A property of continuous bijective maps in the plane.

Question

Take a non-empty open connected bounded set $D$ of $\mathbb{R}^2$ and let $f:D\to D$ be a continuous bijective map. It is well known that $f$ could have no fixed points i. e. $f(p)\not=p$ for all $p\in D$.

I wonder if the following weaker property holds: there is a point $p\in D$ such that $d(f(p),\partial D)=d(p,\partial D)$ where $d(z,\partial D)$ is the euclidean distance from a point $z\in \mathbb{R}^2$ to the boundary of $D$. Any existing reference in the literature will be appreciated.

Answer 1

My thanks to $\bf{ Moishe \ Cohen}$ for pointing out in the comments that my proof is finished upon observing that $g(p)=d(f(p),\partial D)$ is a bounded function on $D$ that converges to $0$ as $p$ approaches $\partial D$; therefore, $g$ attains its maximum for some $p\in D$.

Observe that $d(f(p),\partial D)=\inf\{d(f(p),x)\colon x\in\partial D\}$ is a continuous function of $p$. Hence, $d(f(p),\partial D)-d(p,\partial D)$ is continuous as well. $d(f(p),\partial D)-d(p,\partial D)$ maps into $[-R,R]$ for some $R>0$. Since $D$ is connected, the image of $D$ under $d(f(p),\partial D)-d(p,\partial D)$ is connected, and, hence, an interval as $D$ is bounded.

Thus, if there were no fixed point wrt distance from the boundary, then either $d(f(p),\partial D)<d(p,\partial D)$ or $d(f(p),\partial D)>d(p,\partial D)$ for all $p$ and conversely. We might as well use the equivalent condition $d(f^{-1}(p),\partial D)<d(p,\partial D)$ for $d(f(p),\partial D)>d(p,\partial D)$.

However, $d(f,\partial D)\colon D\to (0,\infty)$ achieves its supremum on $D$ as $d(f,\partial D)$ is bounded on $D$ and converges to $0$ as $p$ approaches $\partial D$. Suppose $p_0$ is a point where $d(f,\partial D)$ achieves its supremum on $D$. Then $d(f(p_0),\partial D)-d(p_0,\partial D)>0$ and $d(f^{-1}(f(p_0)),\partial D)-d(f(p_0),\partial D)<0$. Hence, there exists a point $p\in D$ such that $d(f(p),\partial D)-d(p,\partial D)=0$.