3 Deleted (quasi), on vote of two users and myself.

The Question.

Suppose that f and g are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in Rn) to itself. Does there always exist a point x such that f(x)=g(x)?

If one of the mappings is invertible, then it is just a restatement of the Brower's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension 1.

(No. 1 in http://mathoverflow.net/unanswered today, May 5, 2010, with 58 votes, my rating: nice, easy?? )

[[[ WRONG ]]

• If $x$ is a fix-point of $f$ then $y=g(x)$ is a fix-point of $f$. That is:
If $f(x)=x$ and $y=g(x)$ then $f(y)=y$.

Proof: $f(y)=f(g(x)) = g(f(x))$ ----- by commuting,
$= g(x)$ ---- since $f(x)=x$
$=y$

• There is $x_0$ with $f(x_0)=x_0$. Proof: fix point theorem.

• a) For $n \ge 1$, let $x_n = g(x_{n-1})$.
b) Then $f(x_n) = x_n$ for all $n$ by 2. and 1.

• The closed unit disk is compact, hence $x_{n_k}$ converges to $x$, for some $x$ and some increasing ${n_k}$.

• By 3b. and continuity of $f$, we have $f(x)=x$.
By 3adeleted] .and continuity of $g$, we have $g(x)=x$. [[ WRONG, HERE IS THE MISTAKE ]]

• Thus $f(x)=g(x)$, and moreover, $x$ is a common fix-point of $f$ and $g$.
Works on any compact where a fix point theorem is valid for function $f$.

• 2 added 50 characters in body

The Question.

Suppose that f and g are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in Rn) to itself. Does there always exist a point x such that f(x)=g(x)?

If one of the mappings is invertible, then it is just a restatement of the Brower's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension 1.

(No. 1 in http://mathoverflow.net/unanswered today, May 5, 2010, with 58 votes, my rating: nice, easyeasy?? )

1. If $x$ is a fix-point of $f$ then $y=g(x)$ is a fix-point of $f$. That is:
If $f(x)=x$ and $y=g(x)$ then $f(y)=y$.

Proof: $f(y)=f(g(x)) = g(f(x))$ ----- by commuting,
$= g(x)$ ---- since $f(x)=x$
$=y$

2. There is $x_0$ with $f(x_0)=x_0$. Proof: fix point theorem.

3. a) For $n \ge 1$, let $x_n = g(x_{n-1})$.
b) Then $f(x_n) = x_n$ for all $n$ by 2. and 1.

4. The closed unit disk is compact, hence $x_{n_k}$ converges to $x$, for some $x$ and some increasing ${n_k}$.

5. By 3b. and continuity of $f$, we have $f(x)=x$.
By 3a. and continuity of $g$, we have $g(x)=x$. [[ WRONG, HERE IS THE MISTAKE ]]

6. Thus $f(x)=g(x)$, and moreover, $x$ is a common fix-point of $f$ and $g$.
Works on any compact where a fix point theorem is valid for function $f$.

1

The Question.

Suppose that f and g are two commuting continuous mappings from the closed unit disk (or, if you prefer, the closed unit ball in Rn) to itself. Does there always exist a point x such that f(x)=g(x)?

If one of the mappings is invertible, then it is just a restatement of the Brower's fixed point theorem but I do not know the answer in the general case and would not even dare to guess what it must be. Also, the answer is well-known to be "Yes" in dimension 1.

(No. 1 in http://mathoverflow.net/unanswered today, May 5, 2010, with 58 votes, my rating: nice, easy)

1. If $x$ is a fix-point of $f$ then $y=g(x)$ is a fix-point of $f$. That is:
If $f(x)=x$ and $y=g(x)$ then $f(y)=y$.

Proof: $f(y)=f(g(x)) = g(f(x))$ ----- by commuting,
$= g(x)$ ---- since $f(x)=x$
$=y$

2. There is $x_0$ with $f(x_0)=x_0$. Proof: fix point theorem.

3. a) For $n \ge 1$, let $x_n = g(x_{n-1})$.
b) Then $f(x_n) = x_n$ for all $n$ by 2. and 1.

4. The closed unit disk is compact, hence $x_{n_k}$ converges to $x$, for some $x$ and some increasing ${n_k}$.

5. By 3b. and continuity of $f$, we have $f(x)=x$.
By 3a. and continuity of $g$, we have $g(x)=x$.

6. Thus $f(x)=g(x)$, and moreover, $x$ is a common fix-point of $f$ and $g$.
Works on any compact where a fix point theorem is valid for function $f$.