fixed point roulette
try to escape brouwer
can you beat a theorem?
a map f is a rule that sends every point of a shape to some point. a fixed point is a point the map leaves where it is: f(x) = x.
Brouwer (1911) says some maps always have one, however you design them. the map must be continuous: nearby points go to nearby points, so no tearing and no jumping. and it must send a shape into itself, where the shape is compact (bounded, and it includes its edge) and convex (the straight segment between any two of its points stays inside, so no dents and no holes).
your job: build a map with no fixed point. (you can't. but by level 3 you'll see which condition to break.)
f : [0,1] → [0,1] means f takes any number x from 0 to 1 and returns a number f(x), also from 0 to 1. the blue curve is its graph: height f(x) above each x. on the dashed line y = x, height equals position, so a fixed point is exactly where the curve touches that line. drag the control points and try to make the curve avoid it.
before you press find, look at the two ends. at x = 0 the curve is on or above the line, since f(0) can't be below 0. at x = 1 it is on or below it, since f(1) can't be above 1. a curve drawn without lifting the pen can't get from above to below without touching the line.
now f moves every point of the unit square (all (x, y) with x and y from 0 to 1) to a point of the square. drag the colored dots to set where each corner goes; every point in between follows smoothly, as the warped grid shows. each gray arrow runs from a point to where f sends it. a fixed point is a point whose arrow has length zero. try to give every point a real arrow.
there are no two ends to compare here, so the level 1 argument doesn't carry over, and the 2D proof is much harder. when you press find, the blue trail follows the arrows for a while, then the computer solves f(p) = p directly. following arrows alone can circle the point forever: Brouwer promises the point exists, not a path to it.
same game, different shape. an annulus is a ring: the region between two circles with the same center. the map rotates it by the angle you pick. before you check: which point does a rotation leave in place?
the math
brouwer's fixed point theorem
the 1D proof (IVT)
for f : [0,1] → [0,1] continuous, define g(x) = f(x) − x. then g(0) = f(0) ≥ 0 and g(1) = f(1) − 1 ≤ 0. by the intermediate value theorem (IVT: a continuous function that is ≥ 0 at one end of an interval and ≤ 0 at the other is 0 somewhere in between), g has a zero. that zero is the fixed point.
why continuity matters
define f(x) = x + 1/2 (mod 1). this maps [0,1] to itself, and no point is fixed: every point shifts by 1/2. but f is discontinuous at x = 1/2. without continuity, the curve can "jump" over the diagonal y = x.
why the domain matters
a rotation of the annulus maps it to itself continuously with no fixed point. the center of rotation would be fixed, but it is not in the domain. the annulus has a hole, so it is not convex. convexity is sufficient, not necessary: the theorem holds for any shape that can be continuously deformed, without tearing, into a closed disk (a crescent works fine). the hole is what kills the annulus.
applications
- nash equilibrium. every finite game has a mixed-strategy equilibrium. Nash's 1950 proof uses Brouwer (via Kakutani's generalization to set-valued maps).
- general equilibrium. Arrow and Debreu proved competitive equilibrium exists in any economy satisfying mild conditions. the core step is a fixed-point argument.
- differential equations. the Schauder fixed point theorem extends Brouwer to infinite dimensions, proving existence of solutions to many PDEs.
- the crumpled map theorem. crumple a map of a city and place it on a flat copy. at least one point on the crumpled map sits directly above its real location on the flat one.