Abstract. The present work aims to find the solutions f, g, h, , m : (u, v)  for all x, y, u, v ∈ R without any regularity assumptions on the unknown functions. This equation is a generalization of a functional equation which arises from the characterization of the determinant of symmetric matrices

Kelly B Houston

Prasanna K Sahoo

CiteSeerX

SARAJEVO JOURNAL OF MATHEMATICS
Vol.3 (15) (2007), 61–71
ON A FUNCTIONAL EQUATION RELATED TO THE
DETERMINANT OF SYMMETRIC TWO-BY-TWO
MATRICES
KELLY B. HOUSTON AND PRASANNA K. SAHOO
Abstract. The present work aims to find the solutions f, g, h, `, m :
R2 → R of the functional equation f(ux − vy, uy − vx) = g(x, y) +
h(u, v) + `(x, y) m(u, v) for all x, y, u, v ∈ R without any regularity as-
sumptions on the unknown functions. This equation is a generalization
of a functional equation which arises from the characterization of the
determinant of symmetric matrices.
1. Introduction
Let us define f : R2 → R by
f(x, y) = det
(
x y
y x
)
for all x, y ∈ R. Then, since
det
(
ux− vy uy − vx
uy − vx ux− vy
)
= det
(
x y
y x
)
det
(
u v
v u
)
,
we have the functional equation
f(ux− vy, uy − vx) = f(x, y) f(u, v) (1)
for all x, y, u, v ∈ R. Obviously, f(x, y) = x2 − y2 is a solution of the
functional equation (1). A functional equation similar to (1) was studied in
[2]. In the solved and unsolved problems column of the News Letter of the
European Mathematical Society, the second author [6] posed the following
problem: Determine the general solutions f : R2 → R of the functional
equation
f(ux− vy, uy − vx) = f(x, y) + f(u, v) + f(x, y) f(u, v) (2)
2000 Mathematics Subject Classification. Primary 39B22.
Key words and phrases. Determinant, functional equation, logarithmic function, mul-
tiplicative function.
62 KELLY B. HOUSTON AND PRASANNA K. SAHOO
for all x, y, u, v ∈ R. In [4], among others we gave the general solution of
the functional equation (2) without any regularity assumptions on f . A
generalization of the functional equations (1) and (2) is the following:
f(ux− vy, uy − vx) = g(x, y) + h(u, v) + `(x, y) m(u, v) (3)
for all x, y, u, v ∈ R. Here f, g, h, `, m : R2 → R are unknown functions to
be determined. For an account on the subject of functional equations the
interested reader should refer to [1], [3], [5], [7], [8], and [9].
In this paper, we determine the general solution of the functional equation
(3) without any regularity assumptions on the unknown functions f,g,h, `,m :
R2 → R. Our method of solution is elementary and direct.
2. Preliminary results
A function M : R → R is said to be a multiplicative function if and only
if it satisfies M(xy) = M(x) M(y) for all x, y ∈ R. An identically constant
multiplicative function M is either M = 0 or M = 1.
Let D be an interval in R such that whenever x, y ∈ D, then xy ∈ D.
A function L : D → R is said to be a logarithmic function if and only if
L(xy) = L(x) + L(y) for all x, y ∈ D. Note that if 0 ∈ D, then L = 0.
Lemma 1. Let D ⊆ R be an interval such that if x, y ∈ D, then xy ∈ D.
The general solution f : D2 → R of the functional equation
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) (4)
holding for all x1, x2, y1, y2 ∈ D is given by
f(x, y) = L1(x) + L2(y), (5)
where L1, L2 : D → R are logarithmic functions. If D = R, then f(x, y) ≡ 0
is the only solution to functional equation (4).
Proof. It is easy to check that the solution enumerated in (5) satisfies func-
tional equation (4). Next, we show that (5) is the only solution of (4).
Suppose f is identically a constant, say f ≡ c. Then from (4) we have
c = 0. Hence the identically constant solution of (4) is f(x, y) = 0 for all
x, y ∈ D, which is a solution included in (5).
From now on we assume that f is not identically constant. Let a ∈ D be
a fixed element and f : D2 → R be such that it satisfies (4). Then
f(x, y) = f(x, y) + f(a, a) + f(a, a)− 2f(a, a)
= f(xaa, aay)− 2f(a, a)
= f((xa)a, (ya)a)− 2f(a, a)
= f(xa, a) + f(a, ya)− 2f(a, a)
= f(xa, a)− f(a, a) + f(a, ya)− f(a, a)
ON A FUNCTIONAL EQUATION 63
= L1(x) + L2(y)
where
L1(x) := f(xa, a)− f(a, a)
and
L2(y) := f(a, ya)− f(a, a).
Now we show that L1 and L2 are logarithmic functions in D. Consider
L1(xy) = f(xya, a)− f(a, a)
= f(xya, a) + f(a, a)− 2f(a, a)
= f(xyaa, aa)− 2f(a, a)
= f(xa, a) + f(ya, a)− 2f(a, a)
= f(xa, a)− f(a, a) + f(ya, a)− f(a, a)
= L1(x) + L1(y).
Hence L1 is logarithmic. Similarly, one can show that
L2(xy) = L2(x) + L2(y)
and hence L2 is logarithmic. Thus
f(x, y) = L1(x) + L2(y)
where L1 and L2 are logarithmic functions. ¤
Lemma 2. Let D ⊆ R be an interval such that if x, y ∈ D, then xy ∈ D.
The general solution f : D2 → R of the functional equation
f(x1y2, x2y1) = f(x1, y1) f(y2, x2) (6)
holding for all x1, x2, y1, y2 ∈ D is given by
f(x, y) = M1(x)M2(y), (7)
where M1,M2 : D → R are multiplicative functions.
Proof. It is easy to check that the solution enumerated in (7) satisfies func-
tional equation (6). Next, we will show that (7) is the only solution of (6).
Suppose f is identically a constant, say f ≡ c. Then from (6) we have
c2 − c = 0 for all x, y ∈ D. Hence c = 1 or c = 0. Thus the only identically
constant solutions of (6) are f(x, y) = 1 and f(x, y) = 0 for all x, y ∈ D,
which are solutions included in (7).
From now on we assume that f is not identically constant. Let a ∈ D be
a fixed element and f : D2 → R be such that it satisfies (6) with f(a, a) 6= 0.
Then
f(x, y) = f(x, y) f(a, a) f(a, a) f(a, a)−2
= f(xaa, aay) f(a, a)−2
64 KELLY B. HOUSTON AND PRASANNA K. SAHOO
= f((xa)a, (ya)a) f(a, a)−2
= f(xa, a) f(a, ya) f(a, a)−2
= f(xa, a) f(a, a)−1 f(a, ya) f(a, a)−1
= M1(x)M2(y)
where
M1(x) := f(xa, a) f(a, a)−1
and
M2(y) := f(a, ya) f(a, a)−1.
Now we show that M1 and M2 are multiplicative functions in D. Consider
M1(xy) = f(xya, a) f(a, a)−1
= f(xya, a) f(a, a) f(a, a)−2
= f(xyaa, aa) f(a, a)−2
= f(xa, a) f(ya, a) f(a, a)−2
= f(xa, a) f(a, a)−1 f(ya, a) f(a, a)−1
= M1(x) M1(y).
Hence M1 is multiplicative. Similarly, one can show that
M2(xy) = M2(x)M2(y)
and hence M2 is multiplicative. Thus
f(x, y) = M1(x) M2(y)
where M1 and M2 are multiplicative functions. ¤
Remark 1. Note that if we replace D by a commutative semigroup in
Lemma 1 and Lemma 2, then both of the lemmas are still valid.
Lemma 3. The general solution f, g : R2 → R of the functional equation
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + g(x1, y1) g(x2, y2) (8)
holding for all x1, x2, y1, y2 ∈ R is given by{
f(x, y) = δ2 [M(xy)− 1]
g(x, y) = δ [M(xy)− 1], (9)
where M : R→ R is a multiplicative function and δ is an arbitrary constant.
Proof. It is easy to check that the solution enumerated in (9) satisfies func-
tional equation (8). Next, we will show that (9) is the only solution of (8).
Suppose g is identically constant, say g ≡ −δ. Then from (8) we have
that
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + δ2 (10)
ON A FUNCTIONAL EQUATION 65
for all x1, x2, y1, y2 ∈ R. Now we define a function F : R2 → R by
F (x, y) := f(x, y) + δ2 (11)
for all x, y ∈ R. Then (10) becomes
F (x1y2, x2y1) = F (x1, y1) + F (y2, x2) (12)
for all x1, x2, y1, y2 ∈ R. From Lemma 1, we see that the solution of (12) is
F (x, y) = 0 (13)
for all x, y ∈ R. Hence
f(x, y) = −δ2 (14)
for all x, y ∈ R, which is a solution included in (9).
From now on we assume that g is not identically constant. Interchanging
x1 with y2 and x2 with y1 in (8) and comparing the resulting equations with
(8) we see that
g(x1, y1) g(x2, y2) = g(y1, x1) g(y2, x2) (15)
for all x1, x2, y1, y2 ∈ R. Since g is non-constant, there exists x0, y0 ∈ R such
that g(y0, x0) 6= 0. Letting x2 = x0 and y2 = y0 in (15) we obtain
g(y, x) = α g(x, y) (16)
for all x, y ∈ R where α is a constant and α 6= 0 (since otherwise g ≡ 0).
Setting x2 = y2 = 0 in (8) we obtain
f(0, 0) = f(x1, y1) + f(0, 0) + g(x1, y1) g(0, 0) (17)
for all x1, y1 ∈ R. Thus,
f(x, y) = α0 g(x, y) (18)
for all x, y ∈ R. Again note that α0 6= 0 since otherwise f ≡ 0 and hence
g ≡ 0 by (8). Therefore,
g(x, y) = k f(x, y) (19)
for all x, y ∈ R where the constant k 6= 0.
Now, using (19) in (8) we see that
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + k2 f(x1, y1) f(x2, y2) (20)
for all x1, x2, y1, y2 ∈ R. Using (16) and (19) we see that
k f(x, y) = g(x, y) = α g(y, x) = α k f(y, x)
which gives us
f(x, y) = α f(y, x) (21)
for all x, y ∈ R.
Next, using (20) and (21) we have
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + α k2 f(x1, y1) f(y2, x2) (22)
66 KELLY B. HOUSTON AND PRASANNA K. SAHOO
for all x1, x2, y1, y2 ∈ R. Define F : R2 → R by
F (x, y) := α k2 f(x, y) + 1 (23)
for all x, y ∈ R. Then using (22) and (23) we have
F (x1y2, x2y1) = F (x1, y1) F (y2, x2) (24)
for all x1, x2, y1, y2 ∈ R.
From Lemma 2 we obtain
F (x, y) = M1(x) M2(y) (25)
for all x, y ∈ R. Thus from (19) and (23) we have{
f(x, y) = 1
α k2
[M1(x) M2(y)− 1]
g(x, y) = 1α k [M1(x) M2(y)− 1]
(26)
for all x, y ∈ R.
Substituting (26) back into (8) we see that α = 1 and M1 = M2 = M for
all x ∈ R. Thus if we write δ = 1/ k, (26) becomes{
f(x, y) = δ2 [M(xy)− 1]
g(x, y) = δ [M(xy)− 1] (27)
for all x, y ∈ R, which is a solution to (8). This completes the proof of the
lemma. ¤
Lemma 4. The general solutions f, `, m : R2 → R of the functional equation
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + `(x1, y1) m(x2, y2) (28)
holding for all x1, x2, y1, y2 ∈ R are given by{
f(x, y) ≡ 0
`(x, y) m(u, v) = 0 (29)
and 


f(x, y) = 1k1 k2 [M1(x) M2(y)− 1]
`(x, y) = 1k2 [M1(x) M2(y)− 1]
m(x, y) = 1k1 [M1(y) M2(x)− 1],
(30)
where M1,M2 : R→ R are multiplicative functions and k1 and k2 are arbi-
trary nonzero constants.
Proof. It is easy to check that the solutions enumerated in (29)-(30) satisfy
functional equation (28). Next, we will show that (29)-(30) are the only
solutions of (28).
Suppose `(x, y) m(u, v) = 0. Then (28) becomes
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) (31)
ON A FUNCTIONAL EQUATION 67
for all x1, x2, y1, y2 ∈ R. From (31) and Lemma 1 we see that
f(x, y) = 0 (32)
for all x, y ∈ R, which is a solution to (28).
Now, suppose `(x, y) = m(x, y) for all x, y ∈ R. Then, (28) becomes
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + `(x1, y1) `(x2, y2) (33)
for all x1, x2, y1, y2 ∈ R. From (33) and Lemma 3 we see that


f(x, y) = δ2 [M(xy)− 1]
`(x, y) = δ [M(xy)− 1]
m(x, y) = δ [M(xy)− 1]
(34)
for all x, y ∈ R where M : R → R is a multiplicative function, which is a
solution to (28) and a special case of (30).
Finally, suppose `(x, y) 6= m(x, y) and neither is identically zero. Setting
y2 = x2 = 0 in (28) we see that
f(0, 0) = f(x1, y1) + f(0, 0) + `(x1, y1) m(0, 0) (35)
for all x1, y1 ∈ R. Thus,
`(x, y) = k1 f(x, y) (36)
for all x, y ∈ R where k1 6= 0 (since otherwise ` ≡ 0).
Similarly, setting y1 = x1 = 0 in (28) we see that
f(0, 0) = f(0, 0) + f(y2, x2) + `(0, 0)m(x2, y2) (37)
for all x2, y2 ∈ R. Thus,
m(x, y) = k2 f(y, x) (38)
for all x, y ∈ R where k2 6= 0 (since otherwise m ≡ 0).
Now using (28), (36), and (38) we have
f(x1y2, x2y1) = f(x1, y1) + f(y2, x2) + k1 k2 f(x1, y1) f(y2, x2) (39)
for all x1, x2, y1, y2 ∈ R. Define a function H : R2 → R by
H(x, y) := k1 k2 f(x, y) + 1 (40)
for all x, y ∈ R. Then (39) becomes
H(x1y2, x2y1) = H(x1, y1) H(y2, x2) (41)
for all x1, x2, y1, y2 ∈ R.
From Lemma 2 we get
H(x, y) = M1(x) M2(y) (42)
68 KELLY B. HOUSTON AND PRASANNA K. SAHOO
for all x, y ∈ R. Thus using (36), (38), (40), and (42) we see that



f(x, y) = 1k1 k2 [M1(x) M2(y)− 1]
`(x, y) = 1k2 [M1(x) M2(y)− 1]
m(x, y) = 1k1 [M1(y) M2(x)− 1],
(43)
for all x, y ∈ R where M1,M2 : R → R are multiplicative functions, which
is a solution to (28). ¤
3. Main result
Now we are ready to prove our main result.
Theorem 1. The general solutions f, g, h, `,m : R2 → R of the functional
equation
f(ux− vy, uy − vx) = g(x, y) + h(u, v) + `(x, y) m(u, v) (44)
holding for all x, y, u, v ∈ R are given by


f(x, y) ≡ α1 α2 − β1 − β2
g(x, y) = α1 `(x, y)− β2 + α1 α2
h(x, y) = −β1
`(x, y) is arbitrary
m(x, y) ≡ −α1
(45)



f(x, y) ≡ α1 α2 − β1 − β2
g(x, y) ≡ −β2
h(x, y) = α2 m(x, y)− β1 + α1 α2
`(x, y) ≡ −α2
m(x, y) is arbitrary
(46)
and 


f(x, y) = 1k1k2 [M1(x + y)M2(x− y)− 1] + α1 α2 − β1 − β2
g(x, y) = 1k1k2 [M1(x + y)M2(x− y)− 1]
+ α1k2 [M1(x + y)M2(x− y)− 1]− β2
h(x, y) = 1k1k2 [M1(x + y)M2(x− y)− 1]
+ α2k1 [M1(x− y)M2(x + y)− 1]− β1
`(x, y) = 1k2 [M1(x + y)M2(x− y)− 1]− α2
m(x, y) = 1k1 [M1(x− y)M2(x + y)− 1]− α1,
(47)
where M1,M2 : R → R are multiplicative functions, α1, α2, β2 and β1 are
arbitrary constants, and k1 and k2 are nonzero arbitrary constants.
ON A FUNCTIONAL EQUATION 69
Proof. It is easy to check that the solutions enumerated in (45)-(47) satisfy
functional equation (44). Next, we will show that (45)-(47) are the only
solutions of (44).
First, we define functions F, G,H, L,N : R2 → R by



F (x, y) := f
(x+y
2 ,
x−y
2
)
G(x, y) := g
(x+y
2 ,
x−y
2
)
H(x, y) := h
(x+y
2 ,
x−y
2
)
L(x, y) := `
(x+y
2 ,
x−y
2
)
N(x, y) := m
(x+y
2 ,
x−y
2
)
(48)
for all x, y ∈ R. Using (48) in (44) we see that
F ((x + y)(u− v), (x− y)(u + v))
= G(x + y, x− y) + H(u + v, u− v) + L(x + y, x− y) N(u + v, u− v)
(49)
for all x, y, u, v ∈ R. Letting x1 = x + y, y1 = x− y, x2 = u + v, and y2 =
u− v we have
F (x1y2, x2y1) = G(x1, y1) + H(x2, y2) + L(x1, y1) N(x2, y2) (50)
for all x1, x2, y1, y2 ∈ R.
Setting x2 = y2 = 1 in (50) we have
F (x1, y1) = G(x1, y1) + H(1, 1) + L(x1, y1)N(1, 1) (51)
for all x1, y1 ∈ R. Simplifying we get
F (x1, y1) = G(x1, y1)− α1 L(x1, y1)− β1 (52)
for all x1, y1 ∈ R where α1 and β1 are arbitrary constants.
Similarly, setting x1 = y1 = 1 in (50) we have
F (y2, x2) = G(1, 1) + H(x2, y2) + L(1, 1)N(x2, y2) (53)
for all x2, y2 ∈ R. Simplifying we get
F (y2, x2) = H(x2, y2)− α2 N(x2, y2)− β2 (54)
for all x2, y2 ∈ R where α2 and β2 are arbitrary constants.
Then using (50), (52), and (54) we have
F (x1y2, x2y1)− β = F (x1, y1) + F (y2, x2) + α1 L(x1, y1)
+ α2 N(x2, y2) + L(x1, y1) N(x2, y2) (55)
70 KELLY B. HOUSTON AND PRASANNA K. SAHOO
for all x1, x2, y1, y2 ∈ R where β = β1 + β2. Next, we define functions
F1, L1, N1 : R2 → R by


F1(x, y) := F (x, y) + β − α1 α2
L1(x, y) := L(x, y) + α2
N1(x, y) := N(x, y) + α1
(56)
for all x, y ∈ R. Thus using (55) and (56) we see that
F1(x1y2, x2y1) = F1(x1, y1) + F1(y2, x2) + L1(x1, y1)N1(x2, y2) (57)
for all x1, x2, y1, y2 ∈ R.
By Lemma 4, (48), (52), (54), and (56) we see that


f(x, y) ≡ α1 α2 − β
g(x, y) = α1 `(x, y) + β1 + α1 α2 − β
h(x, y) ≡ β2 − β
`(x, y) is arbitrary
m(x, y) ≡ −α1
(58)



f(x, y) ≡ α1 α2 − β
g(x, y) ≡ β1 − β
h(x, y) = α2 m(x, y)− β + β2 + α1 α2
`(x, y) ≡ −α2
m(x, y) is arbitrary
(59)
and


f(x, y) = 1k1 k2 [M1(x + y) M2(x− y)− 1] + α1 α2 − β
g(x, y) = 1k1 k2 [M1(x + y) M2(x− y)− 1]
+ α1k2 [M1(x + y) M2(x− y)− 1] + β1 − β
h(x, y) = 1k1 k2 [M1(x + y) M2(x− y)− 1]
+ α2k1 [M1(x− y) M2(x + y)− 1]− β + β2
`(x, y) = 1k2 [M1(x + y) M2(x− y)− 1]− α2
m(x, y) = 1k1 [M1(x− y) M2(x + y)− 1]− α1
(60)
for all x, y ∈ R where M1,M2 : R → R are multiplicative functions. Note
that β = β1 + β2. Hence (58) - (60) yield the asserted solutions (45) - (47).
The proof of the theorem is now complete. ¤
Acknowledgment. The authors are very thankful to the referee for valu-
able suggestions. This work was partially supported by a SROP grant from
the Graduate School and an IRIG grant from the Office of the VP for Re-
search, University of Louisville.
ON A FUNCTIONAL EQUATION 71
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(Received: September 8, 2006) K. B. Houston and P. K. Sahoo
(Revised: January 5, 2007) Department of Mathematics
University of Louisville
Louisville, KY, 40292 USA
E–mail: kbhous01@louisville.edu
E–mail: sahoo@louisville.edu


ON A FUNCTIONAL EQUATION RELATED TO THE DETERMINANT OF SYMMETRIC TWO-BY-TWO MATRICES

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ON A FUNCTIONAL EQUATION RELATED TO THE DETERMINANT OF SYMMETRIC TWO-BY-TWO MATRICES

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