# Stability of the parametric fundamental equation of information for
nonpositive parameters

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

In this note we prove that the parametric fundamental equation of information
is stable in the sense of Hyers and Ulam provided that the parameter is
nonpositive. We also prove, as a corollary, that the system of equations that
defines the recursive and semi-symmetric information measures depending on a
nonpositive parameter is stable in a certain sense.

###### Key words and phrases:

Stability, fundamental equation of information, entropy of degree $\alpha$.

###### 1991 Mathematics Subject Classification:

39B82, 39B72

This research has been supported by the Hungarian Scientific Research Fund
(OTKA) Grants NK 68040 and K 62316.

##  1\. Introduction

The basic problem in the stability theory of functional equations is whether
an approximate solution of a functional equation or a system of functional
equations can be approximated by a solution of the equation or the system of
equations in question.

In this paper we prove that the parametric fundamental equation of information

(1.1)
$f(x)+(1-x)^{\alpha}f\left(\frac{y}{1-x}\right)=f(y)+(1-y)^{\alpha}f\left(\frac{x}{1-y}\right)$

is stable in the sense of Hyers and Ulam (see the expository papers Forti [5],
Ger [6], Moszner [10]), provided that $\alpha$ is nonpositive. Equation (1.1)
arises in a natural way in characterizing information measures based on the
properties of $\alpha$–recursivity and semi–symmetry (see Aczél–Daróczy [3]).
In the investigations (1.1) is supposed to hold on

$D=\left\\{(x,y)\in\mathbb{R}^{2}|x,y\in[0,1[,x+y\leq 1\right\\}$

with $f:[0,1]\rightarrow\mathbb{R}$ or only on the interior of $D$,

$D^{\circ}=\left\\{(x,y)\in\mathbb{R}^{2}|x,y,x+y\in]0,1[\right\\},$

with $f:]0,1[\rightarrow\mathbb{R}$ (see Aczél–Daróczy [3], Aczél [1], [2] and
their references). In [9] we proved that (1.1) is stable on $D$, moreover it
is superstable (see Forti [5]) if $\alpha>0$ and $\alpha\neq 1$. The question
of the stability of (1.1) in the exceptional case $\alpha=1$ on $D^{\circ}$
was raised by Székelyhidi [11]. The method we used in [9] can not be applied
neither in this case nor in the case $\alpha\leq 0$ neither on $D$ nor on
$D^{\circ}$. Finally, we should remark that the ideas we use in this paper to
prove the stability of (1.1) on $D^{\circ}$ as well as on $D$ do not work if
$\alpha>0$.

##  2\. The main result

First we prove the following.

######  Theorem 2.1.

Let $\alpha,\varepsilon\in\mathbb{R}$ be fixed, $\alpha\leq 0$ and
$\varepsilon\geq 0$. Suppose that the function $f:]0,1[\rightarrow\mathbb{R}$
satisfies the inequality

(2.1)
$\left|f(x)+(1-x)^{\alpha}f\left(\frac{y}{1-x}\right)-f(y)-(1-y)^{\alpha}f\left(\frac{x}{1-y}\right)\right|\leq\varepsilon$

for all $(x,y)\in D^{\circ}$. Then, in case $\alpha<0$ there exist
$a,b\in\mathbb{R}$ such that

(2.2) $\left|f(x)-\left[ax^{\alpha}+b(1-x)^{\alpha}-b\right]\right|\leq
15\varepsilon,\quad\left(x\in]0,1[\right)$

furthermore, in case $\alpha=0$, there exists a logarithmic function
$l:]0,1[\rightarrow\mathbb{R}$ and $c\in\mathbb{R}$ such that

(2.3) $\left|f(x)-\left[l(1-x)+c\right]\right|\leq
63\varepsilon.\quad\left(x\in]0,1[\right)$

###### Proof.

Define the function $F$ on $]0,+\infty[^{2}$ by

(2.4) $F(u,v)=(u+v)^{\alpha}f\left(\frac{v}{u+v}\right).$

Then

(2.5) $F(tu,tv)=t^{\alpha}F(u,v)\quad\left(t,u,v\in]0,+\infty[\right)$

and

(2.6) $f(x)=F(1-x,x),\quad\left(x\in]0,1[\right)$

furthermore, with the substitutions

$x=\frac{w}{u+v+w},\quad
y=\frac{v}{u+v+w}\quad\left(u,v,w\in]0,+\infty[\right)$

inequality (2.1) implies that

$\begin{array}[]{l}\left|f\left(\frac{w}{u+v+w}\right)+\frac{(u+v)^{\alpha}}{(u+v+w)^{\alpha}}f\left(\frac{v}{u+v}\right)\right.\\\
\left.-f\left(\frac{v}{u+v+w}\right)-\frac{(u+w)^{\alpha}}{(u+v+w)^{\alpha}}f\left(\frac{w}{u+w}\right)\right|\leq\varepsilon\end{array}$

whence, by (2.4)

(2.7)
$\left|F(u+v,w)+F(u,v)-F(u+w,v)-F(u,w)\right|\leq\varepsilon(u+v+w)^{\alpha}$

follows for all $u,v,w\in]0,+\infty[$.

In the next step we define the functions $g$ and $G$ on $]0,+\infty[$ and on
$]0,+\infty[^{2}$, respectively by

(2.8) $g(u)=F(u,1)-F(1,u)$

and

(2.9) $G(u,v)=F(u,v)+g(v).$

We will show that

(2.10) $\left|G(u,v)-G(v,u)\right|\leq
3\varepsilon(u+v+1)^{\alpha}.\quad\left(u,v\in]0,+\infty[\right)$

Indeed, with the substitution $w=1$, inequality (2.7) implies that

(2.11)
$\left|F(u+v,1)+F(u,v)-F(u+1,v)-F(u,1)\right|\leq\varepsilon(u+v+1)^{\alpha}.$

Interchanging $u$ and $v$, it follows from (2.11) that

$\left|-F(u+v,1)-F(v,u)+F(v+1,u)-F(v,1)\right|\leq\varepsilon(u+v+1)^{\alpha}.\quad\left(u,v\in]0,+\infty[\right)$

This inequality, together with (2.11) and the triangle inequality imply that

(2.12) $\left|F(u,v)-F(v,u)-F(u+1,v)-F(u,1)+F(v+1,u)+F(v,1)\right|\leq
2\varepsilon(u+v+1)^{\alpha}$

holds for all $u,v\in]0,+\infty[$. On the other hand, with $u=1$, we get from
(2.7) that

$\left|F(1+v,w)+F(1,v)-F(1+w,v)-F(1,w)\right|\leq\varepsilon(1+v+w)^{\alpha}.$

Replacing here $v$ by $u$ and $w$ by $v$, respectively, we have that

$\left|F(u+1,v)+F(1,u)-F(v+1,u)-F(1,v)\right|\leq\varepsilon(u+v+1)^{\alpha}.\quad\left(u,v\in]0,+\infty[\right)$

Again, the triangle inequality and the definitions (2.8) and (2.9), (2.12)
imply (2.10).

In what follows we will investigate the function $g$. At this point of the
proof we have to distinguish two cases.

In case $\alpha<0$ we will determine the function $g$ by proving that

(2.13) $g(u)=c(u^{\alpha}-1)\quad\left(u\in]0,+\infty[\right)$

with some $c\in\mathbb{R}$. Indeed, (2.10) implies that

$\left|G(tu,tv)-G(tv,tu)\right|\leq
3\varepsilon(tu+tv+1)^{\alpha},\quad\left(t,u,v\in]0,+\infty[\right)$

therefore by (2.5) and (2.9)

$\left|t^{\alpha}F(u,v)+g(tv)-t^{\alpha}F(v,u)-g(tu)\right|\leq
3\varepsilon(tu+tv+1)^{\alpha}\quad\left(t,u,v\in]0,+\infty[\right)$

or

$\left|F(u,v)-F(v,u)-t^{-\alpha}\left(g(tu)-g(tv)\right)\right|\leq
3\varepsilon(u+v+t^{-1})^{\alpha}\quad\left(t,u,v\in]0,+\infty[\right)$

whence

$\lim_{t\rightarrow
0}t^{-\alpha}\left(g(tu)-g(tv)\right)=F(u,v)-F(v,u)\quad\left(t,u,v\in]0,+\infty[\right)$

follows. Particularly, with $v=1$, by (2.8), we have that

(2.14) $g(u)=\lim_{t\rightarrow
0}t^{-\alpha}\left(g(tu)-g(t)\right).\quad\left(u\in]0,+\infty[\right)$

Let now $u,v\in]0,+\infty[$. Then, by (2.14), we obtain that

$\begin{array}[]{rcl}g(uv)&=&\lim_{t\rightarrow
0}t^{-\alpha}\left[g(tuv)-g(t)\right]\\\ &=&\lim_{t\rightarrow
0}\left[(tv)^{-\alpha}\left(g((tv)u)-g(tv)\right)v^{\alpha}+t^{-\alpha}(g(tv)-g(t))\right]\\\
&=&g(u)v^{\alpha}+g(v).\end{array}$

Therefore, $g(u)v^{\alpha}+g(v)=g(v)u^{\alpha}+g(u)$, that is,

$g(u)\left(v^{\alpha}-1\right)=g(v)\left(u^{\alpha}-1\right)\quad\left(u,v\in]0,+\infty[\right)$

which implies (2.13) with $c=g(2)\left(2^{\alpha}-1\right)^{-1}$.

Thus, by (2.6), (2.13), (2.9) and (2.10), we have that

(2.15)
$\begin{array}[]{l}\left|f(x)-c(1-x)^{\alpha}-\left(f(1-x)-cx^{\alpha}\right)\right|\\\
=\left|F(1-x,x)+cx^{\alpha}-\left(F(x,1-x)+c(1-x)^{\alpha}\right)\right|\\\
=\left|G(1-x,x)-G(x,1-x)\right|\leq 3\cdot 2^{\alpha}\varepsilon\end{array}$

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

In the next step we define the functions $f_{0}$ and $F_{0}$ on $]0,1[$ and on
$]0,1[^{2}$ by

(2.16) $f_{0}(x)=f(x)-c\left[(1-x)^{\alpha}-1\right]$

and

(2.17)
$F_{0}(p,q)=f_{0}(p)+p^{\alpha}f_{0}(q)-f_{0}(pq)-(1-pq)^{\alpha}f_{0}\left(\frac{1-p}{1-pq}\right),$

respectively. Then (2.1) and (2.15) imply that

(2.18)
$\left|f_{0}(x)+(1-x)^{\alpha}f_{0}\left(\frac{y}{1-x}\right)-f_{0}(y)-(1-y)^{\alpha}f_{0}\left(\frac{x}{1-y}\right)\right|\leq\varepsilon$

for all $(x,y)\in D^{\circ}$ and

(2.19) $\left|f_{0}(x)-f_{0}(1-x)\right|\leq 3\cdot
2^{\alpha}\varepsilon.\quad\left(x\in]0,1[\right)$

Furthermore, with the substitutions $x=1-p$, $y=pq$ ($p,q\in]0,1[$), (2.18)
implies that

(2.20)
$\left|f_{0}(1-p)+p^{\alpha}f_{0}(q)-f_{0}(pq)-(1-pq)^{\alpha}f_{0}\left(\frac{1-p}{1-pq}\right)\right|\leq\varepsilon$

holds for all $p,q\in]0,1[$. Therefore, due to (2.19) and the triangle
inequality, (2.18) implies that

(2.21) $\left|F_{0}(p,q)\right|\leq\left(1+3\cdot
2^{\alpha}\right)\varepsilon.\quad\left(p,q\in]0,1[\right)$

It can easily be checked that

$\begin{array}[]{l}f_{0}(p)\left[q^{\alpha}+(1-q)^{\alpha}-1\right]-f_{0}(q)\left[p^{\alpha}+(1-p)^{\alpha}-1\right]\\\
=F_{0}(q,p)-F_{0}(p,q)+(1-pq)^{\alpha}\left[F_{0}\left(\frac{1-q}{1-pq},p\right)+f_{0}\left(1-\frac{1-p}{1-pq}\right)-f_{0}\left(\frac{1-p}{1-pq}\right)\right]\end{array}$

holds for all $p,q\in]0,1[$. Thus, by (2.21) and (2.19) we get that

$\begin{array}[]{l}\left|f_{0}(p)\left[q^{\alpha}+(1-q)^{\alpha}-1\right]-f_{0}(q)\left[p^{\alpha}+(1-p)^{\alpha}-1\right]\right|\\\
\leq 2(1+3\cdot 2^{\alpha})\varepsilon+(1-pq)^{\alpha}\left[(1+3\cdot
2^{\alpha})\varepsilon+3\cdot 2^{\alpha}\varepsilon\right],\end{array}$

that is,

$\begin{array}[]{l}\left|f_{0}(p)-\frac{f_{0}(q)}{q^{\alpha}+(1-q)^{\alpha}-1}\left[p^{\alpha}+(1-p)^{\alpha}-1\right]\right|\\\
\leq\frac{2(1+3\cdot 2^{\alpha})+(1-pq)^{\alpha}(1+6\cdot
2^{\alpha})}{q^{\alpha}+(1-q)^{\alpha}-1}\varepsilon.\end{array}\quad\left(p,q\in]0,1[\right)$

Taking into consideration (2.16), with $q=\frac{1}{2}$ with the definitions
$a=f_{0}\left(\frac{1}{2}\right)\left(2^{1-\alpha}-1\right)^{-1}$, $b=a+c$,
this inequality implies that

$\left|f(x)-\left[ax^{\alpha}+b(1-x)^{\alpha}-b\right]\right|\leq\frac{8+6\cdot
2^{\alpha}+2^{-\alpha}}{2^{1-\alpha}-1}\varepsilon.\quad\left(x\in]0,1[\right)$

Since

$\sup_{\alpha<0}\frac{8+6\cdot 2^{\alpha}+2^{-\alpha}}{2^{1-\alpha}-1}=15$

we get (2.2).

In case $\alpha=0$ we will show that there exists a logarithmic function
$l:]0,+\infty[\rightarrow\mathbb{R}$ such that

$\left|g(u)-l(u)\right|\leq 6\varepsilon$

for all $u\in]0,+\infty[$. Indeed, (2.10) yields in this case that

$\left|G(u,v)-G(v,u)\right|\leq
3\varepsilon.\quad\left(u,v\in]0,+\infty[\right)$

Due to (2.5) and (2.9) we obtain that

$\begin{array}[]{rcl}G(tu,tv)&=&F(tu,tv)+g(tv)\\\ &=&F(u,v)+g(tv)\\\
&=&G(u,v)-g(v)+g(tv)\end{array}$

that is,

$G(tu,tv)-G(u,v)=g(tv)-g(v),\quad\left(t,u,v\in]0,+\infty[\right)$

therefore

(2.22) $\begin{array}[]{l}\left|g(tv)-g(v)+g(u)-g(tu)\right|\\\
=\left|G(tu,tv)-G(u,v)-G(tv,tu)+G(v,u)\right|\\\
\leq\left|G(tu,tv)-G(tv,tu)\right|+\left|G(v,u)-G(u,v)\right|\leq
6\varepsilon\end{array}$

for all $t,u,v\in]0,+\infty[$. Now (2.22) with the substitution $u=1$ implies
that

$\left|g(tv)-g(v)-g(t)\right|\leq 6\varepsilon$

holds for all $t,v\in]0,+\infty[$, since obviously $g(1)=0$. This means that
the function $g$ is approximately logarithmic on $]0,+\infty[$. Thus (see e.g.
Forti [5]) there exists a logarithmic function
$l:]0,+\infty[\rightarrow\mathbb{R}$ such that

$\left|g(u)-l(u)\right|\leq 6\varepsilon$

holds for all $u\in]0,+\infty[$.

Furthermore,

(2.23) $\begin{array}[]{l}\left|f(x)-l(1-x)-\left(f(1-x)-l(x)\right)\right|\\\
=\left|F(1-x,x)-l(1-x)-F(x,1-x)+l(x)\right|\\\
=\left|F(1-x,x)+g(x)-g(x)-l(1-x)\right.\\\
-F(x,1-x)+g(1-x)-g(1-x)+l(x)\left.\right|\\\
\leq\left|F(1-x,x)+g(x)-\left(F(x,1-x)+g(1-x)\right)\right|\\\
+\left|g(1-x)-l(1-x)\right|+\left|l(x)-g(x)\right|\\\
=\left|G(1-x,x)-G(x,1-x)\right|\\\
+\left|g(1-x)-l(1-x)\right|+\left|l(x)-g(x)\right|\\\ \leq
3\varepsilon+6\varepsilon+6\varepsilon=15\varepsilon.\end{array}$

As in the first part of the proof define the functions $f_{0}$ and $F_{0}$ on
$]0,1[$ and on $]0,1[^{2}$, respectively, by

$f_{0}(x)=f(x)-l(1-x)$

and

$F_{0}(p,q)=f_{0}(p)+f_{0}(q)-f_{0}(pq)-f\left(\frac{1-p}{1-pq}\right).$

Due to (2.23)

(2.24) $\left|f_{0}(x)-f_{0}(1-x)\right|\leq 15\varepsilon$

holds for all $x\in]0,1[$. Furthermore, inequality (2.1) implies, with the
substitutions $x=1-p$, $y=pq$ ($p,q\in]0,1[$), that

(2.25)
$\left|f_{0}(1-p)+f_{0}(q)-f_{0}(pq)-f_{0}\left(\frac{1-p}{1-pq}\right)\right|\leq\varepsilon$

holds for all $p,q\in]0,1[$. Inequalities (2.24) and (2.25) and the triangle
inequality imply that

(2.26) $\left|F_{0}(p,q)\right|\leq 16\varepsilon$

for all $p,q\in]0,1[$. An easy calculation shows that

$f_{0}(p)-f_{0}(q)=F_{0}(q,p)-F_{0}(p,q)+F_{0}\left(\frac{1-p}{1-pq},p\right)+f_{0}\left(1-\frac{1-p}{1-pq}\right)-f_{0}\left(\frac{1-p}{1-pq}\right)$

therefore,

(2.27) $\begin{array}[]{l}\left|f_{0}(p)-f_{0}(q)\right|\\\
\leq\left|F_{0}(q,p)\right|+\left|F_{0}(p,q)\right|+\left|F_{0}\left(\frac{1-p}{1-pq},p\right)\right|+\left|f_{0}\left(1-\frac{1-p}{1-pq}\right)-f_{0}\left(\frac{1-p}{1-pq}\right)\right|\\\
\leq 3\cdot 16\varepsilon+15\varepsilon=63\varepsilon\end{array}$

holds for all $p,q\in]0,1[$. With the substitution $q=\frac{1}{2}$ (2.27)
implies that

$\left|f_{0}(p)-f_{0}\left(\frac{1}{2}\right)\right|\leq 63\varepsilon.$

Using the definition of the function $f_{0}$, we obtain that

$\left|f(x)-l(1-x)-c\right|\leq 63\varepsilon$

holds for all $x\in]0,1[$, where $c=f_{0}\left(\frac{1}{2}\right)$. Hence
inequality (2.3) holds, indeed. ∎

######  Remark 2.2.

Applying Theorem 2.1 in the case $\varepsilon=0$ we get the general solution
of (1.1) on $D^{\circ}$ (see also Maksa [8]).

##  3\. Two corollaries of the main result

The first corollary says that equation (1.1) is stable on $D$, as well.

######  Theorem 3.1.

Let $\alpha,\varepsilon\in\mathbb{R}$ be fixed, $\alpha\leq 0$,
$\varepsilon\geq 0$. Suppose that the function $f:[0,1]\rightarrow\mathbb{R}$
satisfies inequality (2.1) for all $(x,y)\in D$. Then, in case $\alpha<0$
there exist $a,b\in\mathbb{R}$ such that the function $h_{1}$ defined on
$[0,1]$ by

$h_{1}(x)=\left\\{\begin{array}[]{lcl}0,&\text{if}&x=0\\\
ax^{\alpha}+b(1-x)^{\alpha}-b,&\text{if}&x\in\left.]0,1[\right.\\\
a-b,&\text{if}&x=1\end{array}\right.$

is a solution of (1.1) on $D$ and

(3.1) $\left|f(x)-h_{1}(x)\right|\leq
15\varepsilon,\quad\left(x\in[0,1]\right)$

furthermore, in case $\alpha=0$, there exist $a,b,c\in\mathbb{R}$ such that
the function $h_{2}$ defined on $[0,1]$ by

$h_{2}(x)=\left\\{\begin{array}[]{lcl}a,&\text{if}&x=0\\\
c,&\text{if}&x\in\left.]0,1[\right.\\\ b,&\text{if}&x=1\end{array}\right.$

is a solution of (1.1) on $D$ and

(3.2) $\left|f(x)-h_{2}(x)\right|\leq 63\varepsilon.$

###### Proof.

An easy calculation shows that the functions $h_{1}$ and $h_{2}$ are the
solutions of (1.1) on $D$ in case $\alpha<0$ and in case $\alpha=0$,
respectively. Firstly, we investigate the case $\alpha<0$. Theorem 2.1.
implies that (3.1) holds for all $x\in]0,1[$. Therefore, it is enough to prove
that (3.1) holds for $x=0$ and for $x=1$. It follows from (2.1) with $x=0$,
that $\left((1-x)^{\alpha}-1\right)\left|f(0)\right|\leq\varepsilon$ if
$x\in]0,1[$. Since $\alpha<0$, $f(0)=0$ follows, that is, (3.1) is valid for
$x=0$. Let now $x\in]0,1[$ and $y=1-x$ in (2.1). Then

(3.3)
$\left|f(1-x)-f(x)-f(1)\left((1-x)^{\alpha}-x^{\alpha}\right)\right|\leq\varepsilon.$

Apply (2.2) to $1-x$ instead of $x$. Hence we get that

(3.4) $\left|-f(1-x)+a(1-x)^{\alpha}+bx^{\alpha}-b\right|\leq 15\varepsilon$

Adding the inequalities (3.3), (2.2) and (3.4) up and using the triangle
inequality to obtain that

$\left|a-b-f(x)\right|\cdot\left|(1-x)^{\alpha}-x^{\alpha}\right|\leq
31\varepsilon.\quad\left(x\in]0,1[\right)$

Since $\alpha<0$ we get that $f(1)=a-b$ and so (3.1) holds also for $x=1$.

Now, we fall to deal with the case $\alpha=0$. Let $x\in]0,1[$ and $y=1-x$ in
(2.1), then we obtain that

(3.5) $\left|f(x)-f(1-x)\right|\leq\varepsilon,\quad\left(x\in]0,1[\right)$

furthermore, let us observe that (2.1) does not pose any restriction on the
value of $f(0)$ as well as $f(1)$. Thus $f(0)=a$, $f(1)=b$, where
$a,b\in\mathbb{R}$ are certain constants. Therefore (3.2) holds if $x=0$ or
$x=1$.

In case $x\in]0,1[$, due to Theorem 2.1. there exist a logarithmic function
$l:]0,1[\rightarrow\mathbb{R}$ and $c\in\mathbb{R}$ such that

(3.6) $\left|f(x)-l(1-x)-c\right|\leq 63\varepsilon$

holds for all $x\in]0,1[$. Hence it is enough to prove that the function $l$
is identically zero on $]0,1[$. Indeed, due to (2.3), (3.5) and (3.6)

(3.7) $\begin{array}[]{l}\left|l(1-x)-l(x)\right|\\\
=\left|l(1-x)-f(1-x)+f(1-x)+c\right.\\\ -l(x)+f(x)-f(x)-c\left.\right|\\\
\leq\left|l(1-x)+c-f(x)\right|+\left|f(1-x)-l(x)-c\right|\\\
+\left|f(x)-f(1-x)\right|\\\ \leq 127\varepsilon\end{array}$

holds for all $x\in]0,1[$, using that the function $l$ is logarithmic, the
last inequality can be written as

$\left|l\left(\frac{1-x}{x}\right)\right|\leq
127\varepsilon.\quad\left(x\in]0,1[\right)$

By substitution $x=\frac{1}{1+p}$ ($p\in]0,+\infty[$) into the last
inequality, we obtain that

$\left|l(p)\right|\leq 127\varepsilon$

holds for all $p\in]0,+\infty[$, where we used the fact that every logarithmic
function on $]0,1[$ is uniquely extendable to a logarithmic function on
$]0,+\infty[$.

Thus $l$ is bounded on $]0,+\infty[$. However, the only bounded, logarithmic
function on $]0,+\infty[$ is the identically zero function. Therefore,

$\left|f(x)-c\right|\leq 63\varepsilon$

holds for all $x\in]0,1[$, i.e., (3.2) is proved. ∎

The second corollary concerns the stability of a system of equations.

######  Theorem 3.2.

Let $n\geq 2$ be a fixed positive integer,

$\Gamma^{\circ}_{n}=\left\\{(p_{1},\ldots,p_{n})|p_{i}>0,i=1,\ldots,n,\sum^{n}_{i=1}p_{i}=1\right\\}$

and $(I_{n})$ be the sequence of functions
$I_{n}:\Gamma^{\circ}_{n}\rightarrow\mathbb{R}$ and suppose that there exist a
sequence $(\varepsilon_{n})$ of nonnegative real numbers and a real number
$\alpha<0$ such that

(3.8)
$\left|I_{n}(p_{1},\ldots,p_{n})-I_{n-1}(p_{1}+p_{2},p_{3},\ldots,p_{n})-(p_{1}+p_{2})^{\alpha}I_{2}\left(\frac{p_{1}}{p_{1}+p_{2}},\frac{p_{2}}{p_{1}+p_{2}}\right)\right|\leq\varepsilon_{n-1}$

for all $n\geq 3$ and $(p_{1},\ldots,p_{n})\in\Gamma^{\circ}_{n}$, and

(3.9)
$\left|I_{3}(p_{1},p_{2},p_{3})-I_{3}(p_{1},p_{3},p_{2})\right|\leq\varepsilon_{1}$

holds on $\Gamma^{\circ}_{n}$. Then, in case $\alpha<0$ there exist
$c,d\in\mathbb{R}$ such that

(3.10)
$\begin{array}[]{l}\left|I_{n}\left(p_{1},\ldots,p_{n}\right)-\left(cH^{\alpha}_{n}\left(p_{1},\ldots,p_{n}\right)+d\left(p_{1}^{\alpha}-1\right)\right)\right|\\\
\leq\sum^{n-1}_{k=2}\varepsilon_{k}+15\left(2\varepsilon_{2}+\varepsilon_{1}\right)\left(1+\sum^{n-1}_{k=2}\left(\sum^{k}_{i=1}p_{i}^{\alpha}\right)\right)\end{array}$

for all $n\geq 2$ and $\left(p_{1},\ldots,p_{n}\right)\in\Gamma^{\circ}_{n}$.
Furthermore, in case $\alpha=0$ there exists a logarithmic function
$l:]0,1[\rightarrow\mathbb{R}$ and $c\in\mathbb{R}$ such that

(3.11)
$\begin{array}[]{l}\left|I_{n}\left(p_{1},\ldots,p_{n}\right)-\left(cH^{0}_{n}\left(p_{1},\ldots,p_{n}\right)+l(p_{1})\right)\right|\\\
\sum^{n-1}_{k=2}\varepsilon_{k}+63\left(n-1\right)\left(2\varepsilon_{2}+\varepsilon_{1}\right)\end{array}$

for all $n\geq 2$ and $\left(p_{1},\ldots,p_{n}\right)\in\Gamma^{\circ}_{n}$,
where the convention
$\sum^{1}_{k=2}\varepsilon_{k}=\sum^{1}_{k=2}\left(\sum^{k}_{i=1}p_{i}^{\alpha}\right)=0$
is adapted and

(3.12)
$H^{\alpha}_{n}\left(p_{1},\ldots,p_{n}\right)=\left(2^{1-\alpha}-1\right)^{-1}\left(\sum^{n}_{i=1}p_{i}^{\alpha}-1\right).\quad\left(\left(p_{1},\ldots,p_{n}\right)\in\Gamma^{\circ}_{n}\right)$

###### Proof.

As in Maksa [9], it can be proved that, due to (3.8) and (3.9), for the
function $f$ defined on $]0,1[$ by $f(x)=I_{2}(1-x,x)$ we get that

$\left|f(x)+(1-x)^{\alpha}f\left(\frac{y}{1-x}\right)-f(y)-(1-y)^{\alpha}f\left(\frac{x}{1-y}\right)\right|\leq
2\varepsilon_{2}+\varepsilon_{1}$

for all $(x,y)\in D^{\circ}$, i.e., (2.1) holds with
$\varepsilon=2\varepsilon_{2}+\varepsilon_{1}$. Therefore, applying Theorem
2.1. we obtain (2.2) and (2.3), respectively, with some $a,b,c\in\mathbb{R}$
and a logarithmic function $l:]0,1[\rightarrow\mathbb{R}$ and
$\varepsilon=2\varepsilon_{2}+\varepsilon_{1}$, i.e.,

$\left|I_{2}\left(1-x,x\right)-\left(ax^{\alpha}+b(1-x)^{\alpha}-b\right)\right|\leq
15(2\varepsilon_{2}+\varepsilon_{1}),\quad\left(x\in]0,1[\right)$

in case $\alpha<0$, and

$\left|I_{2}\left(1-x,x\right)-\left(l(1-x)+c\right)\right|\leq
63(2\varepsilon_{2}+\varepsilon_{1})\quad\left(x\in]0,1[\right)$

in case $\alpha=0$.

Therefore (3.10) holds with $c=(2^{1-\alpha}-1)a$, $d=b-a$ in case $\alpha<0$,
and (3.11) holds in case $\alpha=0$, respectively, for $n=2$.

We continue the proof by induction on $n$. Suppose that (3.10) and (3.11)
holds, resp., and for the sake of brevity, introduce the notation

$J_{n}(p_{1},\ldots,p_{n})=\left\\{\begin{array}[]{lcl}cH^{\alpha}_{n}(p_{1},\ldots,p_{n}),&\text{if}&\alpha<0\\\
cH^{0}_{n}(p_{1},\ldots,p_{n})+l(p_{1}),&\text{if}&\alpha=0\end{array}\right.$

for all $n\geq 2$, $(p_{1},\ldots,p_{n})\in\Gamma^{\circ}_{n}$. It can easily
be seen that (3.10) and (3.11) hold on $\Gamma^{\circ}_{n}$ for $J_{n}$
instead of $I_{n}$ ($n\geq 3$) with $\varepsilon_{n}=0$ ($n\geq 2$). Thus, for
all $(p_{1},\ldots,p_{n+1})\in\Gamma^{\circ}_{n+1}$, we get that

$\begin{array}[]{l}I_{n+1}(p_{1},\ldots,p_{n+1})-J_{n+1}(p_{1},\ldots,p_{n+1})\\\
=I_{n+1}(p_{1},\ldots,p_{n+1})-J_{n}(p_{1}+p_{2},p_{3},\ldots,p_{n+1})-(p_{1}+p_{2})^{\alpha}J_{2}\left(\frac{p_{1}}{p_{1}+p_{2}},\frac{p_{2}}{p_{1}+p_{2}}\right)\\\
=I_{n+1}(p_{1},\ldots,p_{n+1})-I_{n}(p_{1}+p_{2},p_{3},\ldots,p_{n+1})-(p_{1}+p_{2})^{\alpha}I_{2}\left(\frac{p_{1}}{p_{1}+p_{2}},\frac{p_{2}}{p_{1}+p_{2}}\right)\\\
+I_{n}(p_{1}+p_{2},p_{3},\ldots,p_{n+1})-J_{n}(p_{1}+p_{2},p_{3},\ldots,p_{n+1})\\\
+(p_{1}+p_{2})^{\alpha}\left(I_{2}\left(\frac{p_{1}}{p_{1}+p_{2}}\right)-J_{2}\left(\frac{p_{1}}{p_{1}+p_{2}},\frac{p_{2}}{p_{1}+p_{2}}\right)\right).\end{array}$

Therefore, if $\alpha<0$, (3.8) (with $n+1$ instead of $n$), (3.10) with $n=2$
and the induction hypothesis (applying to $(p_{1}+p_{2},\ldots,p_{n+1})$
instead of $(p_{1},\ldots,p_{n})$) imply that

$\begin{array}[]{l}\left|I_{n+1}(p_{1},\ldots,p_{n+1})-J_{n+1}(p_{1},\ldots,p_{n+1})\right|\\\
\leq\varepsilon_{n}+\sum^{n-1}_{k=2}\varepsilon_{k}+15(2\varepsilon_{2}+\varepsilon_{1})\left(1+\sum^{n-1}_{k=2}\left(\sum^{k+1}_{i=1}p_{i}\right)^{\alpha}\right)\\\
+15(2\varepsilon_{2}+\varepsilon_{1})(p_{1}+p_{2})^{\alpha}\\\
=\sum^{n}_{k=2}\varepsilon_{k}+15(2\varepsilon_{2}+\varepsilon_{1})\left(1+\sum^{n}_{k=2}\left(\sum^{k}_{i=1}p_{i}\right)^{\alpha}\right),\end{array}$

that is (3.10) holds for $n+1$ instead of $n$.

Finally, if $\alpha=0$, (3.9) (with $n+1$ instead of $n$), (3.11) with $n=2$
and the induction hypothesis (applying to $(p_{1}+p_{2},\ldots,p_{n+1})$
instead of $(p_{1},\ldots,p_{n})$) imply that

$\begin{array}[]{l}\left|I_{n+1}(p_{1},\ldots,p_{n+1})-J_{n+1}(p_{1},\ldots,p_{n+1})\right|\\\
\leq\varepsilon_{n}+\sum^{n-1}_{k=2}\varepsilon_{k}+63(n-1)(2\varepsilon_{2}+\varepsilon_{1})+63(2\varepsilon_{2}+\varepsilon_{1})\\\
=\sum^{n}_{k=2}\varepsilon_{k}+63n(2\varepsilon_{2}+\varepsilon_{1}),\end{array}$

this yields that (3.11) holds for $n+1$ instead of $n$. ∎

######  Remark 3.3.

Applying Theorem 3.2 with the choice $\varepsilon_{n}=0$ for all
$n\in\mathbb{N}$ we get the $\alpha$–recursive, $3$–semisymmetric information
measures.

######  Remark 3.4.

The sequence $H^{\alpha}_{n}$ defined in (3.12) is the entropy of degree
$\alpha$ (see Aczél–Daróczy [3], Havrda–Charvát [7], Daróczy [4], Tsallis
[12]) for $\alpha\leq 0$. Since the sequence $J_{n}$ satisfies (3.8) and (3.9)
with $\varepsilon_{n}=0$ ($n\geq 1$), our theorem says that $(J_{n})$ is
stable in a certain sense.

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