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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?

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

$$S \colon z \mapsto \frac{z-i}{z+i}$$

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.

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