Solution for October 2012

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 $f\left(f(x)\right) = x+1$ for all $x\in \mathbf{ R}$? Does there exist a strictly decreasing function $g$ for which $g\left(g(x)\right) = 2x+1$ for all $x\in \mathbf{ R}$?

The solution:

There was one incorrect submission.

a. No, no function $f$ for which $f\left(f(x)\right) = x+1$ for all $x\in \mathbf{ R}$ can be strictly decreasing. We can use any real number $x$ for a counter example, so we first use $x=f(0)$:

But with $x=0$ we have $f\left(f(0)\right) = 0+1 = 1$, so that

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 $\mathbb{R}$ onto $\mathbb{R}$ 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 $g\left(g(x)\right) = 2x+1$ for all $x\in \mathbf{ R}$. Such a function is
$$g(x) = -\sqrt 2 x - (1+\sqrt 2).$$
Note that $g(x)$ is decreasing because it is the equation of a line with negative slope; moreover,
$$g(g(x)) = -\sqrt 2(-\sqrt 2x -1-\sqrt 2) - (1 + \sqrt 2) = 2x +1,$$
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 $a<0$.
$$g\left(g(x)\right) = g(ax+b) = a(ax+b) + b = a^2x + (ab+b).$$
This quantity equals $2x+1$ only if $a^2=2$, whence $a=-\sqrt 2$, and $ab+b = -\sqrt 2 b + b = 1$, whence $b = \frac1{-\sqrt 2 + 1} = -(\sqrt 2 + 1)$, as claimed.

Further comments. The reader should explore for himself why the preceding calculation fails for $f(x) = ax+b$ when $f\left(f(x)\right) = x+1$. As for possible uniqueness in part (b), both Desprez and Humbert observed that (b) has infinitely many solutions. For all real numbers $a \ne \frac12$,

$$g(x) = \left\{\begin{array}{cc} -2^a(x+1)-1 & \mbox{ if } \; x \le -1 \\
-2^{1-a}(x+1)-1 & \mbox{ if } \; x \ge -1 \end{array}\right.$$

serves as a nonlinear example. Desprez went further and proved that if $a \le 0$, or if $a=1$ and $b \ne 0$, then there are no strictly decreasing functions that satisfy $f\left(f(x)\right) = ax+b$ 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.