Contraction mappings and fixed points
Turn a shrinking distance estimate into a unique fixed point, a convergent iteration, and a usable error certificate.
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The question for today
What you’ll understand
- Check the complete-space, self-map, and uniform contraction hypotheses.
- Derive convergence of iteration from a geometric tail estimate.
- Use both initial-step and latest-step error bounds without confusing indices.
A fixed point is an input left unchanged by a map. Such equations often hide inside optimization, differential equations, and self-consistent physical models. A contraction provides more than a promise that a solution exists: repeatedly applying it from any starting point gives the unique solution, and a geometric estimate certifies the remaining error. The strength of the conclusion comes from three distinct requirements: a complete space, a map that stays in that space, and one contraction factor strictly below one valid for every pair.
Check all three hypotheses
Let T map a nonempty complete metric space X to itself. Suppose there is one number q with 0≤q<1 such that every output pair has distance at most q times the corresponding input distance. Then T is a contraction. The word uniform is essential: showing that each particular pair becomes closer, with no common factor below one, does not establish this hypothesis. A derivative bound can supply the factor for a real differentiable map on an interval, but the interval must also be mapped into itself and be complete under its selected metric.
Iterate . Consecutive distances shrink by powers of q. For q strictly between zero and one, summing these distances through the triangle inequality bounds any long tail by a convergent geometric series. Thus the iterates form a Cauchy sequence, and completeness puts the limit x* in X. A contraction is continuous, so passing to the limit in the iteration equation gives T(x*)=x*. If q=0, the map is constant and reaches its fixed point after one application, provided it is a self-map.
Uniqueness and a posteriori certification
If u and v are both fixed points, their distance is at most q times itself. Since q<1 and distance is nonnegative, that distance must be zero. Thus there is only one fixed point in X. This proof does not require trying different initializations numerically. The iteration converges from every initial point in X because the same geometric estimate applies globally. A contraction on one invariant region does not necessarily describe behavior outside that region, so the theorem's domain should remain explicit when reporting the result.
A useful stopping rule uses the latest observed step rather than the first one. After computing x_n, the remaining distance is at most q/(1−q) times for n at least one. Equivalently, if has already been computed, the error at is at most . These are different indexed statements, not contradictory formulas. They are certificates only when q is a proven global contraction factor. Fitting a convenient ratio to a few early steps does not establish a guarantee.
Failure modes explain the assumptions
On X=(0,1), the map T(x)=x/2 is a self-map with contraction factor one half, but its only possible fixed point is zero, outside X. Completeness fails. On the complete real line, T(x)=x+1 preserves all distances and has no fixed point, so replacing q<1 by q≤1 is invalid. A formula that sends allowed inputs outside the proposed domain fails the self-map assumption before any convergence argument begins. Checking these examples is a way to audit an application rather than merely memorize the theorem.
The quadratic gradient iteration in the lab is a particularly transparent case. For f(x)=x²/2 and step size eta, its update is T(x)=(1−eta)x. On the real line this is a contraction exactly when |1−eta|<1, or 0<eta<2. For eta between one and two, signs alternate while magnitudes shrink, demonstrating that convergence need not be monotone in position. At eta=2 a nonzero initial value alternates forever, and above two the magnitude grows. These explicit boundaries come from the contraction factor, not from the smoothness of the animation.
Work it through
An affine fixed point with an error bound
For T(x)=(x+2)/3 on the real line, start at x_0=0. Find the fixed point and certify the error after four iterations.
The real line is complete, T maps it to itself, and |T(x)−T(y)|=|x−y|/3. Thus q=1/3 is a contraction factor. Solving x=(x+2)/3 gives x*=1.
The first step has length 2/3. The initial-step estimate at n=4 is [(1/3)^4/(1−1/3)]·(2/3)=1/81.
The exact recurrence is x_n=1−3^(−n), obtained by subtracting the fixed point from both sides. Hence x_4=80/81 and its error is exactly 1/81, agreeing with the bound.
The unique fixed point is 1. After four applications the certified and actual errors both equal 1/81, approximately 0.01235.
The living notebook
Explore the model
Interactive lab · optimization
Find the contraction threshold
Set nonzero initial x and compare eta=0.5, 1.5, 2, and 2.5 in the exact quadratic gradient iteration. Predict the factor |1−eta| and convergence before playing or stepping. The model is a scalar linear contraction test, not a generic nonlinear fixed-point solver.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forA factor below one controls the whole iteration; oscillation alone does not imply failure.
Pause & reason
Check your understanding
Your turn
Try it yourself
A contraction has q=0.4. The latest step from to has length 0.003. Bound the current error and decide whether it is certified below 0.001.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use q/(1−q) times the latest completed step, keeping the target iterate at .
Compare with the solution
The current error is at most (0.4/0.6)·0.003=0.002. This does not certify error below 0.001, although the actual error might already be smaller. To certify a strict error below 0.001 with the same q, the latest step should be strictly below 0.001·0.6/0.4=0.0015. A certificate is a sufficient guarantee, so failing it is not evidence that the true error exceeds the requested threshold.
Keep the idea
Take it with you
- A complete nonempty space, a self-map, and one contraction factor below one give a unique fixed point.
- The geometric tail proves convergence and supplies explicit error certificates.
- A small observed step is a rigorous stopping rule only when linked to a valid contraction bound.
Read further
Original sources and deeper explanations to follow at your own pace.
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