Let g :R →R be a continuous function such that lim x →∞ g (x) − x = ∞ and such that the set {x; g(x) = x} is finite and nonempty. Prove that if f :R →R is

4.6.8. Let g :R →R be a continuous function such that lim x →∞ g (x) − x = ∞ and such that the set {x; g(x) = x} is finite and nonempty. Prove that if f :R →R is

continuous and f ◦g=f , then f is constant.

C. Joit¸a, Amer. Math. Monthly, Problem 10818 Solution. (Robin Chapman) Let x 1 ,...,x n with x 1 < ··· < x n denote the val-

ues of x such that g (x) = x. Define intervals I 0 ,...,I n as follows: I 0 = (−∞,x 1 ),

I r = (x r ,x r +1 ) for 1 ≤ r < n, and I n = (x n , ∞). The sign of g(x) − x is constant on each I r . Define ε r = (g(x) − x)/|g(x) − x| for x ∈ I r . Suppose that f : R → R is continuous and f ◦ g = f . We first show that f takes the value f (x n ) on all of I n . Since lim x →∞ (g(x) − x) = ∞, we have ε n = 1 for large x . Choose y 0 >x n . Since g (x n )=x n and g (y 0 )>y 0 , the intermediate value prop- erty yields y 1 with x n <y 1 <y 0 and g (y 1 )=y 0 . Repeating this argument yields

a decreasing sequence (y m ) m ≥0 of elements of I n with g (y m +1 )=y m . Therefore,

f (y m ) = f (g(y m +1 )) = f (y m +1 ). Consequently, f (y 0 ) = f (y m ) for each m. Since (y m ) m ≥0 is a decreasing sequence bounded below by x n , it tends to a limit y ∞ with y ∞ ≥x n . Continuity of g yields g (y ∞ ) = lim m →∞ g (y m ) = lim m →∞ y m −1 =y ∞ . Therefore, y ∞ =x n and f (y ∞ ) = lim m →∞ f (y m ) = f (y 0 ). It follows that f (y) = f (x n ) for all y ∈I n .

168 4 Continuity We now show by descending induction on r that f has value f (x n ) on all of I r .

Suppose that for some r ∈ [0,n − 1], we have f (y) = f (x n ) for y ∈ I s with r < s ≤ n. By continuity, f (y) = f (x n ) for y ∈ [x r +1 , ∞). We divide this into cases accord- ing to whether ε r is 1 or −1. Suppose first that ε r = 1. For y ∈ I r , consider the

sequence (y m ) m ≥0 with y 0 = y and y m +1 = g(y m ). If all the y m lie in I r , then the sequence (y m ) m ≥0 is increasing and bounded above by x r +1 . It tends to a limit y ∞ with y ∞ ≤x r +1 and g (y ∞ )=y ∞ . Therefore, y ∞ =x r +1 . On the other hand, we also have f (y m +1 ) = f (g(y m )) = f (y m ), so the sequence ( f (y m )) m ≥0 is constant. Con-

sequently, f (y) = f (y 0 ) = f (y ∞ ) = f (x r +1 ). Hence f is constant on I r with value

f (x n ). Suppose that ε r = −1. It will be shown that g(I r )⊇I r . For r > 0, this is automatic from the intermediate value property. For r = 0 and y ∈ I 0 , one has g (y) < y < x 0 = g(x 0 ). By the intermediate value property, there exists y ′ ∈ (y,x) with g(y ′ )= y . We now mimic the argument used in the case r = n. Given y ∈ I 0 , there is a sequence (y m ) m ≥0 with y 0 = y and g(y m +1 )=y m for each m, since g (y 0 )<y 0 . By the intermediate value property, there exists y 1 such that g (y 1 )=y 0 . This sequence is increasing and tends to a limit y ∞ with y ∞ ≤x r +1 . Again, g (y ∞ )=y ∞ . Thus, y ∞ =x r +1 . The sequence ( f (y m )) m ≥0 is constant, and so f (y) = f (y 0 ) = f (y ∞ )=

f (x r +1 ) = f (x n ). Again we conclude that f is constant on I r . ⊓ ⊔