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.

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 $\map g b \le 0$

$\exists \xi \in \openint a b: \map g \xi = 0$

Thus:

$\map f \xi = \xi$

$\blacksquare$

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.

$\blacksquare$

This entry was named for Luitzen Egbertus Jan Brouwer.