Brouwer's Fixed Point Theorem/One-Dimensional Version
Theorem
Let $f: \closedint a b \to \closedint a b$ be a real function which is continuous on the closed interval $\closedint a b$.
Then:
- $\exists \xi \in \closedint a b: \map f \xi = \xi$
That is, a continuous real function from a closed real interval to itself fixes some point of that interval.
Proof by Intermediate Value Theorem
As the codomain of $f$ is $\closedint a b$, it follows that the image of $f$ is a subset of $\closedint a b$.
Thus $\map f a \ge a$ and $\map f b \le b$.
Let us define the real function $g: \closedint a b \to \R$ by $\map g x = \map f x - x$.
Then by the Combined Sum Rule for Continuous Real Functions, $\map g x$ is continuous on $\closedint a b$.
But $\map g a \ge 0$ and $\map g b \le 0$.
By the Intermediate Value Theorem, $\exists \xi: \map g \xi = 0$.
Thus $\map f \xi = \xi$.
$\blacksquare$
Proof using Connectedness
By Subset of Real Numbers is Interval iff Connected, $\closedint a b$ is connected.
Aiming for a contradiction, suppose there is no fixed point.
Then $\map f a > a$ and $\map f b < b$.
Let:
- $U = \set {x \in \closedint a b: \map f x > x}$
- $V = \set {x \in \closedint a b: \map f x < x}$
Then $U$ and $V$ are open in $\closedint a b$.
Because $a \in U$ and $b\in V$, $U$ and $V$ are non-empty.
By assumption:
- $U \cup V = \closedint a b$
Thus $\closedint a b$ is not connected, which is a contradiction.
Thus, by Proof by Contradiction, there exists at least one fixed point.
$\blacksquare$
Source of Name
This entry was named for Luitzen Egbertus Jan Brouwer.