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 range 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 $$g: \left[{a \,. \, . \, b}\right] \to \mathbb{R}$$ 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:

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

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:
 * $$t = \frac {f \left({x}\right) \cdot \left({f \left({x}\right) - x}\right)} {\left|{x - f \left({x}\right)}\right|^2}$$

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 =


 * : $$\S 9.16$$