Return to Answer

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and fixed point theory" by Khamsi and Kirk for a brief discussion.

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and fixed point theory" by Khamsi and Kirk for a brief discussion.

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and Fixed Point Theory" by Khamsi and Kirk for a brief discussion.

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and fixed point theory" by Khamsi and Kirk for a brief discussion.

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

I got here via this mathoverflow question. I find I like Dick Palais' simple proof so much that I want to make it simpler. $\newcommand{\from}{\colon}\newcommand{\One}{\mathbb{1}}$

Palais reduces the problem to finding a map $f\from D^\infty \to D^\infty$ without fixed points. Here is an easier proof that such a map exists. Let $T \from D^\infty \to D^\infty$ be the shift map: $T((x_0, x_1, x_2, \ldots)) = (0, x_0, x_1, x_2, \ldots)$. Note that $T$ is continuous and fixes the origin.

Set $\One = (1,0,0,0,\ldots)$. Working with the $L^2$ norm, define $$ f(x) = \sqrt{1 - |x|^2} \cdot \One + T(x). $$ Since $f$ and $T$ agree on $S^\infty$, the map $f$ has no fixed points on the sphere. On the other hand, for all $x \in D^\infty$, the point $f(x)$ lies in $S^\infty$. So $f$ has no fixed points inside the ball. We are done.

Said another way: the Brouwer fixed-point theorem is "obviously" false for $D^\infty$.

EDIT: Problem 36 of the Scottish book, posed by Ulam, asks if $D^\infty$ deformation retracts to its boundary. The book goes on to say that Tychonoff found the required retraction. See page 178 of "Spaces and Fixed Point Theory" by Khamsi and Kirk for a brief discussion.