TITLE: If $f$ is continuous and $f(f(x))+f(x)+x=0$ for all $x$ in $\mathbb R^2$, then $f$ is linear?
QUESTION [6 upvotes]: Let $f\in C^0(\mathbb R^2,\mathbb R^2)$, with $f(f(x))+f(x)+x=0$ for all $x$ in $\mathbb R^2$.
Is $f$ linear ( : $\forall (x,y) \in (\mathbb R^2)^2,f(x+y)=f(x)+f(y)$ ) ?
This problem can be translated to a problem of Geometry of the plane.

REPLY [5 votes]: The computational part of Eric Wofsey's answer can be simplified a little bit ; in particular
it can be done entirely by hand as I show below.
Following this answer, let us start with the three points $p=(0,1),q=(1,0),r=(-1,-1)$. Let $b$ be a $C^1$ bijection $[0,1]\to [0,1]$, to be defined later, and let
$$
\begin{array}{lclcl}
A(t) &=& (1-t)p+tq &=& (t,1-t) \\
B(t) &=& (1-b(t))q+b(t)r &=& (1-2b(t),-b(t)) \\
C(t) &=&  -(A(t)+B(t))  &=& (2b(t)-(t+1),t+b(t)-1). \\
\end{array}
$$
 Then there is a unique bijection $f$, defined on the triangle $pqr$, such that
$f$ cyclically permutes every triple $(A(t),B(t),C(t))$ for $t\in [0,1]$.
As shown in Eric's answer, the only problem is then to find a $b$ such
that f is non-linear and such that the angle $(Or,OC(t))$ varies monotonically
as $t$ goes from $0$ to $1$.
It is easy to see that $f$ is linear iff $b$ is the identity (indeed, $f$ cyclically
permutes $p,q$ and $r$, and this suffices to uniquely define $f$ is $f$ is linear).
A little computation shows that the straight lines $(qr)$ and $(OC(t))$ intersect
at $D(t)=(1-d(t))r+d(t)p$ where
$$
d(t)=\frac{2t-b(t)}{3t-3b(t)+1}
$$
So it suffices to find a $b$ such that the derivative of $d$ is never zero on
$(0,1)$. Another little computation shows that the numerator of $d'$ is
(here $u$ and $v$ denote the numerator and denominator of $d$ respectively)
$$
N(t)=u'(t)v(t)-u(t)v'(t)=(3b(t)-1)b'(t)+2-3b(t)
$$
If we try $b(t)=t^2$, we are lucky : $N(t)=3t^2-2t+2=3(t-\frac{1}{3})^2+\frac{5}{3}$
is never zero.