Questions: Contraction Mapping Theorem

Question 1

A function g on [0, 1] satisfies |g(x) − g(y)| ≤ 0.9|x − y| for all x, y. Starting from x_0 = 0.5, what does the Banach fixed-point theorem guarantee about the sequence x_{n+1} = g(x_n)?

• A. The sequence may or may not converge, depending on the choice of x_0
• B. The sequence converges to a unique fixed point at an exponential rate, regardless of the starting point
• C. The sequence converges but may converge to different fixed points depending on x_0
• D. The sequence converges only if x_0 is already close to the fixed point

Answer: (B)

The conditions are met: g has Lipschitz constant L = 0.9 < 1 on [0, 1] (which is a complete metric space as a closed subset of ℝ). The Banach fixed-point theorem therefore guarantees three things simultaneously: (1) there is exactly one fixed point x* in [0, 1], (2) it is unique, and (3) the iteration converges to x* from ANY starting point in [0, 1] — not just nearby ones. The error after n steps is bounded by L^n / (1 − L) · d(x_1, x_0), shrinking exponentially.

Question 2

A student claims: 'With Lipschitz constant L = 0.999, the iteration x_{n+1} = g(x_n) will barely converge — L is so close to 1 that the theorem barely applies.' Is this correct?

• A. Yes — L must be 0.5 or less for the theorem to guarantee meaningful convergence
• B. No — the theorem guarantees convergence for any L strictly less than 1. L = 0.999 is a valid contraction; convergence is slower (each step reduces error by only 0.1%) but guaranteed
• C. Yes — with L = 0.999, the geometric error bound diverges in practice
• D. No — convergence requires only L ≤ 1, and L = 0.999 qualifies comfortably

Answer: (B)

The theorem's condition is L < 1 (strictly). L = 0.999 satisfies this, so all three guarantees hold: unique fixed point, convergence from any start, exponential error decay. The student is right that convergence is slow — the error shrinks by a factor of 0.999 per step, requiring about 6,900 steps to reduce error by a factor of 10⁶. But 'slow' and 'barely applies' are different claims; the theorem is not a near-miss. Option D is wrong because L = 1 does NOT satisfy the strict inequality — an isometry need not have any fixed point.

Question 3

A contraction mapping on a complete metric space always has exactly one fixed point.

• T. True
• F. False

Answer: True

This is precisely the conclusion of the Banach fixed-point theorem: existence AND uniqueness of the fixed point are both guaranteed. Existence follows from the Cauchy sequence argument (the iterates form a Cauchy sequence whose limit, by completeness, exists in the space, and continuity ensures g maps that limit to itself). Uniqueness follows because two distinct fixed points x* and y* would require d(g(x*), g(y*)) = d(x*, y*), but the contraction condition gives d(g(x*), g(y*)) ≤ L · d(x*, y*) < d(x*, y*) — a contradiction.

Question 4

A function satisfying |g(x) − g(y)| = |x − y| for most x, y (Lipschitz constant L = 1) is a contraction, and the Banach theorem guarantees convergence of iteration to a fixed point.

• T. True
• F. False

Answer: False

False — a contraction requires L strictly less than 1. L = 1 is an isometry (distance-preserving map) and does NOT satisfy the contraction condition. The theorem makes no guarantee when L = 1. For example, g(x) = x + 1 on ℝ has L = 1, no fixed point at all, and iteration diverges. Even a map like g(x) = x (the identity) has L = 1 and every point is a fixed point — but iteration doesn't 'converge' in any useful sense. The strict inequality L < 1 is essential, not a technicality.

Question 5

Why does the Contraction Mapping Theorem require the metric space to be complete? What could go wrong without completeness?

Model answer: Completeness ensures that every Cauchy sequence in the space converges to a point that is also within the space. The iteration x_{n+1} = g(x_n) produces a Cauchy sequence (provable via geometric series from the contraction condition), but without completeness, the limit of this sequence might not exist in the space — it could 'escape' to a missing boundary point. For example, consider g(x) = x/2 on the open interval (0, 1), which is not complete. Starting from x_0 = 0.5, the iteration gives 0.25, 0.125, ..., converging to 0. But 0 is not in (0, 1), so the fixed point exists in ℝ but not in the space where g was defined. Without completeness, the theorem's existence conclusion fails.

Completeness is often called a 'technical' condition by students, but this example shows it is load-bearing. The theorem bundles three guarantees — existence, uniqueness, convergence — and completeness is what makes existence work. In practice, the spaces used in numerical analysis (ℝ^n, closed bounded subsets) are all complete, which is why the theorem is so widely applicable.