Stir your coffee: a point stays still
Tonight's idea
A fixed point of \(f\) is an input it leaves alone: \(f(x) = x\). On a graph it is where the curve meets the diagonal \(y = x\).
Theorem. Every continuous \(f : [0,1] \to [0,1]\) has a fixed point. (Inputs in \([0,1]\), outputs in \([0,1]\), no jumps in the graph.)
Proof. Let \(g(x) = f(x) - x\). Then \(g(0) = f(0) \ge 0\) and \(g(1) = f(1) - 1 \le 0\). A continuous function that is non-negative at one end and non-positive at the other is zero somewhere between: that is the intermediate value theorem (IVT). Where \(g(c) = 0\), \(f(c) = c\).
In two dimensions the same is true and far harder. Brouwer's theorem: a continuous map of a disc or square to itself has a fixed point. Stir a coffee gently, without splashing; when it settles, some point of the surface is exactly where it began. Lay a map of the room on the floor; one point of it lies directly over the place it depicts.
Why not copy the proof? In one dimension \(f(x) - x\) is a number, positive or negative, and the IVT squeezes a zero between the signs. In two it is an arrow, with a direction instead of a sign, and there is no "between". A new idea is needed, such as the winding count of sheet b17.
Watch one: \(f(x) = \cos x\) on \([0,1]\)
Working in radians, \(\cos\) takes \([0,1]\) into \([\cos 1, 1] \approx [0.54, 1]\), inside \([0,1]\), so the theorem applies. Let \(g(x) = \cos x - x\): \(g(0) = 1\), \(g(1) \approx -0.46\). Halve the interval and keep the half where the sign changes:
\[g(0.5) = 0.378,\quad g(0.75) = -0.018,\quad g(0.625) = 0.186,\]
so the fixed point is in \([0.625, 0.75]\). Ten more halvings give \(c \approx 0.7391\). On a calculator, press \(\cos\) repeatedly from any start and the display settles at 0.739085. That convergence is a bonus the theorem does not promise; the theorem only says the point is there.
Your turn
- Check that \(f(x) = 1 - x^2\) satisfies the hypotheses on \([0,1]\), then find its fixed point exactly.
- (a) Find all fixed points of \(f(x) = x^2\) on \([0,1]\). (b) How many does \(f(x) = x^2 + \tfrac14\) have? (c) \(f(x) = x^2 + \tfrac12\) has none. Does this contradict the theorem?
- Every hypothesis matters. Give (a) a continuous \(f : (0,1) \to (0,1)\), on the open interval, with no fixed point; (b) an \(f : [0,1] \to [0,1]\) with one jump and no fixed point.
- A hiker walks up a mountain path on Saturday, leaving the bottom at 6 am and reaching the top at 6 pm. On Sunday she walks down the same path, leaving the top at 6 am and reaching the bottom at 6 pm. Prove there is a time of day at which she was at exactly the same height on both days.
Stretch: opposite points, same temperature
Temperature along the equator is a continuous function \(T(\theta)\) of longitude. Show there are two diametrically opposite points with the same temperature. Hint: let \(g(\theta) = T(\theta) - T(\theta + 180^\circ)\) and compare \(g(0)\) with \(g(180^\circ)\).
Answers
- \(1 - x^2\) is continuous and for \(0 \le x \le 1\) lies between 0 and 1. Fixed point: \(x = 1 - x^2\), so \(x^2 + x - 1 = 0\), \(x = \tfrac{\sqrt5 - 1}{2} \approx 0.618\). The other root is negative. This is the golden ratio minus one.
- (a) \(x^2 = x\) gives \(0\) and \(1\): two fixed points, both at the corners. (b) \(x^2 - x + \tfrac14 = (x - \tfrac12)^2 = 0\): exactly one, at \(x = \tfrac12\), where the parabola touches the diagonal. (c) No. \(x^2 + \tfrac12\) equals \(\tfrac32\) at \(x = 1\), so \(f\) does not map \([0,1]\) into \([0,1]\) and the theorem does not apply. Indeed \(x^2 - x + \tfrac12\) has discriminant \(-1\).
- (a) \(f(x) = x/2\) (or \(x^2\)): \(f(x) = x\) only at \(0\), which is not in \((0,1)\). The proof needed the endpoints to exist. (b) \(f(x) = 1\) for \(x \le \tfrac12\) and \(f(x) = 0\) for \(x > \tfrac12\). A fixed point with \(x \le \tfrac12\) would need \(x = 1\); one with \(x > \tfrac12\) would need \(x = 0\). The graph jumps over the diagonal.
- Let \(u(t)\) be Saturday's height and \(d(t)\) Sunday's, both continuous for \(t\) from 6 to 18, and let the summit height be \(H\). Set \(g(t) = u(t) - d(t)\). Then \(g(6) = 0 - H \lt 0\) and \(g(18) = H - 0 > 0\), so by the IVT \(g(t^*) = 0\) for some \(t^*\): same height at the same time of day. The time need not be noon and need not be unique.
Stretch. \(g(180^\circ) = T(180^\circ) - T(360^\circ) = T(180^\circ) - T(0) = -g(0)\). If \(g(0) = 0\) the points at \(0\) and \(180^\circ\) already match. Otherwise \(g(0)\) and \(g(180^\circ)\) have opposite signs, and the IVT gives a \(\theta\) between them with \(g(\theta) = 0\), that is \(T(\theta) = T(\theta + 180^\circ)\).