Mathematics.

sequences of functions

Pointwise vs. Uniform Convergence

Real Analysis25 minDifficulty4 out of 10

You should know: uniform convergence

Overview

Pointwise convergence of a sequence of functions (fₙ) to f only requires that, at each individual x, the numerical sequence fₙ(x) converges to f(x) — different x's are allowed to converge at wildly different rates. Uniform convergence demands a single rate that works across the whole domain at once. Every uniformly convergent sequence converges pointwise, but the converse is false, and the gap between the two notions matters enormously: pointwise limits can destroy continuity, integrability, and differentiability in ways uniform limits cannot.

Intuition

In pointwise convergence, each point x is allowed its own private 'speed of convergence' — the index N needed to get within ε can differ from point to point, and can even blow up as x approaches some troublesome location (e.g. the edge of the domain). Uniform convergence removes this loophole: a single N must serve every point simultaneously. The classic example fₙ(x) = xⁿ on [0,1] converges pointwise (to 0 for x<1, to 1 at x=1) but not uniformly, precisely because points near x=1 need larger and larger N — there's no uniform bound.

Formal Definition

Definition

Contrasting the two definitions for fₙ: D → ℝ:

Pointwise: ∀x∈D, ∀ε>0, ∃N(ε,x) such that n≥N  ⟹  ∣fn(x)−f(x)∣<ε\text{Pointwise: } \forall x \in D,\ \forall \varepsilon>0,\ \exists N(\varepsilon, x) \text{ such that } n \ge N \implies |f_n(x)-f(x)| < \varepsilon
Pointwise convergence (N may depend on x)
Uniform: ∀ε>0, ∃N(ε) such that ∀x∈D, n≥N  ⟹  ∣fn(x)−f(x)∣<ε\text{Uniform: } \forall \varepsilon>0,\ \exists N(\varepsilon) \text{ such that } \forall x \in D,\ n \ge N \implies |f_n(x)-f(x)| < \varepsilon
Uniform convergence (N depends only on ε)
Uniform convergence  ⟹  Pointwise convergence, but not conversely.\text{Uniform convergence} \implies \text{Pointwise convergence, but not conversely.}
One-way implication

Worked Examples

  1. For fixed x ≠ 0, as n→∞, the denominator dominates, so fₙ(x) → 0. At x=0, fₙ(0)=0 for all n.

    fn(x)→0 pointwise for every xf_n(x) \to 0 \text{ pointwise for every } x

  2. Compute sup over x of |fₙ(x)|: maximize x/(1+nx²) by calculus, giving max at x=1/√n with value 1/(2√n).

    sup⁡x∣fn(x)∣=12n\sup_x |f_n(x)| = \frac{1}{2\sqrt{n}}

  3. Since 1/(2√n) → 0, the supremum tends to 0, so the convergence IS uniform on ℝ (this is a case where pointwise and uniform coincide).

    12n→0\frac{1}{2\sqrt{n}} \to 0

Answer: fₙ → 0 both pointwise and uniformly on ℝ, since the maximum deviation 1/(2√n) → 0.

Practice Problems

Difficulty 4/10

Find the pointwise limit of fₙ(x) = nx(1-x)ⁿ on [0,1] and determine whether convergence is uniform.

Difficulty 3/10

Which statement correctly relates the two convergence notions?

Difficulty 6/10

Prove: if fₙ → f pointwise on [a,b] and each fₙ is continuous but f is discontinuous, then the convergence cannot be uniform.

Quiz

A sequence of continuous functions converges pointwise to a discontinuous function. What can you conclude?
fₙ(x) = xⁿ on [0,1] converges pointwise to a function that is:

Summary

  • Pointwise convergence lets the rate of convergence (N) depend on x; uniform convergence requires one N for the whole domain.
  • Uniform convergence implies pointwise convergence, but the converse fails — e.g. fₙ(x)=xⁿ on [0,1].
  • A discontinuous pointwise limit of continuous functions is decisive proof that the convergence is not uniform.

References

  1. WebsiteWikipedia — Uniform convergence