Pointwise and Uniform Convergence

A sequence of real numbers attaches one number to each $n \in N$ . A sequence of functions attaches one function. For each $n$ we have $f_{n} : S \to R$ , and we ask what it means for the whole list $f_{1}, f_{2}, f_{3}, \dots$ to approach a limit function $f$ . Unlike the numerical case, there is more than one reasonable answer, and the two standard answers behave differently under the operations of calculus.¹

If $f_{n}$ is a sequence of continuous functions and $f_{n} \to f$ in the weaker sense, $f$ can fail to be continuous. Under the stronger sense it cannot. The same split decides whether the limit of the integrals equals the integral of the limit, and whether a power series may be differentiated term by term. Which convergence holds decides each case.

Pointwise convergence

The first notion evaluates at a single point and reduces to a numerical limit.

Because limits of numerical sequences are unique, the pointwise limit function $f$ , when it exists, is unique. Spelling the numerical limit out with its $ϵ$ – $N$ definition gives an equivalent form that exposes the one feature that matters.

The index $N$ is allowed to depend on both $ϵ$ and $x$ . For a fixed $ϵ$ , different inputs may demand different thresholds, and nothing forces those thresholds to have a common bound.

The powers

x^{2 n}

[0, 1]

flatten toward

0

on the interior but stay pinned at

1

when

x = 1

; the pointwise limit is discontinuous there.

A function given by a series is a pointwise limit of its partial sums. Writing

k = 0 \sum \infty x^{k} means f (x) = n \to \infty lim k = 0 \sum n x^{k} = \frac{1}{1 - x},

and the convergence holds only on $(- 1, 1)$ , even though $\frac{1}{1 - x}$ is defined for every $x \neq = 1$ . The partial sums are defined everywhere, but they converge only inside the unit interval.

For $f_{n} (x) = sin (n x)$ there is no pointwise limit on any interval. The values converge at isolated points such as $x = 0$ , but on every interval $[a, b]$ some $x$ makes $sin (n x)$ oscillate without settling.²

Uniform convergence

The second notion removes the dependence of $N$ on $x$ . One threshold must work across the entire domain.

The quantifier order is the whole difference. In pointwise convergence the phrase reads for each $x$ , there is an $N$ ; in uniform convergence it reads there is an $N$ , good for every $x$ . Uniform convergence gives a band of half-width $ϵ$ around the graph of $f$ , and past the index $N$ every graph $f_{n}$ lies inside that band.

Uniform convergence: past some index every

f_{n}

lies inside a band of fixed half-width around

f

; the band width does not shrink or grow with

x

Uniform convergence is the stronger condition. Choosing the single $N$ certainly supplies a working $N$ at each individual $x$ , so the implication is immediate.

The powers $f_{n} (x) = x^{2 n}$ are the standard witness that the converse fails.

Whether uniformity survives can depend on the domain even for a sequence that converges to $0$ at every point.

On a bounded interval the lines

x / n

flatten uniformly to

0

n

grows; on the whole line each still reaches unbounded height, so uniformity holds only after the domain is bounded.

The uniform norm

Uniform convergence becomes a statement about a single sequence of numbers once we measure the size of a function by its largest value.

The subscript names the set over which the supremum runs, so $∥ f ∥_{K}$ for $K \subset S$ uses only the points of $K$ . The quantity $∥ f_{n} - f ∥_{S}$ is the largest vertical gap between the two graphs, and uniform convergence says that gap collapses to $0$ .

The picture is a single measured segment: the tallest gap between $f_{n}$ and $f$ . Uniform convergence drives that one number to zero.

The uniform norm

∥ f_{n} - f ∥_{S}

is the height of the tallest vertical gap between the graphs; uniform convergence means this height goes to

0

The uniform Cauchy criterion

For numerical sequences, Cauchy sequences and convergent sequences coincide because $R$ is complete. The uniform norm gives functions the same internal test: a sequence can be certified without naming its limit.

Proof. Suppose ${f_{n}}$ is uniformly Cauchy. Fix $x \in S$ . Then $∣ f_{m} (x) - f_{k} (x) ∣ \leq ∥ f_{m} - f_{k} ∥_{S}$ , so ${f_{n} (x)}$ is a Cauchy sequence of real numbers and converges; define $f (x) = lim_{n} f_{n} (x)$ . This gives pointwise convergence. To upgrade it, fix $ϵ > 0$ and $N$ with $∥ f_{m} - f_{k} ∥_{S} < ϵ /2$ for $m, k \geq N$ , so $∣ f_{m} (x) - f_{k} (x) ∣ < ϵ /2$ for every $x$ . Holding $m$ fixed and letting $k \to \infty$ sends $∣ f_{m} (x) - f_{k} (x) ∣$ to $∣ f_{m} (x) - f (x) ∣$ , giving $∣ f_{m} (x) - f (x) ∣ \leq ϵ /2 < ϵ$ for all $x$ and all $m \geq N$ . Hence ${f_{n}}$ converges uniformly to $f$ . The converse follows from the triangle inequality: if $f_{n} \to f$ uniformly, choose $N$ with $∣ f_{n} (x) - f (x) ∣ < ϵ /4$ for $n \geq N$ , and then $∣ f_{m} (x) - f_{k} (x) ∣ \leq ∣ f_{m} (x) - f (x) ∣ + ∣ f (x) - f_{k} (x) ∣ < ϵ /2$ for all $x$ and all $m, k \geq N$ ; take the supremum. $□$

The value of the criterion is that it tests the sequence against itself. The Picard iteration produces functions with no closed form, and the uniform Cauchy criterion certifies that their limit exists.

Series of functions and the M-test

A series $\sum_{k = 1}^{\infty} g_{k}$ of functions converges uniformly if its partial sums $s_{n} = \sum_{k = 1}^{n} g_{k}$ do. The uniform Cauchy criterion supplies a test that never mentions the limit and reduces everything to one numerical series.

Each term is dominated by a numerical bound $M_{k}$ , and the convergence of the number series $\sum M_{k}$ is transferred to uniform convergence of the function series.

The M-test caps each term's height by a number

M_{k}

; if the numeric series of caps converges, the function series converges uniformly.

The two convergences at a glance

	Pointwise	Uniform
Quantifier order	for each $x$ , exists $N (ϵ, x)$	exists $N (ϵ)$ , for all $x$
$N$ depends on	$ϵ$ and $x$	$ϵ$ only
Norm statement	$f_{n} (x) \to f (x)$ each $x$	$∥ f_{n} - f ∥_{S} \to 0$
Continuity of limit	can fail	preserved
Cauchy test	not self-contained	uniformly Cauchy $\Leftrightarrow$ converges
Certified by	direct limit	uniform norm bound, M-test

Pointwise convergence is the minimum a limit function must satisfy. Uniform convergence is the hypothesis under which the operations of calculus pass to the limit; the interchange theorems make this precise.

Lebl, §6.1. A solution of a differential equation is often built as a limit of approximate solutions, so the meaning of that limit must be made precise before the operations of calculus can be applied to it. ↩
Lebl, §6.1, Example 6.1.4. That $sin (n x)$ has no limit for some $x$ in every interval is left as an exercise there; the values are equidistributed enough to keep oscillating. ↩
Lebl, §6.1, Example 6.1.11. ↩

Pointwise convergence

Uniform convergence

The uniform norm

The uniform Cauchy criterion

Series of functions and the M-test

The two convergences at a glance

Footnotes