You sample a continuous function at 0, 1/4, 1/2, 3/4 and 1. Every recorded value is zero. Must the whole function be near zero? Draw a tent between each pair of consecutive nodes. The tents can be arbitrarily tall unless you know something beyond continuity of each individual graph.
Now impose a common slope bound in the precise Lipschitz sense: |f(x)−f(y)|≤2|x−y| for all inputs. A tent rising from zero at one endpoint and returning to zero at the next cannot exceed height 1/4. Its midpoint is distance 1/8 from a node, so the allowed change is at most 2·1/8. These are exact geometric constraints, not a conclusion inferred from five measurements.
Grid intervals m
Spacing
Farthest distance to a node
Largest height when L=2
2
1/2
1/4
1/2
4
1/4
1/8
1/4
8
1/8
1/16
1/8
Sketch both the positive tents and their reflections below the axis for m=4. They agree at every node, yet their largest difference is 1/2. This is twice the largest deviation of either function from zero.
Predict
A family is a collection of functions on the same domain. Each member can be continuous without sharing a usable continuity scale with the others. For fₙ(x)=xⁿ on [0,1], every member is continuous and every value lies between 0 and 1. Predict why no single neighborhood of 1 keeps all members within 1/2 of their value at 1.
Choose any y strictly below 1 but as close to 1 as requested. Eventually yⁿ is below 1/2, while fₙ(1)=1. Choosing the function after the neighborhood defeats the proposed common control. In contrast, the constant functions cₙ(x)=n have perfect common continuity control: every difference within one function is zero. Their values are nevertheless not commonly bounded. Continuity control and value bounds address different questions.
Experiment
In the grid explorer, select “One common slope bound” and leave Sample discrepancy e at 0 to inspect the zero-sampled tent family. Before revealing the certificate, predict the maximum possible intersample height from the displayed Lipschitz constant L and number of equal intervals m. Then compare the single-function height L/(2m) with the discrepancy L/m between opposite tents. Change the grid density while retaining the same L, and explain why halving the spacing halves both bounds.
Can matching samples hide a gap?
Place m+1 sample nodes at 0, 1/m, …, 1. The nearest-node distance is at most δ=1/(2m). These prescribed functions form opposite triangular tents between nodes: the diagram uses their formulas, not interpolation justified by sample data alone.
At every sample node, f=e/2 and g=−e/2. Predict the largest difference between nodes. A Lipschitz bound L means |f(x)−f(y)|≤L|x−y| for every pair of inputs, and likewise for g.
Solid line: f. Dashed line: g. Squares: observed grid nodes. Circles: midpoint values known from the specified formulas. Overlapping sample markers mean equal values, not missing data.
Inspect all nodes and exact tent breakpoints
All piecewise-linear breakpoints; decimals rounded.
Input
f
g
Sample node?
0
0
0
Yes
0.125
0.25
-0.25
No: midpoint
0.25
0
0
Yes
0.375
0.25
-0.25
No: midpoint
0.5
0
0
Yes
0.625
0.25
-0.25
No: midpoint
0.75
0
0
Yes
0.875
0.25
-0.25
No: midpoint
1
0
0
Yes
Compare m=4 and m=32 first with fixed L=2, then with fixed-height tents. Finally set e=0.1 and explain which part of the bound is sampling error and which comes from the common continuity control. Controls are temporary and award no completion.
Next choose “Fixed-height tents; slopes grow.” Each individual tent still reaches height 1 above its node value as m increases, while its slope bound becomes 2m. The pairwise gap stays 2+e. This changed family does not have a fixed common Lipschitz bound. Return to the bounded family and set e=0.1: the largest pairwise gap becomes 0.1+L/m, while each graph’s node value is shifted by half the sample discrepancy.
The functions in this experiment are defined by their formulas. The shared Lipschitz constant is supplied and can be proved; it is not estimated from the zero samples. For an unknown measured function, the same sample table would not establish that assumption. A narrow, steeper tent could agree with every sample and violate the proposed bound.
For a separate paper calculation, suppose two functions both have Lipschitz constant 3 and their discrepancies at the quarter-grid nodes are at most 1/20. Predict a bound over the entire interval. The nearest-node distance is 1/8, so the bound is 1/20+2·3·1/8=4/5. The two factor-3 terms control movement from each function’s sample to its value between nodes.
Understand
Let K be a metric space with distance d, and F a family of real-valued functions on K. Equicontinuity at every point means: for every ε>0 and every x∈K there exists δ>0 such that, for every f∈F and y∈K, d(x,y)<δ implies |f(x)−f(y)|<ε. The radius may depend on x and ε, but not on the chosen function. This differs from merely requiring each member to be continuous, where the radius may depend on f as well.
Uniform equicontinuity moves the choice of δ earlier: for every ε>0 there exists δ>0 working for every f∈F and every x,y∈K. The domain points no longer affect the radius. Some texts use “equicontinuous” for this stronger version; the present lesson keeps the two quantifier orders explicit.
On a compact K, the first property implies the second. Fix ε>0. At each center z choose r_z>0 so every family member changes by less than ε/2 within distance r_z of z. The balls of radius r_z/2 cover K. Choose finitely many of them and let δ be half the smallest corresponding full radius. If d(x,y)<δ, some selected half-radius ball contains x. The triangle inequality then puts y inside its full-radius ball. For every f, both f(x) and f(y) differ from the center value by less than ε/2, so |f(x)−f(y)|<ε. The same δ works for every member and both input points. If K is empty, the conclusion is vacuous and no minimum of an empty list is needed.
Now assume additionally pointwise boundedness: for each x there is a finite M_x bounding |f(x)| for all f∈F. Use equicontinuity with tolerance 1 to cover K by neighborhoods on which every f differs from its center value by less than 1. Extract finitely many centers z₁,…,zᵣ. Then every function value is bounded in magnitude by 1+max(M_z₁,…,M_zᵣ). This is a common bound over both functions and points. The finite collection of center bounds is the step that pointwise boundedness alone does not provide.
A single anchor does not replace pointwise boundedness on an arbitrary compact domain. On K consisting of 0 and 1, set fₙ(0)=0 and fₙ(1)=n. Radius 1/2 gives uniform equicontinuity, because inputs within that distance coincide. The anchor 0 is bounded while values at the isolated point 1 are unbounded.
For a quantitative grid certificate, let f and g both be L-Lipschitz. Suppose each input x lies within δ of a grid node z, and all node discrepancies are at most ε. Then
Taking the supremum gives the whole-domain certificate. The covering radius δ, the sample error ε and the common regularity bound L are separate inputs.
Apply
Prove sharpness, not merely plausibility. Let G be the equally spaced grid with m intervals, and let dist(x,G) be the distance to its nearest node. The triangle inequality gives dist(x,G)≤|x−y|+dist(y,G); reversing x and y proves this distance function is 1-Lipschitz. Thus h(x)=L dist(x,G) is L-Lipschitz. It vanishes at the nodes and reaches L/(2m) at every midpoint. Any other L-Lipschitz function with zero node values has magnitude at most this height, by comparison to a nearest node. The pair h and −h attains discrepancy L/m, so the two-function factor of two cannot generally be removed.
Change the domain to (0,1) and consider the single function 1/x. At each point it is continuous, so the singleton family is equicontinuous there and pointwise bounded. Yet inputs 1/n and 1/(n+1) approach one another while their outputs differ by 1. There is no uniform equicontinuity and no global value bound. The compactness hypothesis was doing substantive work.
Finally, contrast the isolated-anchor example with a globally L-Lipschitz family on [0,1] satisfying |f(0)|≤M. Here |f(x)|≤M+Lx≤M+L. The stronger global inequality connects the anchor directly to every point; no false general anchor theorem is needed.
The next extraction argument will combine common control with convergence on finitely many or countably many selected inputs. Today’s bounds explain how sampled agreement can extend to a whole domain, but they do not themselves choose a convergent subsequence or prove that an arbitrary family is compact.
Apply · explain · revise
Practice and reasoning
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For real functions on a metric space K, write the quantifiers for equicontinuity at every point and for uniform equicontinuity. Classify the families f_n(x)=x^n on [0,1] and c_n(x)=n on [0,1]. Discuss both common continuity control and pointwise boundedness.
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Let K be a compact metric space and F an equicontinuous family of real functions on K in the pointwise sense. Prove uniform equicontinuity by a finite-ball argument. Do not assume that nearby x,y belong to a neighborhood centered at x with an already uniform radius.
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Assume a family F of real functions on compact metric K is equicontinuous and pointwise bounded: for each x, sup over f∈F of |f(x)| is finite. Prove a common finite bound for all f and x. Explain why a bound at just one anchor point is insufficient on arbitrary compact spaces.
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Let f,g:[0,1]→R both be L-Lipschitz, with L≥0. A finite grid G has the property that each x is within distance δ of some grid point, and |f(z)−g(z)|≤ε for every z∈G. Prove a sup-norm bound. Apply it to L=3, grid {0,1/4,1/2,3/4,1}, and ε=1/20. Why does the sample discrepancy alone not bound the sup norm?
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For m≥1 an integer let G_m={j/m:0≤j≤m}, and define h(x)=L dist(x,G_m) on [0,1], where L≥0. Prove h is L-Lipschitz, find its exact maximum, and prove the best possible bound for an L-Lipschitz function that vanishes on G_m. Then compare h and −h.
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On K=(0,1), consider the singleton family containing f(x)=1/x. Prove it is equicontinuous at each point and pointwise bounded, but neither uniformly equicontinuous nor commonly bounded. Compare with a family of L-Lipschitz functions on [0,1] satisfying |f(0)|≤M, and prove the latter has a common bound.
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Reflect and continue
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