Set of fixed points is convex.

Question

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?

Answer 1

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 $z$.

Answer 2

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 $f(x_t)=x_t$.