Advanced book on Mathematics Olympiad

Real Analysis 507

g(c+ 2 h 0 )>g(c+h 0 )>g

(

c+

h 0

2

)

>···>g

(

c+

h 0

2 n

)

>···.

Passing to the limit, we obtain thatg(c+ 2 h) > g(c), contradicting the maximality of
c. The contradiction proves that our initial assumption was false, and the conclusion
follows.

390.From the given condition, it follows thatfis one-to-one. Indeed, iff(x)=f(y),
thenf(f(x))=f(f(y)),sobx=by, which impliesx=y. Becausefis continuous
and one-to-one, it is strictly monotonic.
We will show thatfhas a fixed point. Assume by way of contradiction that this is
not the case. So eitherf(x)>xfor allx,orf(x)<xfor allx. In the first casefmust
be strictly increasing, and then we have the chain of implications

f(x)>x⇒f(f(x)) > f(x)⇒af (x)+bx > f (x)⇒f(x) <

bx

1 −a

,

for allx∈R. In particular,f( 1 )< 1 −ba<1, contradicting our assumption.
In the second case the simultaneous inequalitiesf(x)<xandf(f(x)) < f(x)
show thatfmust be strictly increasing again. Again we have a chain of implications

f(x)<x⇒f(f(x)) < f(x)⇒f(x) > af(x)+bx⇒f(x) >

bx

1 −a

,

for allx∈R. In particular,f(− 1 )>− 1 −ba>−1, again a contradiction.
In conclusion, there exists a real numbercsuch thatf(c)=c. The condition
f(f(c))=af (c)+bcimpliesc=ac+bc; thusc(a+b− 1 )=0. It follows that
c=0, and we obtainf( 0 )=0.

Remark.This argument can be simplified if we use the fact that a decreasing monotonic
function onRalways has a unique fixed point. (Prove it!)
(45th W.L. Putnam Mathematical Competition, 2002, proposed by T. Andreescu)

391.Being continuous on the closed interval[ 0 , 1 ], the functionfis bounded and has
a maximum and a minimum. LetMbe the maximum andmthe minimum. Then
m
2 n≤

f(xn)

2 n ≤

M
2 n, which implies that the series is absolutely convergent and its limit is a
number in the interval[m, M].
Leta∈( 0 , 1 )andmaandMabe the minimum and the maximum offon[ 0 ,a].If
α∈[ 0 ,a]is such thatf(α)=Ma, then

Ma=f(α)=

∑∞

n= 1

f(αn)

2 n

≤Ma

∑∞

n= 1

1

2 n

=Ma,

whence we must have equality in the above inequality, sof(αn) = Ma. Since
limn→∞αn = 0, it follows thatMamust equal limx→ 0 f(x)= f( 0 ). Similarly,