS>
(Date: May 7, 2023)

###### Abstract.

The present paper aims to survey known results and to point out the wealth of
rather important open problems that are out there.

###### Key words and phrases:

higher order convexity, function with positive differences, spectral
decomposition of real symmetric matrices

###### 2000 Mathematics Subject Classification:

Primary 26A51, 39B62; Secondary 26D15, 15A20

##  1\. Introduction

Higher order convexity was introduced by Hopf [13] and Popoviciu [26], [28],
who defined it in terms of divided differences of a function. Assuming $f$ a
real-valued function defined on a real interval $I,$ the divided differences
of order $0,1,\ldots,n$ associated to a family $x_{0},x_{1},\ldots,x_{n}$ of
$n+1$ distinct points are respectively defined by the formulas

$\displaystyle[x_{0};f]$ $\displaystyle=f(x_{0})$
$\displaystyle[x_{0},x_{1};f]$
$\displaystyle=\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}$ $\displaystyle...$
$\displaystyle[x_{0},x_{1},...,x_{n};f]$
$\displaystyle=\frac{[x_{1},x_{2},...,x_{n};f]-[x_{0},x_{1},...,x_{n-1};f]}{x_{n}-x_{0}}$
$\displaystyle={\displaystyle\sum\nolimits_{j=0}^{n}}\frac{f(x_{j})}{\prod\nolimits_{k\neq
j}\left(x_{j}-x_{k}\right)}.$

Notice that all these divided differences are invariant under the permutation
of points $x_{0},x_{1},...,x_{n}.$ As a consequence, we may always assume that
$x_{0}<x_{1}<\cdots<x_{n}.$

A function $f$ is called $n$-_convex_(respectively __ $n$-_concave_) if all
divided differences $[x_{0},x_{1},\ldots,x_{n};f]$ are nonnegative
(respectively nonpositive). In particular,

  * •

the convex functions of order 0 are precisely the nonnegative functions;

  * •

the convex functions of order 1 are the nondecreasing functions;

  * •

the convex functions of order 2 are nothing but the usual convex functions
since in this case for all $x_{0}<x_{1}<x_{2}$ in $I,$

$[x_{0},x_{1},x_{2};f]=\frac{\frac{f(x_{0})-f(x_{1})}{x_{0}-x_{1}}-\frac{f(x_{1})-f(x_{2})}{x_{1}-x_{2}}}{x_{0}-x_{2}}\geq
0,$

that is,
$\left(x_{2}-x_{0}\right)f(x_{1})\leq\left(x_{2}-x_{1}\right)f(x_{0})+\left(x_{1}-x_{0}\right)f(x_{2}).$

While the properties of the above three classes of $n$-convex functions are
well understood, only few relevant results are known in the case where $n\geq
3.$ Besides the work of Popoviciu (see [26], [27], [28], [29] and [30]) we
should mention here the contribution of Bennett [1], Bessenyei and Páles [2],
Boas and Widder [4], Bojanić and Roulier [5], Brady [6], Bullen [7], [8],
Kuczma [16], Marinescu and Monea [17], Pecarić and his collaborators [15],
[14], [19], [25], Rajba [31], Szostok [37] and Wasowicz [40].

This paper is aimed to present an overview of the present state of art
concerning the $3$-convex functions, to add some new results, and to single
out some open problems which seem of interest. The reason to restrict
ourselves to this particular case is two-fold: it offers a convenient
framework to illustrate the richness of the class of 3-convex functions and
also a context that keeps the technical aspects still simple and intuitive.

For the convenience of the reader, some very basic facts are recalled in
Section 2.

Section 3 is mostly dedicated to the identity of 3-convex functions with the
functions having positive differences up to order 3. See Theorem 2 below. This
result, already known to Popoviciu, was rarely mentioned by the various books
dedicated to convex functions, except that by Kuczma [16], that also includes
a full proof. Our approach here combines results due to Hopf [13], Popoviciu
[26], Boas and Widder [4] and Bennett [1] (to cite them according to their
apparition). An important role is played by the fact that on an open interval
the 3-convex functions are precisely the differentiable functions whose
derivatives are convex functions. This offers the possibility to deduce
results for the 3-convex functions from know results for the usual convex
functions and vice-versa. See Theorem 3 and Theorem 4 and the comments that
accompany them in Section 4.

Section 5 is devoted to an overview of the Hermite-Hadamard inequality in the
context of continuous 3-convex functions. The central result is the remarkable
extension obtained by Bessenyei and Páles [2], [3], that covers the general
case of Borel probability measures on a compact interval $[a,b]$. The Hermite-
Hadamard inequality for the usual convex functions is centered around the role
played by the barycenter $p$ and the extremal points. As $f(p)=\delta_{p}(f),$
in the $3$-convexity variant, the role of $\delta_{p}$ is taken by a discrete
probability measure supported at two points, one of them being inside the
interval. Naturally, this raises the interesting problem how looks the analog
of Choquet’s theory in the framework of $3$-convex functions. Several open
problems in this connection are mentioned at the end of Section 5.

Section 6 present a radiography of a recent result due Ressel [32], concerning
the connection between the functions which are continuous, nondecreasing,
concave and 3-convex and the Hornich-Hlawka inequality. See Theorem . We show
that this result is actually the juxtaposition of two distinct results
covering complementary domains of the variables, one involving the properties
of continuity and 3-convexity, while the other involving only the properties
of monotonicity and concavity. See Lemma 4, and Lemma 5 below.

The paper ends with a section discussing the extension of the entire theory to
the case of functions taking values in an ordered Banach space.

##  2\. Preliminaries

A function $f$ defined on an interval $I$ is $3$-_convex_ if for every
quadruple $x_{0}<x_{1}<x_{2}<x_{3}$ of elements in $I,$

$[x_{0},x_{1},x_{2},x_{3};f]=\frac{f(x_{0})}{(x_{0}-x_{1})(x_{0}-x_{2})(x_{0}-x_{3})}-\frac{f(x_{1})}{(x_{0}-x_{1})(x_{1}-x_{2})(x_{1}-x_{3})}\\\
+\frac{f(x_{2})}{(x_{0}-x_{2})(x_{1}-x_{2})(x_{2}-x_{3})}-\frac{f(x_{3})}{(x_{0}-x_{3})(x_{1}-x_{3})(x_{2}-x_{3})}\geq
0,$

equivalently,

(2.1)
$\displaystyle(x_{2}-x_{0})(x_{3}-x_{0})(x_{3}-x_{2})f(x_{1})+(x_{1}-x_{0})(x_{2}-x_{0})(x_{2}-x_{1})f(x_{3})$
$\displaystyle\geq(x_{2}-x_{1})(x_{3}-x_{1})(x_{3}-x_{2})f(x_{0})+(x_{1}-x_{0})(x_{3}-x_{0})(x_{3}-x_{1})f(x_{2}).$

When the points $x_{0},x_{1},x_{2},x_{3}$ are equidistant, that is, when
$x_{1}=x_{0}+h,$ $x_{2}=x_{0}+2h,$ $x_{3}=x_{0}+3h$ for some $h>0,$ the last
inequality becomes

$f(x_{0}+3h)-3f(x_{0}+2h)+3f(x_{0}+h)-f(x_{0})\geq 0,$

equivalently,

(2.2) $f(x_{0})+3f\left(\frac{x_{0}+2x_{3}}{3}\right)\leq
3f\left(\frac{2x_{0}+x_{3}}{3}\right)+f(x_{3}).$

If $f$ is $n$-times differentiable, then a repeated application of Lagrange’s
mean value theorem yields the existence of a point
$\xi\in\left(\min_{k}x_{k},\max_{k}x_{k}\right)$ such that

$[x_{0},x_{1},...,x_{n};f]=\frac{f^{(n)}(\xi)}{n!}.$

As a consequence, one obtains the sufficiency part of the following practical
criterion of $n$-convex.

######  Lemma 1.

Suppose that $f$ is a continuous function defined on an interval $I$ which is
$3$-times differentiable on the interior of $I.$ Then $f$ is $3$-convex if and
only if its third derivative is nonnegative.

The necessity part is also immediate by using the standard formulas for
derivatives via iterated differences,

$f^{\prime\prime\prime}(x_{0})=\lim_{h\rightarrow
0+}\frac{f(x_{0}+3h)-3f(x_{0}+2h)+3f(x_{0}+h)-f(x_{0})}{h^{3}}.$

In connection with Lemma 1, it is worth mentioning a result due to Hopf [13],
p. 24, and Popoviciu [26], p. 48, which asserts that every $3$-convex function
$f$ defined on an open interval is differentiable and $f^{\prime}$ is convex.
This can be easily turned into a characterization of 3-convexity in the
framework of continuous functions. See Theorem 2 below.

Lemma 1 allows us to notice the existence of a large variety of $3$-convex
functions:

$\displaystyle x/(x+1),~{}1-e^{-\alpha x}\text{ (for }\alpha>0),\text{ }$
$\displaystyle\log(1+x),\text{ }-x\log x\,,\ (x-1)/\log x,\text{
}x^{\alpha}\text{ }(\text{for }\alpha\in(0,1]\cup[2,\infty)),$
$\displaystyle-x^{2}+\sqrt{x},~{}\sinh,~{}\cosh,~{}-\log\left(\Gamma(x)\right).$

Also the primitive of any continuous convex function is a $3$-convex function.

The function $1-\left(x-3\right)+\frac{\left(x-3\right)^{3}}{6}$ is continuous
and $3$-convex on $\mathbb{R}_{+}$ but not $n$-convex for any
$n\in\left\\{0,1,2\right\\}.$

The polynomials with positive coefficients and the exponential are $n$-convex
for every $n\geq 0.$

######  Remark 1.

All polynomials of degree less than or equal to $2$ are both $3$-convex and
$3$-concave. These functions together with the finite sums
$\sum\nolimits_{i=1}^{m}c_{i}\left(\left(x-a_{i}\right)_{+}\right)^{2}$ with
positive coefficients represent the building blocks of any $3$-convex
function. See Popoviciu _[28]_, pp. $29$-$30$ $($and also _[5]_ and _[38]_$)$.

The continuous 3-convex functions on interval $I$ constitute a convex cone in
the vector space $C(I),$ of all continuous functions on $I.$ This cone is
closed under convolution, but not under usual product.

The extremal properties of $3$-convex functions differ from those of convex
functions.

The maximum (or minimum) of two $3$-convex function is not necessarily a
$3$-convex function; consider the case of the functions $-x$ and $x.$ Also, an
interior critical point of a $3$-convex function is not necessarily a point of
minimum. It can be a point of inflection (the case of the cubic function
$x^{3})$ or a point of maximum (the case of the function $-x^{2}+\sqrt{x}).$

The following approximation theorem due to Popoviciu __[27]__(see also [12],
Theorem 1.3.1 $(i)$, p. 20) allows us to reduce the reasoning with $n$-convex
functions to the case where they are also differentiable.

######  Theorem 1.

$($Popoviciu’s approximation theorem$)$ If a continuous function
$f:[0,1]\rightarrow\mathbb{R}$ is $k$-convex, then so are the Bernstein
polynomials associated to it,

$B_{n}(f)(x)=\sum_{i=0}^{n}\binom{n}{i}x^{i}(1-x)^{n-i}f\left(\frac{i}{n}\right).$

Moreover, by the well-known property of simultaneous uniform approximation of
a function and its derivatives by the Bernstein polynomials and their
derivatives, it follows that $B_{n}(f)$ and any derivative (of any order) of
it, converge uniformly to $f$ and to its derivatives, correspondingly.

Using a change of variable, one can easily see that the approximation theorem
extends to functions defined on compact intervals $[a,b]$ with $a<b.$

######  Corollary 1.

If $f:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ is a continuous $3$-convex
function which is also nondecreasing and concave, then the same properties
hold for $f^{\alpha}$ if $\alpha\in(0,1].$

###### Proof.

According to Theorem 1, we may reduce the proof to the case where the involved
function is of class $C^{3}$, in which case the conclusion follows from Lemma
1. ∎

######  Remark 2.

An important class of functions $f:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$
which are continuous, nondecreasing concave and $3$-convex is that of
Bernstein functions. Recall that a function
$\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ is called Bernstein if it is
continuous on $\mathbb{R}_{+},$ indefinitely differentiable on $(0,\infty)$
and

$(-1)^{n+1}f^{(n)}\geq 0\text{\quad for all }n\geq 1.$

Their theory is exposed in the monograph of Schilling, Song, and Vondraček
_[34]_. According to Corollary 3.8, p. 28, in this monograph, the composition
of two Bernstein functions is also a Bernstein function $($which is not the
case for the class of $3$-convex functions$)$.

##  3\. Some characterizations of 3-convexity

The _difference operator_ $\Delta_{h}$ $($of step size $h\geq 0)$ associates
to each function $f$ defined on an interval $I$ the function $\Delta_{h}f$
defined by

$\left(\Delta_{h}f\right)(x)=f(x+h)-f(x),$

for all $x$ such that the right-hand side formula makes sense. Notice that no
restrictions are necessary if $I=\mathbb{R}^{+}$ or $I=\mathbb{R}.$

The difference operators are linear and commute to each other,

$\Delta_{h_{1}}\Delta_{h_{2}}=\Delta_{h_{2}}\Delta_{h_{1}}.$

They also verify the following property of invariance under translation:

$\Delta_{h}\left(f\circ T_{a}\right)=\left(\Delta_{h}f\right)\circ T_{a},$

where $T_{a}$ is the translation defined by the formula $T_{a}(x)=x+a.$

######  Lemma 2.

If $n$ is a positive integer, then the following formula holds:

$\Delta_{h_{1}}\Delta_{h_{2}}\cdots\Delta_{h_{n}}f(x)={\displaystyle\sum\limits_{\varepsilon_{1},...,\varepsilon_{n}\in\left\\{0,1\right\\}}}\left(-1\right)^{n-\left(\varepsilon_{1}+\cdots+\varepsilon_{n}\right)}f\left(x+\varepsilon_{1}h_{1}+\cdots+\varepsilon_{n}h_{n}\right).$

The proof is immediate, by mathematical induction.

A function $f$ has _positive differences of order_ $n\geq 1$ if

$\Delta_{h_{1}}\Delta_{h_{2}}\cdots\Delta_{h_{n}}f(x)\geq 0,$

whenever the left-hand size is well defined. To outline a parallel to
$n$-convexity, we say that a function $f$ has positive differences of order 0
if $f\geq 0.$

Notice that a function has positive differences of first order if it is
nondecreasing and it has positive differences of second order if it is convex
(a simple exercise left to the reader). As was noticed by Popoviciu [28] (at
the beginning of Section 24, p. 49), this remark works in the general case of
$n$-convex continuous functions. A detailed proof can be found in the book of
Kuczma [16]; see Theorem 15.6.1, p. 440. More comments are available in [24].
The next result concerns the case $3$-convex functions.

######  Theorem 2.

If $f:[0,A]\rightarrow\mathbb{R}$ is a continuous function then the following
assertions are equivalent:

$(i)$ $f$ is $3$-convex;

$(ii)$ $\Delta_{h}\Delta_{h}\Delta_{h}f(x)=f(x+3h)-3f(x+2h)+3f(x+h)-f(x)\geq
0$ for all $x\in[0,A)$ and $h>0$ such that $x+3h\leq A.$

$(iii)$ $f$ has positive differences of order $3$, that is, it verifies the
inequality

$\Delta_{x}\Delta_{y}\Delta_{z}f(t)=f\left(x+t\right)+f\left(y+t\right)+f\left(z+t\right)+f\left(x+y+z+t\right)\\\
-f\left(x+y+t\right)-f\left(y+z+t\right)-f\left(z+x+t\right)-f(t)\geq 0$

whenever $x,y,z,t\in[0,A]$ and $x+y+z+t\leq A;$

$(iv)~{}f$ is differentiable on $(0,A)$ and its derivative $f^{\prime}$ is a
convex function.

###### Proof.

The implication $(i)\Longrightarrow(ii)$ is straightforward. The fact that
$(i)\Longleftrightarrow(ii)$ under the presence of continuity is stated by
Popoviciu in his book [28] (at the beginning of Section 24, p. 49).

The implication $(i)\Longrightarrow(iv)$ was noticed both by Hopf [13] and
Popoviciu [26] (in the general case of $n$-convex functions).

The implications $(ii)\Longrightarrow(iii)\Longrightarrow(iv)$ are covered by
the paper of Boas and Widder [4] (see Lemma 1 and the Theorem, p. 497).

The implication $(iv)\Longrightarrow(i)$ was noticed by Bennett ([1],
Proposition 1), who used the identity

$\displaystyle[a,b,c,d;f]$
$\displaystyle=\frac{1}{\left(b-a\right)\left(c-a\right)\left(d-a\right)}\int_{a}^{b}f^{\prime}(t)dt$
$\displaystyle-\frac{c+d-a-b}{\left(c-a\right)\left(c-b\right)\left(d-a\right)\left(d-b\right)}\int_{b}^{c}f^{\prime}(t)dt$
$\displaystyle+\frac{1}{\left(d-a\right)\left(d-b\right)\left(d-c\right)}\int_{c}^{d}f^{\prime}(t)dt,$

for all $a<b<c<d.$ ∎

It is worth mentioning that the property of $3$-convex functions of having
positive differences up to order 3 can be also deduced by adapting the
argument used by Popoviciu [29] for a weaker variant of it.

We shall need the following special case of the Hardy-Littlewood-Pólya
inequality of majorization (see [23], Theorem 4.1.3, p. 186):

######  Lemma 3.

If __ $g:\left[a,b\right]\rightarrow\mathbb{R}$ is a convex function and __
$c$ and $d$ are two points in $\left[a,b\right]$ such that $a+b=c+d$, then

$g(c)+g(d)\leq g(a)+g(b).$

Popoviciu’s alternative argument for the implication $(i)\Longrightarrow(iii)$
in Theorem 2 works as follows: According to Popoviciu’s approximation theorem,
we may restrict to the case where $f$ is a $3$-convex function of class
$C^{\infty}.$ Then fix arbitrarily $y,z,t$ in $[0,A]$ such that $y+z+t<A$ and
consider the function

$\displaystyle F(x)$
$\displaystyle=f\left(x+t\right)+f\left(y+t\right)+f\left(z+t\right)+f\left(x+y+z+t\right)$
$\displaystyle-f\left(x+y+t\right)-f\left(y+z+t\right)-f\left(z+x+t\right)-f(t)$

defined on the interval $[0,A-y-z-t].$ This function is also of class
$C^{\infty}$ and

$F^{{}^{\prime}}\left(x\right)=f^{{}^{\prime}}\left(x+t\right)+f^{{}^{\prime}}\left(x+y+z+t\right)-f^{{}^{\prime}}\left(x+y+t\right)-f^{{}^{\prime}}\left(z+x+t\right).$

From Lemma 1 we infer that $F^{\prime}$ is a convex function, so that
$F^{\prime}\geq 0$ according to Lemma 3. Therefore

$F(x)\geq F(0)=0$

for all $x,$ a fact which is equivalent to the assertion $(iii).$ The proof is
done.

The connection of Theorem 2 __ $(iii)$ with the Hornich-Hlawka inequality will
be discussed in Section 6. The next section is devoted to the applications of
Theorem 2 __ $(iv).$

##  4\. Applications of Theorem 2 __ $(iv)$

Theorem 2 __ $(iv)$ easily allows us to deduce results for the usual convex
functions from those for the 3-convex functions (and vice-versa). Two examples
are exhibited below. The first one is a refinement of the Jensen inequality.

######  Theorem 3.

A continuous function $f$ defined on an interval $I$ is convex if and only if

$\frac{1}{\left|J\right|}\int_{J}f(x)\mathrm{d}x\geq\frac{1}{\left|K\right|}\int_{K}f(x)\mathrm{d}x,$

whenever $K\subset J$ are two compact subintervals of $I$ with the same
midpoint.

###### Proof.

Suppose that $f$ is convex and that the two intervals under attention are
$J=[a,b]$ and $K=[a+\varepsilon,b-\varepsilon]$ (where
$\varepsilon\in(0,(b-a)/2))$. Every primitive $F$ of $f$ is a $3$-convex
function of class $C^{1}$ so, according to the formula (2.1) (applied to the
points $a<a+\varepsilon<b-\varepsilon<b),$ it verifies the inequality

$F(b)-F(a)\geq\frac{b-a}{b-a-2\varepsilon}\left[F(b-\varepsilon)-F(a+\varepsilon)\right].$

Thus

$\displaystyle\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x$
$\displaystyle=\frac{1}{b-a}\int_{a}^{b}F^{\prime}(x)\mathrm{d}x=\frac{F(b)-F(a)}{b-a}$
$\displaystyle\geq\frac{1}{b-a-2\varepsilon}\left[F(b-\varepsilon)-F(a+\varepsilon)\right]$
$\displaystyle=\frac{1}{b-a-2\varepsilon}\int_{a+\varepsilon}^{b-\varepsilon}f(x)\mathrm{d}x$

and the proof of the necessity part is done.

For the sufficiency part, notice that $f$ verifies the condition

$\frac{1}{\left|I_{n}\right|}\int_{I_{n}}f(x)dx\searrow
f\left(\frac{a+b}{2}\right),$

whenever $I_{0}=[a,b]\supset I_{1}\supset I_{2}\supset\cdots$ is a sequence of
nested compact subintervals of $I$ that shrink to $(a+b)/2,$ supposed to be
their common midpoint. Therefore $f$ is a continuous function such that

$\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x\geq f\left(\frac{a+b}{2}\right)$

whenever $a<b$ in $I.$ As it is well known, this property implies that $f$ is
a convex function. See [23], Exercise 1, p. 63. ∎

######  Remark 3.

In the same manner one can prove that a continuous function $f$ defined on an
interval $I$ is convex if and only if

$\frac{1}{2\varepsilon}\int_{a}^{a+\varepsilon}f(x)\mathrm{d}x+\frac{1}{2\varepsilon}\int_{b-\varepsilon}^{b}f(x)\mathrm{d}x\geq\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x,$

whenever $a<b$ in $I$ and $\varepsilon\in(0,(b-a)/2).$ This represents a
refinement of the right-hand side of the Hermite-Hadamard inequality $($see
_[23]_, Section $1.10$, pp. $59-64)$.

Combining Theorem 3 and Remark 3 one obtains double inequalities such as

$\frac{4}{b-a}\int_{a}^{\left(3a+b\right)/4}f(x)\mathrm{d}x+\frac{4}{b-a}\int_{(a+3b)/4}^{b}f(x)\mathrm{d}x\\\
\geq\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x\geq\frac{3}{b-a}\int_{\left(2a+b\right)/3}^{(a+2b)/3}f(x)\mathrm{d}x,$

for every convex function $f:[a,b]\rightarrow\mathbb{R}.$ Continuity of $f$ is
not necessary. See [23], Proposition 1.1.3, p. 3.

Similarly, one can use the theory of convex functions to characterize the
3-convex functions.

######  Theorem 4.

Suppose that $f:I\rightarrow\mathbb{R}$ is a function continuous on $I$ and of
class $C^{1}$ on the interior of $I.$ Then the following assertions are
equivalent:

$(i)$ $f$ is $3$-convex;

$(ii)$ $f$ verifies the inequality

(4.1) $f^{\prime}\left(\frac{a+b}{2}\right)\leq\frac{f(b)-f(a)}{b-a},$

for all points $a<b$ in $\operatorname{int}I;$

$(ii)$ $f$ verifies the inequality

(4.2)
$\frac{f(b)-f(a)}{b-a}\leq\frac{1}{2}\left(\frac{f^{\prime}(a)+f^{\prime}(b)}{2}+f^{\prime}\left(\frac{a+b}{2}\right)\right),$

for all points $a<b$ in $\operatorname{int}I$.

###### Proof.

$(i)\Longrightarrow(ii)$ According to Theorem 2, $f$ is differentiable on
$(a,b)$ and $f^{\prime}$ is a convex function on this interval. Taking into
account the Jensen inequality (see [23], Corollary 1.7.4, p. 43), for every
$\varepsilon\in(0,(b-a)/2),$ we have

$f^{\prime}\left(\frac{a+b}{2}\right)\leq\frac{1}{b-a}\int_{a+\varepsilon}^{b-\varepsilon}f^{\prime}(x)\mathrm{d}x=\frac{f(b-\varepsilon)-f(a+\varepsilon)}{b-a}$

and the inequality (4.1) follows by passing to the limit as
$\varepsilon\rightarrow 0+.$

$(ii)\Longrightarrow(i)$ Indeed, the inequality (4.1) assures the fulfillment
of the inequality Jensen inequality by the function $f^{\prime\prime}$ on
every compact interval included in $\operatorname{int}I,$ which is known to
imply the convexity of $f^{\prime}.$ See [23], Exercise 1, p. 63.

$(i)\Longrightarrow(iii)$ The proof of inequality (4.2) can be done in the
same manner by using Remark 1.10.5, p. 61, in [23]. The implication
$(iii)\Longrightarrow(i)$ follows from Exercise 2, p. 63, loc. cit.) ∎

######  Example 1.

Applying Theorem _4_ in the case of the function $f(x)=\log(1+x),$ $x\geq 0,$
one obtains the double inequality

$\frac{b-a}{1+\frac{a+b}{2}}\leq\log\frac{1+b}{1+a}\leq\left(\frac{b-a}{4}\right)\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{2}{1+\frac{a+b}{2}}\right),$

valid for all $0\leq a\leq b.$ This provides a rational estimate of
$\log(1+x),$ better than the estimate offered by Maclaurin’s expansion.

######  Example 2.

If $f:[a,b]\rightarrow\mathbb{R}$ is a $3$-times differentiable function with
$M=\sup_{x\in[a,b]}f^{\prime\prime\prime}(x)<\infty,$ then $Mx^{3}/6-f$ is a
continuous $3$-convex function. For example, in the case of the sine function,
this works for $M=1,$ which implies that

$\frac{\left(a+b\right)^{2}}{2}-\cos\frac{a+b}{2}\leq\frac{a^{2}+ab+b^{2}}{6}-\frac{\sin
b-\sin a}{b-a}\\\ \leq\frac{1}{4}\left(\frac{a^{2}+ab+b^{2}}{2}-\cos a-\cos
b-2\cos\left(\frac{a+b}{2}\right)\right).$

Similarly, when $m=\inf_{x\in[a,b]}f^{\prime\prime\prime}(x)>-\infty,$ then
the function $f-mx^{3}/6$ is $3$-convex and the conclusion of Theorem _4
_applies to it.

Another application of Theorem 2 refers to the ”support” of a $3$-convex
functions.

If $f$ is a continuous $3$-convex function defined on an interval $I$, then
$f^{\prime}$ is a convex function on $\operatorname{int}I$ and we can apply to
it the theory of subdifferentiability of convex functions. According to
[23]_,_ Theorem $1.6.2$, p. $36$, the subdifferential $\partial f^{\prime}(a)$
$($of $f^{\prime}$ at a point $a$ interior to $I)$ equals the interval
$[f_{-}^{\prime\prime}(a),f_{+}^{\prime\prime}(a)]$ and

$f^{\prime}(x)\geq f^{\prime}(a)+y(x-a)\text{\quad for all
}x\in\operatorname{int}I\text{ and }y\in\partial f^{\prime}(a).$

Notice that $\partial f^{\prime}(a)=\left\\{f^{\prime\prime}(a)\right\\}$ when
$f$ is twice differentiable at $a$.

Therefore

(4.3) $f(x)\geq
f(a)+(x-a)f^{\prime}(a)+\frac{\left(x-a\right)^{2}}{2}f_{+}^{\prime\prime}(a)\text{\quad
if }x\in I\text{ and }x\geq a$

and

(4.4) $f(x)\leq
f(a)+(x-a)f^{\prime}(a)+\frac{\left(x-a\right)^{2}}{2}f_{-}^{\prime\prime}(a)\text{\quad
if }x\in I\text{ and }x\leq a.$

As a consequence,

$f(x)=\sup\left\\{f(a)+(x-a)f^{\prime}(a)+\frac{\left(x-a\right)^{2}}{2}y:a\in\operatorname{int}I,\text{
}a<x,\text{ }y\in\partial f^{\prime}(a)\right\\}\ \ \ \ $

for all $x\in I$ different from the left endpoint of $I.$

Geometrically,
$g=f(a)+(x-a)f^{\prime}(a)+\frac{\left(x-a\right)^{2}}{2}f_{+}^{\prime\prime}(a)$
$($respectively
$y=f(a)+(x-a)f^{\prime}(a)+\frac{\left(x-a\right)^{2}}{2}f_{-}^{\prime\prime}(a))$
represents the unique parabola tangent to the graph of $f$ at $a$ and which
has the same right-hand $($left-hand$)$ derivative of second order at $a.$
These ”tight tangent parabolas” represent analogs of the line supports from
the theory of convex functions. Notice that a tight tangent parabola at a
point $a$ is above the graph of $f$ on $I\cap(-\infty,a]$ and under the graph
on $I\cap[a,\infty).$

######  Problem 1.

Suppose that $f$ is a function continuous on $I$ and differentiable on
$\operatorname{int}I,$ which admit at every point $a\in\operatorname{int}I$ a
tight tangent parabola. Is $f$ necessarily $3$-convex?

The answer to this problem seems to be positive as suggests Bullen’s analogue
for $3$–convex functions, of the familiar fact that the graph of a convex
function lies always beneath its chords.

######  Theorem 5.

A continuous function $f$ is $3$-convex on $[a,b]$ if and only if for every
quadratic function $Q$ that agrees with $f$ at $\alpha$, $\beta$ and $\gamma$
$($where $a<\alpha<\beta<\gamma<b)$ we have

$Q(t)\geq f(t)\text{\quad if }x\in[a,\alpha]\cup[\beta,\gamma]$

and

$f(t)\geq Q(t)\text{\quad if }x\in[\alpha,\beta]\cup[\gamma,b].$

See [7], Theorem $5$, p. $85$ and also Theorem $10$, p. $88$.

The existence of tight tangent parabolas raises naturally the problem of an
analogue of Fenchel duality in the case of continuous $3$-convex functions.

######  Problem 2.

Does there exist an analogue of Fenchel conjugate in the case of continuous
$3$-convex functions?

In connection with the last problem, notice that a function which is both
convex and $3$-convex, may have a Fenchel conjugate which is not 3-convex. See
the case of the exponential function, whose Fenchel conjugate is the function
$f^{\ast}(x)=x\log x-x$ for $x>0$ and $f^{\ast}(0)=0.$

######  Remark 4.

The support-type properties of $3$-convex functions were investigated also by
Wasowicz who proved the following result: If $f:[a,b]\rightarrow\mathbb{R}$ is
a $3$-convex function, then for every point $c\in(a,b)$ there exist quadratic
functions $p$ and $q$ such that

$p(a)=f(a),\text{ }p(c)=f(c)\text{ and }p\leq f\text{ on }[a,b]$

and

$q(c)=f(c),\text{ }p(b)=f(b)\text{ and }q\geq f\text{ on }[a,b].$

See _[40]_, Corollaries $10$ and $11.$

Another application of Theorem 2 __ $(iv)$ is provided by the inequality
(5.4), proved in the next section.

##  5\. The Hermite-Hadamard inequality in the context of 3-convex functions

According to Choquet’s theory, the meaning of the Hermite-Hadamard inequality
for convex functions on intervals is that of a double estimate for the
integral mean of every convex function $f:[a,b]\rightarrow\mathbb{R}$ with
respect to a Borel probability measure $\mu$ on $[a,b]$, precisely, __

(5.1)
$\,f\left(\operatorname*{bar}(\mu)\right)\leq\,\int_{a}^{b}\,f(x)\mathrm{d}\mu\leq\frac{b-\operatorname*{bar}(\mu)}{b-a}\cdot
f(a)+\frac{\operatorname*{bar}(\mu)-a}{b-a}\cdot f(b)\,.$

Here $\operatorname*{bar}(\mu)$ represents the barycenter of $\mu,$ that is,
the unique point $p$ in $[a,b]$ such that

(5.2) $f(p)=\int_{a}^{b}\,f(x)\,\mathrm{d}\mu(x)\,$

for every continuous affine function $f:[a,b]\rightarrow\mathbb{R}$. One can
easily check that
$\operatorname*{bar}(\mu)=\int_{a}^{b}x\,\mathrm{d}\mu(x)\,,$ the moment of
the first order of $\mu.$

See [23] (and also [20] and [22]).

When $\mu$ is the an absolutely continuous probability measure of the form
$wdx,$ where the weight $w\geq 0$ is continuous and symmetric about the
vertical line $x=(a+b)/2,$ that is,

$w(x)=w(a+b-x)\text{\quad for all }x\in[a,b],$

then $\operatorname*{bar}(\mu)=\left(a+b\right)/2$ and the inequality (5.1)
becomes

(5.3)
$\,f\left(\frac{a+b}{2}\right)\leq\,\int_{a}^{b}\,f(x)w(x)\mathrm{d}x\leq\frac{f(a)+f(b)}{2}\,,$

This is Fejér’s variant of the classical Hermite-Hadamard inequality, also
known as the Hermite-Hadamard-Fejér inequality. See [23], the remark after
Exercise $6$, p. $64.$

######  Remark 5.

The inequality _( 5.1)_ also works outside the framework of probability
measure, for examples for the so called Hermite-Hadamard measures, an example
being $3(x^{2}-1/6)\mathrm{d}x$ on $[-1,1].$ For details, see _[11]_ and
_[23]_, Section _7.5_ , pp. _322-324_.

Theorem 2 $(iv)$ allows us to derive from the Hermite-Hadamard-Fejér
inequality some consequences for the 3-convex functions.

For this, consider the case of a differentiable $3$-convex function
$f:[a,b]\rightarrow\mathbb{R}$ and of a continuous real weight $w$ which
admits a primitive $W\geq 0,$ symmetric about the vertical line $x=(a+b)/2$.
Three such examples are: 1) $w(x)=a+b-2x$ on $[a,b]$ (with the primitive
$W(x)=(x-a)(b-x))$; 2) $2nx^{2n-1}$ on $[-a,a]$ (with the primitive $x^{2n});$
3) $\cos x$ on $[0,\pi]$ (with the primitive $\sin x).$

Then

$\displaystyle\int_{a}^{b}\,f(x)w(x)\mathrm{d}x$
$\displaystyle=fW|_{a}^{b}-\int_{a}^{b}\,f^{\prime}(x)W(x)\mathrm{d}x$
$\displaystyle=f(b)W(b)-f(a)W(a)-\int_{a}^{b}\,f^{\prime}(x)W(x)\mathrm{d}x,$

and the Hermite-Hadamard-Fejér inequality leads to

(5.4)
$\displaystyle-\frac{f^{\prime}(a)+f^{\prime}(b)}{2}\int_{a}^{b}\,W(x)\mathrm{d}x$
$\displaystyle\leq\int_{a}^{b}\,f(x)w(x)\mathrm{d}x-\left(f(b)W(b)-f(a)W(a)\right)\leq-f^{\prime}\left(\frac{a+b}{2}\right)\int_{a}^{b}\,W(x)\mathrm{d}x.$

In the particular case where $w(x)=a+b-2x$ and $W(x)=(x-a)(b-x)$ we have

$\int_{a}^{b}W(x)\mathrm{d}x=\allowbreak\frac{1}{6}\left(b-a\right)^{3}$

and the inequality (5.4) becomes

$\frac{(b-a)^{3}}{6}f^{\prime}\left(\frac{a+b}{2}\right)\leq\int_{a}^{b}\,f(x)(2x-a-b)\mathrm{d}x\leq\frac{(b-a)^{3}}{12}\left(f^{\prime}(a)+f^{\prime}(b)\right).$

We pass now to the existence of an analogue of the Hermite-Hadamard inequality
for the $3$-convex functions.

As in the case of usual convex function it is useful to consider the following
$3$-convex ordering on the set $\operatorname*{Prob}([a,b],$ of all Borel
probability measures on $[a,b]:$

$\nu\prec_{3cvx}\mu\text{ if and only
if}\int_{a}^{b}f(x)\,\mathrm{d}\nu(x)\leq\int_{a}^{b}f(x)\,\mathrm{d}\mu(x)\text{
}\\\ \text{for all continuous and 3-convex functions
}f:[a,b]\rightarrow\mathbb{R}.$

Some important necessary and sufficient conditions for higher order convex
ordering are available in the papers of Denuit, Lefevre and Shaked [9], Rajba
[31] and Szostok [37].

The relation $\prec_{3cvx}$is indeed a partial order relation. Clearly, it is
transitive and reflexive; the fact that $\nu\prec_{3cvx}\mu$ and
$\mu\prec_{3cvx}\nu$ imply $\mu=\nu$ comes from the fact that the linear space
generated by the continuous 3-convex functions is dense in $C([a,b])$.

######  Remark 6.

For every $\mu\in\operatorname*{Prob}([a,b]$ one can choose a minimal Borel
probability measure $\lambda$ such that $\lambda\prec_{3cvx}\mu$ $($and the
same is true for the maximal measures$)$. Indeed, $\operatorname*{Prob}([a,b]$
can be identified with a weak star convex and compact subset of the dual space
of $C\left([a,b]\right),$ precisely with $\left\\{x^{\ast}\in
C\left([a,b]\right):x^{\ast}\geq 0\text{ and }x^{\ast}(1)=1\right\\},$ so that
every net of measures minorizing $\mu$ admit a convergent subnet in the weak
star topology. Thus the existence of minimal Borel probability measures
majorized by $\mu$ follows from Zorn’s lemma.

######  Remark 7.

Given a Borel probability measure $\mu$ on $[a,b]$ whose support includes more
than two points, _no_ Dirac measure $\delta_{p}$ can be found such that

$\delta_{p}(f)=f(p)\leq\int_{a}^{b}f\left(x\right)\mathrm{d}\mu(x)$

for all continuous $3$-convex functions. Indeed, checking this for the
functions $\pm x$ and $\pm x^{2}$ we should have

$p=\int_{a}^{b}x\mathrm{d}\mu(x)\text{ and
}p^{2}=\int_{a}^{b}x^{2}\mathrm{d}\mu(x),$

which is not possible because the equality occurs in the Cauchy-Schwarz
inequality if and only if one the two functions involved is a scalar multiple
of the other.

The last two remarks lead naturally to the problem of characterizing the
minimal Borel probability measures with respect to the ordering
$\prec_{3cvx}.$

######  Problem 3.

Given a Borel probability measure $\mu$ on $[a,b],$ find numbers
$\lambda,\mu,\nu\in[0,1]$ such that for every continuous $3$-convex function
$f:[a,b]\rightarrow\mathbb{R}$ we have

(3J) $\lambda f((1-\mu)a+\mu b)+(1-\lambda)f((1-\nu)a+\nu
b)\leq\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}\mu(x).$

A result due to Bessenyei and Páles [3], Theorem 3.4, combined with a careful
inspection of the argument of Lemma 3.2 in [37], shows that this problem has a
unique solution, provided that the support of $\mu$ contains at least 3
points. In the particular case when $\mu$ equals
$\left(1/(b-a)\right)\mathrm{d}x,$ this solution corresponds to

$\lambda=1/4,\text{ }\mu=0\text{ and }\nu=2/3,$

and their result reads as follows:

######  Theorem 6.

For a continuous function $f:I$ $\rightarrow$ $\mathbb{R}$, the following
statements are equivalent:

$(i)$ $f$ is $3$-convex;

$(ii)~{}$for all $a,b\in I$ with $a<b,$

$\frac{1}{4}f(a)+\frac{3}{4}f\left(\frac{a+2b}{3}\right)\leq\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x;$

$(iii)~{}$for all $a,b\in I$ with $a<b,$

$\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x\leq\frac{3}{4}f\left(\frac{2a+b}{3}\right)+\frac{1}{4}f(b).$

In what follows we will refer to the extremal probability measures
$\frac{1}{4}\delta_{a}+\frac{3}{4}\delta_{\left(a+3b\right)/3}$ and
$\frac{3}{4}\delta_{(2a+b)/3}+\frac{1}{4}\delta_{b}$ as the $3$-_condensation_
of $(1/(b-a))\mathrm{d}x$__ and respectively the $3$-_dispersion_ of
$(1/(b-a))\mathrm{d}x.$The non symmetric form of these probability measures
seems to be a consequence of Bullen’s Theorem 5.

Notice that

$\displaystyle f\left(\frac{a+b}{2}\right)$
$\displaystyle\leq\frac{1}{4}f(a)+\frac{3}{4}f\left(\frac{a+2b}{3}\right)\leq\frac{1}{b-a}\int_{a}^{b}f(x)\mathrm{d}x$
$\displaystyle\leq\frac{3}{4}f\left(\frac{2a+b}{3}\right)+\frac{1}{4}f(b)\leq\frac{f(a)+f(b)}{2}$

in the case functions $f$ which are both convex and 3-convex.

Theorem 6 outlines the following property of rigidity of the continuous
$3$-convex functions:

######  Remark 8.

If $f:[a,b]\rightarrow\mathbb{R}$ is a continuous $3$-convex function such
that $f(a)\geq 0$ and $f\left(\frac{a+2b}{3}\right)\geq 0$, then its integral
mean value is also greater than or equal to $0$. This imposes that the values
of $f$ in the interval $[\left(a+2b\right)/3,b]$ cannot be ”too” negative
$($though it can be negative as shows the case of the function
$-x^{2}+\sqrt{x}$ defined on $[0,3/2]).$

Similarly, if $f\left(\frac{2a+b}{3}\right)\leq 0$ and $f(b)\leq 0,$ then the
integral mean value of $f$ is also less than or equal to $0$ $($and the values
of $f$ in the interval $[a,\left(2a+b\right)/3]$ cannot be ”too” positive$)$.

There are numerous open problems related to Theorem 6 which seems of interest.

######  Problem 4.

What is the statistical meaning of the $3$-condensation __ of
$(1/(b-a))\mathrm{d}x?$ The same in the case of the $3$-dispersion of
$(1/(b-a))\mathrm{d}x.$

######  Problem 5.

Is any Fejér analog of Theorem _6_?

######  Problem 6.

Does the relation $\mu\prec_{3cvx}\nu$ admit a characterization à la Sherman
_[35],_ when $\mu$ and $\nu$ are discrete probability measures?

######  Problem 7.

Does Theorem _6_ admit an extension to the context of signed measures $($as is
the case of the Hermite-Hadamard measures for convex functions$)$?

##  6\. 3-convexity and the Hornich-Hlawka functional inequality

A straightforward consequence of Theorem 2 is the fact that every nonnegative
and continuous $3$-convex function verifies the Hornich-Hlawka functional
inequality:

######  Proposition 1.

If $f:\left[0,A\right]\rightarrow\mathbb{R}$ is a continuous $3$-convex
function, then

(6.1) $f\left(x\right)+f\left(y\right)+f\left(z\right)+f\left(x+y+z\right)\geq
f\left(x+y\right)+f\left(y+z\right)+f\left(z+x\right)+f\left(0\right)$

for all points $x,y,z\in[0,A]$ such that $x+y+z\leq A;$ if in addition
$f(0)\geq 0,$ then

(6.2) $f\left(x\right)+f\left(y\right)+f\left(z\right)+f\left(x+y+z\right)\geq
f\left(x+y\right)+f\left(y+z\right)+f\left(z+x\right).$

The Hornich-Hlawka functional inequality is not characteristic to the 3-convex
functions. Indeed, as was noticed by Sendov and Zitikis [36], Theorem 4.2 (see
also [24], Theorem 7), the Hornich-Hlawka functional inequality (HH2) also
works in the framework of completely monotone functions. Recall that a
function $f:[0,\infty)\rightarrow\mathbb{R}_{+}$ is _completely monotone_ if
it is continuous on $[0,\infty)$, indefinitely differentiable on $(0,\infty)$
and

$(-1)^{n}f^{(n)}(x)\geq 0\text{\quad for all }x>0\text{ and }n\geq 0.$

Some simple examples are $e^{-x},$ $1/(1+x),~{}$and
$\left(1/x\right)\log(1+x).$ Notice that every completely monotone function is
nonincreasing, convex and 3-concave.

The result of Proposition 1 can be considerably improved by adding additional
hypothesis.

######  Theorem 7.

Suppose that $f:[0,A]\rightarrow\mathbb{R}$ is a continuous $3$-convex
function which is also nondecreasing and concave. Then

$f\left(\left|x\right|\right)+f\left(\left|y\right|\right)+f\left(\left|z\right|\right)+f\left(\left|x+y+z\right|\right)\geq
f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f(0)$

for all $x,y,z\in[-A,A]$ with
$\left|x\right|+\left|y\right|+\left|z\right|\leq A.$

Theorem 7 is implicit in a paper due to Ressel, who formulated his result in
terms of differences of higher order. See [32],Theorem 1 and formula (5). For
the convenience of the reader we will include here a full argument.

Notice first that the inequality stated in Theorem 7 is invariant under the
permutations of the elements $x,y,z$ and also to the symmetry
$(x,y,z)\rightarrow(-x,-y,-z).$ As a consequence, the proof of Theorem 7 can
be reduced to the following two cases:

Case $\mathbf{1}$: the elements $x,y$ and $z$ have the same sign, in which
case we may reduce ourselves to the situation where

$x\geq y\geq z\geq 0;$

Case $\mathbf{2}$: two of the elements $x,y,z$ are nonnegative, while the
third is nonpositive, in which case the proof reduces to the situation where

$x\geq y\geq 0\geq z.$

Case $1$ is covered by the assertion $(iii)$ of Theorem 2. Case $2$ can be
split into four subcases:

Case $\mathbf{2a}$: $x\geq y\geq 0\geq z$ and $\left|z\right|\geq x+y;$

Case $\mathbf{2b}$: $x\geq y\geq 0\geq z$ and $x+y\geq\left|z\right|\geq x,y;$

Case $\mathbf{2c}$: $x\geq y\geq 0\geq z$ and $x\geq\left|z\right|\geq y;$

Case $\mathbf{2d}$:  $z\leq 0\leq y\leq x$ and $x\geq y\geq\left|z\right|.$

The assertion of Theorem 7 in Case $2a$ makes the objective of Lemma 4, while
the other cases (Case $2b$, Case $2c$ and Case $2d$) are settled by Lemma 5.

######  Lemma 4.

Suppose that $f:[0,A]\rightarrow\mathbb{R}$ is a continuous function such that

$f\left(x\right)+f\left(y\right)+f\left(z\right)+f\left(x+y+z\right)\geq
f\left(x+y\right)+f\left(y+z\right)+f\left(z+x\right)+f(0)$

for all $x,y,z\geq 0$ with $x+y+z\leq A$. Then $f$ also verifies the
functional inequality

$f\left(\left|x\right|\right)+f\left(\left|y\right|\right)+f\left(\left|z\right|\right)+f\left(\left|x+y+z\right|\right)\geq
f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f(0)$

for all triplets $x,y,z\in[-A,A]$ of which two elements are nonnegative and
their sum does not exceed the absolute value of the third element.

###### Proof.

It suffices to consider the case where $x,y\geq 0\geq z$ and
$x+y\leq\left|z\right|$. Then, $\left|z\right|-x=\left|x+z\right|,$
$\left|z\right|-y=\left|y+z\right|$ and
$\left|z\right|-x-y=\left|x+y+z\right|$. According to our hypothesis, applied
to $x,~{}y$ and $\left|z\right|-x-y,$ we have

$f(x)+f(y)+f\left(\left|z\right|-x-y\right)+f\left(\left|z\right|\right)\geq
f(x+y)+f\left(\left|z\right|-y\right)+f\left(\left|z\right|-x\right)+f(0),$

equivalently,

$f\left(\left|x\right|\right)+f\left(\left|y\right|\right)+f\left(\left|z\right|\right)+f\left(\left|x+y+z\right|\right)\\\
\geq
f\left(\left|y+z\right|\right)+f\left(\left|x+z\right|\right)+f\left(\left|x+y\right|\right)+f(0).$

∎

######  Lemma 5.

If $f:[0,A]\rightarrow\mathbb{R}$ is a nondecreasing and concave function,
then

$f\left(\left|x\right|\right)+f\left(\left|y\right|\right)+f\left(\left|z\right|\right)+f\left(\left|x+y+z\right|\right)\geq
f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f\left(0\right),$

for all $x,y,z\in[-A,A]$ such that $z\leq 0\leq y\leq x.$

When the domain of $f$ is $\mathbb{R}_{+}$ and $f$ is nonnegative, then the
property of concavity implies the property of being nondecreasing. See [23],
Exercise 4, p. 31.

###### Proof.

The range $z\leq 0\leq y\leq x$ can be split into the following cases:

Case $2b:$ $z\leq 0\leq y\leq x$ and $x,y\leq\left|z\right|\leq x+y.$ Then
$0\leq x+y+z\leq\left|z\right|\leq x+y,$ so by Lemma 3 it follows that

$f(x+y)+f(0)\leq f(\left|z\right|)+f(x+y+z).$

On the other hand, $\left|x+z\right|=\left|z\right|-x,$
$\left|y+z\right|=\left|z\right|-y$ and $\left|x+y+z\right|=x+y+z.$ Since $f$
is nondecreasing, we have $f(\left|z\right|-x)\leq f(y)$ and
$f(\left|z\right|-y)\leq f(\left|x\right|)$ . Therefore

$f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f(0)\\\
=f(x+y)+f(0)+f(\left|z\right|-y)+f(\left|z\right|-x)\\\ \leq
f(x+y+z)+f(\left|z\right|)+f(x)+f(y)\\\
=f(\left|z\right|)+f(\left|x+y+z\right|)+f(\left|x\right|)+f(\left|y\right|)$

Case $2c:$ $z\leq 0\leq y\leq x$ and $y\leq\left|z\right|\leq x.$ Applying
Lemma 3 for $0\leq y\leq x\leq x+y$ and taking into account that $f$ is
nondecreasing we obtain

$f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f(0)\\\
=f(x+y)+f(0)+f(\left|z\right|-y)+f(x-\left|z\right|)\\\ \leq
f(x)+f(y)+f(\left|z\right|)+f(x+y+z)\\\
=f(\left|x\right|)+f(\left|y\right|)+f(\left|z\right|)+f(\left|x+y+z\right|).$

Case $2d:$ $z\leq 0\leq y\leq x$ and $\left|z\right|\leq y\leq x.$ Similar to
Case $2c$. Applying Lemma 3 for $0\leq\left|z\right|\leq
x+y-\left|z\right|\leq x+y$ and using the fact that $f$ is nondecreasing we
obtain

$f\left(\left|x+y\right|\right)+f\left(\left|y+z\right|\right)+f\left(\left|z+x\right|\right)+f(0)\\\
=f(x+y)+f(0)+f(y-\left|z\right|)+f(x-\left|z\right|)\\\ \leq
f\left(x+y-\left|z\right|\right)+f\left(\left|z\right|\right)+f(y-\left|z\right|)+f(x-\left|z\right|)\\\
\leq f(x+y+z)+f(\left|z\right|)+f(y)+f(x)\\\
=f(\left|x\right|)+f(\left|y\right|)+f(\left|z\right|)+f(\left|x+y+z\right|).$

The proof of Lemma 5 is now complete. ∎

Lemma 5, fails in the case of nondecreasing and $3$-convex functions which are
not concave. To check this, consider the restriction of the cubic function
$x^{3}$ to $[0,\infty)$ and the triplet $x=y=1$ and $z=-1.$

Some example illustrating Theorem 7 and Corollary 1 in the case of the
Bernstein functions $x/(1+x)$, $\log(1+x)$ and the identity of $[0,\infty))$
are indicated below:

  1. $(RHH)$

the _rational form of the Hornich-Hlawka inequality,_

$\frac{\left|x\right|^{\alpha}}{1+\left|x\right|^{\alpha}}+\frac{\left|y\right|^{\alpha}}{1+\left|y\right|^{\alpha}}+\frac{\left|z\right|^{\alpha}}{1+\left|z\right|^{\alpha}}+\frac{\left|x+y+z\right|^{\alpha}}{1+\left|x+y+z\right|^{a}}\\\
\geq\frac{\left|x+y\right|^{a}}{1+\left|x+y\right|^{a}}+\frac{\left|y+z\right|^{a}}{1+\left|y+z\right|^{a}}+\frac{\left|z+x\right|^{a}}{1+\left|z+x\right|^{a}};$

  2. $(MHH)$

the _multiplicative form of the Hornich-Hlawka inequality,_

$(1+\left|x\right|)(1+\left|y\right|)(1+\left|z\right|)(1+\left|x+y+z\right|)\allowbreak\\\
\geq(1+\left|x+y\right|)(1+\left|y+z\right|)(1+\left|z+x\right|);$

  3. $(HH^{\alpha})$

the _fractional power form of the Hornich-Hlawka inequality_ :

$\left|x\right|^{\alpha}+\left|y\right|^{\alpha}+\left|z\right|^{\alpha}+\left|x+y+z\right|^{\alpha}\geq\left|x+y\right|^{\alpha}+\left|y+z\right|^{\alpha}+\left|z+x\right|^{\alpha}.$

Here $\alpha\in(0,1]$ is a parameter.

The natural analogue of Theorem 7, for more that 3 numbers does not hold.
Indeed, according to a comment made by Freudenthal in connection with the
Hornich-Hlawka inequality, the function

$\sum\nolimits_{i=1}^{4}\left|x_{i}\right|-\sum\limits_{1\leq i<j\leq
4}\left|x_{i}+x_{j}\right|+\sum\limits_{1\leq i<j<k\leq
4}\left|x_{i}+x_{j}+x_{k}\right|-\left|x_{1}+x_{2}+x_{3}+x_{4}\right|$

takes both positive and negative values as the variables $x_{1},...,x_{4}$ run
over $\mathbb{R}.$

However, an extension of Theorem 7 to the case of $n>3$ variables is still
possible by using an inductive scheme due to Vasić and Adamović [39]. We state
here a slightly modified version of their result as appeared in [19], Theorem
2, p. 528:

######  Theorem 8.

Suppose that $\varphi$ is a real-valued function defined on a commutative
additive semigroup $\mathcal{S}$ such that

$\sum\limits_{k=1}^{3}{\varphi\left(x_{k}\right)}+{\varphi}\left({\sum\limits_{k=1}^{3}}x_{k}\right)\gtrless\sum\limits_{1\leq{i}<{j}\leq
3}{\varphi\left({x{{}_{i}+x}}_{j}\right)}$

for all $x_{1},x_{2},x_{3}\in\mathcal{S}.$ Then for each pair
$\left\\{k,n\right\\}$ of integers with $2\leq k<n$ we also have

$\binom{n-2}{k-1}\sum\limits_{k=1}^{n}{\varphi\left(x_{k}\right)}+\binom{n-2}{k-2}{\varphi}\left({\sum\limits_{k=1}^{n}}x_{k}\right)\gtrless\sum\limits_{1\leq{i_{1}}<...<{i_{k}}\leq
n}{\varphi\left({\sum\limits_{j=1}^{k}{x{{}_{i_{j}}}}}\right),}$

whenever $x_{1},...,x_{n}\in\mathcal{S}.$

This result yields the following generalization of Theorem 7:

######  Theorem 9.

If $f:\mathbb{R}_{+}\rightarrow E$ is a continuous $3$-convex function which
is also nondecreasing and concave, then

$\binom{n-2}{k-1}\sum\limits_{k=1}^{n}{f\left(\left|x_{k}\right|\right)}+\binom{n-2}{k-2}{f}\left(\left|{\sum\limits_{k=1}^{n}}x_{k}\right|\right)\\\
\geq\sum\limits_{1\leq{i_{1}}<...<{i_{k}}\leq
n}{f\left(\left|{\sum\limits_{j=1}^{k}{x{{}_{i_{j}}}}}\right|\right)+}\binom{n-1}{k}f(0),$

for all pairs $\left\\{k,n\right\\}$ of integers with $2\leq k<n$ and all
strings $x_{1},...,x_{n}$ of real numbers.

###### Proof.

Apply Theorem 8 to $f(\left|\cdot\right|)-f(0),$ taking into account the last
assertion of 7 and the formula

$\binom{n-2}{k-1}+\binom{n-2}{k-2}=\binom{n-1}{k-1}.$

∎

##  7\. The case of vector-valued functions

The concept of $n$-convexity can be extended in a straightforward way to the
case of functions with values in an ordered Banach space by using the same
definition based on divided differences.

Recall that an _ordered Banach space_ is any Banach space $E$ endowed with the
ordering $\leq$ associated to a closed convex cone $E_{+}$ via the formula

$x\leq y\ \text{if and only if }y-x\in E_{+},~{}$

such that

$E=E_{+}-E_{+},\text{\quad}\left(-E_{+}\right)\cap E_{+}=\left\\{0\right\\},$

and

$0\leq x\leq y\ \text{in}~{}E~{}\text{implies
}\left\|x\right\|\leq\left\|y\right\|.$

The basic facts concerning the theory of ordered Banach spaces are made
available by the book of Schaefer and Wolff [33]. See [21] for a short
overview centered on two important particular cases: $\mathbb{R}^{n},$ the
$n$-dimensional Euclidean space endowed with the coordinate-wise ordering, and
$\operatorname*{Sym}(n,\mathbb{R)}$ the ordered Banach space of all $n\times
n$-dimensional symmetric matrices with real coefficients endowed with the
operator norm

$\left\|A\right\|=\sup_{\left\|x\right\|\leq 1}\left|\langle
Ax,x\rangle\right|$

and the Löwner ordering,

$A\leq B\text{ if and only if }\langle
A\mathbf{x},\mathbf{x}\rangle\leq\langle B\mathbf{x},\mathbf{x}\rangle\text{
for all }\mathbf{x}\in\mathbb{R}^{n}.$

Here the operator norm can be replaced by any Schatten norm, in particular
with the _Frobenius norm_ ,

$\left\|A\right\|_{F}=\left(\sum_{i=1}^{n}\sum_{j=1}^{n}a_{ij}^{2}\right)^{1/2},$

provided that $A=\left(a_{ij}\right)_{i,j=1}^{n}.$ The Frobenius norm is
associated to the trace inner product

$\langle A,B\rangle=\operatorname*{trace}(AB).$

The positive cone of $\mathbb{R}^{n}$ is the first orthant
$\mathbb{R}_{+}^{n},$ while the positive cone of
$\operatorname*{Sym}(n,\mathbb{R)}$ is the set
$\operatorname*{Sym}^{+}(n,\mathbb{R)}$ consisting of all positive semi-
definite matrices.

######  Remark 9.

The study of vector-valued functions can be reduced to that of real-valued
functions. Indeed, in any ordered Banach space $E,$ any inequality of the
form$\ u\leq v$ is equivalent to $x^{\ast}(u)\leq x^{\ast}(v)$ for all
$x^{\ast}\in E_{+}^{\ast}$. See _[21]_.

As a consequence, a function $f:I\rightarrow E$ is respectively nondecreasing,
convex or $n$-convex if and only if $x^{\ast}\circ f$ has this property
whenever $x^{\ast}\in E^{\ast}$ is a positive functional. For
$E=\mathbb{R}^{n},$ this remark concerns the components of $f$.

Remark 9 easily yields that most of the results in the preceding sections
extends verbatim to the vector-valued framework. In particular, so are Theorem
2, Proposition 1, Theorem 7, Theorem 4 and Theorem 9.

Combining Remark 9 with Lemma 1 one obtains the following practical test of
$3$-convexity for the vector-valued differentiable functions:

######  Theorem 10.

Suppose that $f$ is a continuous function defined on an interval $I$ and
taking values in an ordered Banach space $E$. If $f$ is three times
differentiable on the interior of $I,$ then $f$ is a $3$-convex function if
and only if $f^{\prime\prime\prime}\geq 0.$

An example illustrating Theorem 10 is provided by the function

$f:\mathbb{R}_{+}\rightarrow\operatorname*{Sym}(n,\mathbb{R)},\text{\quad}f(t)=-e^{-tA},$

associated to a positive semi-definite matrix
$A\in\operatorname*{Sym}(n,\mathbb{R)}.$ This function is of class
$C^{\infty}$ and its first three derivatives are given by the formulas

$f^{\prime}(t)=Ae^{-tA},\text{\quad}f^{\prime\prime}(t)=-A^{2}e^{-tA},\text{\quad}f^{\prime\prime\prime}(t)=A^{3}e^{-tA}.$

This shows that $f$ is nondecreasing, concave and 3-convex (according to the
ordering of $\operatorname*{Sym}(n,\mathbb{R))}.$ The fact that $A^{3}e^{-tA}$
is positive semidefinite follows from the fact that the product of positive
semi-definite matrices that commute with each other is a matrix of the same
type.

According to Theorem 7,

$e^{-\left|r\right|A}+e^{-\left|s\right|A}+e^{-\left|t\right|A}+e^{-\left|r+s+t\right|A}\leq
I+e^{-\left|r+s\right|A}+e^{-\left|s+t\right|A}+e^{-\left|t+r\right|A}$

for all $r,s,t\in\mathbb{R}.$ Here $I$ is the identity matrix. In the
1-dimensional case, this reduces to the inequality

$e^{-\alpha\left|x\right|}+e^{-\alpha\left|y\right|}+e^{-\alpha\left|z\right|}+e^{-\alpha\left|x+y+z\right|}\leq
1+e^{-\alpha\left|x+y\right|}+e^{-\alpha\left|y+z\right|}+e^{-\alpha\left|x+z\right|},$

which works for all $x,y,z\in\mathbb{R}$ and $\alpha>0$. This last inequality
can be extended to the framework of real symmetric matrices:

######  Theorem 11.

Suppose that $f:[0,\infty)\rightarrow\mathbb{R}$ is a continuous $3$-convex
function which is also nondecreasing and concave. Then

$f\left(\left|A\right|\right)+f\left(\left|B\right|\right)+f\left(\left|C\right|\right)+f\left(\left|A+B+C\right|\right)\\\
\geq
f\left(\left|A+B\right|\right)+f\left(\left|B+C\right|\right)+f\left(\left|C+A\right|\right)+f(0)I_{n},$

whenever $A,B,C$ are three real symmetric matrices of order $n$ that commute
with each other.

Here $\left|A\right|=\left(A^{2}\right)^{1/2}$ denotes the modulus of $A$ and
$I_{n}$ is the unit matrix of order $n$.

###### Proof.

Notice first that every finite family of self-adjoint matrices that commute
with each other admits an orthonormal basis consisting of vectors that are
eigenvectors of each these matrices. See Mirsky [18], Theorem 10.6.8, p. 322\.
This reduces the proof of the theorem to the case where all the matrices $A,$
$B$ and $C$ are diagonal. Or, if

$A=\left(\begin{array}[c]{ccc}\lambda_{1}(A)&&0\\\ &\ddots&\\\
0&&\lambda_{n}(A)\end{array}\right),$

then

$f(A)=\left(\begin{array}[c]{ccc}f\left(\lambda_{1}(A)\right)&&0\\\
&\ddots&\\\ 0&&f\left(\lambda_{n}(A)\right)\end{array}\right),$

so that the conclusion of the theorem follows from Theorem 7. ∎

######  Remark 10.

Theorem _11_ also works in the context of commuting self-adjoint compact
operators defined on an infinite dimensional Hilbert space provided that
$f:[0,\infty)\rightarrow\mathbb{R}$ is continuous, nondecreasing, concave
$3$-convex and $f(0)=0.$ We do not know whether the commutativity condition is
necessary or not.

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# The Rank of the Cartier operator on Picard Curves

Vahid Nourozi And Farhad Rahmati∗ Faculty of Mathematics and Computer Science,
Amirkabir University of Technology
(Tehran Polytechnic), 424 Hafez Ave., Tehran 15914, Iran<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

For an algebraic curve $\mathcal{X}$ defined over an algebraically closed
field of characteristic $p>0$, $a$-number $a(\mathcal{X})$ is the dimension of
the space of exact holomorphic differentials on $\mathcal{X}$. We computed the
$a$-number for a family of certain Picard curves using the action of the
Cartier operator on $H^{0}(\mathcal{X},\Omega^{1})$.

###### Key words and phrases:

$a$-number; Cartier operator; Super-singular Curves; Picard Curves.

* Corresponding author

##  1\. Introduction

Let $\mathcal{X}$ be a geometrically irreducible, projective, and non-singular
algebraic curve defined over the finite field $\mathbb{F}_{\ell}$ of order
$\ell$. Let $\mathcal{X}(\mathbb{F}_{\ell})$ denote the set of
$\mathbb{F}_{\ell}$-rational points for $\mathcal{X}$. In studying curves over
finite fields, determining the size of $\mathcal{X}(\mathbb{F}_{\ell})$ is a
fundamental problem. The exact primary result here is the Hasse-Weil bound,
which asserts that

$\mid\\#\mathcal{X}(\mathbb{F}_{\ell})-(\ell+1)\mid\leq 2g\sqrt{\ell},$

where $g=g(\mathcal{X})$ is the genus of $\mathcal{X}$.

In this paper, we consider Picard curves of genus $3$ over
$\mathbb{F}_{q^{2}}$ of characteristic $p>3$. Let $\mathcal{X}$ be a Picard
curve. Then $\mathcal{X}$ can be defined by an affine equation of the form

$y^{3}=f(x),$ (1.1)

where $f(x)$ is a polynomial of degree $4$, without multiple roots. It is easy
to see that the maximal curve $\mathcal{X}$ is super-singular because all the
slopes of its Newton polygon are equal to $1/2$. This implies that the
Jacobian $X:=\mbox{Jac}(\mathcal{X})$ has no $p$-torsion points over
$\bar{\mathbb{F}}_{p}$. A relevant invariant of the $p$-torsion group scheme
of the Jacobian of the curve is the $a$-number.

Consider multiplication by $p$-morphism $[p]:X\rightarrow X$, a finite flat
morphism of degree $p^{2g}$. These factors were $[p]=V\circ F$. Here,
$F:X\rightarrow X^{(p)}$ is the relative Frobenius morphism originating from
the $p$-power map on the sheaf structure, and the Verschiebung morphism
$V:X^{(p)}\rightarrow X$ is the dual of $F$. The multiplication-by-$p$ kernel
on $X$ is defined as $X[p]$. An important invariant is the $a$-number
$a(\mathcal{X})$ of curve $\mathcal{X}$ defined by

$a(\mathcal{X})=\mbox{dim}_{\mathbb{\bar{F}}_{p}}\mbox{Hom}(\alpha_{p},X[p]),$

where $\alpha_{p}$ is the kernel of the Frobenius endomorphism in the group
scheme $\mbox{Spec}(k[X]/(X^{p}))$. Another definition for the $a$-number is

$a(\mathcal{X})=\mbox{dim}_{\mathbb{F}_{p}}(\mbox{Ker}(F)\cap\mbox{Ker}(V)).$

A few results on the rank of the Cartier operator (especially $a$-number) of
curves are introduced by Kodama and Washio [9], González [6], Pries and Weir
[18], Yui [21]. Besides that, Vahid talked about the rank of the Cartier of
the maximal curves in [16], the hyperelliptic curve in [13], and the maximal
function fields of $\mathbb{F}_{q^{2}}$ in [12, 17]. Vahid also wrote about
these topics in his [15] dissertation.

In Section 3, we prove that the $a$-number of the Picard curves with Equation
(1.1) is $0$ for $p\equiv 1$ mod $3$, and is $1$ for $p\equiv 2$ mod $3$ , see
Theorem 3.1. The proofs directly use the Cartier operator action on
$H^{0}(\mathcal{X},\Omega^{1})$. Finally, we provide an example of the
$a$-number of Picard curves obtained from the Hasse-Witt matrix.

##  2\. The Cartier operator

Let $k$ be an algebraically closed field of characteristic $p>0$. Let
$\mathcal{X}$ be a curve defined over $k$. The Cartier operator is a
$1/p$-linear operator acting on the sheaf
$\Omega^{1}:=\Omega^{1}_{\mathcal{X}}$ of differential forms on $\mathcal{X}$
with positive characteristics, $p>0$.

Let $K=k(\mathcal{X})$ be the function field of curve $\mathcal{X}$ of genus
$g$ defined over $k$. The separating variable for $K$ is an element $x\in
K\setminus K^{p}$. We can write any function $f\in K$ uniquely in form

$f=f^{p}_{0}+f^{p}_{1}x+\cdots+f^{p}_{p-1}x^{p-1},$

where $f_{i}\in K$, for $i=0,\cdots,p-1$.

######  Definition 2.1.

(Cartier Operator). Let $\omega\in\Omega_{K/K_{p}}$. There exist
$f_{0},\cdots,f_{p-1}$ such that
$\omega=(f^{p}_{0}+f^{p}_{1}x+\cdots+f^{p}_{p-1}x^{p-1})dx$. The Cartier
operator $\mathfrak{C}$ is defined by

$\mathfrak{C}(\omega):=f_{p-1}dx.$

This definition does not depend on the choice of $x$ (see [19, Proposition
1]).

We refer the reader to [19, 2, 3, 20] for proofs of the following statements.

######  Proposition 2.2.

(Global Properties of $\mathfrak{C}$). For all $\omega\in\Omega_{K/K_{q}}$ and
all $f\in K$,

  * •

$\mathfrak{C}(f^{p}\omega)=f\mathfrak{C}(\omega)$;

  * •

$\mathfrak{C}(\omega)=0\Leftrightarrow\exists h\in K,\omega=dh$;

  * •

$\mathfrak{C}(\omega)=\omega\Leftrightarrow\exists h\in K,\omega=dh/h$.

######  Remark 2.3.

Moreover, one can easily show that

$\mathfrak{C}(x^{j}dx)=\left\\{\begin{array}[]{ccc}0&\mbox{if}&\hskip
11.38092ptp\nmid j+1\\\ x^{s-1}dx&\mbox{if}&\hskip
11.38092ptj+1=ps.\end{array}\right.$

If $\mbox{div}(\omega)$ is effective, then differential $\omega$ is
holomorphic. The set $H^{0}(\mathcal{X},\Omega^{1})$ of holomorphic
differentials is a $g$-dimensional $k$-vector subspace of $\Omega^{1}$, such
that $\mathfrak{C}(H^{0}(\mathcal{X},\Omega^{1}))\subseteq
H^{0}(\mathcal{X},\Omega^{1})$. If $\mathcal{X}$ is a curve, then the
$a$-number of $\mathcal{X}$ equals the dimension of the kernel of the Cartier
operator $H^{0}(\mathcal{X},\Omega^{1})$ (or equivalently, the dimension of
the space of the exact holomorphic differentials on $\mathcal{X}$) (see [10,
5.2.8]).

The following theorem is based on Gorenstein; see [7, Theorem 12].

######  Theorem 2.4.

A differential $\omega\in\Omega^{1}$ is holomorphic if and only if it has the
form $(h(x,y)/F_{y})dx$, where $H:h(X,Y)=0$ is a canonical adjoint.

######  Theorem 2.5.

[11] With the above assumptions,

$\mathfrak{C}(h\dfrac{dx}{F_{y}})=(\dfrac{\partial^{2p-2}}{\partial
x^{p-1}\partial y^{p-1}}(F^{p-1}h))^{\frac{1}{p}}\dfrac{dx}{F_{y}}$

for any $h\in K(\mathcal{X})$.

The differential operator $\nabla$ is defined by

$\nabla=\dfrac{\partial^{2p-2}}{\partial x^{p-1}\partial y^{p-1}},$

has the property

$\nabla(\sum_{i,j}c_{i,j}X^{i}Y^{j})=\sum_{i,j}c_{ip+p-1,jp+p-1}X^{ip}Y^{jp}.$
(2.1)

##  3\. The $a$-number of Picard Curves

This section considers the Picard curves over a finite field with $q^{2}$
elements of characteristic $p>3$. Let $\mathcal{X}$ be the Picard curve of
genus $3$ over $\mathbb{F}_{q^{2}}$. Then $\mathcal{X}$ can be defined by an
affine equation of the form

$y^{3}=f(x),$ (3.1)

where $f(x)$ is a polynomial of degree $4$ without multiple roots. From
Theorem 2.4, one can find a basis for the space
$H^{0}(\mathcal{X},\Omega^{1})$ of holomorphic differentials on $\mathcal{X}$,
namely

$\mathcal{B}=\Bigg{\\{}z_{i}=\dfrac{h_{i}dx}{f_{y}}\Bigg{\\}},$

where $h_{1}=3,h_{2}=3x$ and $h_{3}=3y$ for $1\leq i\leq 3$. Therefore,
$\mathcal{B}=\\{\frac{dx}{y^{2}},\frac{xdx}{y^{2}},\frac{dx}{y}\\}$.

######  Theorem 3.1.

Suppose that $p\not\equiv 0$ mod $3$ and $f(x)$ have a non-zero constant-
coefficient, then

  * 1.

If $p\equiv 1$ mod $3$, then $a(\mathcal{X})=0$.

  * 2.

If $p\equiv 2$ mod $3$, then $a(\mathcal{X})=1$.

###### Proof.

  * 1.

Let us compute the image of $\mathfrak{C}(\omega)$ for any
$\omega\in\mathcal{B}$. Taking $p=3r+1$ and $r\in\mathbb{Z}$. Since $p-1=3r$.
We get $2p-2=6r$. So

$\begin{array}[]{ccccccc}\mathfrak{C}(xdx/y^{2})&=&\mathfrak{C}(y^{2p-2}y^{-2p}xdx)=y^{-2}\mathfrak{C}(xf(x)^{2r}dx)\\\
&=&y^{-2}\sum_{i=0}^{2r}c_{i}\mathfrak{C}(x^{i+1}dx)=c_{r}dx\neq 0.\\\
\end{array}$
$\begin{array}[]{ccccccc}\mathfrak{C}(dx/y^{2})&=&\mathfrak{C}(y^{2p-2}y^{-2p}dx)=y^{-2}\mathfrak{C}(f(x)^{2r}dx)\\\
&=&y^{-2}\sum_{i=0}^{2r}c_{i}\mathfrak{C}(x^{i}dx)=c_{r}dx\neq 0.\\\
\end{array}$

Now we let $p-1=3r$,

$\begin{array}[]{ccccccc}\mathfrak{C}(dx/y)&=&\mathfrak{C}(y^{p-1}y^{-p}dx)=y^{-1}\mathfrak{C}(f(x)^{r}dx)\\\
&=&y^{-1}\sum_{i=0}^{r}c_{i}\mathfrak{C}(x^{i}dx)=c_{r}dx\neq 0.\\\
\end{array}$

Thus, $a(\mathcal{X})=0$.

  * 2.

We assume that $p=3r+2$ and $r\in\mathbb{Z}$. We get $p-2=3r$. Hence

$\begin{array}[]{ccccccc}\mathfrak{C}(xdx/y^{2})&=&\mathfrak{C}(y^{p-2}y^{-p}xdx)=y^{-1}\mathfrak{C}(xf(x)^{r}dx)\\\
&=&y^{-1}\sum_{i=0}^{r}c_{i}\mathfrak{C}(x^{i+1}dx)=c_{r}dx\neq 0.\\\
\end{array}$
$\begin{array}[]{ccccccc}\mathfrak{C}(dx/y^{2})&=&\mathfrak{C}(y^{p-2}y^{-p}dx)=y^{-1}\mathfrak{C}(f(x)^{r}dx)\\\
&=&y^{-1}\sum_{i=0}^{r}c_{i}\mathfrak{C}(x^{i}dx)=c_{r}dx\neq 0.\\\
\end{array}$

Finally, from Remark 2.3 we have

$\begin{array}[]{ccccccc}\mathfrak{C}(dx/y)&=&\mathfrak{C}(y^{p-1}y^{-p}dx)=y^{-1}\mathfrak{C}(f(x)^{\frac{3r-1}{3}}dx)\\\
&=&y^{-1}\sum_{i=0}^{\frac{3r-1}{3}}c_{i}\mathfrak{C}(x^{i}dx)=0.\\\
\end{array}$

Therefore, $a(\mathcal{X})=1$.

Where the coefficients $c_{i}\in k$ are obtained from the expansion

$y^{p-1}=f(x)^{(p-1)/2}=\sum_{i=0}^{N}c_{i}x^{i}\hskip
19.91684pt~{}\mbox{with}~{}N=\frac{p-1}{2}.$

∎

In [4], Estrada Sarlabous showed that if
$F^{p-1}=\sum_{i,j}c_{i,j}x^{i}y^{j}$, then the Hasss-Wit matrix of
$\mathcal{B}$ is

$H=\left(\begin{array}[]{cccc}c_{p-1,p-1}&c_{2p-1,p-1}&c_{p-1,2p-1}\\\
c_{p-2,p-1}&c_{2p-2,p-1}&c_{p-2,2p-1}\\\
c_{p-1,p-2}&c_{2p-1,p-2}&c_{p-1,2p-2}\end{array}\right).$

Since $F=y^{3}-f(x),c_{i,j}=0$ for $j\not\equiv 0$ mod $3$. Thus, we have

  * 1.

$p\equiv 1$ mod $3$. Then

$H=\left(\begin{array}[]{cccc}c_{p-1,p-1}&c_{2p-1,p-1}&0\\\
c_{p-2,p-1}&c_{2p-2,p-1}&0\\\ 0&0&c_{p-1,2p-2}\end{array}\right).$ (3.2)

where $c_{p-1,p-1}$, $c_{2p-1,p-1}$, $c_{p-2,p-1},c_{2p-2,p-1}$ are the
respective coefficients of $x^{p-1},x^{2p-1}$, $x^{p-2}$ and $x^{2p-2}$ in
$f(x)^{(2p-2)/3}$ mod $p$. $c_{p-1,2p-2}$ is the coefficient of $x^{p-1}$ in
$f(x)^{(p-1)/3}$ mod $p$.

  * 2.

$p\equiv 2$ mod $3$. Then

$H=\left(\begin{array}[]{cccc}0&0&c_{p-1,2p-1}\\\ 0&0&c_{p-2,2p-1}\\\
c_{p-1,p-2}&c_{2p-1,p-2}&0\end{array}\right).$ (3.3)

where $c_{p-1,2p-1}$ and $c_{p-2,2p-1}$ are the coefficients of $x^{p-1}$ and
$x^{p-2}$ in $f(x)^{(p-2)/3}$ mod $p$, and $c_{p-1,p-2}$, and $c_{2p-1,p-2}$
are the coefficients of $x^{p-1}$ and $x^{2p-1}$ in $f(x)^{(2p-1)/3}$ mod $p$.

######  Example 3.2.

Consider the Picard curves with the following equation:

$y^{3}=x^{4}+1.$

We calculate the a-number with the Hasse-Wit matrix of this curve with $p=5$
and $p=13$. Firstly, suppose that $p=5$ in this case from equation (3.3), we
have

$H=\left(\begin{array}[]{cccc}0&0&1\\\ 0&0&0\\\ 3&0&0\end{array}\right).$

So $\mbox{rank}(H)=2$. From the definition of $a$-number, it is clear that
$a(\mathcal{X})=g(\mathcal{X})-\mbox{rank}(H)$. Therefore, $a(\mathcal{X})=1$.
Finally, assuming that $p=13$ in this case from equation (3.2), we have

$H=\left(\begin{array}[]{cccc}4&0&0\\\ 0&2&0\\\ 0&0&4\end{array}\right).$

So $\mbox{rank}(H)=3$, hence $a(\mathcal{X})=0$. In addition, we achieve this
result from MAGMA computation [1].

### Acknowledgements.

This paper was written while Vahid Nourozi was visiting Unicamp (Universidade
Estadual de Campinas) supported by TWAS/Cnpq (Brazil) with fellowship number
$314966/2018-8$. We are extremely grateful to the referee for their valuable
comments and suggestions, which led us to find correct references for many
assertions and improve the exposition.

## References

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# On monogenity of certain pure number fields defined by $x^{p^{r}}-m$

Hamid Ben Yakkou and Lhoussain El Fadil  Faculty of Sciences Dhar El Mahraz,
P.O. Box 1794 Atlas-Fez , Sidi Mohamed Ben Abdellah University, Fez– Morocco
<EMAIL_ADDRESS>
<EMAIL_ADDRESS>
###### Abstract.

Let $K=\mathbb{Q}(\alpha)$ be a pure number field generated by a complex root
$\alpha$ a monic irreducible polynomial $F(x)=x^{p^{r}}-m$, with $m\neq 1$ is
a square free rational integer, $p$ is a rational prime integer, and $r$ is a
positive integer. In this paper, we study the monogenity of $K$. We prove that
if $\nu_{p}(m^{p}-m)=1$, then $K$ is monogenic. But if $r\geq p$ and
$\nu_{p}(m^{p}-m)>p$, then $K$ is not monogenic. Some illustrating examples
are given.

###### Key words and phrases:

Power integral basis, Theorem of Ore, prime ideal factorization

###### 2010 Mathematics Subject Classification:

11R04, 11R16, 11R21

##  1\. Introduction

Let $K$ be a number field defined by a monic irreducible polynomial
$f(x)\in\mathbb{Z}[x]$ and $\mathbb{Z}_{K}$ its ring of integers. It is well
know that the ring $\mathbb{Z}_{K}$ is a free $\mathbb{Z}$-module of rank
$n=[K:\mathbb{Q}]$. Thus the abelian group $\mathbb{Z}_{K}/\mathbb{Z}[\alpha]$
is finite. Its cardinal order is called the index of $\mathbb{Z}[\alpha]$, and
denoted $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Kummer showed that for a
rational prime integer, if $p$ does not divide
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$, then the factorization of
$p\mathbb{Z}_{K}$ can be derived directly from the factorization of
$\overline{F(x)}$ in $\mathbb{F}_{p}[x]$. In $1878$, Dedekind gave a criterion
to tests wither $p$ divides or not $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ (see
[5, Theorem 6.1.4] and [23]). The index $i(K)$ of the field $K$ is $i(K)=\
gcd\ \\{ind(\theta)\ |\ \theta\in\mathbb{Z}_{K}\mbox{ generates }K\ \\}$. A
rational prime integer $p$ dividing $i(K)$ is called a prime common index
divisor of $K$. The ring $\mathbb{Z}_{K}$ is said to have a power integral
basis if it has a $\mathbb{Z}$-basis $(1,\theta,\cdots,\theta^{n-1})$ for some
$\theta\in\mathbb{Z}_{K}$. That means $\mathbb{Z}_{K}=\mathbb{Z}[\theta]$ or
$\mathbb{Z}_{K}$ is mono-generated as a ring, with a single generator
$\theta$. In such a case, the field $K$ is said to be monogenic and not
monogenic otherwise. The problem of checking the monogenity of the field $K$
is called a problem of Hasse [12, 14, 18, 19, 20]. The problem of testing the
monogenity of number fields and constructing power integral bases have been
intensively studied, this last century, mainly by Gaál, Nakahara, Pethö, and
their research groups (see for instance [3, 13, 15, 22]). In [6], El Fadil
gave conditions for the existence of power integral bases of pure cubic fields
in terms of the index form equation. In [12], Funakura, studied the integral
basis in pure quartic fields. In [16], Gaál and Remete, calculated the
elements of index $1$ (with coefficients with absolute value $<10^{1000}$ in
the integral basis) of pure quartic fields generated by $m^{\frac{1}{4}}$ for
$1<m<10^{7}$ and $m\equiv 2,3\ \mbox{\rm(mod }{4})$. In [1], Ahmad, Nakahara,
and Husnine proved that if $m\equiv 2,3\ \mbox{\rm(mod }{4})$ and
$m\not\equiv\mp 1\ \mbox{\rm(mod }{9})$, then the sextic number field
generated by $m^{\frac{1}{6}}$ is monogenic. They also showed in [3], that if
$m\equiv 1\ \mbox{\rm(mod }{4})$ and $m\not\equiv\mp 1\ \mbox{\rm(mod }{9})$,
then the sextic number field generated by $m^{\frac{1}{6}}$ is not monogenic.
Also, Hameed and Nakahara [4], proved that if $m\equiv 1\ \mbox{\rm(mod
}{4})$, then the octic number field generated by $m^{1/8}$ is not monogenic,
but if $m\equiv 2,3\ \mbox{\rm(mod }{4})$, then it is monogenic. In [17], Gaál
and Remete, by applying the explicit form of the index forms, they obtained
new results on monogenity of the number fields generated by $m^{\frac{1}{n}}$,
where $3\leq n\leq 9$. While Gaál’s and Pethö’s techniques are based on the
index calculation, Nakahara’s methods are based on the existence of power
relative integral bases of some special sub-fields. The goal of this paper is
to study the monogenity of pure number fields of degree $p^{n}$ for every
rational prime integer $p$ and $n$ is a natural integer. Our proposed results
generalize [2, Theorem 2.1], where pure number fields of degree $2^{n}$ are
previously studied. Our method is based on prime ideal factorization and is
similar to that used in [7, 8, 9].

##  2\. Main results

Let $K$ be the number field defined by a complex root $\alpha$ of a monic
irreducible polynomial $F(x)=x^{p^{r}}-m$, with $m\neq 1$ is a square free
rational integer.

######  Theorem 2.1.

If $\nu_{p}(m^{p}-m)=1$, then $\mathbb{Z}[\alpha]$ is the ring of integers of
$K$; $K$ is monogenic and $\alpha$ generates a power integral basis.

######  Corollary 2.2.

For $p=2$, if $m\equiv 2$ or $3$, then $\mathbb{Z}[\alpha]$ is integrally
closed for every natural integer $r$. So that Theorem 2.1 generalizes [2,
Theorem 2.1], where $p=2$ is previously considered.

######  Theorem 2.3.

Assume that $p$ is an odd prime integer and $p$ does not divide $m$. If
$\nu_{p}(m^{p-1}-1)>p$ and $r\geq p$, then $K$ is not monogenic.

######  Corollary 2.4.

Assume that $p=3$.

  1. (1)

If $m\not\equiv\mp 1\ \mbox{\rm(mod }{9})$, then $K$ is monogenic.

  2. (2)

If $r\geq 3$ and $m\equiv\mp 1\ \mbox{\rm(mod }{81})$, then $K$ is not
monogenic.

######  Theorem 2.5.

Assume that $p=2$.

  1. (1)

If $r=2$ and $m\equiv 1\ \mbox{\rm(mod }{16})$, then the pure quartic number
field $K$ is not monogenic.

  2. (2)

If $r\geq 3$ and $m\equiv 1\ \mbox{\rm(mod }{32})$, then $K$ is not monogenic.

##  3\. Proofs

We start by recalling some fundamental notions on Newton polygon’s techniques.
For more details, we refer to [10, 11]. For any prime integer $p$ and for any
monic polynomial $\phi\in\mathbb{Z}[x]$ whose reduction is irreducible in
$\mathbb{F}_{p}[x]$, let $\mathbb{F}_{\phi}$ be the field
$\frac{\mathbb{F}_{p}[x]}{(\overline{\phi})}$. For any monic polynomial
$f(x)\in\mathbb{Z}[x]$, upon the euclidean division by successive powers of
$\phi$, we expand $f(x)$ as $f(x)=\sum_{i=0}^{l}a_{i}(x)\phi(x)^{i}$, called
the $\phi$-expansion of $f(x)$ (for every $i$, $deg(a_{i}(x))<deg(\phi)$). The
$\phi$-Newton polygon of $f(x)$ with respect to $p$, is the lower boundary
convex envelope of the set of points $\\{(i,\nu_{p}(a_{i}(x))),\,a_{i}(x)\neq
0\\}$ in the euclidean plane, which we denote by $N_{\phi}(f)$. For every
$i\neq j=0,\dots,l$, let $a_{i}=a_{i}(x)$ and
$\mu_{ij}=\frac{\nu_{p}(a_{i})-\nu_{p}(a_{j})}{i-j}\in\mathbb{Q}$. Then we
obtain the following integers $0=i_{0}<i_{1}<\dots<i_{r}=l$ satisfying
$i_{j+1}=\mbox{ max }\\{i=i_{j}+1,\dots
l,\,\mu_{i_{j}i_{j+1}}\leq\mu_{i_{j}i}\\}$. For every $j=1,\dots r$, let
$S_{j}$ be the segment joining the points $A_{j-1}=(i_{j-1},\nu(a_{i_{j-1}}))$
and $A_{j}=(i_{j},\nu(a_{i_{j}}))$ in the euclidean plane. The segments
$S_{1},\dots,S_{r}$ are called the sides of the polygon $N_{\phi}(f)$. For
every $j=1,\dots,r$, the rational number
$\lambda_{j}=\frac{\nu_{p}(a_{i_{j}})-\nu_{p}(a_{i_{j-1}})}{i_{j}-i_{j-1}}\in\mathbb{Q}$
is called the slope of $S_{j}$, $l(S_{j})=i_{j}-i_{j-1}$ is its length, and
$h(S_{j})=-\lambda_{j}l(S_{j})$ is its height. In what follows
$\nu(a_{i_{j}})=\nu(a_{i_{j-1}})+l(S_{j})\lambda_{j}$. The $\phi$-Newton
polygon of $f$, is the process of joining the segments $S_{1},\dots,S_{r}$
ordered by the increasing slopes, which can be expressed as
$N_{\phi}(f)=S_{1}+\dots+S_{r}$. Notice that $N_{\phi}(f)=S_{1}+\dots+S_{r}$
is only a notation and not the sum in the euclidean plane. For every side $S$
of the polygon $N_{\phi}(f)$, $l(S)$ is the length of its projection to the
$x$-axis and $h(S)$ is the length of its projection to the $y$-axis. The
principal part of $N_{\phi}(f)$, denoted $N_{\phi}^{+}(f)$, is the part of the
polygon $N_{\phi}(f)$, which is determined by joining all sides of negative
slopes.

For every side $S$ of $N_{\phi}(f)$, with initial point $(s,u_{s})$ and length
$l$, and for every $i=0,\dots,l$, we attach the following residual coefficient
$c_{i}\in\mathbb{F}_{\phi}$ as follows:

$c_{i}=\left\\{\begin{array}[]{ll}0,&\mbox{ if }(s+i,{\it u_{s+i}})\mbox{ lies
strictly above }S,\\\ \left(\dfrac{a_{s+i}(x)}{p^{{\it u_{s+i}}}}\right)\,\,\
\mbox{\rm(mod }{(p,\phi(x))}),&\mbox{ if }(s+i,{\it u_{s+i}})\mbox{ lies on
}S.\end{array}\right.$

where $(p,\phi(x))$ is the maximal ideal of $\mathbb{Z}[x]$ generated by $p$
and $\phi$. That means if $(s+i,{\it u_{s+i}})\mbox{ lies on }S$, then
$c_{i}=\overline{\dfrac{a_{s+i}(\beta)}{p^{{\it u_{s+i}}}}}$, where $\beta$ is
a root of $\phi$.

Let $\lambda=-h/e$ be the slope of $S$, where $h$ and $e$ are two positive
coprime integers, and let $d=l/e$ be the degree of $S$. Notice that, the
points with integer coordinates lying in $S$ are exactly
$(s,u_{s}),(s+e,u_{s}+h),\cdots,(s+de,u_{s}+dh)$. Thus, if $i$ is not a
multiple of $e$, then $(s+i,u_{s+i})$ does not lie in $S$, and so, $c_{i}=0$.
Let
$f_{S}(y)=t_{0}y^{d}+t_{1}y^{d-1}+\cdots+t_{d-1}y+t_{d}\in\mathbb{F}_{\phi}[y]$,
called the residual polynomial of $f(x)$ associated to the side $S$, where for
every $i=0,\dots,d$, $t_{i}=c_{ie}$.

######  Remark.

Note that if $\nu(a_{s}(x))=0$, $\lambda=0$, and $\phi=x$, then
$\mathbb{F}_{\phi}=\mathbb{F}_{p}$ and for every $i=0,\dots,l$,
$c_{i}=\overline{{a_{s+i}}}\ \mbox{\rm(mod }{p})$. Thus this notion of
residual coefficient generalizes the reduction modulo a maximal ideal and
$f_{S}(y)\in\mathbb{F}_{p}[y]$ coincides with the reduction of $f(x)$ modulo
the maximal ideal $(p)$.

Let $N_{\phi}^{+}(f)=S_{1}+\dots+S_{r}$ be the principal $\phi$-Newton polygon
of $f$ with respect to $p$. We say that $f$ is a $\phi$-regular polynomial
with respect to $p$, if for every $i=1,\dots,r$, $f_{S_{i}}(y)$ is square free
in $\mathbb{F}_{\phi}[y]$. We say that $f$ is a $p$-regular polynomial if $f$
is a $\phi_{i}$-regular polynomial with respect to $p$ for every
$i=1,\dots,t$, for some monic polynomials $\phi_{1},\dots,\phi_{t}$ with
$\overline{f(x)}=\prod_{i=1}^{t}\overline{\phi_{i}}^{l_{i}}$ is the
factorization of $\overline{f(x)}$ in $\mathbb{F}_{p}[x]$.

The theorem of Ore plays a fundamental key for proving our main Theorems:
Let $\phi\in\mathbb{Z}[x]$ be a monic polynomial, with $\overline{\phi(x)}$ is
irreducible in $\mathbb{F}_{p}[x]$. As defined in [11, Def. 1.3], the
$\phi$-index of $f(x)$, denoted by $ind_{\phi}(f)$, is deg$(\phi)$ times the
number of points with natural integer coordinates that lie below or on the
polygon $N_{\phi}^{+}(f)$, strictly above the horizontal axis, and strictly
beyond the vertical axis (see Figure 1).

$0$$S_{1}$$S_{2}$$S_{3}$ Figure 1.  $N_{\phi}^{+}(f)$.

Now assume that $\overline{f(x)}=\prod_{i=1}^{t}\overline{\phi_{i}}^{l_{i}}$
is the factorization of $\overline{f(x)}$ in $\mathbb{F}_{p}[x]$, where every
$\phi_{i}\in\mathbb{Z}[x]$ is monic, $\overline{\phi_{i}(x)}$ is irreducible
in $\mathbb{F}_{p}[x]$, $\overline{\phi_{i}(x)}$ and $\overline{\phi_{j}(x)}$
are coprime when $i\neq j=1,\dots,t$. For every $i=1,\dots,t$, let
$N_{\phi_{i}}^{+}(f)=S_{i1}+\dots+S_{ir_{i}}$ be the principal part of the
$\phi_{i}$-Newton polygon of $f$ with respect to $p$. For every
$j=1,\dots,r_{i}$, let
$f_{S_{ij}}(y)=\prod_{s=1}^{s_{ij}}\psi_{ijk}^{a_{ijs}}(y)$ be the
factorization of $f_{S_{ij}}(y)$ in $\mathbb{F}_{\phi_{i}}[y]$. Then we have
the following theorem of Ore (see [11, Theorem 1.7 and Theorem 1.9], [10,
Theorem 3.9], and [21]):

######  Theorem 3.1.

(Theorem of Ore)

  1. (1)

$\nu_{p}(ind(f))=\nu_{p}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))\geq\sum_{i=1}^{t}ind_{\phi_{i}}(f)$
and equality holds if $f(x)$ is $p$-regular; every $a_{i,j,s}=1$.

  2. (2)

If every $a_{i,j,s}=1$, then

$p\mathbb{Z}_{K}=\prod_{i=1}^{r}\prod_{j=1}^{r_{i}}\prod_{s=1}^{s_{ij}}\mathfrak{p}^{e_{ij}}_{i,j,s},$

where $e_{ij}$ is the ramification index of the side $S_{ij}$ and
$f_{ijs}=\mbox{deg}(\phi_{i})\times\mbox{deg}(\psi_{ijs})$ is the residue
degree of $\mathfrak{p}_{ijs}$ over $p$.

######  Corollary 3.2.

Under the hypothesis of Theorem 3.1, if for every $i=1,\dots,t$, $l_{i}=1$ or
$N_{\phi_{i}}(f)=S_{i}$ has a single side of height $1$, then
$\nu_{p}(ind(f))=0$.

The following lemma allows to evaluate the $p$-adic valuation of the binomial
coefficient $\binom{p^{r}}{j}$.

######  Lemma 3.3.

Let $p$ be a rational prime integer and $r$ be a positive integer. Then
$\nu_{p}\left(\binom{p^{r}}{j}\right)=r-\nu_{p}(j)$ for any integer
$j=1,\dots,p^{r}-1$.

###### Proof.

Since for any natural number $m$, $\nu_{p}(m!)=\sum_{t=1}^{m}\nu_{p}(t)$, we
have
$\nu_{p}\left(\binom{p^{r}}{j}\right)=\nu_{p}(p^{r}!)-\nu_{p}((p^{r}-j)!)-\nu_{p}(j!)=\sum_{t=1}^{p^{r}}\nu_{p}(t)-\sum_{t=1}^{p^{r}-j}\nu_{p}(t)-\sum_{t=1}^{j}\nu_{p}(t)$.
Thus
$\nu_{p}\left(\binom{p^{r}}{j}\right)=\sum_{t=p^{r}-j+1}^{p^{r}}\nu_{p}(t)-\sum_{t=1}^{j}\nu_{p}(t)=\nu_{p}(p^{r})+\sum_{t=1}^{j-1}\nu_{p}(p^{r}-t)-\sum_{t=1}^{j}\nu_{p}(t)$.
As for every $t=1,\dots,j-1<p^{r}$, then $\nu_{p}(p^{r}-t)=\nu_{p}(t)$. Hence
$\nu_{p}\left(\binom{p^{r}}{j}\right)=r-\nu_{p}(j)$. ∎

###### Proof.

of Theorem 2.1.
Since $\triangle(f)=\mp p^{rp^{r}}m^{p^{r}-1}$, then by the formula linking
$\triangle(f)$, $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ and $d_{K}$,
$\mathbb{Z}[\alpha]$ is integrally closed if and only if $q$ does not divide
$(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$ for every rational prime integer $q$
dividing $pm$. Let $q$ be a rational prime dividing $m$, then
$f(x)\equiv\phi^{p^{r}}(\ mod\ q)$, where $\phi=x$. As $m$ is a square free
integer, the $\phi$-principal Newton polygon with respect to $\nu_{q}$,
$N_{\phi}^{+}(f)=S$ has a single side of degree $1$. Thus $f_{S}(y)$ is
irreducible over $\mathbb{F}_{\phi}\simeq\mathbb{F}_{q}$. By Theorem 3.1, we
get $\nu_{q}((\mathbb{Z}_{K}:\mathbb{Z}[\alpha]))=ind_{\phi}(f)=0$; $q$ does
not divide $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. Now, we deal with $p$, when
$p$ does not divide $m$. In this case, let $\nu=v_{p}(m^{p-1}-1)$. Since
$f(x)=(x-m+m)^{p^{r}}-m=(x-m)^{p^{r}}+\sum_{j=1}^{p^{r}-1}\binom{p^{r}}{j}m^{p^{r}-j}(x-m)^{j}+m^{p^{r}}-m$,
then by Lemma 3.3, $f(x)\equiv\phi^{p^{r}}(mod\ p)$, where $\phi=x-m$. If
$\nu=1$, then $N_{\phi}^{+}(f)=S$ has a single side of height $1$ and by
Theorem 3.1, $p$ does not divide $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. But if
$\nu\geq 2$, then $p$ divides $(\mathbb{Z}_{K}:\mathbb{Z}[\alpha])$. ∎

For the proof of Theorem 2.3, Corollary 2.4, and Theorem 2.5, we need the
following Lemma, which is an immediate consequence of a theorem of Kummer.

######  Lemma 3.4.

Let $p$ be rational prime integer and $K$ be a number field. For every
positive integer $f$, let $P_{f}$ be the number of distinct prime ideals of
$\mathbb{Z}_{K}$ lying above $p$ with residue degree $f$ and $N_{f}$ be the
number of monic irreducible polynomials of $\mathbb{F}_{p}[x]$ of degree $f$.
If $P_{f}>N_{f}$, then $p$ is a common index divisor of $K$.

###### Proof.

of Theorem 2.3.
Under the hypothesis of Theorem 2.3, $\overline{F(x)}=(x-m)^{p^{r}}$ in
$\mathbb{F}_{p}[x]$ and
$F(x)=(x-m+m)^{p^{r}}-m=(x-m)^{p^{r}}+\sum_{j=1}^{p^{r}-1}\binom{p^{r}}{j}m^{p^{r}-j}(x-m)^{j}+m^{p^{r}}-m$.
Let $\phi=x-m$ and $\nu=\nu_{p}(m^{p^{r}}-m)$. As
$m^{p^{r}}-m=(m^{p}-m)(m^{t}+\dots+1)$, where $t=p^{r}-1$, if $m\equiv 1\
\mbox{\rm(mod }{p^{\nu}})$, then $m^{t}+\dots+1\equiv-1\ \mbox{\rm(mod }{p})$,
and so, $\nu_{p}(m^{t}+\dots+1)=0$. Thus
$\nu_{p}(m^{p-1}-1)=\nu_{p}(m^{p}-m)=\nu_{p}(m^{p^{r}}-m)$. It follows that if
$\nu>p$ and $r\geq p$, then by Lemma 3.3,
$N_{\phi}^{+}(f)=S_{1}+\dots+S_{t-p+1}+\dots+S_{t}$ has $t$-distinct sides of
degree $1$ each one, with $t\geq p+1$. More precisely, $S_{t-i}$ has
$(p^{r-i-1},i+1)$ as the first point and $(p^{r-i},i)$ as the last point for
every $i=0,\dots,p-1$, and $S_{1}$ has $(0,\nu)$ as the first point and
$(1,r)$ or $(p^{r-\nu+1},\nu-1)$ as the last point according to $\nu>r$ or
$r\geq\nu$. It follows by Theorem 3.1 that, there are $t$ distinct prime
ideals of $\mathbb{Z}_{K}$ lying above $p$,with residue degree $1$ each one.
As $t=P_{1}>p=N_{1}$, by Lemma 3.4, $p$ is a common index divisor of $K$ and
$K$ is not monogenic. ∎

###### Proof.

of Theorem 2.5.
Let $m$ be an odd rational integer. Then $\overline{F(x)}=(x-1)^{p^{r}}$ in
$\mathbb{F}_{p}[x]$ and
$F(x)=(x-1+1)^{p^{r}}-m=(x-1)^{p^{r}}+\sum_{j=1}^{p^{r}-1}\binom{p^{r}}{j}(x-1)^{j}+1-m$.
Let $\phi=x-1$ and $\nu=\nu_{p}(1-m)$. It follows that:

  1. (1)

If $r=2$ and $\nu\geq 4$, then by Lemma 3.3,
$N_{\phi}^{+}(f)=S_{1}+S_{2}+S_{3}$ has $3$-distinct sides, with degree $1$
each side. Thus there are $3$-distinct prime ideals of $\mathbb{Z}_{K}$ lying
above $2$ with residue degree $1$ each one. So, by Lemma 3.4, $2$ is a common
index divisor of $K$.

  2. (2)

If $r\geq 3$ and $\nu\geq 5$, then by Lemma 3.3,
$N_{\phi}^{+}(f)=S_{1}+S_{2}+\dots+S_{t}$ has $t$-distinct sides, with $t\geq
4$ and degree $S_{i}$ equals $1$ for every $i=2,\dots,t$. Thus there are at
least $t-1$-distinct prime ideals of $\mathbb{Z}_{K}$ lying above $2$ with
residue degree $1$ each one. As $t-1\geq 3$, then by Lemma 3.4, $2$ is a
common index divisor of $K$.

∎

######  Examples.

  1. (1)

If
$\overline{m}\not\in\\{\overline{1},\overline{18},\overline{19},\overline{30},\overline{31},\overline{48}\\}\
\mbox{\rm(mod }{49})$, then the pure number field
$\mathbb{Q}(\sqrt[7^{r}]{m})$ is monogenic for every positive integer $r$.

  2. (2)

If $m^{p-1}\equiv 1\ \mbox{\rm(mod }{p^{p+1}})$, then for every natural
integer $r\geq p$ the pure field $\mathbb{Q}({\alpha})$ is not monogenic. In
particular, if $m^{4}\equiv 1\ \mbox{\rm(mod }{5^{6}})$, then the pure field
$\mathbb{Q}(\sqrt[5^{r}]{m})$ is not monogenic, for every natural integer
$r\geq 5$.

  3. (3)

If $m\equiv 1\ \mbox{\rm(mod }{32})$, then the octic number field
$\mathbb{Q}(\sqrt[8]{m})$ is not monogenic .

  4. (4)

More general, if $m\equiv 1\ \mbox{\rm(mod }{32})$, then for every natural
integer $r\geq 2$, $\mathbb{Q}({\alpha})$ is not monogenic, where $\alpha$ is
a complex root of $x^{2^{r}}-m$.

  5. (5)

If $\nu_{3}(m^{2}-1)\geq 4$, the field $\mathbb{Q}(\sqrt[27]{m})$ is not
monogenic. ( see figure $2$ ).

  6. (6)

More general, if $m\equiv\mp 1\ \mbox{\rm(mod }{81})$, then the field
$\mathbb{Q}({\alpha})$ is not monogenic, where $\alpha$ is a complex root of
$x^{3^{r}}-m$ and $r\geq 3$ is a natural integer.

$0$$1$$3$$3^{2}$$3^{3}$$1$$2$$3$$v$$S_{1}$$S_{2}$$S_{3}$$S_{4}$ Figure 2.  The
$(x-m)$-Newton polygon for $F(x)=x^{27}-m$ at $p=3$ and $v\geq 4$.

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