real-valued function f(x)

The Problem:
	We say that a real-valued function f(x) of the real variable x is strictly decreasing if f(a)>f(b) whenever a<b . Does there exist a strictly decreasing function f for which f(f(x))=x+1 for all x∈R ? Does there exist a strictly decreasing function g for which g(g(x))=2x+1 for all x∈R ?

The solution:

a. No, no function f for which f(f(x))=x+1 for all x∈R can be strictly decreasing. We can use any real number x for a counter example, so we first use x=f(0) :

f(f(f(0)))=f(0)+1.

(1)

But with x=0 we have f(f(0))=0+1=1 , so that

f(f(f(0)))=f(1).

(2)

Comparing equations (1) and (2) we deduce that f(1)=f(0)+1 ;
that is, f(1)>f(0) , whence f cannot be decreasing, let alone strictly decreasing.

An alternative argument was given by a few correspondents who know some basic analysis: Because the function that takes x to x+1 , namely f(f(x)) , is bijective, it follows that f(x) must also be bijective. From a course in real analysis we know that a decreasing bijection that maps R onto R must have a fixed point (namely the value of x where its graph intersects the line y=x ), which would also be a fixed point of f(f(x))=x+1 . But this map does not have a fixed point (since x=x+1 has no real solution), a contradiction.

b. Yes, there exists a strictly decreasing function g for which g(g(x))=2x+1 for all x∈R . Such a function is

g(x)=−2√x−(1+2√).

Note that g(x) is decreasing because it is the equation of a line with negative slope; moreover,

g(g(x))=−2√(−2√x−1−2√)−(1+2√)=2x+1,

as desired.
In fact, we easily see that this is the only linear function that does the job: a decreasing linear function must satisfy g(x)=ax+b for real numbers a and b with a<0 .

g(g(x))=g(ax+b)=a(ax+b)+b=a2x+(ab+b).

This quantity equals 2x+1 only if a2=2 , whence a=−2√ , and ab+b=−2√b+b=1 , whence b=1−2√+1=−(2√+1) , as claimed.

Further comments. The reader should explore for himself why the preceding calculation fails for f(x)=ax+b when f(f(x))=x+1 . As for possible uniqueness in part (b), both Desprez and Humbert observed that (b) has infinitely many solutions. For all real numbers a≠12 ,

g(x)={−2a(x+1)−1−21−a(x+1)−1 if x≤−1 if x≥−1

serves as a nonlinear example. Desprez went further and proved that if a≤0 , or if a=1 and b≠0 , then there are no strictly decreasing functions that satisfy f(f(x))=ax+b for all x ; otherwise there are infinitely many such functions. He showed, furthermore, how to construct all of them: take an arbitrary continuous strictly decreasing function on an appropriate interval, then use the functional equation f(ax+b)=af(x)+b to extend that function to the real line.


math central