# On the diameter of a super-order-commuting graph

Janko Bračič  Faculty of Natural Sciences and Engineering, University of
Ljubljana, Aškerčeva c. 12, SI-1000 Ljubljana, Slovenia<EMAIL_ADDRESS>lj.si and  Bojan Kuzma  1University of Primorska, Glagoljaška 8, SI-6000
Koper, Slovenia and 2IMFM, Jadranska 19, SI-1000 Ljubljana, Slovenia
<EMAIL_ADDRESS>
###### Abstract.

We answer a question about the diameter of an order-super-commuting graph on a
symmetric group by studying the number-theoretical concept of $d$-complete
sequences of primes in arithmetic progression.

###### Key words and phrases:

Symmetric group, Alternating group, Order relation, Super graphs, Commuting
graphs, Prime numbers, d-complete sequence

###### 2020 Mathematics Subject Classification:

Primary: 05C25, Secondary: 05C69, 11B13, 11A67, 11N13.

##  1\. Introduction

In a recent paper [5] the authors studied the properties of super graphs on
finite groups. These graphs are based on already well-studied commuting,
power, and enhanced power graphs (we refer to [3] for a nice survey and the
current development), but with an equivalence relation thrown in. More
precisely, if $\sim$ is an equivalent relation on a graph $\Gamma$, then
$\sim$-super-$\Gamma$ has the same vertex set as $\Gamma$ however, its edge
set is enlarged, whereby distinct $x,y$ form an edge if there exist $u\sim x$
and $v\sim y$ with either $u=v$ or $(u,v)\in E(\Gamma)$. In particular, with
the order relation on a finite group $\mathcal{G}$ (i.e., $x\sim y$ if $x$ and
$y$ have the same order), the order-super-commuting graph,
$\Delta^{o}(\mathcal{G})$, of a group $\mathcal{G}$ is a simple graph
$\Delta^{o}(\mathcal{G})$ with the vertex set equal to $\mathcal{G}$ and where
two disjoint vertices $x,y$ form an edge if there exists commuting $u,v$ with
$|u|=|x|$ and $|v|=|y|$ (here, we also allow $u=v$, so in particular, each
conjugacy class forms an induced complete graph). We should caution that in
[5] the central elements and in particular, the identity also belong to the
vertex set of the commuting graph, but we do not allow the loops. Notice that
this contrasts with a similar definition in some of the existing literature
[1, 7], where the central elements are removed. It was shown in [5] that, for
$n\geq 4$, the identity, that is, the center of the group, is the only
dominant vertex of $\Delta^{o}(\mathcal{S}_{n})$, the order-super-commuting
graph of the symmetric group $\mathcal{S}_{n}$ on $n$ elements. By deleting
all the dominant vertices, one obtains the reduced graph,
$\Delta^{o}(\mathcal{S}_{n})^{\ast}$. This is connected if and only if neither
$n-1$ nor $n$ is a prime number; moreover, if it is disconnected, then it has
exactly two components, and if it is connected, then its diameter is bounded
above by $3$, see [5, Proposition 4.9 and Theorem 4.11]. Whether its diameter
is $3$ or smaller was not determined for all values of $n\geq 4$, but it was
shown [5, Proposition 4.12] that the following holds.

######  Proposition 1.1.

Let $n\geq 4$. If neither $n$ nor $n-1$ is a prime number, then
$\operatorname{diam}\Delta^{o}(\mathcal{S}_{n})^{\ast}=3$ if and only if there
exist nonempty disjoint subsets $\mathscr{T}_{1},\mathscr{T}_{2}$ consisting
of primes smaller or equal to $n$, such that, for some positive integers
$\alpha_{p}$ and $\beta_{q}$ we have

$M^{\alpha}_{\mathscr{T}_{1}}:=\sum_{p\in\mathscr{T}_{1}}p^{\alpha_{p}}\in\\{n-1,n\\},\qquad
M^{\beta}_{\mathscr{T}_{2}}:=\sum_{q\in\mathscr{T}_{2}}q^{\beta_{q}}\leq n,$

and $p+M^{\beta}_{\mathscr{T}_{2}}>n$, for every $p\in\mathscr{T}_{1}$.∎

With this proposition, the authors managed to prove that the diameter is $3$
if $n$ or $n-1$ is a nontrivial power of a prime or a sum of two prime powers,
with both primes distinct and greater or equal to $5$. The latter provided
that the strong Goldbach conjecture is true, immediately yields that
$\operatorname{diam}\Delta^{o}(\mathcal{S}_{n})^{\ast}=3$, for every even
integer $n\geq 4$. This paper aims to give a complete solution, without
relying on the strong Goldbach conjecture and proves that
$\operatorname{diam}\Delta^{o}(\mathcal{S}_{n})^{\ast}=3$ for every $n\geq 4$
such that neither $n$ nor $n-1$ is a prime number. Our main ingredient is the
fact that the sequences consisting of primes congruent to $\pm 1$ modulo $4$
are complete (see Theorem 2.5 and its consequences).

An infinite sequence of distinct positive integers
$\\{a_{n};\;n\in\mathbb{N}\\}$ is called complete if every sufficiently large
positive integer is a sum of distinct $a_{i}$ (sometimes such sequences are
called weakly complete, while the term complete sequence is reserved for the
case when every integer is a sum of distinct $a_{i}$). Erdős and Lewin [6]
call a complete sequence $d$-complete if every sufficiently large integer is a
sum of distinct $a_{i}$ such that no one divides the other. In [6], there are
given several examples of $d$-complete sequences. For instance, it is proved
that, for positive integers $p$ and $q$, the sequence
$\\{p^{a}q^{b};\;a,b\in\mathbb{N}\\}$ is $d$-complete if and only if
$\\{p,q\\}=\\{2,3\\}$. Bruckman [2] has proved that the sequence
$\mathscr{P}=\\{2,3,5,\ldots\\}$ of all prime numbers is $d$-complete. We
adjust Bruckman’s proof and show that the sequences of all prime numbers
congruent to $1$, respectively to $3$, modulo $4$ are $d$-complete.

##  2\. Results

###  2.1. Generating polynomials

Let $1\leq q_{1}<q_{2}<\dots$ be a sequence of integers. For every
$n\in\mathbb{N}$, let

$f_{n}(x)=\bigl{(}1+x^{q_{1}}\bigr{)}\cdots\bigl{(}1+x^{q_{n}}\bigr{)}.$

It is clear that $f_{n}(x)$ is a polynomial of degree
$S_{n}=q_{1}+\dots+q_{n}$. Denote the coefficient of $f_{n}(x)$ at power
$x^{m}$ by $\gamma_{m}(n)$; we also let $\gamma_{m}(n)=0$ if $m\geq S_{n}+1$.
Then

$f_{n}(x)=\sum_{m=0}^{S_{n}}\gamma_{m}(n)x^{m}.$

Since

$\begin{split}\sum_{m=0}^{S_{n+1}}\gamma_{m}(n+1)x^{m}&=f_{n+1}(x)=f_{n}(x)\cdot(1+x^{q_{n+1}})=\biggl{(}\sum_{m=0}^{S_{n}}\gamma_{m}(n)x^{m}\biggr{)}\bigl{(}1+x^{q_{n+1}}\bigr{)}\\\
&=\sum_{m=0}^{S_{n}}\gamma_{m}(n)x^{m}+\sum_{m=0}^{S_{n}}\gamma_{m}(n)x^{m+q_{n+1}}\end{split}$

the comparison of the coefficients gives

(2.1) $\gamma_{m}(n+1)=\left\\{\begin{array}[]{cl}\gamma_{m}(n),&\quad 0\leq
m<q_{n+1};\\\ \gamma_{m}(n)+\gamma_{m-q_{n+1}}(n),&\quad q_{n+1}\leq m\leq
S_{n};\\\ \gamma_{m-q_{n+1}}(n),&\quad S_{n}<m\leq S_{n+1}.\end{array}\right.$

It follows from (2.1) that $\gamma_{m}(n+1)\geq\gamma_{m}(n)\geq 0$. On the
other hand, let $m\geq 0$ be arbitrary but fixed. Let $n\in\mathbb{N}$ be such
that $m<q_{n+1}$. Then, by (2.1), $\gamma_{m}(n+1)=\gamma_{m}(n)$. Since
$m<q_{n+1}<q_{n+2}$ we have $\gamma_{m}(n+2)=\gamma_{m}(n+1)$ and therefore
$\gamma_{m}(n+2)=\gamma_{m}(n)$. By induction, $\gamma_{m}(k)=\gamma_{m}(n)$,
for all $k>n$. Thus, we may define
$\Gamma_{m}=\max\\{\gamma_{m}(n);\;n\in\mathbb{N}\\}$. Note that
$\Gamma_{m}>0$ if and only if there exist $n$ such that $\gamma_{m}(n)>0$
which is equivalent to the fact that there exist distinct sequence members
$q_{j_{1}},\ldots,q_{j_{\ell}}$ where $\ell\geq 1$, such that
$m=q_{j_{1}}+\cdots+q_{j_{\ell}}$.

######  Theorem 2.1.

Let $1\leq q_{1}<q_{2}<\cdots$ be a sequence of integers with partial sums
$S_{n}=q_{1}+\dots+q_{n}$. Consider

$f_{n}(x)=(1+x^{q_{1}})\cdots(1+x^{q_{n}})=\sum_{m=0}^{S_{n}}\gamma_{m}(n)x^{m}.$

If there exist positive integers $k_{0}$ and $n_{0}$ such that

$2k_{0}+q_{n+1}\leq S_{n},\quad\text{for}\;\;n\geq
n_{0},\qquad\text{and}\qquad\gamma_{m}(n_{0})\geq
1,\quad\text{for}\;\;k_{0}\leq m\leq S_{n_{0}}-k_{0},$

then every integer $n\geq k_{0}$ is a sum of (one or more) different members
of $\\{q_{1},q_{2},\dots\\}$.

###### Proof.

We will prove by induction that, for $n\geq n_{0}$, we have

$\gamma_{m}(n)\geq 1\quad\text{if}\;\;k_{0}\leq m\leq S_{n}-k_{0}.$

It is obvious that this will imply $\Gamma_{m}\geq 1$, for all $m\geq k_{0}$,
and the statement will follow.

Let $\mathscr{N}\subseteq\mathbb{N}$ denote the set of all integers $n\geq
n_{0}$ such that $\gamma_{m}(n)\geq 1$ if $k_{0}\leq m\leq S_{n}-k_{0}$. By
the hypothesis, $n_{0}\in\mathscr{N}$. Assume that $n\in\mathscr{N}$. If
$k_{0}\leq m\leq S_{n}-k_{0}$, then $\gamma_{m}(n+1)\geq\gamma_{m}(n)\geq 1$,
by the inductive hypothesis. Similarly, if $m$ is such that $k_{0}\leq
m-q_{n+1}\leq S_{n}-k_{0}$, then $\gamma_{m-q_{n+1}}(n)\geq 1$, again by the
inductive hypothesis, and therefore $\gamma_{m}(n+1)\geq 1$, by (2.1).
Clearly, $k_{0}\leq m-q_{n+1}\leq S_{n}-k_{0}$ is equivalent to
$k_{0}+q_{n+1}\leq m\leq S_{n}+q_{n+1}-k_{0}=S_{n+1}-k_{0}$. Also, by the
assumptions, $k_{0}+q_{n+1}\leq S_{n}-k_{0}$ so that the intersection of the
intervals $[k_{0},S_{n}-k_{0}]$ and $[k_{0}+q_{n+1},S_{n+1}-k_{0}]$ is
nonempty. We conclude that $\gamma_{m}(n+1)\geq 1$, for all $m$ such that
$k_{0}\leq m\leq S_{n+1}-k_{0}$. Hence, $n+1\in\mathscr{N}$. ∎

We next show that there always exists $k_{0}$ which satisfies the first
condition in Theorem 2.1, provided that the sequence of integers grows at most
exponentially.

######  Lemma 2.2.

Let $1\leq q_{1}<q_{2}<\dots$ be a sequence of integers satisfying
$q_{n+1}<2q_{n}$ and let $k_{0}\geq 0$ be a given integer. Then,
$S_{n}-q_{n+1}\geq 2k_{0}$, for every $n\geq q_{1}+2k_{0}+1$.

###### Proof.

Notice first that $q_{i}<q_{i+1}\leq 2q_{i}-1$, so that $q_{i}-q_{i+1}\geq
q_{i}-(2q_{i}-1)=-q_{i}+1$. Then, proceeding backward, we get

$\begin{split}S_{i}&-q_{i+1}=q_{1}+q_{2}+\cdots+(q_{i}-q_{i+1})\geq
q_{1}+q_{2}+\cdots+(q_{i-1}-q_{i})+1\geq\\\ &\geq
q_{1}+q_{2}+\cdots+(q_{i-2}-q_{i-1})+1+1\geq\dots\geq q_{1}-q_{2}+(i-2)\geq-
q_{1}+(i-1).\end{split}$

Hence, with $i=q_{1}+2k_{0}+1$ we get that $S_{i}-q_{i+1}\geq 2k_{0}$. This
was the start of the induction. To finish the proof, let $\mathscr{M}$ denote
the set of all integers $n\geq q_{1}+2k_{0}+1$ such that the statement of the
lemma is valid. If $n\in\mathscr{M}$, then

$S_{n+1}-q_{n+2}=S_{n}-q_{n+1}+2q_{n+1}-q_{n+2}\geq S_{n}-q_{n+1}\geq 2k_{0},$

and therefore $n+1\in\mathscr{M}$. ∎

###  2.2. Prime numbers congruent $1$, respectively $3$, modulo $4$

Let $d$ be a positive integer and let $1\leq r<d$ be such that $\gcd(d,r)=1$.
The celebrated Dirichlet’s Theorem says that there are infinitely many prime
numbers congruent to $r$ modulo $d$. Let
$\mathscr{P}(d,r)=\\{q_{d,r}(1)<q_{d,r}(2)<q_{d,r}(3)<\cdots\\}$ be the
sequence of all prime numbers congruent to $r$ modulo $d$ and let $S_{d,r}(n)$
denote the sum of the first $n$ prime numbers in $\mathscr{P}(d,r)$. In what
follows, we are interested in $\mathscr{P}(4,1)$ and $\mathscr{P}(4,3)$.

By [4, Theorem 1], for every $x\geq 887$, sets
$(x,1.040x]\cap\mathscr{P}(4,1)$ and $(x,1.040x]\cap\mathscr{P}(4,3)$ are
nonempty, that is, there exist prime numbers congruent to $1$ and to $3$
modulo $4$ in the interval $(x,1.040x]$. Using this and a straightforward
verification of the tables of prime numbers congruent to $1$, respectively to
$3$, modulo $4$ gives the following lemma.

######  Lemma 2.3.

For $x\geq 7$, the interval $(x,2x]$ contains a prime number congruent to $1$
modulo $4$ as well as a prime number congruent to $3$ modulo $4$.∎

In the following table, we list the first $15$ prime numbers congruent to $1$,
respectively to $3$, modulo $4$, and their partial sums.

$n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$q_{4,1}(n)$ | 5 | 13 | 17 | 29 | 37 | 41 | 53 | 61 | 73 | 89 | 97 | 101 | 109 | 113 | 137
$S_{4,1}(n)$ | 5 | 18 | 35 | 64 | 101 | 142 | 195 | 256 | 329 | 418 | 515 | 616 | 725 | 838 | 975
$q_{4,3}(n)$ | 3 | 7 | 11 | 19 | 23 | 31 | 43 | 47 | 59 | 67 | 71 | 79 | 83 | 103 | 107
$S_{4,3}(n)$ | 3 | 10 | 21 | 40 | 63 | 94 | 137 | 184 | 243 | 310 | 381 | 460 | 543 | 656 | 753
Table 1.

######  Lemma 2.4.

(a) If $n\geq 10$, then $S_{4,1}(n)-q_{4,1}(n+1)\geq 244$.

(b) If $n\geq 8$, then $S_{4,3}(n)-q_{4,3}(n+1)\geq 112$.

###### Proof.

(a) Let $\mathscr{M}$ denote the set of all integers $n\geq 10$ such that the
statement (a) of the lemma is valid. The table shows that
$S_{4,1}(10)-q_{4,1}(10)=418-97=321$, so $10\in\mathscr{M}$. To prove the
inductive step, let $n\in\mathscr{M}$. Then, $n\geq 10>7$ so by Lemma 2.3,

$\displaystyle S_{4,1}(n+1)-q_{4,1}(n+2)$
$\displaystyle=S_{4,1}(n)-q_{4,1}(n+1)+2q_{4,1}(n+1)-q_{4,1}(n+2)$
$\displaystyle\geq S_{4,1}(n)-q_{4,1}(n+1)\geq 244,$

and therefore $n+1\in\mathscr{M}$.

(b) Let now $\mathscr{M}$ denote the set of all integers $n\geq 8$ such that
the statement (b) of the lemma is valid. The table shows that
$S_{4,3}(8)-q_{4,3}(8)=184-59=125$, so $8\in\mathscr{M}$. The rest proceeds as
above. ∎

######  Theorem 2.5.

(a) For every integer $m\geq 122$, there exist distinct prime numbers
$p_{j_{1}},\ldots,p_{j_{k}}$ $\in\mathscr{P}(4,1)$ ($k\geq 1$) such that
$m=p_{n_{j_{1}}}+\cdots+p_{j_{k}}$.

(b) For every integer $m\geq 56$, there exist distinct prime numbers
$q_{j_{1}},\ldots,q_{j_{\ell}}\in\mathscr{P}(4,3)$ ($\ell\geq 1$) such that
$m=q_{j_{1}}+\cdots+q_{j_{\ell}}$.

###### Proof.

(a) Let $(k_{0},n_{0})=(122,13)$. By Lemma 2.4, $S_{4,1}(n)-q_{4,1}(n+1)\geq
2k_{0}=244$ for all $n\geq n_{0}$. Also, a direct, though tedious, computation
shows that the generating polynomial

$\prod_{i=1}^{13}(1+x^{q_{4,1}(i)})=(1+x^{5})(1+x^{13})\cdots(1+x^{101})(1+x^{109})=\sum_{i=0}^{S_{4,1}(13)}\gamma_{m}(13)x^{m}$

satisfies $\gamma_{m}(13)\geq 1$ for all $m$ in interval
$[122,603]=[122,725-122]=[k_{0},S_{4,1}(n_{0})-k_{0}]$ (we remark that
$\gamma_{121}(13)=\gamma_{604}(13)=0$). The rest follows from Theorem 2.1.

(b) Let now $(k_{0},n_{0})=(56,12)$. By Lemma 2.4,
$S_{4,3}(n)-q_{4,3}(n+1)\geq 2k_{0}=112$ for all $n\geq n_{0}$. Also, a direct
computation shows that the generating polynomial

$\prod_{i=1}^{12}(1+x^{q_{4,3}(i)})=(1+x^{3})(1+x^{7})\cdots(1+x^{71})(1+x^{79})=\sum_{i=0}^{S_{4,3}(12)}\gamma_{m}(12)x^{m}$

satisfies $\gamma_{m}(12)\geq 1$ for all $m$ in interval
$[56,404]=[56,460-56]=[k_{0},S_{4,3}(n_{0})-k_{0}]$ (here,
$\gamma_{55}(12)=\gamma_{405}(12)=0$). The rest again follows from Theorem
2.1. ∎

######  Corollary 2.6.

For every integer $m\geq 122$, there exist $k,l\in\mathbb{N}$ and distinct
prime numbers $p_{1},\ldots,p_{k}\in\mathscr{P}(4,1)$ and
$q_{1},\ldots,q_{l}\in\mathscr{P}(4,3)$ such that
$m=p_{1}+\cdots+p_{k}=q_{1}+\cdots+q_{l}.$∎

Table 2 shows that in addition to the above corollary, every integer which
does not belong to the union
$[1,17]\cup[19,21]\cup[23,28]\cup[31,33]\cup[35,36]\cup[38,40]\cup[43,45]\cup[48,49]\cup[51,52]\cup[55,57]\cup\cup\\{60,62,65,68,69,77,80,81,85,93,121\\}$
can also be written as a sum of different primes from the class
$\mathscr{P}(4,1)$ as well as the sum of different primes from the class
$\mathscr{P}(4,3)$.

18=5+13=7+11 | 22=5+17=3+19 | 29=29=3+7+19
---|---|---
30=13+17=7+23 | 34=5+29=3+31 | 37=37=3+11+23
41=41=3+7+31 | 42=5+37=11+31 | 46=5+41=3+43
47=5+13+29=47 | 50=13+37=3+47 | 53=53=3+7+43
54=13+41=7+47 | 58=5+53=11+47 | 59=5+13+41=59
61=61=3+11+47 | 63=5+17+41=3+7+11+19+23 | 64=5+13+17+29=3+7+11+43
66=5+61=7+59 | 67=13+17+37=67 | 70=17+53=3+67
71=5+13+53=71 | 72=5+13+17+37=3+7+19+43 | 73=73=3+11+59
74=13+61=3+71 | 75=5+17+53=3+7+11+23+31 | 76=5+13+17+41=3+7+19+47
78=5+73=7+71 | 79=5+13+61=79 | 82=29+53=3+79
83=5+17+61=83 | 84=5+13+29+37=3+7+31+43 | 86=13+73=3+83
87=5+29+53=3+7+11+19+47 | 88=5+13+17+53=3+7+11+67 | 89=89=3+7+79
90=17+73=7+83 | 91=5+13+73=3+7+11+23+47 | 92=5+17+29+41=3+7+11+71
94=5+89=11+83 | 95=5+17+73=3+7+11+31+43 | 96=5+13+17+61=3+7+19+67
97=97=3+11+83 | 98=37+61=19+79 | 99=5+41+53=3+7+11+19+59
100=5+13+29+53=3+7+11+79 | 101=101=3+19+79 | 102=5+97=19+83
103=5+37+61=3+7+11+23+59 | 104=5+17+29+53=3+7+11+83 | 105=5+13+17+29+41=3+19+83
106=5+101=23+83 | 107=5+13+89=3+7+11+19+67 | 108=5+13+17+73=3+7+19+79
109=109=3+23+83 | 110=13+97=31+79 | 111=5+17+89=3+7+11+19+71
112=5+13+41+53=3+7+19+83 | 113=113=3+31+79 | 114=5+109=31+83
115=5+13+97=3+7+11+23+71 | 116=5+13+37+61=3+7+23+83 | 117=5+13+17+29+53=3+31+83
118=5+113=47+71 | 119=5+13+101=3+7+11+19+79 | 120=5+13+29+73=3+7+31+79
Table 2.

By using also other residuum classes of primes we can get additional integers
that can be expressed as a sum of different primes from two disjoint subsets
(Table 3).

16=3+13=5+11 | 19=19=3+5+11 | 20=3+17=7+13
---|---|---
23=23=3+7+13 | 24=5+19=7+17 | 26=3+23=7+19
28=5+23=11+17 | 31=31=3+5+23 | 32=3+29=13+19
33=3+7+23=5+11+17 | 35=3+13+19=5+7+23 | 36=5+31=7+29
38=7+31=3+5+11+19 | 39=3+5+31=7+13+19 | 40=3+37=11+29
43=43=3+11+29 | 44=3+41=7+37 | 45=3+11+31=5+17+23
48=5+43=7+41 | 49=3+5+41=7+11+31 | 51=3+5+43=7+13+31
52=5+47=11+41 | 55=3+5+47=7+11+37 | 56=13+43=19+37
57=3+7+47=5+11+41 | 60=13+47=17+43 | 62=19+43=3+5+7+47
65=3+19+43=5+13+47 | 68=31+37=3+5+13+47 | 69=3+19+47=5+23+41
77=3+31+43=7+23+47 | 80=37+43=3+5+31+41 | 81=3+31+47=11+29+41
85=5+37+43=7+31+47 | 93=3+43+47=23+29+41 | 121=31+43+47=3+11+29+37+41
Table 3.

Notice that Tables 2 and 2 together with Corollary 2.6 show that all integers
except those in $\\{1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,17,21,25,27\\}$ can be
written as a sum of distinct primes in at least two ways such that no prime
appears in two different sums.

######  Corollary 2.7.

Let $n\geq 4$. If $n$ and $n-1$ are not prime numbers, then
$\operatorname{diam}\Delta^{o}(\mathcal{S}_{n})^{\ast}=3$.

###### Proof.

By Corollary 2.6 and the tables above, we see that for every nonprime number
$n\geq 16$, except for $n=21,25,27$, there are disjoint subsets
$\mathscr{T}_{1}$ and $\mathscr{T}_{2}$ of the sequence of primes
$\mathscr{P}$, neither containing the prime $2$, such that
$n=\sum\limits_{p\in\mathscr{T}_{1}}p=\sum\limits_{q\in\mathscr{T}_{2}}q$. For
those integers, the claim follows directly from Proposition 1.1. The remaining
integers $n$, for which neither $n$ nor $n-1$ is a prime number, are
$n=9,10,15,21,25,27$. Except for $n=15,21$, they are all either powers of a
prime number or are immediate successors of powers of a prime number and the
claim follows from [5, Corollary 4.13] ($n=15$ is also considered there). If
$n=21$, then $n-1=20=3+17=7+13$ and Proposition 1.1 again is applicable. ∎

### Acknowledgment

The authors were partially supported by the Slovenian Research and Innovation
Agency through the research programs P2-0268, P1-0285, and research projects
N1-0210, N1-0296, and J1-50000.

## References

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  * [2] P. S. Bruckman, The Primes Are a D-Complete Sequence, Pi Mu Epsilon Journal, 10, no. 10 (1999), 769–771.
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# Entropy functions and functional equations

Eszter Gselmann  Institute of Mathematics
University of Debrecen
P. O. Box: 12.
Debrecen
H–4010
Hungary<EMAIL_ADDRESS>
###### Abstract.

The purpose of this note is to give the general solution of two functional
equations connected to the Shannon entropy and also to the Tsallis entropy. As
a result of this, we present the regular solution of these equations, as well.
Furthermore, we point out that the regularity assumptions used in previous
works can substantially be weakened.

###### Key words and phrases:

functional equation, entropy, Shannon entropy, Tsallis entropy

###### 1991 Mathematics Subject Classification:

Primary 39B22; Secondary 94A17.

##  1\. Introduction and preliminaries

Since the celebrated paper of Claude E. Shannon (see [10]) appeared, the
information theory has become an extensive branch of mathematics. Furthermore,
it is known that information measures can be characterized via functional
equations. Concerning this, the reader can consult the two basic monographs
Aczél–Daróczy [1] and Ebanks–Sahoo–Sander [4].

Although the characterization problem of information measures nearly comes to
the end, from time to time one can meet new functional equations from this
area. A possible explanation for this is that the Shannon entropy and also the
entropy of degree alpha (or Tsallis entropy) has been re–discovered by
physicists and engineers, see Daróczy [3] and Tsallis [12].

The aim of this note is to give the general solution of two functional
equations connected to the notion of the Shannon entropy and also that of the
Tsallis entropy. More precisely, in the second section we will firstly solve
the equation

$f(xy)+f((1-x)y)-f(y)=\left(f(x)+f(1-x)\right)y^{q},$

which is supposed to hold for the unknown function
$f:]0,1]\rightarrow\mathbb{R}$ for all $x\in]0,1[$ and $y\in]0,1]$, where
$q\in\mathbb{R}$ is a fixed parameter.

For the unknown function $f:]0,1]\rightarrow\mathbb{R}$ the equation

$f(xy)=\left(\frac{x^{\alpha}+x^{\beta}}{2}\right)f(y)+\left(\frac{y^{\alpha}+y^{\beta}}{2}\right)f(x)$

will also be solved which is assumed to hold for all $x,y\in]0,1]$, where
$\alpha,\beta\in\mathbb{R}$ are fixed parameters.

These two functional equations were solved in Sharma–Taneja [11] and also in
Furuichi [5] under the assumptions that the unknown function is _nonnegative_
and _differentiable_ and they called the solutions of these equations _entropy
functions_ – to this alludes the title of the present work. First we give the
general solution of these equations and then we will point out that the
regularity suppositions (that is, nonnegativity and differentiability) can
_essentially_ be weakened to get the same result as that of [5, 11].

In what follows some preliminary definitions and results will be listed,
mainly from the theory of functional equation, these results can also be found
in Kuczma [8].

######  Definition 1.1.

Let $I\subset\mathbb{R}$ and $\mathscr{A}=\left\\{(x,y)|x,y,x+y\in
I\right\\}.$ A function $a:I\rightarrow\mathbb{R}$ is called _additive on
$\mathscr{A}$_ if

(1.1) $a\left(x+y\right)=a\left(x\right)+a\left(y\right)$

holds for all pairs $(x,y)\in\mathscr{A}$.
Consider the set $\mathscr{I}=\left\\{(x,y)|x,y,xy\in I\right\\}.$ We say that
$\mu:I\rightarrow\mathbb{R}$ is _multiplicative on $\mathscr{I}$_ if the
functional equation

(1.2) $\mu\left(xy\right)=\mu\left(x\right)\mu\left(y\right)$

is fulfilled for all $(x,y)\in\mathscr{I}$.
A function $\ell:I\rightarrow\mathbb{R}$ is called _logarithmic on
$\mathscr{I}$_ if it satisfies the functional equation

(1.3) $\ell\left(xy\right)=\ell\left(x\right)+\ell\left(y\right)$

for all $(x,y)\in\mathscr{I}$.

Henceforth, for all $n\geq 2$ we define the set $D_{n}$ by

$D_{n}=\left\\{(x_{1},\ldots,x_{n})\in\mathbb{R}^{n}|x_{1},\ldots,x_{n},\sum_{i=1}^{n}x_{i}\in]0,1[\right\\}.$

As we wrote above, we will also determine the regular solutions of equations
$(\ast)$ and $(\ast\ast)$. To do this, the following regularity theorems will
be applied.

######  Lemma 1.2.

Let $a:]0,1[\rightarrow\mathbb{R}$ be an additive function on the set $D_{2}$
and assume that

  1. (i)

$a$ is bounded above or below on subset of $]0,1[$ that has positive Lebesgue
measure;

  2. (ii)

or $a$ is Lebesgue measurable.

Then there exists $c\in\mathbb{R}$ such that

$a(x)=cx$

holds for all $x\in]0,1[$.

######  Lemma 1.3.

Let $\ell:]0,1[\rightarrow\mathbb{R}$ be a logarithmic function on the set

$\widetilde{\mathscr{L}}=\left\\{(x,y)|x,y,xy\in]0,1[\right\\}$

and assume that

  1. (i)

$\ell$ is bounded above or below on subset of $]0,1[$ that has positive
Lebesgue measure;

  2. (ii)

or $\ell$ is Lebesgue measurable.

Then there exists $c\in\mathbb{R}$ such that

$\ell(x)=\ln(x)$

holds for all $x\in]0,1[$.

We also mention that in case $a:]0,1[\rightarrow\mathbb{R}$ is an additive
function on the set $D_{2}$, then it can be uniquely extended to function
$\widetilde{a}:\mathbb{R}\rightarrow\mathbb{R}$ which is additive on
$\mathbb{R}$ (cf. Kuczma [8]). For the sake of simplicity we will always bear
in mind this fact, and the extension of the function in question will always
be denoted by the same character.

The notion of derivations will also be utilized in the next section, see
Kuczma [8].

######  Definition 1.4.

An additive function $a:\mathbb{R}\rightarrow\mathbb{R}$ is termed to be a
_real derivation_ , if it also fulfills the equation

$d(xy)=xd(y)+yd(x)$

for all $x,y\in\mathbb{R}$.

From this definition immediately follows that every real derivation vanishes
at the rationals. Additionally, something more is true. Namely, every real
derivation is identically zero on the set of algebraic numbers (over the
rationals). Furthermore, if a real derivation is Lebesgue measurable or
bounded above or below on the set that has positive Lebesgue measure, then it
is identically zero. Therefore, it can be seen that the non–trivial real
derivations can be very irregular. Although it is surprising, there exists non
identically zero real derivation, see Theorem 14. 2. 2 in Kuczma [8].

The following lemma was proved in [6].

######  Lemma 1.5.

Suppose that the function $\varphi:]0,+\infty[\rightarrow\mathbb{R}$ is such
that

$\varphi(xy)=x\varphi(y)+y\varphi(x)\qquad\left(x,y\in]0,1[\right)$

and the function $g:]0,1[\rightarrow\mathbb{R}$ defined by

$g(x)=\varphi(x)+\varphi(1-x)\qquad\left(x\in]0,1[\right)$

is Lebesgue measurable or it is bounded (above and below) on a subset of
$]0,1[$ that has positive Lebesgue measure. Then there exist $c\in\mathbb{R}$
and a real derivation $d:\mathbb{R}\rightarrow\mathbb{R}$ such that

$\varphi(x)=cx\ln(x)+d(x)$

is fulfilled for any $x\in]0,1[$.

During the proof of our main theorem concerning equation $\left(\ast\right)$
we will apply a result concerning the so–called cocycle equation, see
Jessen–Karpf–Thorup [7]. In the proof however this equation (i.e., the cocycle
equation) will not be satisfied on the whole domain but only on a restricted
one. Therefore we will apply a result of Ng [9], in which the author solves
the cocycle equation on a restricted domain, see also Aczél–Ng [2] and
Ebanks–Sahoo–Sander [4].

######  Theorem 1.6 (Ng [9]).

Let $\mu:]0,1[\rightarrow\mathbb{R}$ be a given multiplicative function and
$G:D_{2}\rightarrow\mathbb{R}$ be a function. Then the general solution of the
system of functional equations

(1.4) $G(x,y)=G(y,x);\qquad\left((x,y)\in D_{2}\right)$ (1.5)
$G(x,y)+G(x+y,z)=G(y,z)+G(x,y+z);\qquad\left((x,y,z)\in D_{3}\right)$

and

(1.6) $G(tx,ty)=\mu(t)G(x,y)\qquad\left(t\in]0,1[,(x,y)\in D_{2}\right)$

is given by in case $\mu(x)=x$,

(1.7) $G(x,y)=\varphi(x)+\varphi(y)-\varphi(x+y),\qquad\left((x,y)\in
D_{2}\right)$

where $\varphi:]0,+\infty[\rightarrow\mathbb{R}$ is such that

$\varphi(xy)=y\varphi(x)+x\varphi(y),\qquad\left(x,y\in]0,+\infty[\right)$

otherwise there exists $c\in\mathbb{R}$ such that

$G(x,y)=c\left[\mu(x)+\mu(y)-\mu(x+y)\right]$

is fulfilled for all $(x,y)\in D_{2}$.

##  2\. Main results

In this section we will find the general solutions of equations
$\left(\ast\right)$ and $\left(\ast\ast\right)$. After this, the regular
solutions of these equations will be presented. Furthermore, it will be
pointed out that the regularity assumptions of [5] and [11] can be
substantially weakened. Moreover, in some cases these suppositions can even be
omitted.

######  Theorem 2.1.

Let $q\in\mathbb{R}$ be arbitrarily fixed, then the function
$f:]0,1]\rightarrow\mathbb{R}$ fulfills equation

(2.1) $f(xy)+f((1-x)y)-f(y)=\left(f(x)+f(1-x)\right)y^{q}$

for all $x\in]0,1[$ and $y\in]0,1]$, in case $q\neq 1$, if and only if, there
exist $c\in\mathbb{R}$ and an additive function
$a:\mathbb{R}\rightarrow\mathbb{R}$ such that

$f(x)=\left\\{\begin{array}[]{rcl}a(x)+cx^{q},&\text{ if }&x\in]0,1[\\\
0,&\text{ if }&x=1\end{array}\right.$

furthermore, in case $q=1$, if and only if there exists an additive function
$a:\mathbb{R}\rightarrow\mathbb{R}$ and a function
$\varphi:]0,+\infty[\rightarrow\mathbb{R}$ such that

$\varphi(xy)=x\varphi(y)+y\varphi(x)\qquad\left(x,y\in]0,+\infty[\right)$

and

$f(x)=\left\\{\begin{array}[]{rcl}a(x)+\varphi(x),&\text{ if }&x\in]0,1[\\\
0,&\text{ if }&x=1\end{array}\right.$

is fulfilled.

###### Proof.

Assume that the function $f:]0,1]\rightarrow\mathbb{R}$ fulfills equation
(2.1). With the substitution $y=1$ we immediately get that $f(1)=0$. Therefore
it is enough to restrict ourselves to the interval $]0,1[$. Let $(u,v)\in
D_{2}$ and let us replace in equation (2.1) $x$ by $\dfrac{u}{u+v}$ and $y$ by
$(u+v)$, respectively. In this case we obtain that

(2.2)
$f(u)+f(v)-f(u+v)=\left[f\left(\frac{u}{u+v}\right)+f\left(\frac{v}{u+v}\right)\right](u+v)^{q}$

holds for all $(u,v)\in D_{2}$.

Define the functions $\mathscr{C}_{f}$ and $\mathscr{R}_{f}$ on the set
$D_{2}$ by

$\mathscr{C}_{f}(u,v)=f(u)+f(v)-f(u+v)\qquad\left((u,v)\in D_{2}\right)$

and

$\mathscr{R}_{f}(u,v)=\left[f\left(\frac{u}{u+v}\right)+f\left(\frac{v}{u+v}\right)\right](u+v)^{q}.\qquad\left((u,v)\in
D_{2}\right)$

With this notations equation (2.2) yields that

$\mathscr{C}_{f}(u,v)=\mathscr{R}_{f}(u,v).\qquad\left((u,v)\in D_{2}\right)$

Let us observe that the function $\mathscr{R}_{f}$ is $q$–homogeneous. Indeed,
for all $t\in]0,1[$ and $(u,v)\in D_{2}$

$\mathscr{R}_{f}(tu,tv)=\left[f\left(\frac{tu}{tu+tv}\right)+f\left(\frac{tv}{tu+tv}\right)\right](tu+tv)^{q}\\\
=t^{q}\left[f\left(\frac{u}{u+v}\right)+f\left(\frac{v}{u+v}\right)\right](u+v)^{q}=t^{q}\mathscr{R}_{f}(u,v).$

This implies that the function $\mathscr{C}_{f}$ is also a $q$–homogeneous
function. Furthermore, the function $\mathscr{C}_{f}$ is symmetric and also
fulfills the cocycle equation. All in all, this means that the function
$\mathscr{C}_{f}$ satisfies equations (1.4), (1.5) and (1.6) with the
multiplicative function $\mu(t)=t^{q}$. Thus by Theorem 1.6, in case $q\neq 1$
there exists $c\in\mathbb{R}$ such that

(2.3)
$\mathscr{C}_{f}(x,y)=c\left[x^{q}+y^{q}-(x+y)^{q}\right],\qquad\left((x,y)\in
D_{2}\right)$

and in case $q=1$ there exists a function
$\varphi:]0,+\infty[\rightarrow\mathbb{R}$ such that

$\varphi(xy)=x\varphi(y)+y\varphi(x)\qquad\left(x,y\in]0,+\infty[\right)$

and

(2.4)
$\mathscr{C}_{f}(x,y)=\varphi(x)+\varphi(y)-\varphi(x+y)\qquad\left((x,y)\in
D_{2}\right)$

is satisfied.

Firstly, we deal with the case $q\neq 1$. Define the function
$\widetilde{f}:]0,1[\rightarrow\mathbb{R}$ by

$\widetilde{f}(x)=f(x)-cx^{q},\qquad\left(x\in]0,1[\right)$

then equation (2.3) yields that the function $\widetilde{f}$ is additive on
the set $D_{2}$. Regarding the function $f$ this shows that there exists an
additive function $a:\mathbb{R}\rightarrow\mathbb{R}$ such that

$f(x)=cx^{q}+a(x)$

holds for any $x\in]0,1[$. Since $f(1)=0$, function $f$ has the form which had
to be proved.

If $q=1$, then let us define the function
$\widetilde{f}:]0,1[\rightarrow\mathbb{R}$ by

$\widetilde{f}(x)=f(x)-\varphi(x).\qquad\left(x\in]0,1[\right)$

Equation (2.4) yields that the function $\widetilde{f}$ is an additive
function on $D_{2}$. Concerning the function $f$ from this we get that there
exists an additive function $a:\mathbb{R}\rightarrow\mathbb{R}$ such that

$f(x)=\varphi(x)+a(x)$

is fulfilled for any $x\in]0,1[$, where the function
$\varphi:]0,+\infty[\rightarrow\mathbb{R}$ satisfies

$\varphi(xy)=x\varphi(y)+x\varphi(y).\qquad\left(x,y\in]0,+\infty[\right)$

Since $f(1)=0$ has to hold, the function $f$ is of the form

$f(x)=\left\\{\begin{array}[]{rcl}a(x)+\varphi(x),&\text{ if }&x\in]0,1[\\\
0,&\text{ if }&x=1\end{array}\right.$

that had to be proved. The converse direction is an easy computation. ∎

The following corollary contains the regular solutions of equation (2.1).

######  Corollary 2.2.

Let $q\in\mathbb{R}$ be arbitrary and suppose that the function
$f:]0,1]\rightarrow\mathbb{R}$ satisfies equation (2.1) for all $x\in]0,1[$
and $y\in]0,1]$. If $q\neq 1$ assume further that one of the following is
true.

  1. (i)

$f$ is bounded above or below on a subset of $]0,1[$ that has positive
Lebesgue measure;

  2. (ii)

$f$ is Lebesgue measurable.

Then there exist $c,c^{\ast}\in\mathbb{R}$ such that

$f(x)=\left\\{\begin{array}[]{rcl}c^{\ast}x+cx^{q},&\text{ if }&x\in]0,1[\\\
0,&\text{ if }&x=1\end{array}\right.$

In case $q=1$ suppose additionally that one of the statements below hold.

  1. (i)

$f$ is bounded (above and below) on a subset of $]0,1[$ that has positive
Lebesgue measure;

  2. (ii)

$f$ is Lebesgue measurable.

Then there exist $c,c^{\ast}\in\mathbb{R}$ such that

$f(x)=\left\\{\begin{array}[]{rcl}cx\ln(x)+c^{\ast}x,&\text{ if }&x\in]0,1[\\\
0,&\text{ if }&x=1\end{array}\right.$

is fulfilled.

###### Proof.

Firstly, we investigate the case $q\neq 1$. From Theorem 2.1 we obtain that

$f(x)=a(x)+cx^{q}$

holds for all $x\in]0,1[$ and $f(1)=0$. If we rearrange this, it follows that

$a(x)=f(x)-cx^{q}.\qquad\left(x\in]0,1[\right)$

By our assumptions $f$ is bounded above or below on a subset of $]0,1[$ that
has positive Lebesgue measure, or $f$ is a Lebesgue measurable function.

This implies that the additive function $a$ fulfills condition (i) or (ii) of
Lemma 1.2. From this, we obtain that there exists a constant
$c^{\ast}\in\mathbb{R}$ such that $a(x)=c^{\ast}x$. This implies however that

$f(x)=c^{\ast}x+cx^{q}$

holds for all $x\in]0,1[$ and $f(1)=0$.

Secondly, we assume that $q=1$. From Theorem 2.1, we obtain that $f(1)=0$ and

$f(x)=\varphi(x)+a(x),\qquad\left(x\in]0,1[\right)$

where $\varphi:]0,+\infty[\rightarrow\mathbb{R}$ fulfills equation

$\varphi(xy)=x\varphi(y)+y\varphi(x)\qquad\left(x,y\in]0,+\infty[\right)$

and $a:\mathbb{R}\rightarrow\mathbb{R}$ is an additive function. Define the
function $g:]0,1[\rightarrow\mathbb{R}$ by

$g(x)=f(x)+f(1-x)-a(1).\qquad\left(x\in]0,1[\right)$

In this case the function $g$ is Lebesgue measurable or bounded on a subset of
$]0,1[$ that has positive Lebesgue measure. Furthermore,

$g(x)=f(x)+f(1-x)-a(1)=\varphi(x)+a(x)+\varphi(1-x)+a(1-x)-a(1)=\varphi(x)+\varphi(1-x),$

where we used that $a$ is an additive function. All in all, this implies that
the function $g$ satisfies the assumptions of Lemma 1.5. Thus there exist
$c\in\mathbb{R}$ and a real derivation $d:\mathbb{R}\rightarrow\mathbb{R}$
such that

$\varphi(x)=cx\ln(x)+d(x).\qquad\left(x\in]0,1[\right)$

Concerning the function $f$ this yields that

$f(x)=cx\ln(x)+d(x)+a(x),\qquad\left(x\in]0,1[\right)$

or if we rearrange this,

$f(x)-cx\ln(x)=d(x)+a(x).\qquad\left(x\in]0,1[\right)$

By our assumptions the function $f$ is bounded on a subset of $]0,1[$ that has
positive Lebesgue measure or it is Lebesgue measurable. Furthermore, the
function $d(x)+a(x)$ is a sum of two additive function, that is, this function
is also additive, which is bounded on a subset of $]0,1[$ with positive
Lebesgue measure or it is Lebesgue measurable. In view of Lemma 1.2, there
exists $c^{\ast}\in\mathbb{R}$ such that $d(x)+a(x)=c^{\ast}x$ holds for all
$x\in\mathbb{R}$. Thus

$f(x)=cx\ln(x)+c^{\ast}x$

holds for all $x\in]0,1f$ and $f(1)=0$. ∎

######  Remark.

Under the assumptions of the previous corollary, in case for the function
$\lim_{x\to 1-}f(x)$ exists and $\lim_{x\to 1-}f(x)=f(1)$, then in case $q\neq
1$,

$f(x)=c^{\ast}(x-x^{q})\qquad\left(]0,1]\right)$

is fulfilled with some $c^{\ast}\in\mathbb{R}$. Furthermore, in case $q=1$,

$f(x)=cx\ln(x)\qquad\left(x\in]0,1]\right)$

is satisfied with a certain $c\in\mathbb{R}$.

At this point of the paper we turn to deal with equation
$\left(\ast\ast\right)$. Before this, we present a more general equation. Thus
the solutions of the above mentioned functional equation will be showed as a
corollary of the following result. Additionally, the regular solutions of
$\left(\ast\ast\right)$ will be dealt with, as well.

The following lemma was proved by E. Vincze in 1962 for commutative groups.
Although $\left(]0,1],\cdot\right)$ is not a group, only a semigroup, we
remark that the method used in Satz 5 in Vincze [13] is appropriate for
commutative semigroups, as well.

######  Lemma 2.3.

Let $g:]0,1]\rightarrow\mathbb{R}$ be a given function, and
$f:]0,1]\rightarrow\mathbb{R}$ be such that

(2.5) $f(xy)=g(y)f(x)+g(x)f(y)$

holds for all $x,y\in]0,1]$.

If $g$ is a multiplicative function, then

$f(x)=g(x)\ell(x),\qquad\left(x\in]0,1]\right)$

where $\ell:]0,1]\rightarrow\mathbb{R}$ is a logarithmic function, and in case
$g$ is not a multiplicative function, that is, there exist
$t_{1},t_{2}\in]0,1]$ such that $g(t_{1}t_{2})\neq g(t_{1})g(t_{2})$, then

$f(x)=\frac{\left[g(t_{1}x)-g(t_{1})g(x)\right]f(t_{1})}{g(t_{1}t_{2})-g(t_{1})g(t_{2})}$

holds for all $x\in]0,1]$.

######  Corollary 2.4.

Let $\alpha,\beta\in\mathbb{R}$ be arbitrary, $f:]0,1]\rightarrow\mathbb{R}$
be a function for which

(2.6)
$f(xy)=\left(\frac{x^{\alpha}+x^{\beta}}{2}\right)f(y)+\left(\frac{y^{\alpha}+y^{\beta}}{2}\right)f(x)$

holds for any $x,y\in]0,1]$. Then we have the following two possibilities.

  1. (i)

if $\alpha\neq\beta$, then there exist $c\in\mathbb{R}$ such that

$f(x)=c\left(x^{\alpha}-x^{\beta}\right);\qquad\left(x\in]0,1]\right)$

  2. (ii)

if $\beta=\alpha$, then there exist a logarithmic function
$\ell:]0,1]\rightarrow\mathbb{R}$ such that

$f(x)=cx^{\alpha}\ell(x)$

holds for all $x\in]0,1]$.

###### Proof.

Firstly, let us suppose that $\alpha\neq\beta$ . In this case the function
$g:]0,1]\rightarrow\mathbb{R}$ defined by
$g(x)=\frac{1}{2}\left(x^{\alpha}+x^{\beta}\right)$ is not a multiplicative
function. Furthermore, with this notation, from (2.6) equation (2.5) follows.
Thus, by Lemma 2.3,

$f(x)=\frac{\left[g(t_{1}x)-g(t_{1})g(x)\right]f(t_{1})}{g(t_{1}t_{2})-g(t_{1})g(t_{2})}$

holds for all $x\in]0,1]$, where $t_{1},t_{2}\in]0,1]$ are arbitrarily fixed.
After using the form of the function $g$, we get that

$f(x)=\frac{f(t_{1})}{t_{2}^{\alpha}-t_{2}^{\beta}}\left(x^{\alpha}-x^{\beta}\right),\qquad\left(x\in]0,1]\right)$

that is, there exists a constant $c\in\mathbb{R}$ such that

$f(x)=c\left(x^{\alpha}-x^{\beta}\right)$

is satisfied for all $x\in]0,1]$.

Secondly, assume that $\alpha=\beta$. In this case the function
$g:]0,1]\rightarrow\mathbb{R}$ defined by $g(x)=x^{\alpha}$ is a
multiplicative function. Additionally, let us observe that with this notations
(2.6) becomes to equation (2.5). Therefore, due to Lemma 2.3, there exist
$c\in\mathbb{R}$ and a logarithmic function $\ell:]0,1]\rightarrow\mathbb{R}$
such that

$f(x)=cx^{\alpha}\ell(x)$

is fulfilled for all $x\in]0,1]$.

Finally, a facile computation shows the correctness of the converse direction.
∎

From this corollary we can effortlessly get the regular solutions of equation
(2.6). Let us observe that in case $\alpha\neq\beta$, the solutions of this
equation are regular already. Therefore in this case the regularity
assumptions (that is, the nonnegativity and the differentiability) are
superfluous in the papers [5] and [11]. Furthermore, if $\alpha=\beta$, then
the above mentioned regularity suppositions can be significantly weakened.
Namely, making use of Lemma 1.3 the following statement can be proved.

######  Corollary 2.5.

Let $\alpha\in\mathbb{R}$ be arbitrarily fixed and assume that the function
$f:]0,1]\rightarrow\mathbb{R}$ fulfills equation

(2.7) $f(xy)=y^{\alpha}f(x)+x^{\alpha}f(y)$

for all $x,y\in]0,1]$. Suppose further that one of the following statements is
true.

  1. (i)

$f$ is bounded above or below on a subset of $]0,1[$ that has positive
Lebesgue measure;

  2. (ii)

$f$ is Lebesgue measurable.

Then there exists $c\in\mathbb{R}$ such that

$f(x)=cx^{\alpha}\ln(x).\qquad\left(x\in\mathbb{R}\right)$

## Acknowledgments

The author is indebted to the anonymous referee. His/her suggestions improved
the manuscript significantly.
This research has been supported by the Hungarian Scientific Research Fund
(OTKA) Grant NK 814 02 and by the TÁMOP 4.2.1./B-09/1/KONV-2010-0007 project
implemented through the New Hungary Development Plan co-financed by the
European Social Fund and the European Regional Development Fund.

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