Continuous Solutions of a Functional Equation Involving the Harmonic and Arithmetic Means Rebecca Whitehead and Bruce Ebanks (faculty advisor) Department of Mathematics and Statistics Mississippi State University PO Drawer MA Mississippi State, MS 39762
[email protected] May 17, 2012
Let
f:
R+ × R+ → R
Abstract be a continuous solution to the functional equation
f a, b > 0. The aim f (a, b) = φ(ab) where φ :
for all
a + b 2ab , 2 a+b
= f (a, b)
of this paper will be to prove that
R+ → R
f
is of the form
is an arbitrary continuous function.
The
procedure used to prove the theorem will be to develop a sequence related to the arithmetic and harmonic means and show that it converges to a limit. Then the continuous function of the convergent sequence will also converge. It will be shown that that
ta = x
f ta, at and
the function
1
t a
is actually a function of the variable
= y,
f (x, y)
then
a
and
t
are found to be
a=
t alone. q
x, y > 0 such √ and t = xy . Thus √ single variable xy . If
x y
is equal to a continuous function of the
Introduction
A functional equation is an equation in which one or more functions appearing there are unknown. These functions are implicitly de ned by the equation, and the job of the solver is to determine their explicit forms. Functional equations can involve functions of one or more variables. The functional equation discussed in this paper is related to the arithmetic and harmonic means a+b A(a, b) = 2 1
and
2ab , a+b respectively, where a, b > 0. In particular, the functional equation discussed in this paper is related to the Gaussian arithmetic-geometric mean iteration. Starting with a = a0 > 0 and b = b0 > 0, the recursive relationship for terms an and bn of the Gaussian arithmetic-geometric mean iteration are de ned by H(a, b) =
an+1 =
an + b n 2
bn+1 =
p
and
an b n .
The limits have been shown to have the property lim bn = lim an as n → ∞. The limit M (a, b) = lim bn = lim an as n → ∞ is called the Gaussian arithmetic-geometric mean of a and b. This limit has been shown (see [2], pages 222-223) to satisfy the functional equation a+b √ , ab . M (a, b) = M 2 The functional equation discussed in this paper is a variant in which we replace the geometric mean by the harmonic mean.
2
Problem of Haruki and Rassias
In 1995, Haruki and Rassias [1] posed the following problem. They observed that any function of the form f (x, y) = φ(xy) satis es the functional equation a + b 2ab f , = f (a, b), (2.1) 2 a+b and they asked whether all continuous solutions of this functional equation must be of this form. This question was again posed in the 1998 book of Sahoo and Riedel [2] (see pages 230-231). We answer this question in the a rmative. That is, we will prove the following. Theorem 2.1. A function equation
R
+
→
R
(2.1),
f:
if and only if
R+ × R+ → R
f (a, b) = φ(ab)
.
2
is a continuous solution of the functional for an arbitrary continuous function
φ:
3
A Key Lemma
The proof of our theorem hinges on the convergence of a certain sequence of real numbers. To motivate our consideration of this sequence, suppose t and x are two positive real numbers and let a = tx and b = xt . Then equation (2.1) becomes
t tx + 2t2 t x , = f tx, f t 2 x tx + x
which simpli es to
f
tx2 + t 2tx , 2 2x x +1
t = f tx, . x
(3.1)
Let h : R+ × R+ → R be the continuous function de ned by t h(x, t) := f tx, x
(3.2)
for all x, t > 0. Then from (3.1) we obtain 2 x +1 h , t = h(x, t) 2x
(3.3)
for all x, t > 0. Our plan is to iterate this functional equation, so we consider the function s : R+ → + R de ned by x2 + 1 , s(x) := 2x where x > 0. Now consider the sequence {s(n) (x)} = {x, s(x), s(s(x)), s(3) (x), . . . } for n ≥ 0. That is, we de ne s(0) (x) := x, s(1) (x) := s(x), and s(n+1) (x) = s(s(n) (x)) for each positive integer n. Lemma 3.1. For each 1. 2.
x>0
s(n) (x) ≥ 1, and (n) s (x) is a monotone
and for all positive integers
n,
we have
non-increasing sequence.
We prove part 1 of the lemma by induction. For the case n = 1, we note that (x − 1) ≥ 0 for any x > 0. Expanding, we have x2 − 2x + 1 ≥ 0, and so
Proof.
2
s(x) =
x2 + 1 ≥1 2x
for all x > 0. 3
Now let x > 0 and suppose that s(m) (x) ≥ 1 for some positive integer m. Then, mimicking the argument above, from (s(m) (x) − 1)2 ≥ 0 it follows that (s(m) (x))2 − 2s(m) (x) + 1 ≥ 0, and therefore
s
(m+1)
(s(m) (x))2 + 1 (x) = ≥ 1. 2s(m) (x)
Thus, by the Principal of Mathematical Induction, s(n) (x) ≥ 1 for all x > 0 and for all positive integers n. We also prove part 2 by induction. For the case n = 1, s(x)2 + 1 (2) s(x) − s (x) = s(x) − 2s(x) 2 (s(x)) − 1 = 2s(x) (s(x) − 1)(s(x) + 1) = 2s(x) By the proof of part 1, we know that s(x) ≥ 1 for all x > 0. Thus the displayed quantity is nonnegative, and this yields s(x) ≥ s(2) (x) for x > 0. Now let x > 0 and suppose that s(m) (x) ≥ s(m+1) (x) for some positive integer m. Then we calculate (m+1) (s (x))2 + 1 (m+1) (m+2) (m+1) s (x) − s (x) = s (x) − 2s(m+1) (x) (s(m+1) (x))2 − 1 = 2s(m+1) (x) (s(m+1) (x) − 1)(s(m+1) (x) + 1) . = 2s(m+1) (x) Since for x > 0 we have s(n) (x) ≥ 1 for all n ≥ 1 by part 1, the displayed quantity is nonnegative. Hence s(m+1) (x) ≥ s(m+2) (x) for all x > 0. By the Principal of Mathematical Induction, s(n) (x) ≥ s(n+1) (x) for all x > 0 and for all positive integers n. This completes the proof of the lemma.
4
Proof of our Theorem
Since the sequence s(n) (x) is monotone non-increasing and bounded below by 1, it converges to a limit L ≥ 1. Letting n → ∞ in the equation s
(n+1)
(s(n) (x))2 + 1 (x) = 2s(n) (x) 4
we obtain
L2 + 1 L= (n) 2L Therefore L = 1, and the series s (x) converges to 1. Now we rewrite equation (3.3) as h(s(x), t) = h(x, t) and iterate: h(s(x), t) = h(x, t), h(s(2) (x), t) = h(s(x), t), h(s(3) (x), t) = h(s(2) (x), t), and so on. Thus by iteration we have
h(x, t) = h(s(n) (x), t), for all x, t > 0 and for all positive integers n. Since the function h is continuous, letting n → ∞ in this last equation gives
h(x, t) = h(1, t), a function of only one variable. Let us de ne the function φ : R+ → R by √ φ(t) := h(1, t),
(4.1)
for all t > 0. The function φ inherits continuity from the function h. The equation (3.2) can now be written as t f tx, = h (x, t) = φ(t2 ), x for all x, t > 0. Now given a, b > 0, we can nd t, x > 0 such that tx = a and r √ a Namely, we choose x = and t = ab. Then we have b t f (a, b) = f tx, = φ(t2 ) = φ(ab), x
t x
= b.
where φ is an arbitrary continuous function. This completes the proof of our theorem. As a nal observation, we note that all values of solution functions f are determined by their values on the diagonal. By equations (3.2) and (4.1), we see that √ √ √ f (a, b) = φ(ab) = h(1, ab) = f ( ab, ab).
References [1] H. Haruki and Th. Rassias, A new analogue of Gauss's functional equation, J. Math. Math. Sci. 18 (1995), 749756.
Int.
[2] P.K. Sahoo and T. Riedel, Mean Value Theorems and Functional Equations, World Scienti c Publishing Co. Pte. Ltd., Singapore, 1998. 5
