Brouwer's Fixed Point Theorem

= One-Dimensional Version =

Theorem
Let $f: \left[{a \,. \, . \, b}\right] \to \left[{a \,. \, . \, b}\right]$ be a real function which is continuous on the closed interval $\left[{a \,. \, . \, b}\right]$.

Then:
 * $\exists \xi \in \left[{a \, . \, . \, b}\right]: f \left({\xi}\right) = \xi$.

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 $\left[{a \,. \, . \, b}\right]$, it follows that the image of $f$ is a subset of $\left[{a \, . \, . \, b}\right]$.

Thus $f \left({a}\right) \ge a$ and $f \left({b}\right) \le a$.

Let us define the real function $g: \left[{a \,. \, . \, b}\right] \to \R$ by $g \left({x}\right) = f \left({x}\right) - x$.

Then by the Combination Theorem for Functions, $g \left({x}\right)$ is continuous on $\left[{a \,. \, . \, b}\right]$.

But $g \left({a}\right) \ge 0$ and $g \left({b}\right) \le 0$.

By the Intermediate Value Theorem, $\exists \xi: g \left({\xi}\right) = 0$.

Thus $f \left({\xi}\right) = \xi$.

= General Case =

Theorem
Any smooth map $f$ of the closed unit ball $B^n \subset \R^n$ into itself must have a fixed point:

$\forall f \in C^\infty (B^n \to B^n): \exists x \in B^n: f(x)=x$

Proof
Suppose there exists such a map $f$ of the unit ball to itself without fixed points.

Since $f \left({x}\right) \ne x$, the two points $x$ and $f \left({x}\right)$ are distinct and there is a unique straight line on which they both lie.

Call this line $L$ and let $h \left({x}\right) = \partial B^n \cap L$.

If $x \in \partial B^n$, then $h \left({x}\right) = x$ and $h$ restricts to the identity on $\partial B^n$.

Since $x$ is in the line segment between $f \left({x}\right)$ and $h \left({x}\right)$, one may write the vector $h \left({x}\right) - f \left({x}\right)$ as a multiple $t$ times the vector $x - f \left({x}\right)$, where $t \ge 1$.

Hence $h \left({x}\right) = tx + \left({1 - t}\right) f \left({x}\right)$.

Since $f$ is smooth, the smoothness of $t$ with respect to $x$ implies the smoothness of $h$.

Taking the dot product of both sides of this formula and noting that $\left|{h \left({x}\right)}\right| = 1$, we have:
 * $t^2 \left|{x - f \left({x}\right)}\right|^2 + 2 t f \left({x}\right) \cdot \left({x - f \left({x}\right)}\right) + \left|{f \left({x}\right)}\right|^2 - 1 = 0$.

Applying the quadratic formula gives:
 * $\displaystyle t = \frac {f \left({x}\right) \cdot \left({f \left({x}\right) - x}\right)} {\left|{x - f \left({x}\right)}\right|^2}$

an expression for $t$ in smooth terms of $x$.

Hence $h$ is a smooth retract of a compact manifold onto its boundary, which contradicts the Retraction Theorem.

= Sources =


 * : $\S 9.16$