Convex feasibility and projection geometry
Derive projection certificates, analyze alternating projections, and separate feasibility from nearest-point optimization.
On this page
The question for today
What you’ll understand
- Verify a projection through its normal-cone inequality.
- Explain Fejér decrease and finite-dimensional convergence of alternating projections.
- Distinguish an intersection point, projection onto the intersection and an infeasibility certificate.
Many constraints are easy to enforce separately and awkward to enforce simultaneously. Projecting onto each set in turn seems a natural repair procedure. Convex geometry explains both why this can work and what it does not promise. Projection onto a closed convex set has a global variational certificate and a firm distance bound. Alternating projections can find a common point under finite-dimensional closedness, convexity and nonempty-intersection assumptions, but that common point need not be the closest one to the original input. A small residual also needs careful interpretation when feasibility is uncertain.
The nearest-point certificate
For a nonempty closed convex set C in a real Hilbert space, the projection p of x is the unique point minimizing squared distance to x. It is characterized by for every z in C. Necessity follows by differentiating squared distance along the feasible segment from p to z. Sufficiency follows by expanding the squared distance to any z: the cross term has the correct sign, so no feasible point is closer. This certificate proves a global nearest-point statement without searching the whole set numerically.
Projection is the proximal map of the indicator and the resolvent of the normal cone. Its firm nonexpansiveness therefore follows from the preceding lesson, but the variational inequality also gives a useful Pythagorean inequality relative to every feasible z. Orthogonal projection onto an affine subspace gives equality in the associated orthogonality relation; a curved set or a halfspace generally yields an inequality. Confusing these cases can produce an incorrect geometric proof. Closedness matters for attainment, and convexity matters for uniqueness and the global normal certificate.
Alternating projections decrease distance to the intersection
Now take nonempty closed convex C and D in , with a nonempty intersection, and iterate x_{k+1}=P_C(P_Dx_k). For any fixed z in the intersection, applying the Pythagorean inequality twice shows that distances to z never increase and that the squared correction lengths have a finite sum. This is Fejér monotonicity with respect to the intersection. The corrections therefore tend to zero. The iterates stay bounded, so finite-dimensional compactness gives a cluster point; vanishing corrections and closedness place it in both sets.
Once a cluster point lies in the intersection, Fejér monotonicity forces the full sequence to converge to it: the distances to that particular point are nonincreasing and have a subsequence tending to zero. This proves norm convergence in the stated finite-dimensional setting. It gives neither a universal convergence rate nor a guarantee that the limit equals projection of the starting point onto C∩D. Computing that nearest point is a different optimization task; methods such as Dykstra’s algorithm add correction variables rather than simply alternating the unmodified projections.
Residuals, infeasibility and stopping
If the intersection is empty, the preceding argument has no feasible reference z. Alternating points may approach a best-approximation pair, may fail to attain an inter-set distance, or may escape to infinity. Two disjoint closed convex sets can even have distance zero: take the epigraph of exp(x) in the plane and the lower halfspace y≤0. Their vertical gap can be arbitrarily small as x tends to negative infinity, yet they share no finite point. Thus a small distance between two repaired points is not an exact feasibility certificate.
A practical stopping rule should measure the original constraints and say whether its tolerance is absolute or relative to chosen units. To certify infeasibility one needs additional information, such as a strict separating inequality with a verified gap; a long sequence of unsuccessful projections is insufficient. For equality subspaces, projection can often be computed by solving a linear system, whose numerical conditioning also matters. The abstract nearest-point operator assumes exact evaluation. Approximate projections require an error-controlled convergence argument, especially when errors accumulate instead of diminishing along the iteration.
Work it through
Alternating two lines
Let C be the x-axis and D the line y=x in R². Starting at x₀=(2,0), compute two full cycles of P_C P_D and identify the limit.
Projection of (a,b) onto D is ((a+b)/2,(a+b)/2), obtained by minimizing (t−a)²+(t−b)² over t. Projection onto C keeps the first coordinate and sets the second to zero.
The first D projection sends (2,0) to (1,1), and the subsequent C projection gives x₁=(1,0). The second cycle passes through (1/2,1/2) and yields x₂=(1/2,0).
Inductively x_k=(2^{1-k},0). The limit is (0,0), the unique point in C∩D. Here it also equals the projection onto the intersection because that intersection is a singleton.
Each full cycle halves the norm in this example. The rate comes from the angle between these particular lines and is not a universal rate for arbitrary convex sets.
The living notebook
Explore the model
Interactive lab · optimization
Inspect the same scalar contraction factor
Pause and Reset, then set step size η=0.5 and initial x₀=2. Before each Step predict the sequence 2,1,0.5,0.25, matching the x-coordinate after full cycles in the two-line example. For a fresh comparison Reset and set η=0.5,x₀=−2 to see the symmetric sequence. The instrument computes scalar quadratic gradient descent; its matching factor illustrates the derived sequence and does not perform the two intermediate geometric projections.
- 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 simple contraction can encode a particular projection sequence, while its geometric assumptions must still be proved separately.
Pause & reason
Check your understanding
Your turn
Try it yourself
Project x=(3,2) onto the triangle C={(u,v):u≥0,v≥0,u+v≤2}. Prove your answer using the projection inequality rather than only a sketch.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
First project onto the boundary line u+v=2, and check whether that point also satisfies both nonnegativity constraints.
Compare with the solution
Subtract 3/2 from each coordinate to obtain p=(3/2,1/2), which is feasible and lies on u+v=2. The residual is x−p=(3/2,3/2). For every z=(u,v) in C, its inner product with z−p is (3/2)(u+v−2)≤0. The variational certificate therefore proves p is the unique nearest point in C. The squared distance is (3/2)²+(3/2)²=9/2. The initial line projection works here because it satisfies the remaining constraints; that check cannot be omitted in general.
Keep the idea
Take it with you
- Projection is characterized by a global normal inequality.
- Alternating projections converge under the stated finite-dimensional feasible-intersection assumptions.
- A feasible limit, a nearest-point solution and an infeasibility certificate are different deliverables.
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