Find all functions $f\colon \Bbb R\to \Bbb R$ such that $f(1-f(x)) = x$ for all $x \in \Bbb R$

Question

Find all functions $f\colon \Bbb R\to \Bbb R$ such that $f(1-f(x)) = x$ for all $x \in \Bbb R$.

This is a question from the national olympiad in Germany 2018.

All i could do is to try with some linear functions like $f(x) = x+1$

I am very good at calculus but I don't know how to approach the problems where I have to find a particular function. Is there a specific algorithm I should use?

Answer 1

This is a partial answer.

Any solution of the functional equation must be bijective.

Injectivity is easy: $$f(x)=f(y) \quad \Rightarrow \quad 1-f(x)=1-f(y) \quad \Rightarrow \quad f(1-f(x))=f(1-f(y)) \quad \Rightarrow \quad x=y$$

Surjectivity is even easier: that's because $x=f(1-f(x))= f(\mathrm{something})$.

If we look for "nice" functions, we could start by looking for continuous ones. However, there is no continuous function which solves $f(1-f(x))=x$.

Indeed, if $f$ is continuous, then $f$ must be monotone (that's because it's bijective!). Suppose that $f$ is increasing. Then, for $x<y$ you have $$x<y \quad \Rightarrow \quad f(x)<f(y) \quad \Rightarrow \quad 1-f(x) > 1-f(y) \quad \Rightarrow \quad f(1-f(x)) > f(1-f(y))$$ which implies $x>y$. A similar contradiction shows that $f$ cannot be decreasing.

Thus, there is no continuous solution.

Answer 2

As requested within the comment section I will provide the solution as it was derived by a teacher of mine. He struggled for himself and went back and forth several times until he finally got the solution.

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First of all we tried to figure out whether the function is invertible or not hence the property $f(f^{-1}(x))=x$ will be quite useful here. By making the assumption $x_1\ne x_2$ and $f(x_1)=f(x_2)$ - in other words the function is not invertible - we get

$$1-f(x_1)=1-f(x_2)\Leftrightarrow f(1-f(x_1))=f(1-f(x_2))$$

and by the functinal equation we obtain $x_1=x_2$ which is a contradiction thus the function is invertible.

Now we applied $f^{-1}$ to the functional equation and so we get

$$f^{-1}(x)=1-f(x)\Leftrightarrow \frac{f(x)+f^{-1}(x)}2=\frac12$$

which states that the arithmetic mean of the function and its inverse equals $1/2$ for all $x$. From hereon my teacher reasoned that this implies the function $f$ is symmetric to the line $y=1/2$. Hence $f$ is defined for all $x$ we can reason that there are points of the graph of $f$ within the I. quadrant and the IV. quadrant as well. Now assume a point $P$ on the graph of $f$ within the I. quadrant (below the line $y=1/2$ otherwise you have to work within the IV. quadrant). Reflect the point on the line $y=1/2$ and $P'$ will lie on $f^{-1}$. Mirroring again this time on $y=x$ leads to the point $P''$ which again lies on $f$. Therefore a rotation of nighty degree around the point $S(1/2,1/2)$reflects the graph of $f$ on itself. By doing this procedure again we will end up on the graph of $f$ aswell. Therefore $f$ is also symmetric towards the point $S$.

Thus consider the points $x_1,x_2$ which are symmetric towards $1/2$ ($1/2(x_1+x_2)=1/2$) which implies

$$\frac12-f(x_2)=-\frac12+f(x_2)\Leftrightarrow f(x_1)+f(x_2)=1 \Leftrightarrow f(x)+f(1-x)=x$$

Additonally $S$ itself is part of $f$. Hence

$$f\left(\frac12\right)=f^{-1}\left(\frac12\right)\Leftrightarrow f\left(\frac12\right)=1-f\left(\frac12\right)\Leftrightarrow f\left(\frac12\right)=\frac12$$

Form hereon the main question remains: are there any functions which satisfy this functional equation; the answer is yes. My teacher constructed a quite weird one but howsoever I will just go on.

Starting with the two cases

$$ f(x)=\begin{cases}\frac12x+\frac34&,\text{for }\frac12<x\le1\\-2x+\frac52&,\text{for }1<x\le\frac54\end{cases} $$

which satisfy the condition since

$$f_1(1-f_2(x))=\frac12\left(1-\left(-2x+\frac52\right)\right)+\frac34=x$$

where $f_1(x)$ denotes the first branch of the function $f$ and $f_2$ respectively the second one. The crucial point is that adding $1$ to the negative of the one branch automatically results in the other branch. By continuing these relations he obtained

$$ y_n=\begin{cases}\frac12x+\frac34&,\text{for }\frac32-\frac1{4^n}<x\le\frac32-\frac1{2\cdot4^n}\\-2x+\frac52&,\text{for }\frac32-\frac1{2\cdot4^n}<x\le\frac32-\frac1{4^{n+1}}\end{cases} $$

wiht $n\in\mathbb{N}$. I am not quite sure about this function but it should also provide the solution function on the left side of $S$ by plugging in values of $n\in\mathbb{Z}$. So far so good. I will upload the original notes somewhere so that you can look at them for yourself and understand what I maybe missed right now. Refering to your pseudonym and since you are asking for the solution to a German olympiad question I guess you are capable of reading german.

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Here the promised original notes side $1$ and side $2$.

Answer 3

with $g=1-f$, the equation is simply $$g(g(x))=1-x.$$ Noting the importance of $(y,x)=(1/2,1/2)$ on the graph of $f$(and $g$) from the other answers, set $h(x)=g(x+1/2)-1/2$, so that $h(0)=0$. Then we see that

$$h(h(x)) = h\big(g(x+1/2)-1/2\big) = g(g(x+1/2)) -1/2 = 1-(x+1/2)-1/2 = -x,$$

So $$h(h(x)) = -x.$$

It is known that this equation has a solution on $\mathbb R$. Moreover, the solutions to both problems are in one-to-one correspondence via the above transformations. Undoing these transformations, we see that $$ g(x)=h(x-1/2)+1/2,$$ and finally $$ f(x) = 1/2 - h(x-1/2).$$ By modifying the graph from this great answer by alex.jordan, I produced a graph of a solution $f$. In $\color{green}{light\ green}$ is the graph of $g=1-f$, and you can verify that the graph is correct by following a trajectory like the $\color{lightblue}{light\ blue}$ one ($\color{lightblue}x\mapsto \color{green}g(\color{lightblue}x) \mapsto f(\color{green}g(\color{lightblue}x)) \overset{!}{=} \color{lightblue}x $.)