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A remark on a logarithmic functional equation
J. Math. Anal. Appl. 336 (2007) 745–748 www.elsevier.com/locate/jmaa
Note
A remark on a logarithmic functional equation ✩ Jae-Young Chung School of Mathematics, Informatics and Statistics, Kunsan National University, Kunsan 573-701, Republic of Korea Received 30 August 2006 Available online 6 March 2007 Submitted by Richard M. Aron
Abstract We revisit the logarithmic functional equation of Heuvers and Kannappan [K.J. Heuvers, Pl. Kannappan, A third logarithmic functional equation and Pexider generalizations, Aequationes Math. 70 (2005) 117–121] and give a simple proof of the result and discuss the locally integrable solutions of the equation. © 2007 Elsevier Inc. All rights reserved. Keywords: Logarithmic functional equation; Distributions
1. On the results of Heuvers and Kannappan In [2] K.J. Heuvers and Pl. Kannappan showed that the general solution f, g, h : (0, ∞) → R of the Pexider-logarithmic functional equation f (x + y) − g(xy) = h(1/x + 1/y),
x, y > 0,
(1.1)
has the form f (x) = c1 + c2 + L(x),
(1.2)
g(x) = c1 + L(x),
(1.3)
h(x) = c2 + L(x),
(1.4)
where L : (0, ∞) → R satisfies the logarithmic functional equation L(xy) = L(x) + L(y).
(1.5)
✩ This work was supported by the Korean Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2005-015-C00026). E-mail address: [email protected].
0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.02.072
746
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
Here we give a simple proof of the result. Set xy = t,
x+y xy
= s to get
f (ts) = g(t) + h(s)
(1.6)
for ts 2 4. Let t, s, u > 0, ts 2 u2 4 and tsu2 4. Then we have by (1.6) f (tsu) = g(t) + h(su), f (tsu) = g(ts) + h(u), so that g(ts) − g(t) = h(su) − h(u) := α(s).
(1.7)
Hence g(ts) = g(t) + α(s) for all t, s > 0. Thus there is a logarithmic function L such that g(t) = c1 + L(t),
α(s) = L(s).
(1.8)
Then we have by (1.7) and (1.8) h(su) = h(u) + L(s) for all s, u > 0. Thus we have h(s) = c2 + L(s). These g and h in (1.1) yields (1.2). This completes the proof. Remark. As a direct consequence of the above result it is obtained that every solution f : (0, ∞) → R of the equation f (x + y) − f (xy) = f (1/x + 1/y),
x, y > 0,
satisfies the logarithmic functional equation f (xy) = f (x) + f (y),
x, y > 0.
2. Regular solutions of the equation In this section using methods of the Schwartz distributions in [1] we consider the locally integrable solution f, g, h : (0, ∞) → C of Eq. (1.1). Theorem 2.1. Every locally integrable solution f, g, h of Eq. (1.1) has the form f (z) = c1 + c2 + a ln z,
z > 0,
(2.1)
g(z) = c1 + a ln z,
z > 0,
(2.2)
h(z) = c2 + a ln z,
z > 0,
(2.3)
where c1 , c2 , a ∈ C.
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
747
Proof. We denote by S, P , R the functions S(x, y) = x +y, P (x, y) = xy, R(x, y) = 1/x +1/y. Every locally integrable function on (0, ∞) can be viewed as a distribution (a continuous linear functional on the space Cc∞ ((0, ∞)) of smooth functions defined on (0, ∞) with bounded supports, equipped with a suitable topology) and Eq. (1.1) can be read as f ◦ S − g ◦ P = h ◦ R,
(2.4)
where ◦ denotes the pullback of distributions. Applying ∂1 − ∂2 in (2.4) we have (x − y)(g ◦ P ) =
x2 − y2 (h ◦ R), x2y2
which implies g ◦ P =
x +y (h ◦ R) x2y2
(2.5)
on the space Cc∞ (V ), where V = {(x, y): x > y > 0}. Let J : V → U := {(s, t): st 2 > 4, t > 0} be a diffeomorphism defined by J (x, y) = (xy, 1/x + 1/y). Then, since (S ◦ J −1 )(s, t) = s, (R ◦ J −1 )(s, t) = t, taking pullback of (2.5) by J −1 , we have zg (z) ◦ P1 = zh (z) ◦ P2 (2.6) as distributions in D (U ), where P1 (s, t) = s, P2 (s, t) = t. Applying ∂1 in (2.6) (or localize and apply tensor product of test functions) it follows that zg (z) = zh (z) := a
(2.7)
are the same constants. Thus it follows that g = c1 + a ln z,
(2.8)
h = c2 + a ln z
(2.9)
in where a, c1 , c2 ∈ C. Choose ψ ∈ Cc∞ ((0, ∞)) such that the sense of Schwartz distributions, ∞ ψ(y) dy = 1. For given ϕ ∈ Cc ((0, ∞)), applying φ(x, y) = ϕ(x + y)ψ(y) in (2.4) and using (2.8) and (2.9), we have (2.10) f, ϕ = (c1 + c2 + a ln z)ϕ(z) dz for all ϕ ∈ Cc∞ ((0, ∞)). Thus it follows from (2.8), (2.9) and (2.10) that f (z) = c1 + c2 + a ln z,
a.e. z > 0,
(2.11)
g(z) = c1 + a ln z,
a.e. z > 0,
(2.12)
h(z) = c2 + a ln z,
a.e. z > 0.
(2.13)
Equations (2.11)–(2.13) imply that (2.1)–(2.3) hold for all z in a subset Ω ⊂ (0, ∞) with m(Ω c ) = 0. For given z > 0, let p, q : (0, ∞) → R by p(t) = t + z/t, q(t) = 1/t + t/z. Then since m(p −1 (Ω)c ∪ q −1 (Ω)c ) = 0, we have p −1 (Ω) ∩ q −1 (Ω) = ∅. Thus we can choose x, y > 0, so that xy = z and x + y, 1/x + 1/y ∈ Ω. It follows from (1.1), (2.2) and (2.3) that g(z) = f (x + y) − h(1/x + 1/y) = c1 + a ln(xy) = c1 + a ln z, which means that Eq. (2.2) holds for all z > 0. Similarly, let p : (0, z) → R by p(t) = 1/t + 1/(z − t). Then we have p −1 (Ω) = ∅. Thus we can choose x, y > 0, so that x + y = z and 1/x + 1/y ∈ Ω. Similarly we can verify that Eq. (2.3) holds for all z > 0, and then, from (1.1) Eq. (2.1) follows. This completes the proof. 2
748
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
Acknowledgments The author would like to express his sincere gratitude to the referee for improving the paper (in particular, the proof in Section 1). This work was supported by the Korean Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2005-015-C00026).
References [1] J.A. Baker, Distributional methods for functional equations, Aequationes Math. 62 (2001) 136–142. [2] K.J. Heuvers, Pl. Kannappan, A third logarithmic functional equation and Pexider generalizations, Aequationes Math. 70 (2005) 117–121.
Note
A remark on a logarithmic functional equation ✩ Jae-Young Chung School of Mathematics, Informatics and Statistics, Kunsan National University, Kunsan 573-701, Republic of Korea Received 30 August 2006 Available online 6 March 2007 Submitted by Richard M. Aron
Abstract We revisit the logarithmic functional equation of Heuvers and Kannappan [K.J. Heuvers, Pl. Kannappan, A third logarithmic functional equation and Pexider generalizations, Aequationes Math. 70 (2005) 117–121] and give a simple proof of the result and discuss the locally integrable solutions of the equation. © 2007 Elsevier Inc. All rights reserved. Keywords: Logarithmic functional equation; Distributions
1. On the results of Heuvers and Kannappan In [2] K.J. Heuvers and Pl. Kannappan showed that the general solution f, g, h : (0, ∞) → R of the Pexider-logarithmic functional equation f (x + y) − g(xy) = h(1/x + 1/y),
x, y > 0,
(1.1)
has the form f (x) = c1 + c2 + L(x),
(1.2)
g(x) = c1 + L(x),
(1.3)
h(x) = c2 + L(x),
(1.4)
where L : (0, ∞) → R satisfies the logarithmic functional equation L(xy) = L(x) + L(y).
(1.5)
✩ This work was supported by the Korean Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2005-015-C00026). E-mail address: [email protected].
0022-247X/$ – see front matter © 2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.02.072
746
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
Here we give a simple proof of the result. Set xy = t,
x+y xy
= s to get
f (ts) = g(t) + h(s)
(1.6)
for ts 2 4. Let t, s, u > 0, ts 2 u2 4 and tsu2 4. Then we have by (1.6) f (tsu) = g(t) + h(su), f (tsu) = g(ts) + h(u), so that g(ts) − g(t) = h(su) − h(u) := α(s).
(1.7)
Hence g(ts) = g(t) + α(s) for all t, s > 0. Thus there is a logarithmic function L such that g(t) = c1 + L(t),
α(s) = L(s).
(1.8)
Then we have by (1.7) and (1.8) h(su) = h(u) + L(s) for all s, u > 0. Thus we have h(s) = c2 + L(s). These g and h in (1.1) yields (1.2). This completes the proof. Remark. As a direct consequence of the above result it is obtained that every solution f : (0, ∞) → R of the equation f (x + y) − f (xy) = f (1/x + 1/y),
x, y > 0,
satisfies the logarithmic functional equation f (xy) = f (x) + f (y),
x, y > 0.
2. Regular solutions of the equation In this section using methods of the Schwartz distributions in [1] we consider the locally integrable solution f, g, h : (0, ∞) → C of Eq. (1.1). Theorem 2.1. Every locally integrable solution f, g, h of Eq. (1.1) has the form f (z) = c1 + c2 + a ln z,
z > 0,
(2.1)
g(z) = c1 + a ln z,
z > 0,
(2.2)
h(z) = c2 + a ln z,
z > 0,
(2.3)
where c1 , c2 , a ∈ C.
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
747
Proof. We denote by S, P , R the functions S(x, y) = x +y, P (x, y) = xy, R(x, y) = 1/x +1/y. Every locally integrable function on (0, ∞) can be viewed as a distribution (a continuous linear functional on the space Cc∞ ((0, ∞)) of smooth functions defined on (0, ∞) with bounded supports, equipped with a suitable topology) and Eq. (1.1) can be read as f ◦ S − g ◦ P = h ◦ R,
(2.4)
where ◦ denotes the pullback of distributions. Applying ∂1 − ∂2 in (2.4) we have (x − y)(g ◦ P ) =
x2 − y2 (h ◦ R), x2y2
which implies g ◦ P =
x +y (h ◦ R) x2y2
(2.5)
on the space Cc∞ (V ), where V = {(x, y): x > y > 0}. Let J : V → U := {(s, t): st 2 > 4, t > 0} be a diffeomorphism defined by J (x, y) = (xy, 1/x + 1/y). Then, since (S ◦ J −1 )(s, t) = s, (R ◦ J −1 )(s, t) = t, taking pullback of (2.5) by J −1 , we have zg (z) ◦ P1 = zh (z) ◦ P2 (2.6) as distributions in D (U ), where P1 (s, t) = s, P2 (s, t) = t. Applying ∂1 in (2.6) (or localize and apply tensor product of test functions) it follows that zg (z) = zh (z) := a
(2.7)
are the same constants. Thus it follows that g = c1 + a ln z,
(2.8)
h = c2 + a ln z
(2.9)
in where a, c1 , c2 ∈ C. Choose ψ ∈ Cc∞ ((0, ∞)) such that the sense of Schwartz distributions, ∞ ψ(y) dy = 1. For given ϕ ∈ Cc ((0, ∞)), applying φ(x, y) = ϕ(x + y)ψ(y) in (2.4) and using (2.8) and (2.9), we have (2.10) f, ϕ = (c1 + c2 + a ln z)ϕ(z) dz for all ϕ ∈ Cc∞ ((0, ∞)). Thus it follows from (2.8), (2.9) and (2.10) that f (z) = c1 + c2 + a ln z,
a.e. z > 0,
(2.11)
g(z) = c1 + a ln z,
a.e. z > 0,
(2.12)
h(z) = c2 + a ln z,
a.e. z > 0.
(2.13)
Equations (2.11)–(2.13) imply that (2.1)–(2.3) hold for all z in a subset Ω ⊂ (0, ∞) with m(Ω c ) = 0. For given z > 0, let p, q : (0, ∞) → R by p(t) = t + z/t, q(t) = 1/t + t/z. Then since m(p −1 (Ω)c ∪ q −1 (Ω)c ) = 0, we have p −1 (Ω) ∩ q −1 (Ω) = ∅. Thus we can choose x, y > 0, so that xy = z and x + y, 1/x + 1/y ∈ Ω. It follows from (1.1), (2.2) and (2.3) that g(z) = f (x + y) − h(1/x + 1/y) = c1 + a ln(xy) = c1 + a ln z, which means that Eq. (2.2) holds for all z > 0. Similarly, let p : (0, z) → R by p(t) = 1/t + 1/(z − t). Then we have p −1 (Ω) = ∅. Thus we can choose x, y > 0, so that x + y = z and 1/x + 1/y ∈ Ω. Similarly we can verify that Eq. (2.3) holds for all z > 0, and then, from (1.1) Eq. (2.1) follows. This completes the proof. 2
748
J.-Y. Chung / J. Math. Anal. Appl. 336 (2007) 745–748
Acknowledgments The author would like to express his sincere gratitude to the referee for improving the paper (in particular, the proof in Section 1). This work was supported by the Korean Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2005-015-C00026).
References [1] J.A. Baker, Distributional methods for functional equations, Aequationes Math. 62 (2001) 136–142. [2] K.J. Heuvers, Pl. Kannappan, A third logarithmic functional equation and Pexider generalizations, Aequationes Math. 70 (2005) 117–121.