Advanced book on Mathematics Olympiad

Real Analysis 583

f(eiπ/^3 )=β,

f(e−iπ/^3 )= ̄α+ 1 − ̄αβ,

whereβis an arbitrary complex parameter.

(20th W.L. Putnam Competition, 1959)

537.Successively, we obtain

f(− 1 )=f

(

−

1

2

)

=f

(

−

1

3

)

= ··· =nlim→∞f

(

−

1

n

)

=f( 0 ).

Hencef(x)=f( 0 )forx∈{ 0 ,− 1 ,−^12 ,...,−n^1 ,...}.

Ifx = 0 ,− 1 ,...,−n^1 ,...,replacingxby 1 +xxin the functional equation, we obtain

f

(

x

1 +x

)

=f

( x

1 +x

1 − 1 +xx

)

=f(x).

And this can be iterated to yield

f

(

x

1 +nx

)

=f(x), n= 1 , 2 , 3 ....

Becausefis continuous at 0 it follows that

f(x)=nlim→∞f

(

x

1 +nx

)

=f( 0 ).

This shows that only constant functions satisfy the functional equation.

538.Plugging inx=t,y= 0 ,z=0 gives

f(t)+f( 0 )+f(t)≥ 3 f(t),

orf( 0 )≥f(t)for all real numberst. Plugging inx= 2 t,y= 2 t,z=− 2 tgives

f(t)+f( 0 )+f( 0 )≥ 3 f( 0 ),

orf(t)≥f( 0 )for all real numberst. Hencef(t)=f( 0 )for allt,sof must be

constant. Conversely, any constant functionfclearly satisfies the given condition.

(Russian Mathematical Olympiad, 2000)

539.No! In fact, we will prove a more general result.

Proposition.LetSbe a set andg:S→Sa function that has exactly two fixed points

{a, b}and such thatg◦ghas exactly four fixed points{a, b, c, d}. Then there is no

functionf:S→Ssuch thatg=f◦f.