# Only one fixed point for $f:\bar{\mathbb{D}}\rightarrow\bar{\mathbb{D}}$ on the boundary.

We know for Brouwer theorem that $f$ (continuous bijective function) have a fixed point. My questions are:

1) Is there a function with only one fixed point $x_0\in Int(\bar{\mathbb{D}})$ (open disk)?

2) Is there a function with only one fixed point $x_0\in \partial(\bar{\mathbb{D}})$?

For (1) is true since a rotation by any angle $\theta\not=2k\pi$. But for (2) I can't find $f$, I think that for any function with one fixed point on the boundary always have at least on fixed point in the open disk $\mathbb{D}$. Maybe for the continuity of $\partial f$ (the restricion of $f$ to $\partial \mathbb{D}$).

Any ideas or suggestions or maybe a counterexample?

• How about a constant function? – Harald Hanche-Olsen Nov 19 '14 at 16:42
• All point are fixed point, I need only one on the boundary. – Donyarley Nov 19 '14 at 16:44
• I'm confused. If $f(x)=x_0$ for all $x$, how can $f(x)=x$ when $x\ne x_0$? – Harald Hanche-Olsen Nov 19 '14 at 16:49
• Perhaps I was not clear explaining, the function must be continuous bijective function. This the case non trivial. – Donyarley Nov 19 '14 at 17:00
• Do you know Möbius transformations? – Daniel Fischer Nov 19 '14 at 17:26

$$S \colon z \mapsto \frac{z-i}{z+i}$$
maps the upper half-plane biholomorphically to the unit disk. Now it is easy to see bijections of the closed upper half-plane having exactly one fixed point on the boundary, all translations $T_a \colon z \mapsto z+a$ with $a\in \mathbb{R}\setminus \{0\}$ have $\infty$ as their only fixed point. Conjugating such a translation with $T$ then gives us a homeomorphism (even a holomorphic one) of the closed unit disk with itself having exactly one fixed point on the boundary.
Explicitly, $S^{-1}(z) = i\frac{1+z}{1-z}$, then
$$S\circ T_a \circ S^{-1} \colon z \mapsto \frac{a+(2i-a)z}{2i+a-az}$$
is, for every $a \in \mathbb{R}\setminus \{0\}$, an automorphism of the unit disk with $1$ as its only fixed point.