Pointwise and uniform convergence

Try

For f_n(x)=x^n on [0,1], each fixed input below one eventually produces a small output, while the input one always produces one. The pointwise limit is therefore zero on [0,1) and one at the endpoint. The inputs are fixed before n increases. A moving input can behave differently: x_n=(1/2)^(1/n) approaches one while f_n(x_n)=1/2 forever. This moving-point observation reveals the distinction between pointwise and uniform convergence.

Predict

Predict whether removing the endpoint repairs uniform convergence. On [0,1), the pointwise limit becomes the continuous zero function, but the same moving inputs remain admissible. Their errors stay at 1/2. Thus deleting the endpoint changes the limit function without creating uniform convergence. Restricting instead to [0,r] for a fixed r<1 succeeds because every output is bounded by r^n. Domain choice is part of a convergence claim, not an optional plotting preference.

Experiment

Use the live experiment to compare pointwise traces with the largest error over the displayed domain.

Live experiment: control every point with one index

Pointwise convergence fixes x before choosing a sufficiently late index. Uniform convergence asks for one index that works across the whole domain. Compare the curves with the pointwise limit, then inspect the exact supremum error; a small error at a few sample points is not enough.

Function sequence xⁿ on [0,1]xⁿ on [0,0.8]xⁿ/n on [0,1]nx/(1+n²x²) on [0,1]Index n 1248163264128Tolerance ε 0.10.050.01Reset experiment

sup |fₙ−f| = 1. This value is not attained: it is approached as x tends to 1 from below, while the error at 1 itself is 0. It is not strictly below the selected tolerance 0.1.

For each fixed x<1, xⁿ→0, while fₙ(1)=1 for all n. For every finite n, xⁿ can be arbitrarily close to 1 with x<1, so the supremum error remains 1. The pointwise limit jumps at 1; convergence is not uniform.

No index can make the supremum error tend to zero for this family. Increasing the plot resolution or inspecting more fixed points does not remove the moving or endpoint obstruction.

Compare xⁿ on the two domains, then inspect the moving peak at n=128. Explain the order of “for every x” and “there exists N.” Curves use 200 intervals plus the exact moving maximizer; rounded display values are not a proof. Settings are temporary.

Requirement	Order of choices
Pointwise convergence	Fix x, then ε, then choose N that may depend on both
Uniform convergence	Fix ε, then choose one N for every x
Disproof of uniformity	Fix an error threshold and find late indices with violating points

A finite menu illustrates these quantifiers but does not test every point or every tail. The mathematical argument must provide an all-input estimate or a construction that defeats any proposed cutoff.

Understand

Formally, f_n→f pointwise on D means that for each x∈D and ε>0 there is N such that n≥N implies |f_n(x)−f(x)|<ε. Uniform convergence moves the choice of x after N: for each ε>0 there is one N such that all n≥N and all x∈D satisfy that bound.

To prove uniform convergence of x^n on [0,r], take arbitrary ε>0. Geometric convergence supplies N with r^n<ε for all n≥N. Since 0≤x^n≤r^n, this same N controls every input. To disprove it on [0,1), choose ε₀=1/4. For any N choose n≥N and x_n as above. The error 1/2 exceeds ε₀, exactly contradicting the required quantifiers. No assertion about an attained maximum at the excluded endpoint is needed.

Apply

Repair “all the graphs look close, so convergence is uniform.” A graph samples finitely many inputs and indices. Narrow features can move between samples, and a cutoff that works at selected points need not control the domain. Conversely, uniform convergence does not require monotone improvement at every point or every index; it requires eventual control for each tolerance.

For f_n(x)=x/n, the error on [−M,M] is at most M/n, so convergence is uniform on every fixed bounded interval. On R, choosing x=n leaves error one. Being uniform on every compact interval is called locally uniform convergence; it does not automatically become uniform on the entire unbounded line.

Continue

Optional depth asks how uniform convergence changes when the output is vector-valued. Replace absolute value by a norm; the same quantifier structure remains. Completeness of the output space becomes important when constructing a limit from Cauchy information.

Save one positive proof with a point-independent estimate and one negative proof with moving witnesses. Explicitly state the domain in both. Next, the supremum norm will package the all-input error as one number and make uniform convergence resemble ordinary convergence in a new space whose points are functions.