Set of fixed points is convex. Problem: Suppose that $f:\mathbb{R}^n \to \mathbb{R}^n$ has Lipschitz constant $1$. Is it true that the set of fixed points of $f$ is convex?
My attempt: I managed to show that if $x,y$ are two fixed points of $f$ and $z  = (1-\lambda)x+\lambda y$ for $\lambda \in (0,1)$, then $||f(z)-x||=||z-x||$ and $||f(z)-y||=||z-y||$. But I'm not quite sure what to do after this. Is there any hint or suggestion?
 A: So $f(z)$ is on the sphere around $x$ passing through $z$, and is also on the sphere around $y$ passing through $z$. These two spheres "kiss" at $z$ so that $f(z)$ must be $z$.
A: Let $x_0,x_1\in\mathbb R^n$, $x_0\ne x_1$.
Assume $f(x_0)=x_0$ and $f(x_1)=x_1$, and $x_t=tx_1+(1-t)x_0$, $t\in (0,1)$. We need to show that $f(x_t)=x_t$.
Assume that $f(x_t)\ne x_t$.
Then
$$
\|x_1-x_0\|=\|f(x_1)-f(x_0)\|\le \|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|\\ \le \|x_t-x_0\|+\|x_t-x_1\|
=t\|x_1-x_0\|+(1-t)\|x_1-x_0\|=\|x_1-x_0\|.
$$
Hence
$$
\|f(x_1)-f(x_0)\|= \|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|   \tag{1}
$$
Meanwhile
$$
\|f(x_t)-f(x_0)\|\le\|x_t-x_0\|=t\|x_1-x_0\|. \tag{2}
$$
Similarly
$$
\|f(x_t)-f(x_1)\|\le (1-t)\|x_1-x_0\|. \tag{3}
$$
Adding (2) and (3) we obtain
$$
\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\| \le t\|x_1-x_0\|+(1-t)\|x_1-x_0\|=\|x_1-x_0\|
$$
and hence
$$
\|x_1-x_0\|=\|f(x_1)-f(x_0)\|\le 
\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\| \le \|x_1-x_0\|
$$
which implies that
$$
\|x_1-x_0\|=\|f(x_t)-f(x_0)\|+\|f(x_t)-f(x_1)\|=\|f(x_t)-x_0\|+\|f(x_t)-x_1\|
$$
and we obtain that (2) and (3) are equalities.
Lemma. If $\,a,b\in\mathbb R^n\,$ and $\,\|a+b\|=\|a\|+\|b\|$, then $a=\lambda c$ and $b=\mu c$, for some $c\in\mathbb R^n$ and $\lambda,\mu\ge 0$. In particular, if
$\|a-b\|=\|a-c\|+\|c-b\|$, then $c=sa+(1-s)b$, for some $s\in [0,1]$.
This Lemma implies that
$$
f(x_t)=sx_1+(1-s)x_0, \quad\text{for some $s\in [0,1]$}
$$
and due to (2) or (3), $s=t$, and hence $f(x_t)=x_t$.
