paper1.dvi


The general solutions of a functional equation
related to information theory

Prem Nath and Dhiraj Kumar Singh

Department of Mathematics, University of Delhi
Delhi – 110007, INDIA

E-mail: dksingh@maths.du.ac.in
dhiraj426@rediffmail.com

Abstract. The general solutions of a functional equation, containing two

unknown functions, and related to a functional equation characterizing the

Shannon entropy and the entropy of degree α, are obtained.

Keywords: Functional equations, continuous solutions, Lebesgue measurable

solutions, the Shannon entropy, the nonadditive entropy of degree α, multi-

plicative functions, additive functions.



2 Prem Nath and Dhiraj Kumar Singh

1. Introduction

For n = 1, 2, 3, . . ., let

Γn =

{

(p1, . . . , pn) : pi ≥ 0, i = 1, . . . , n;
n
∑

i=1

pi = 1

}

denote the set of all n-component complete discrete probability distributions

with nonnegative elements and let F : I → R, R denoting the set of all real

numbers and I =
{

x ∈ R : 0 ≤ x ≤ 1
}

, the unit closed interval.

The functional equation

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

F (pi) +
∑̀

j=1

F (qj)(1.1)

with (p1, . . . , pk) ∈ Γk and (q1, . . . , q`) ∈ Γ` was first studied by T.W. Chaundy

and J.B. Mcleod [4]. They proved that if (1.1) holds for integers k = 2, 3, . . . and

` = 2, 3, . . . and F is continuous on I , then F is of the form

F (p) = c p log
2
p , 0 ≤ p ≤ 1(1.2)

where c is an arbitrary real constant and 0 log
2
0 = 0. Later on, J. Aczél and

Z. Daróczy [1] proved the same by assuming k = ` = 2, 3, . . .. Z. Daróczy [5]

obtained the Lebesgue measurable solutions of (1.1) by fixing k = 3, ` = 2 and

assuming F (1) = 0. Gy. Maksa [13] obtained the solutions of (1.1) by fixing

k = 3, ` = 2 but assuming F to be bounded on a subset, of I , of positive

Lebesgue measure.

If F

(

1

2

)

=
1

2
, then (1.2) gives c = −1 and then (1.2) reduces to

F (p) = − p log
2
p(1.3)

for all p ∈ I .



The general solutions of a functional equation ... 3

For any probability distribution (p1, . . . , pm) ∈ Γm ,

Hm(p1, . . . , pm) = −
m
∑

i=1

pi log2 pi(1.4)

is known as the Shannon entropy [15] of the probability distribution

(p1, . . . , pm) ∈ Γm and the sequence Hm : Γm → R, m = 1, 2, . . . is known

as the sequence of the Shannon entropies.

A generalization of the Shannon entropy with which we shall be concerned in

this paper is (with H αm : Γm → R, m = 1, 2, 3, . . .)

Hαm(p1, . . . , pm) = (1−2
1−α)−1

(

1−
m
∑

i=1

pαi

)

, α > 0, α 6= 1, 0α := 0, α∈R.(1.5)

The entropies (1.5) are due to J. Havrda and F. Charvat [7].

The axiomatic characterization of the entropies (1.5) leads to the study of the

functional equation

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

F (pi) +
∑̀

j=1

F (qj) + λ
k
∑

i=1

F (pi)
∑̀

j=1

F (qj)(1.6)

where (p1, . . . , pk) ∈ Γk , (q1, . . . , q`) ∈ Γ` and λ = 2
1−α − 1, α ∈ R. Clearly,

(1.6) reduces to (1.1) if λ = 0.

By taking λ = 21−α − 1, α 6= 1, α ∈ R, 0α := 0, the continuous solutions

of (1.6) were obtained by M. Behara and P. Nath [3] for all positive integers

k = 2, 3, . . . ; ` = 2, 3, . . . . Later on PL. Kannappan [10] and D.P. Mittal [14]

also obtained the continuous solutions of (1.6) for λ 6= 0 and k = 2, 3, . . .;

` = 2, 3, . . . . For fixed integers k ≥ 3 and ` ≥ 2, L. Losonczi [11] obtained the

measurable solutions of (1.6). Also, PL. Kannappan [8] obtained the Lebesgue

measurable solutions of both (1.1) and (1.6) for fixed integers k ≥ 3, ` ≥ 3.



4 Prem Nath and Dhiraj Kumar Singh

It seems that L. Losonczi and Gy. Maksa [12] are the first to obtain the general

solutions of (1.6) in both cases, namely λ 6= 0 and λ = 0, by fixing integers k

and `, k ≥ 3 and ` ≥ 3.

There are several generalizations of (1.6), with λ ∈ R, containing at least two

unknown functions. Below we list only three important generalizations of (1.6),

namely,

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

H(pi) +
∑̀

j=1

H(qj) + λ
k
∑

i=1

H(pi)
∑̀

j=1

H(qj)(1.7)

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

F (pi) +
∑̀

j=1

H(qj) + λ
k
∑

i=1

F (pi)
∑̀

j=1

H(qj)(1.8)

and

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

G(pi) +
∑̀

j=1

H(qj) + λ
k
∑

i=1

G(pi)
∑̀

j=1

H(qj).(1.9)

The object of this paper is to investigate the general solutions of the functional

equation (1.7) for fixed integers k ≥ 3 and ` ≥ 3. The corresponding results for

the functional equations (1.8) and (1.9) have also been investigated by the authors

and shall be presented elsewhere in our subsequent research work.

The process of finding the general solutions of (1.7) requires a detailed study

of the following two functional equations :

k
∑

i=1

∑̀

j=1

g(piqj) =
k
∑

i=1

g(pi)
∑̀

j=1

g(qj) + `(k − 1) g(0)(1.10)

and

k
∑

i=1

∑̀

j=1

f (piqj) =
k
∑

i=1

h(pi)
∑̀

j=1

h(qj)(1.11)

where f : [0, 1] → R, g : [0, 1] → R and h : [0, 1] → R.



The general solutions of a functional equation ... 5

The functional equation (1.10) is, indeed, a generalization of the multiplica-

tivetype functional equation

k
∑

i=1

∑̀

j=1

g(piqj) =
k
∑

i=1

g(pi)
∑̀

j=1

g(qj)(1.12)

whose importance in information theory is well-known (see L. Losonczi and

Gy. Maksa [12]). The functional equation (1.6), for λ 6= 0, can be written in

the multiplicative form (1.12) by defining g : I → R as g(x) = λ F (x) + x for all

x ∈ I . Likewise, each of the functional equations (1.7), (1.8) and (1.9), for λ 6= 0,

can also be written in the corresponding multiplicative forms. This is precisely

the reason for paying attention to the functional equations (1.7) to (1.9).

2. The general solutions of functional equation (1.10)

Before investigating the general solutions of (1.10) for fixed integers k and

`, k ≥ 3, ` ≥ 3, we need some definitions and results already existing in the

literature (see [12]). Let

∆ =
{

(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ x + y ≤ 1
}

.

In other words, ∆ denotes the unit closed triangle in

R
2 = R × R =

{

(x, y) : x ∈ R, y ∈ R
}

.

A mapping a : R → R is said to be additive if it satisfies the equation

a(x + y) = a(x) + a(y)(2.1)

for all x ∈ R, y ∈ R.

A mapping a : I → R, I = [0, 1] is said to be additive on the triangle ∆ if it

satisfies (2.1) for all (x, y) ∈ ∆.



6 Prem Nath and Dhiraj Kumar Singh

A mapping m : [0, 1] → R is said to be multiplicative if m(0) = 0, m(1) = 1

and m(xy) = m(x) m(y) for all x ∈ ]0, 1[, y ∈ ]0, 1[.

Now we state :

Lemma 1. Let Ψ : I → R be a mapping which satisfies the functional

equation

n
∑

i=1

Ψ(pi) = c(2.2)

for all (p1, . . . , pn) ∈ Γn ; c a given constant and n ≥ 3 a fixed integer. Then

there exists an additive mapping a : R → R such that

Ψ(p) = a(p) + Ψ(0) , 0 ≤ p ≤ 1(2.3)

where

a(1) = c − n Ψ(0) .(2.4)

Conversely, if (2.4) holds, then the mapping Ψ : I → R, defined by (2.3), satisfies

the functional equation (2.2).

This lemma appears on p-74 in [12].

Lemma 2. Every mapping a : I → R, I = [0, 1], additive on the unit triangle

∆, has a unique additive extension to the whole of R.

Note. This unique additive extension to the whole of R will also be denoted

by the symbol a but now a : R → R.

For Lemma 2, See Theorem (0.3.7) on p-8 in [2] or Z. Daróczy and L. Losonczi

[6]. Now we prove :



The general solutions of a functional equation ... 7

Theorem 1. Let k ≥ 3, ` ≥ 3 be fixed integers and g : [0, 1] → R be a

mapping which satisfies the functional equation (1.10) for all (p1, . . . , pk) ∈ Γk

and (q1, . . . , q`) ∈ Γ` . Then g is of the form

g(p) = a(p) + g(0)(2.5)

where a : R → R is an additive function such that a(1) satisfies the equation

a(1) + k` g(0) = [a(1) + k g(0)][a(1) + ` g(0)] + `(k − 1) g(0)(2.6)

or

g(p) = M (p) − A(p) + g(0)(2.7)

where A : R → R is an additive function with

A(1) = ` g(0)(2.8)

and M : [0, 1] → R is a mapping such that

M (0) = 0(2.9)

M (1) = g(1) + (` − 1) g(0)(2.10)

and

M (pq) = M (p) M (q) for all p ∈ ]0, 1[ , q ∈ ]0, 1[ .(2.11)

Proof. Let us put p1 = 1, p2 = . . . = pk = 0 in (1.10). We obtain

[1 − g(1) − (k − 1)g(0)]
∑̀

j=1

g(qj) = 0.(2.12)



8 Prem Nath and Dhiraj Kumar Singh

Case 1. 1 − g(1) − (k − 1) g(0) 6= 0. Then (2.12) reduces to

∑̀

j=1

g(qj) = 0.(2.13)

Hence, by Lemma 1, g is of the form (2.5) in which a : R → R is an additive

mapping such that a(1) = − ` g(0) satisfies the equation (2.6).

Case 2. 1 − g(1) − (k − 1) g(0) = 0.

The functional equation (1.10) may be written in the form

∑̀

j=1

[

k
∑

i=1

g(piqj) − g(qj)
k
∑

i=1

g(pi)

]

= `(k − 1) g(0).

Hence, by Lemma 1,

k
∑

i=1

g(piq) − g(q)
k
∑

i=1

g(pi)(2.14)

= A1(p1, . . . , pk, q) −
1

`
A1(p1, . . . , pk, 1) + (k − 1) g(0)

where A1 : Γk ×R → R is additive in the second variable. The substitution q = 0

in (2.14) gives

A1(p1, . . . , pk, 1) = ` g(0)

[

k
∑

i=1

g(pi) − 1

]

.(2.15)

Let x ∈ [0, 1], (r1, . . . , rk) ∈ Γk . Put q = xrt , t = 1, . . . , k in (2.14); add the

resulting k equations; and use the additivity of A1 . We get

k
∑

i=1

k
∑

t=1

g(pirtx) −
k
∑

i=1

g(pi)
k
∑

t=1

g(xrt)(2.16)

= A1(p1, . . . , pk, x) −
k

`
A1(p1, . . . , pk, 1) + k(k − 1) g(0).



The general solutions of a functional equation ... 9

Now put q = x, p1 = r1, . . . , pk = rk in (2.14). We obtain

k
∑

t=1

g(xrt) = g(x)
k
∑

t=1

g(rt) + A1(r1, . . . , rk, x)(2.17)

−
1

`
A1(r1, . . . , rk, 1) + (k − 1) g(0).

From (2.16) and (2.17), it follows that

k
∑

i=1

k
∑

t=1

g(pirtx) − g(x)
k
∑

i=1

g(pi)
k
∑

t=1

g(rt) − k(k − 1) g(0)(2.18)

= (k − 1) g(0)
k
∑

i=1

g(pi) + A1(r1, . . . , rk, x)
k
∑

i=1

g(pi)

−
1

`
A1(r1, . . . , rk, 1)

k
∑

i=1

g(pi) + A1(p1, . . . , pk, x)

−
k

`
A1(p1, . . . , pk, 1).

The left hand side of (2.18) does not undergo any change if we interchange pi and

ri, i = 1, . . . , k . So, the right hand side of (2.18) must also remain unchanged on

interchanging pi and ri , i = 1, . . . , k . Consequently, we obtain

A1(p1, . . . , pk, x)

[

k
∑

t=1

g(rt)− 1

]

−
1

`
A1(p1, . . . , pk, 1)

[

k
∑

t=1

g(rt) − k

]

(2.19)

+ (k − 1) g(0)
k
∑

t=1

g(rt)

= A1(r1, . . . , rk, x)

[

k
∑

i=1

g(pi)−1

]

−
1

`
A1(r1, . . . , rk, 1)

[

k
∑

i=1

g(pi)−k

]

+ (k − 1) g(0)
k
∑

i=1

g(pi).

Now we divide our discussion into two cases depending upon whether
k
∑

t=1

g(rt)−1

vanishes identically on Γk or does not vanish identically on Γk .



10 Prem Nath and Dhiraj Kumar Singh

Case 2.1.

k
∑

t=1

g(rt) − 1 vanishes identically on Γk . Then

k
∑

t=1

g(rt) = 1

for all (r1, . . . , rk) ∈ Γk . By using Lemma 1, it follows that g is of the form (2.5)

in which a(1) = 1 − k g(0) satisfies the equation (2.6).

Case 2.2.

k
∑

t=1

g(rt) − 1 does not vanish identically on Γk .

In this case, there exists a probability distribution (r∗
1
, . . . , r∗k) ∈ Γk such that

k
∑

t=1

g(r∗t ) − 1 6= 0.(2.20)

Putting r1 = r
∗

1
, . . . , rk = r

∗

k in (2.19), making use of (2.20) and (2.15); and

performing necessary calculations, it follows that

A1(p1, . . . , pk, x) = A(x)

[

k
∑

i=1

g(pi) − 1

]

(2.21)

where A : R → R is such that

A(x) =

[

k
∑

t=1

g(r∗t ) − 1

]

−1

A1(r
∗

1
, . . . , r∗k, x)(2.22)

From (2.22) it is easy to conclude that A : R → R is additive as the mapping

x 7→ A1(r
∗

1
, . . . , r∗k, x) is additive. Also, putting x = 1 in (2.22) and making use

of (2.15) by taking pi = r
∗

i , i = 1, . . . , k ; (2.8) follows. Also, from (2.14), (2.15),

(2.21) and (2.8), it follows that

k
∑

i=1

[g(piq) + A(piq) − g(0)] − [g(q) + A(q) − g(0)](2.23)

×

k
∑

i=1

[g(pi)+A(pi) − g(0)] + [g(q)+A(q) − g(0)](` − k) g(0) = 0.



The general solutions of a functional equation ... 11

Define a mapping M : I → R, I = [0, 1], as

M (p) = g(p) + A(p) − g(0)(2.24)

for all p ∈ I . Then, (2.23) reduces to the equation

k
∑

i=1

[M (piq) − M (q) M (pi) + (` − k) g(0) M (q) pi] = 0.(2.25)

Hence, by Lemma 1,

M (pq) − M (q) M (p) + (` − k) g(0) M (q) p = E1(p, q) −
1

k
E1(1, q)(2.26)

where E1 : R × [0, 1] → R is additive in its first variable.

Since A(0) = 0 and A(1) = ` g(0), (2.9) and (2.10) follow from (2.24). Also,

putting p = 0 in (2.26) and making use of (2.9), it follows that

E1(0, q) = 0(2.27)

for all q , 0 ≤ q ≤ 1. Consequently,

E1(1, q) = 0(2.28)

for all q , 0 ≤ q ≤ 1. Now, (2.26) reduces to

M (pq) − M (p) M (q) = E1(p, q) − (` − k) g(0) M (q) p(2.29)

for all p ∈ [0, 1] and q ∈ [0, 1].

Since M (1) = g(1) + (` − 1) g(0), from now onwards, we divide our discus-

sion into two subcases, depending upon whether g(1) + (` − 1) g(0) = 1 or

g(1) + (` − 1) g(0) 6= 1.

Case 2.2.1. g(1) + (` − 1) g(0) = 1.

In this case, 1 = g(1) + (` − 1) g(0) = g(1) + (k − 1) g(0) + (` − k) g(0).



12 Prem Nath and Dhiraj Kumar Singh

Since g(1)+(k −1)g(0) = 1, it follows that (`−k) g(0) = 0. Then, (2.29) reduces

to

M (pq) − M (p) M (q) = E1(p, q)(2.30)

where E1 : R × [0, 1] → R is additive in the first variable and 0 ≤ p ≤ 1,

0 ≤ q ≤ 1. The left hand side of (2.30) is symmetric in p and q . Hence,

E1(p, q) = E1(q, p) for all p ∈ [0, 1], q ∈ [0, 1]. Consequently, E1 is also additive in

second variable. Also, we may suppose that E1(p, ·) has been extended additively

to the whole of R and this extension is unique by Lemma 2.

From (2.30), as on p-77 in [12], it follows that

M (pqr) − M (p) M (q) M (r) = E1(pq, r) + M (r) E1(p, q)(2.30a)

= E1(qr, p) + M (p) E1(q, r)

for all p, q , r in [0,1]. Now, we prove that E1(p, q) = 0 for all p, q , 0 ≤ p ≤ 1,

0 ≤ q ≤ 1. If possible, suppose there exist p∗ and q∗ , 0 ≤ p∗ ≤ 1, 0 ≤ q∗ ≤ 1,

such that E1(p
∗, q∗) 6= 0. Then, from (2.30a),

M (r) =
[

E1(p
∗

, q
∗)
]

−1
[

E1(q
∗

r, p
∗) + M (p∗) E1(q

∗

, r) − E1(p
∗

q
∗

, r)
]

from which it is easy to conclude that M is additive. Now, making use of (2.8),

(2.10), (2.20), (2.24), the condition g(1) + (` − 1) g(0) = 1; and the additivity of

A and M , we have

1 6=
k
∑

t=1

g(r∗t ) = M (1) − A(1) + k g(0) = 1

a contradiction. Hence E1(p, q) = 0 for all p and q , 0 ≤ p ≤ 1, 0 ≤ q ≤ 1. Thus,

(2.30) reduces to M (pq) = M (p) M (q) for all p and q , 0 ≤ p ≤ 1, 0 ≤ q ≤ 1.

So, M is a nonconstant multiplicative function. Hence, from (2.24), it follows

that g is of the form (2.7).



The general solutions of a functional equation ... 13

Case 2.2.2. g(1) + (` − 1) g(0) 6= 1.

Since the values of M at 0 and 1 are given by (2.9) and (2.10), our next task

is to get some information about M (r) when 0 < r < 1. For this purpose, we

proceed as follows:

Let p, q , r be in ]0,1[. Now, from (2.29), one can derive

M (pqr) − M (p) M (q) M (r)(2.31)

= E1(r, pq)−(`−k) g(0) M (pq) r+M (r)
[

E1(p, q)−(`−k) g(0) M (q) p
]

= E1(rq, p)−(`−k) g(0) M (p) rq+M (p)
[

E1(r, q)−(`−k) g(0) M (q) r
]

.

Now, we prove that E1(p, q) − (` − k) g(0) M (q) p = 0 for all p, q , 0 < p < 1,

0 < q < 1. If possible, suppose there exist p∗ ∈ ]0, 1[ and q∗ ∈ ]0, 1[ such that

E1(p
∗, q∗) − (` − k) g(0) M (q∗) p∗ 6= 0. Then, from (2.31), it follows that for all

r ∈ ]0, 1[,

M (r) =
[

E1(p
∗

, q
∗) − (` − k) g(0) M (q∗) p∗

]

−1

(2.32)

×
[

E1(rq
∗

, p
∗) − (` − k) g(0) M (p∗) rq∗

+ M (p∗)
{

E1(r, q
∗) − (` − k) g(0) M (q∗) r

}

− E1(r, p
∗q∗)

+ (` − k) g(0) M (p∗q∗) r
]

.

Now we prove that M : [0, 1] → R is additive on ∆, that is,

M (x + y) = M (x) + M (y)(2.33)

for all 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ x + y ≤ 1.

If x = 0, 0 ≤ y ≤ 1 or y = 0, 0 ≤ x ≤ 1, then (2.33) holds trivially.

If 0 < x < 1, 0 < y < 1, 0 < x + y < 1, then (2.33) follows from (2.32).

Now consider the case when 0 < x < 1, 0 < y < 1 but x + y = 1. In this

case, let us choose q = 1 and p = x + y in (2.29) and use the additivity of E1



14 Prem Nath and Dhiraj Kumar Singh

with respect to first variable. We obtain

M (x + y)
{

1 − g(1) − (` − 1) g(0)
}

=
{

M (x) + M (y)
}{

1 − g(1) − (` − 1) g(0)
}

.

Since g(1) + (` − 1) g(0) 6= 1, (2.33) follows. Thus, M is additive on the

triangle ∆. Now, making use of (2.8), (2.10), (2.20), (2.24), the condition

g(1) + (k − 1) g(0) = 1; and the additivity of A and M , we have

1 6=
k
∑

t=1

g(r∗t ) = M (1) − A(1) + k g(0) = 1

a contradiction. Hence E1(p, q) − (` − k) g(0) M (q) p = 0 for all p, q, 0 < p < 1,

0 < q < 1. Thus, (2.29) reduces to (2.11). But in this case M is not multiplicative

because M (1) = g(1) + (` − 1) g(0) 6= 1. Hence from (2.24), the solution (2.7)

follows.

Note. It is easy to verify that (2.5); subject to the condition (2.6), satisfies

(1.10). However (2.7) also satisfies (1.10). In this case we need to use (2.25) in

addition to (2.8) to (2.11).

3. The general solutions of functional equation (1.11)

Now we prove :

Theorem 2. Let k ≥ 3, ` ≥ 3 be fixed integers and f : I → R, h : I → R,

I = [0, 1], be mappings which satisfy the functional equation (1.11) for all

(p1, . . . , pk) ∈ Γk and (q1, . . . , q`) ∈ Γ` . Then any general solution of (1.11)

is of the form
{

f (p) = b(p) + f (0)

h(p) = B(p) + h(0)
(3.1)



The general solutions of a functional equation ... 15

subject to the condition

b(1) + k` f (0) = [B(1) + k h(0)][B(1) + ` h(0)];(3.2)

or






f (p) = [h(1) + (k − 1) h(0)]2 a(p) + A∗(p) + f (0)

h(p) = [h(1) + (k − 1) h(0)] a(p) + h(0)
(3.3)

subject to the condition

[h(1) + (k − 1)h(0)]2a(1) + A∗(1) + k` f (0)(3.3a)

=
{

[h(1)+(k−1)h(0)]a(1)+k h(0)
}{

[h(1)+(k−1)h(0)]a(1)+` h(0)
}

or






































f (p) = [h(1) + (k − 1) h(0)]2[M (p) − A(p)] + A∗(p) + f (0)

h(p) = [h(1) + (k − 1) h(0)][M (p) − A(p)] + h(0)

A(1) =
` h(0)

h(1) + (k − 1) h(0)
,

A∗(1) = `
{

[h(1) + (k − 1) h(0)] h(0) − k f (0)
}

(3.4)

where A∗ : R → R, A : R → R, B : R → R , a : R → R, b : R → R are

additive functions; f (0) and h(0) are arbitrary constants; and M : [0, 1] → R is

a mapping which satisfies (2.9), (2.11) and

M (1) =
h(1) + (` − 1) h(0)

h(1) + (k − 1) h(0)
(3.5)

with h(1) + (k − 1) h(0) 6= 0 in (3.3), (3.3a), (3.4) and (3.5).

To prove this theorem, we need to prove some Lemmas :

Lemma 3. If a mapping f : I → R satisfies the functional equation

k
∑

i=1

∑̀

j=1

f (piqj) = 0(3.6)



16 Prem Nath and Dhiraj Kumar Singh

for all (p1, . . . , pk) ∈ Γk , (q1, . . . , q`) ∈ Γ` , k ≥ 3, ` ≥ 3 fixed integers; then

f (p) = b(p) + f (0)(3.7)

where b : R → R is an additive function with b(1) = − k` f (0).

Proof. Choose q1 = 1, q2 = . . . = q` = 0. Then, equation (3.6) reduces to

k
∑

i=1

f (pi) = − k(` − 1) f (0).

Hence, by Lemma 1,

f (p) = b(p) −
1

k
b(1) − (` − 1) f (0)(3.8)

for all p, 0 ≤ p ≤ 1, b : R → R being any additive function with

b(1) = − k` f (0). Putting this value of b(1) in (3.8), (3.7) readily follows.

Lemma 4. Under the assumptions stated in the statement of Theorem 2, the

following conclusions hold :

f (p) = [h(1)+(k−1)h(0)]h(p)+A∗(p)−[h(1)+(k−1)h(0)]h(0)+f (0)(3.9)

[h(1) + (k − 1) h(0)]
k
∑

i=1

∑̀

j=1

h(piqj) −
k
∑

i=1

h(pi)
∑̀

j=1

h(qj)(3.10)

= `(k − 1) h(0)[h(1) + (k − 1) h(0)]

[h(1) + (` − 1) h(0)]
k
∑

i=1

∑̀

j=1

h(piqj) −
k
∑

i=1

h(pi)
∑̀

j=1

h(qj)(3.11)

= k(` − 1) h(0)[h(1) + (` − 1) h(0)]

where A∗ : R → R is an additive function.



The general solutions of a functional equation ... 17

Proof. Putting p1 = 1, p2 = . . . = pk = 0 in (1.11), we obtain

∑̀

j=1

{

f (qj) − [h(1) + (k − 1) h(0)] h(qj)
}

= − `(k − 1) f (0).(3.12)

Hence, by Lemma 1 (changing q to p),

f (p) = [h(1) + (k − 1) h(0)] h(p) + A∗(p) −
1

`
A∗(1) − (k − 1) f (0)(3.13)

for all p, 0 ≤ p ≤ 1, A∗ : R → R being an additive function with

A∗(1) = `
{

[h(1) + (k − 1) h(0)] h(0) − k f (0)
}

.(3.14)

From equations (3.13) and (3.14), equation (3.9) follows.

From (3.9) and (3.14), it is easy to see that

k
∑

i=1

∑̀

j=1

f (piqj) = [h(1) + (k − 1) h(0)]
k
∑

i=1

∑̀

j=1

h(piqj)(3.15)

− `(k − 1) [h(1) + (k − 1) h(0)] h(0).

From (1.11) and (3.15), we get (3.10). The proof of (3.11) is similar and hence

omitted.

Proof of Theorem 2. We divide our discussion into three cases :

Case 1.

k
∑

i=1

h(pi) vanishes identically on Γk , that is,

k
∑

i=1

h(pi) = 0(3.16)

for all (p1, . . . , pk) ∈ Γk . Then, (1.11) reduces to (3.6). So, f is of the form (3.7)

for all p, 0 ≤ p ≤ 1. Also applying Lemma 1 to (3.16), we obtain

h(p) = B(p) −
1

k
B(1)(3.17)

for all p, 0 ≤ p ≤ 1, B : R → R being an additive function with B(1) = − k h(0).



18 Prem Nath and Dhiraj Kumar Singh

Now (3.17) reduces to

h(p) = B(p) + h(0).(3.18)

Equations (3.7), (3.18), together with the condition (3.2), constitute the solution

(3.1) of (1.11).

Case 2.
∑̀

j=1

h(qj) vanishes identically on Γ` . In this case, we also get the

solution (3.1), subject to the condition (3.2); of (1.11). The proof is omitted as

it is similar to that in case 1.

Case 3. Neither
k
∑

i=1

h(pi) vanishes identically on Γk nor
∑̀

j=1

h(qj) vanishes

identically on Γ` . Then, there exist a (p
∗

1
, . . . , p∗k) ∈ Γk and a (q

∗

1
, . . . , q∗` ) ∈ Γ`

such that
k
∑

i=1

h(p∗i ) 6= 0 and
∑̀

j=1

h(q∗j ) 6= 0; and consequently

k
∑

i=1

h(p∗i )
∑̀

j=1

h(q∗j ) 6= 0 .(3.19)

Now, we prove that h(1) + (k − 1) h(0) 6= 0. If possible, suppose

h(1) + (k − 1) h(0) = 0. Then (3.10) reduces to the equation

k
∑

i=1

h(pi)
∑̀

j=1

h(qj) = 0

valid for all (p1, . . . , pk) ∈ Γk and (q1, . . . , q`) ∈ Γ` . In particular,
k
∑

i=1

h(p∗i )
∑̀

j=1

h(q∗j ) = 0 contradicting (3.19). Hence h(1) + (k − 1) h(0) 6= 0.

Similarly, making use of(3.11), we can prove that h(1) + (` − 1) h(0) 6= 0.

Let us consider the case when h(1) + (k − 1) h(0) 6= 0. In this case, let us

define a mapping g : [0, 1] → R as

g(x) = [h(1) + (k − 1) h(0)]−1h(x)(3.20)

for all x ∈ [0, 1]. Then, with the aid of (3.20), (3.10) reduces to the functional



The general solutions of a functional equation ... 19

equation (1.10). Also, from (3.20), it is easy to see that g(1) + (k − 1) g(0) = 1.

Consequently, from the discussion, carried out under this case, in the proof of

theorem 1, it follows that g is of the form (2.5), subject to the condition (2.6); and

(2.7). From equations (2.5), (2.7), (3.9) and (3.20), the solutions (3.3) subject to

the condition (3.3a); and (3.4) of functional equation (1.11) follow. The details

are omitted for the sake of brevity.

4. The general solutions of functional equation (1.7) when λ 6= 0

In this section we prove the following :

Theorem 3. Let k ≥ 3, ` ≥ 3 be fixed integers and F : I → R,

H : I → R, I = [0, 1], be mappings which satisfy the functional equation (1.7) for

all (p1, . . . , pk) ∈ Γk and (q1, . . . , q`) ∈ Γ` . Then, any general solution of (1.7) is

of the form

F (p) =
b(p) + λ F (0) − p

λ
, H(p) =

B(p) + λ H(0) − p

λ
(4.1)

subject to the condition

b(1) + λk` F (0) = [B(1) + λk H(0)][B(1) + λ` H(0)](4.2)

or














F (p) =
[λ (H(1) + (k − 1) H(0)) + 1]2 a(p) + A∗(p) + λ F (0) − p

λ

H(p) =
[λ (H(1) + (k − 1) H(0)) + 1] a(p) + λ H(0) − p

λ

(4.3)

subject to the condition

[λ (H(1) + (k − 1) H(0)) + 1]2 a(1) + A∗(1) + λk` F (0)(4.3a)

=
{

[λ (H(1) + (k − 1) H(0)) + 1] a(1) + λk H(0)
}

×
{

[λ (H(1) + (k − 1) H(0)) + 1] a(1) + λ` H(0)
}



20 Prem Nath and Dhiraj Kumar Singh

or










































































F (p) =





[λ (H(1)+(k − 1) H(0))+1]2[M (p)−A(p)]

+ A∗(p)+λ F (0)−p





λ

H(p) =
[λ(H(1) + (k − 1) H(0)) + 1][M (p)−A(p)] + λ H(0)−p

λ

A(1) =
λ` H(0)

[λ (H(1) + (k − 1) H(0))]
,

A∗(1) = λ`
{

[λ (H(1) + (k − 1) H(0)) + 1] H(0) − k F (0)
}

(4.4)

where A∗ : R → R, A : R → R, B : R → R, a : R → R, b : R → R are additive

functions; M : [0, 1] → R satisfies (2.9), (2.11) and

M (1) =
λ (H(1) + (` − 1) H(0)) + 1

λ (H(1) + (k − 1) H(0)) + 1
(4.5)

with [λ (H(1) + (k − 1) H(0)) + 1] 6= 0 in (4.3), (4.3a), (4.4) and (4.5).

Proof. Let us write (1.7) in the multiplicative form

k
∑

i=1

∑̀

j=1

[λ F (piqj) + piqj] =
k
∑

i=1

[λ H(pi) + pi]
∑̀

j=1

[λ H(qj) + qj].(4.6)

Define the mappings f : I → R, h : I → R as

f (x) = λ F (x) + x, h(x) = λ H(x) + x(4.7)

for all x ∈ I . Then, (4.6) reduces to the functional equation (1.11) whose

solutions are given by (3.1) subject to the condition (3.2); (3.3) subject to (3.3a);

and (3.4) in which A∗ : R → R, A : R → R, B : R → R, a : R → R , b : R → R

are additive functions; and M : [0, 1] → R is a mapping which satisfies (2.9),

(2.11) and (3.5). Now making use of (4.7) and (3.1) subject to the condition

(3.2); (3.3) subject to the condition (3.3a); and (3.4); the required solutions (4.1)



The general solutions of a functional equation ... 21

subject to the condition (4.2); (4.3) subject to the condition (4.3a) and (4.4)

follow. The details are omitted.

5. The general solutions of functional equation (1.7) when λ = 0

If λ = 0, then (1.7) reduces to the functional equation

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

H(pi) +
∑̀

j=1

H(qj)(5.1)

where k ≥ 3, ` ≥ 3 are fixed integers and (p1, . . . , pk) ∈ Γk , (q1, . . . , q`) ∈ Γ` .

The substitutions p1 = 1, p2 = . . . = pk = 0 in (5.1) yield

∑̀

j=1

[F (qj) − H(qj)] = H(1) + (k − 1) H(0) − `(k − 1) F (0).(5.2)

Hence, by Lemma 1,

F (p) = H(p)+A∗
1
(p)−

1

`
A∗

1
(1)+

1

`

{

H(1)+(k−1)H(0)−`(k−1)F (0)
}

(5.3)

where A∗
1

: R → R is additive with

A∗
1
(1) = H(1) + (k + ` − 1) H(0) − k` F (0).(5.4)

From (5.3) and (5.4), we obtain

k
∑

i=1

∑̀

j=1

F (piqj) =
k
∑

i=1

∑̀

j=1

H(piqj) + H(1) − (k − 1)(` − 1) H(0).(5.5)

From (5.1) and (5.5), we obtain

k
∑

i=1

∑̀

j=1

H(piqj) =
k
∑

i=1

H(pi)+
∑̀

j=1

H(qj)−
{

H(1)−(k−1)(`−1)H(0)
}

.(5.6)

Define H1 : [0, 1] → R as

H1(x) = H(x) −
{

H(1) − (k − 1)(` − 1) H(0)
}

x(5.7)



22 Prem Nath and Dhiraj Kumar Singh

for all x ∈ [0, 1]. Then, equation (5.6) reduces to

k
∑

i=1

∑̀

j=1

H1(piqj) =
k
∑

i=1

H1(pi) +
∑̀

j=1

H1(qj).(5.8)

Putting p1 = q1 = 1 and p2 = . . . = pk = q2 = . . . = q` = 0 in (5.8), we obtain

H1(1) = (k − 1)(` − 1) H1(0). Define H2 : [0, 1] → R as

H2(x) = H1(x) − H1(0) − [H1(1) − H1(0)] x(5.9)

for all x ∈ [0, 1]. Then

k
∑

i=1

∑̀

j=1

H2(piqj) =
k
∑

i=1

H2(pi) +
∑̀

j=1

H2(qj)(5.10)

where H2(1) = H2(0) = 0, and (p1, . . . , pk) ∈ Γk , (q1, . . . , q`) ∈ Γ` , k ≥ 3, ` ≥ 3

fixed integers. Theorem 2 (p-78 in [12]) may now be written as :

Theorem 4. Let k ≥ 3, ` ≥ 3 be fixed integers. The mapping H2 : [0, 1] → R

with H2(1) = 0, H2(0) = 0, defined in (5.9) is a solution of (5.10) if and only if

H2(p) =

{

a(p) + D(p, p) if 0 < p ≤ 1

0 if p = 0
(5.11)

where a : R → R is additive; D : R × ]0, 1] → R is additive in the first variable

and there exists a function E : R × R → R, additive in both variables such that

E(1, 1) = a(1) and, moreover,

D(pq, pq) − D(pq, p) − D(pq, q) = E(p, q) if 0 < p ≤ 1, 0 < q ≤ 1.(5.12)

Making use of corollary 3 on p-81 in [12], it follows that

H1(p) =







c + c (k` − k − `) p + a(p) + D(p, p) if 0 < p ≤ 1

c if p = 0
(5.13)

where c = H1(0) is an arbitrary real constant, a : R → R is additive,

D : R × ]0, 1] → R is as described above in Theorem 4. Now from (5.7) and



The general solutions of a functional equation ... 23

(5.13), we obtain

H(p) =







c (1 − p) + d1 p + a(p) + D(p, p) if 0 < p ≤ 1

c if p = 0
(5.14)

where c = H(0) and d1 = H(1) are arbitrary real constants, a : R → R is

additive function; D : R × ]0, 1] → R as described above in Theorem 4. Now from

(5.3), (5.4) and (5.14), we obtain

F (p) =







d0 + d1 p − c p + a(p) + A
∗

1
(p) + D(p, p) if 0 < p ≤ 1

d0 if p = 0
(5.15)

where c = H(0), d0 = F (0), d1 = H(1) are arbitrary real constants; a : R → R,

A∗
1

: R → R are additive functions with A∗
1
(1) given by (5.4); D : R × ]0, 1] → R

as described above in Theorem 4. Thus, we have proved the following:

Theorem 5. Let k ≥3, `≥3 be fixed integers. The mappings F : [0, 1] → R,

H : [0, 1] → R satisfy the equation (5.1) if and only if F and H are respectively

of the forms (5.15) and (5.14) with A∗
1
(1) given by (5.4) and D as described above

in Theorem 4.

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