Brouwer's Fixed Point Theorem/One-Dimensional Version

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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

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 $g \left({x}\right) = \map f x - x$.

Then by the Combined Sum Rule for Continuous 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, $\left[{a \,.\,.\, b}\right]$ is connected.

Aiming for a contradiction, suppose there is no fixed point.

Then $f \left({a}\right) > a$ and $f \left({b}\right) < b$.

Let:

$U = \left\{ {x \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) > x}\right\}$
$V = \left\{ {x \in \left[{a \,.\,.\, b}\right]: f \left({x}\right) < x}\right\}$

Then $U$ and $V$ are open in $\left[{a \,.\,.\, b}\right]$.

Because $a \in U$ and $b\in V$, $U$ and $V$ are non-empty.

By assumption:

$U \cup V = \left[{a \,.\,.\, b}\right]$

Thus $\left[{a \,.\,.\, b}\right]$ 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.


Sources