Try
Consider the constant functions fₙ(x)=1/n on [0,1]. Every graph is flat, every value lies between 0 and 1, and all functions share the strongest possible continuity control: changing x does not change the value at all. Nevertheless, the family consisting of precisely these functions is not compact in uniform distance. The graphs approach the zero function, which was left out.
Uniform distance measures the largest vertical difference over the whole domain. Here the distance from fₙ to zero is exactly 1/n. Write down the functions with values 1,1/2,1/3 and the missing limit, and ask which collection includes all its uniform limits.
| Collection | Uniform limit of fₙ | Included? | Maximum of −f(0) |
|---|
| Constants 1/n only | zero function | No | None; supremum is 0 |
| Same constants plus zero | zero function | Yes | 0, attained by zero |
The enlarged collection is the closure: all functions that can be uniformly approximated by members of the original family. A family is relatively compact when its closure is compact. This is not the same as saying the family itself is compact. For optimization, that distinction decides whether the best value belongs to an admissible function.
Predict
Now take all functions on [0,1] that start at height zero and satisfy |f(x)−f(y)|≤2|x−y|. Predict whether this family has an escaping sequence. The anchor and the common slope bound imply |f(x)|≤2. Moreover, if the horizontal distance is less than ε/2, the vertical difference is less than ε for every member. These are uniform controls on the entire family.
The inequality here is a Lipschitz condition; no derivative is assumed to exist. Piecewise linear tents, including their corners, are allowed. If the anchor is removed, arbitrary constant functions are allowed too. Their oscillation in x is still zero, but the constants can drift to arbitrarily large heights. Continuity control alone does not control absolute height.
Predict what happens if the equality f(0)=0 is replaced by the strict condition f(0)>0 while retaining a common size bound. Constants 1/n again approach a function that violates the strict condition. A uniform limit can preserve a non-strict inequality or equality while violating a strict inequality.
Experiment
Use three candidate graphs on [0,1]: zero, x, and the tent t(x)=1/2−|x−1/2|. All start at zero and are 1-Lipschitz. For each, compute J(f)=f(1/2) minus the integral of f from 0 to 1. This functional compares the midpoint height with the average height over the interval.
Check the three calculations
For zero, both terms vanish, so J=0. For f(x)=x, midpoint height and integral are both 1/2, so J=0 again. The tent has height 1/2 and area 1/4, giving J=1/4. Comparing these three examples does not by itself prove that no other admissible function does better. Later we obtain an upper bound for the entire family and identify the tent as an attaining function.
Now change a graph by at most δ vertically everywhere. Its midpoint value changes by at most δ, and its integral changes by at most δ because the domain has length one. Therefore J changes by at most 2δ. This quantitative continuity estimate is the mechanism that will transfer a maximizing sequence’s values to its uniform limit.
Understand
Work on a nonempty compact metric space K. The real continuous functions C(K) have distance ||f−g||∞, the supremum of their pointwise absolute differences. Every individual continuous function is bounded and uniformly continuous on K. A family is pointwise bounded if, for each x, all its values at x form a bounded set. Equicontinuity means a continuity radius can be chosen independently of the function, at each fixed point; on compact K it yields a uniform family modulus. The preceding extraction theorem establishes that pointwise boundedness and equicontinuity give a uniformly convergent subsequence from every sequence in the family.
First prove the converse. Suppose the uniform closure of a nonempty family F is compact. Cover that closure by uniform balls of radius 1 and choose a finite subcover with continuous centers h₁,…,h_s. Every f∈F lies within 1 of some center. Consequently ||f||∞≤1+max over i of ||h_i||∞, a finite bound. This proves more than pointwise boundedness.
Given ε>0, use a new finite net of uniform radius ε/4. Each of its finitely many centers is uniformly continuous. Choose one positive δ that makes every center vary by less than ε/4 whenever d(x,y)<δ. If f is within ε/4 of center h_i, then
|f(x)−f(y)| ≤ |f(x)−h_i(x)| + |h_i(x)−h_i(y)| + |h_i(y)−f(y)| < 3ε/4.
Thus the entire family is uniformly equicontinuous. Finite nets matter: taking a minimum of infinitely many positive radii need not give a positive radius. The empty family satisfies these properties vacuously and has compact empty closure.
Now make the forward conclusion about the closure explicit. Let F be pointwise bounded and equicontinuous, and let gₙ be a sequence in its closure. For nonempty F choose fₙ∈F with ||gₙ−fₙ||∞<1/n. Apply the preceding extraction theorem to fₙ, obtaining strictly increasing indices n_j and a uniform limit f∈C(K). Then
||g at index n_j − f||∞ ≤ 1/n_j + ||f at index n_j − f||∞ → 0.
The limit lies in the closure because it is a uniform limit of members of F. We have proved sequential compactness of the closure, not just extracted a subsequence from its smaller subset. The equivalence between compactness and sequential compactness for metric spaces now gives compactness. That metric-space theorem is an explicit reused ingredient, not a consequence of merely observing several convergent graphs.
Combining the two directions gives the relative compactness criterion. To conclude that F itself is compact, additionally show that it is closed in C(K). For example, uniform convergence preserves a fixed anchor f(x₀)=c and each fixed Lipschitz or Hölder inequality: pass to the limit at x₀, or at each pair x,y. The common bound supplies relative compactness; passing constraints to the limit supplies closedness. These are separate jobs.
Apply
For the anchored 1-Lipschitz family in the experiment, every function has absolute value at most 1. The family is nonempty, equicontinuous and closed, hence compact. The earlier estimate |J(f)−J(g)|≤2||f−g||∞ proves continuity of J. Its values are bounded, so its supremum S is a finite real number. Choose fₙ with J(fₙ)>S−1/n. Compactness gives a subsequence converging uniformly to a function f still in the family. Continuity gives J(f)=S. This is the full attainment argument.
We can now identify the maximum directly. For every t, the Lipschitz inequality gives f(1/2)−f(t)≤|t−1/2|. Integrating yields J(f)≤1/4. The tent from the experiment makes this inequality an equality for every t, so it attains the maximum 1/4. A derivative calculation was unnecessary.
For integral constraints, distinguish equality from strict inequality. On [0,1], |∫f−∫g|≤||f−g||∞, so a condition such as ∫f=1 is preserved by uniform limits. With a size bound 3 and a fixed Hölder bound H|x−y|^α, where H≥0 and 0<α≤1, constant 1 is feasible and the constrained family is compact. Replacing the integral equality by ∫f<1 loses closedness: constants 1−1/n are admissible but converge to the excluded constant 1. Always check nonemptiness too; a closed collection of inconsistent constraints has no maximizing function.
Continue
Return to the uniform subsequence extraction proof when establishing relative compactness. Then check every proposed constraint under uniform limits, and prove continuity of the objective in that same norm. A pointwise-continuous expression is not automatically a continuous functional under every topology.
Keep the logical chain visible: hypotheses control subsequences; closure keeps their limits; a continuous functional transfers objective values. Neither a finite graph gallery nor a formal supremum establishes attainment. The six tasks ask for general finite-net proofs, a missing-limit counterexample, closed and nonclosed constraints, and a concrete optimizer with a valid bound.
