core29 min studyLesson 176 in Mathematics

Pointwise and uniform convergence

Read the quantifier order that distinguishes local agreement from a simultaneous error guarantee.

Before you begin

On this page

The question for today

What you’ll understand

Translate pointwise and uniform convergence into explicit tolerance statements.
Use supremum errors to prove uniform convergence or construct moving witnesses against it.
Explain continuity preservation through a three-term error decomposition.

Suppose every location in a simulated profile eventually becomes accurate. That sounds like convergence, but it leaves a crucial ambiguity: may each location demand a different waiting time, or can one cutoff guarantee accuracy everywhere? Pointwise and uniform convergence encode these alternatives. The distinction determines which properties pass to a limit and whether a finite approximation can be trusted across a whole domain. Learning to move the quantifiers correctly is more useful than memorizing isolated counterexamples.

Who chooses the waiting time?

Pointwise convergence fixes an input first and then asks for late enough indices to meet a tolerance at that input. The cutoff may depend on both the tolerance and the location. Uniform convergence asks for one cutoff depending only on the tolerance, valid simultaneously for every input. This reversal is not a stylistic change. It excludes a region of stubborn error that moves or narrows as the index grows, even if no fixed input remains inaccurate forever.

For bounded errors, the supremum norm packages the simultaneous requirement into a scalar sequence. The supremum need not be attained: a domain may omit the very boundary where error is largest. To disprove uniform convergence, find a fixed positive tolerance and, for arbitrarily late indices, an input at which that tolerance is violated. The witness input is allowed to depend on the index. Insisting on one fixed witness would miss precisely the moving-error phenomenon.

f_n\to f\ \text{uniformly on }E\quad\Longleftrightarrow\quad\sup_{x\in E}|f_n(x)-f(x)|\longrightarrow0

Equivalently, for every ε>0 there is one N such that n≥N and x∈E imply |fₙ(x)−f(x)|<ε. The cutoff is independent of x.↔ Scroll wide equations horizontally.

A boundary layer defeats uniform convergence

On the closed unit interval, the powers xⁿ tend to zero for x<1 and stay one at x=1. Their pointwise limit therefore jumps at the right endpoint. The functions are individually continuous, so this already suggests the convergence cannot be uniform. More directly, values just below one remain near one for each fixed n, making the supremum error from the pointwise limit equal to one despite every fixed interior error tending to zero.

Restricting the same functions to a shorter closed interval from zero to r, with r<1, changes the answer. The supremum error is rⁿ and tends to zero. Even deleting the endpoint one without imposing a positive gap does not suffice: on the half-open interval from zero to one the supremum is still one. Uniform convergence depends on the entire domain, not merely on an algebraic formula or whether the limiting function happens to be continuous there.

0\le r<1:\quad\sup_{0\le x\le r}x^n=r^n\to0,\qquad\sup_{0\le x<1}x^n=1

A supremum can equal one without any input attaining it. This is why excluding the endpoint does not produce uniform convergence on [0,1).↔ Scroll wide equations horizontally.

Continuity survives through a common approximation

To prove that a uniform limit of continuous functions is continuous at a chosen point, allocate the desired error among three terms. First choose one approximating function uniformly close to the limit. Next use continuity of that fixed approximant to control its values near the chosen point. The two approximation errors and the middle continuity error add to the target tolerance. The approximating index must be chosen before the nearby point, which is exactly what uniform convergence allows.

This argument works on any metric domain; compactness is unnecessary for preserving continuity under uniform convergence. Compactness becomes relevant to other conclusions, such as turning continuity into uniform continuity. In the quadratic lab, send the coefficient a toward zero on a fixed bounded input interval. Predict how the largest value shrinks compared with the value at one selected position. The model exposes the difference between a pointwise reading and a whole-domain bound without pretending to display the xⁿ boundary layer.

Work it through

A moving witness locates the missing uniform bound

On [0,1), show that $f_n(x)=x^n$ tends pointwise to zero but not uniformly.

Fix x in [0,1). Since its absolute value is below one, the geometric sequence xⁿ tends to zero. This proves pointwise convergence at every allowed input.
For each n choose $x_n=2^{-1/n}$ , which lies strictly between zero and one. Its nth power equals 1/2, so this moving input retains a fixed error.
If uniform convergence held, one cutoff would force every later error below, for example, 1/4 at every input. The inputs xₙ violate that condition at every n, giving a contradiction.

The result

The sequence converges pointwise to zero on [0,1), but the error cannot be bounded uniformly by a quantity tending to zero.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

One location versus a domain-wide error budget

The controls vary a and the inspected x for y=ax². On the displayed bounded interval [−R,R], predict the maximum error from zero as |a|R² and compare it with the selected point’s value. This exact quadratic family illustrates a uniform bound; it does not simulate xⁿ or arbitrary function sequences.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA common bound over the domain is stronger than accuracy at one selected input.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Uniform convergence requires one cutoff for every input at the chosen tolerance. Allowing a different cutoff for each input gives pointwise convergence instead.

Your turn

Try it yourself

Prove that $g_n(x)=x^2/n$ converges uniformly to zero on [−3,3], and explain why the same claim fails on the whole real line.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Bound the largest squared input before choosing a cutoff.

Compare with the solution

On [−3,3], the supremum of |gₙ| is 9/n, so choosing an integer N>9/ε controls every input for all n≥N. On the real line, for each n the function is unbounded. Taking x=√n gives error one at arbitrarily late indices, contradicting any eventual uniform error below one half. The same formula thus has different convergence behavior on different domains.

Keep the idea

Take it with you

Uniform convergence changes the order of the tolerance and input quantifiers.
Moving witness inputs can disprove a simultaneous bound.
Uniform limits preserve continuity, but the domain still controls the error norm.

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