Pointwise vs Uniform Convergence

Function Sequence

Index n 1

Epsilon band ε 0.150

Convergence Test

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sup|fₙ−f| (this n)

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Smallest N for this ε

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How it works

A sequence of functions f_n: D → ℝ converges pointwise to f if, at every fixed x, the number sequence f_n(x) → f(x). It converges uniformly if the worst-case gap over the whole domain vanishes:

Pointwise: ∀x ∀ε>0 ∃N(x, ε): n≥N ⟹ |fₙ(x)−f(x)| < ε
Uniform:   ∀ε>0 ∃N(ε): n≥N ⟹ |fₙ(x)−f(x)| < ε  for ALL x at once
Equivalently: uniform  ⟺  sup_x |fₙ(x) − f(x)| → 0

The surface is a "waterfall" of the whole family: x runs across, n runs into the depth, height is f_n(x), and colour is the error |f_n(x) − f(x)| (blue = small, red = large). The translucent band hugs the limit function f(x) at height ±ε; the highlighted curve is the current slice n, with green/red instanced markers showing where that slice sits inside or outside the band.

xⁿ on [0,1] — each f_n is continuous, but the pointwise limit is 0 on [0,1) and jumps to 1 at x=1: discontinuous. Since a uniform limit of continuous functions must be continuous, this sequence cannot converge uniformly — and indeed sup|f_n−f| stays exactly 1 for every n, no matter how large.
sin(nx)/n on [0,2π] — |f_n(x)| ≤ 1/n everywhere, so the sup-norm itself shrinks to 0. One N works for every x simultaneously: uniform convergence to f = 0.
n·x·(1−x)ⁿ on [0,1] — a "travelling bump" that slides toward x = 0 as n grows. At every fixed x the value still → 0 (pointwise), and the limit function 0 is perfectly continuous — yet the bump's peak height settles near 1/e ≈ 0.368 forever, so the sup-norm never shrinks: not uniform, even though continuity alone didn't rule it out.

This is the central subtlety of real analysis courses: pointwise limits can lose properties (continuity, integrability, differentiability) that every term in the sequence had; uniform convergence is exactly the extra condition that preserves them.

Low error (near limit f) High error (far from f) ε-band around f(x)

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