Find all functions f :R + →R + such that for all x ,y∈R + ,

4.6.5. Find all functions f :R + →R + such that for all x ,y∈R + ,

f (x) f (y f (x)) = f (x + y) .

International Mathematics Competition for University Students, 1999 Solution. We first assume that f (x) > 1 for some x ∈ R + . Setting y = x/ f (x) − 1,

we obtain the contradiction f (x) = 1. Hence f (x) ≤ 1 for each x ∈ R + , which implies that f is a decreasing function.

(x) = 1 for some x ∈ R + , then f (x + y) = f (y) for each y ∈ R , and by the monotonicity of f , it follows that f ≡ 1.

If f

Let now f (x) < 1 for each x ∈ R + . Then f is strictly decreasing, in particular one-to-one. By the equalities

f (x) f (y f (x)) = f (x + y) = f (y f (x) + x + y(1 − f (x))) = f (y f (x)) f

we deduce that x = (x + y(1 − f (x))) f (y f (x)). Setting x = 1, z = x f (1), and

a = 1 − f (1)/ f (1), we obtain f (z) = 1/1 + az. Combining the two cases, we conclude that f

(x) = 1/1 + ax for each x ∈ R + , where a ≥ 0. Conversely, a direct verification shows that the functions of this form

satisfy the initial equality. ⊓ ⊔

4.6.6. Let f : [−1,1]→R

be a continuous function such that f (2x 2 −1) = 2x f (x) ,

for all x ∈ [−1,1] . Prove that f equals zero identically.

Solution. For any real number t that is not an integer multiple of π we define

g (t) = f (cost)/sint. It follows that g(t + π ) = g(t). Moreover, by hypothesis,

f (2 cos 2 t − 1)

2 cost f (cost)

g (2t) =

= g(t).

sin 2t

sin 2t

In particular,

g (1 + n π /2 k ) = g(2 k +n π ) = g(2 k ) = g(1).

4.6 Functional Equations and Inequalities 167 By the continuity of f we deduce that g is continuous on its domain of definition.

But the set

{1 + n π /2 k ;n , k ∈ Z}

is dense in R. This implies that g should be constant on its domain. But g is odd, so g (t) = 0 for every t that is not an integer multiple of π . Therefore f (x) = 0, ∀x ∈ (−1,1). Taking now x = 0 and x = 1 in the functional equation in the hypoth- esis, we deduce that f (−1) = f (1) = 0. ⊓ ⊔

4.6.7. Let a and b be real numbers in (0, 1/2) and f :R →R a continuous function such that f ( f (x)) = a f (x)+bx , for all x . Prove that there exists a real constant c such that f (x) = cx .

Solution. We first observe that if f (x) = f (y) then x = y, so f is one-to-one. So, by the continuity of f , we deduce that f is strictly monotone. Moreover, f cannot have a finite limit L as x → + ∞. Indeed, in this case we have f ( f (x)) − a f (x) = bx, and the left-hand side is bounded, while the right-hand side is unbounded. Similarly,

f cannot have a finite limit as x → − ∞. Next, since f is monotone, we deduce that

f is onto. Let x 0 be arbitrarily chosen and x n +1 = f (x n ), for all n > 0 and x n =f −1 √ √ −1 (x n ) if n < 0. Let r 1 = (a + a 2 + 4b)/2 and r 2 = (a − a 2 + 4b)/2 be the roots of x 2 − ax − b = 0, so r 1 >0>r 2 and |r 1 | > |r 2 |. Thus, there exist c 1 ,c 2 ∈ Z such that x n =c 1 r n +c 2 r 1 n 2 for all n ∈ Z. Let us assume that f is increasing. If c 1 n is dominated by r n 2 , provided that n < 0 is small enough. In this case we have 0 < x n <x n +2 , contradiction, since

f (x n ) > f (x n +2 ). It follows that c 2 = 0, so x 0 =c 1 and x 1 =c 1 r 1 . Hence f (x) = r 1 x for all x. Analogously, if f is decreasing then f (x) = r 2 x . ⊓ ⊔