Coffee without the Cup
Suppose f is a continuous function that maps the closed unit disk on R 2 to itself. Then, Brouwer's fixed-point theorem tells us that there is a fixed point x 0 in the closed disk which is mapped to itself, i.e. f ( x 0 ) = x 0 .
In a similar spirit, let g be a continuous function that maps the open unit disk to itself. Must g have a fixed point?
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3 solutions
Thanks for the constructive solution! Will any value of a work?
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Yes, any nonzero value of a ∈ D works. The Mobius transformation maps the unit circle to itself, and also maps the open unit disc to itself. Try solving f ( z ) = z to see that the only fixed points have modulus 1 , so do not lie in the open disc.
First, think of the 1-D analog of the problem: a continuous function f mapping the open interval ( 0 , 1 ) to itself. One of the simplest such f which does not have a fixed point is f ( x ) = 1 − exp ( − 1 − x 1 ) , 0 < x < 1 . The above can be easily verified by either calculus, or by simply drawing pictures.
Next, lift this example to 2-D. Representing the points in polar coordinates, consider the continuous function g defined on the open disc as follows g ( r , θ ) = ( 1 − exp ( − 1 − r 1 ) , θ ) , 0 ≤ r < 1 , 0 ≤ θ ≤ 2 π . Hence, this function has no fixed point on the open disc as well.
Note: The construction above is general and extends to any n -dimensional open ball.
The simplest continuous function from ( 0 , 1 ) to itself with no fixed point is surely x ↦ x 2 .
More to the point, this solution is faulty; this function is not continuous at the centre r = 0 , since lim r → 0 g ( r , θ ) = ( 1 − e − 1 , θ ) takes different values for different values of θ .
The short version of the proof is as follows:
1) Fixed point theorems don't work for open intervals
2) An open circle is equivalent to an open square, which can be seen as a stack of open intervals
In what sense are they equivalent? Do you mean homeomorphisms?
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Yes, an open square can map onto an open circle. So, first imagine a simple vector field in an open square, a "wind" blowing from one side of the square to the other, and then map this onto an open circle. Then there are no interior singularities. This only works if both are open. Maybe later I can come up with an illustrative graphic?
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An illustrative graphic would be great!
Consider the open disc as the set D of complex numbers of modulus less than 1 . For any nonzero a ∈ D , the Möbius transformation f ( z ) = 1 + a z z + a Is a continuous bijection from D to itself with no fixed point.