Theory

Sequences and series of functions

Pointwise convergence and uniform convergence of a sequence of functions, the theorems on passing continuity, integrals and derivatives to the limit, and total convergence for series of functions.

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Sequences of functions and pointwise convergence
Uniform convergence

Sequences of functions and pointwise convergence

1. Sequence of functions

A sequence of functions is a family $(f_{n})_{n \in N}$ of functions $f_{n} : A \to R$ , all defined on the same set $A \subseteq R$ . For each fixed $x \in A$ , the values $f_{1} (x), f_{2} (x), f_{3} (x), \dots$ form a numerical sequence. Examples: $f_{n} (x) = x^{n}$ on $[0, 1]$ , $f_{n} (x) = \frac{sin ( n x )}{n}$ on $R$ . The chapter answers two questions: in what sense $f_{n}$ can tend to a limit function $f$ , and which properties of $f_{n}$ (continuity, integral, derivative) pass to the function $f$ .

2. Pointwise convergence

The sequence $(f_{n})$ converges pointwise to $f$ on $A$ if for every $x \in A$ the numerical sequence $f_{n} (x)$ converges to $f (x)$ , that is, if

n \to + \infty lim f_{n} (x) = f (x) for every x \in A .

In the $ε$ definition: for every $x \in A$ and every $ε > 0$ there exists $N$ such that $∣ f_{n} (x) - f (x) ∣ < ε$ for every $n \geq N$ ; the index $N$ depends on $ε$ and also on the point $x$ . Pointwise convergence is studied with the methods for limits of sequences from Mathematical Analysis 1, one point at a time, and it determines the candidate limit $f$ : every other notion of convergence in the chapter compares $f_{n}$ with this function.

The sequence xⁿ on [0, 1]

$f_{n} (x) = x^{n}$ on $[0, 1]$ . For $0 \leq x < 1$ the geometric sequence $x^{n}$ tends to $0$ ; for $x = 1$ we have $f_{n} (1) = 1$ for every $n$ . The pointwise limit is therefore

f (x) = {01 if 0 \leq x < 1 if x = 1

As $n$ increases the graph drops toward the $x$ -axis on the whole interval $[0, 1)$ , but every curve rises back up to the point $(1, 1)$ .

y = x^{3}

y = x^{3}

: the curve is already close to

0

on half the interval, but it still reaches

(1, 1)

y = x^{15}

y = x^{15}

: almost the whole graph is flattened onto the

x

-axis, then it rises sharply toward

(1, 1)

Warning

The pointwise limit of continuous functions need not be continuous. Each $f_{n} (x) = x^{n}$ is continuous on $[0, 1]$ , but the pointwise limit equals $0$ on $[0, 1)$ and $1$ at the point $1$ , so it is discontinuous. Pointwise convergence does not preserve continuity; preserving it requires the stronger notion introduced below.

y = 0

y = x^{2}

y = x^{12}

y = x^{2}

and

y = x^{12}

with the pointwise limit (in color): it equals

0

[0, 1)

, on the

x

-axis, and

1

at the point

x = 1

Uniform convergence

The sequence $(f_{n})$ converges uniformly to $f$ on $A$ if for every $ε > 0$ there exists $N$ such that

∣ f_{n} (x) - f (x) ∣ < ε for every n \geq N and for every x \in A .

The difference from pointwise convergence lies in the order of the requirements: the same index $N$ works for all points $x \in A$ simultaneously. Geometrically, for $n \geq N$ the graph of $f_{n}$ is contained in the tube of radius $ε$ around the graph of $f$ , that is, in the strip between $f - ε$ and $f + ε$ . Uniform convergence implies pointwise convergence; the counterexamples below show that the converse does not hold.

y = \sin x + \dfrac{1}{5} \sin \left(5 x\right)

y = \sin x

Strip between

f - ε

and

f + ε

with

f (x) = sin x

(dashed) and

ε = \frac{3}{10}

: the graph of

f_{5} (x) = sin x + \frac{sin ( 5 x )}{5}

lies entirely inside the strip, because

∣ f_{5} (x) - f (x) ∣ \leq \frac{1}{5}

for every

x

Sup criterion

$(f_{n})$ converges uniformly to $f$ on $A$ if and only if

M_{n} = x \in A sup ∣ f_{n} (x) - f (x) ∣ n \to + \infty 0.

The equivalence follows from the definition, since the condition $∣ f_{n} (x) - f (x) ∣ < ε$ for every $x \in A$ implies $M_{n} \leq ε$ and, conversely, the condition $M_{n} < ε$ implies $∣ f_{n} (x) - f (x) ∣ < ε$ for every $x \in A$ . The criterion turns the study of uniform convergence into a Mathematical Analysis 1 problem: for each $n$ one looks for the maximum of $∣ f_{n} - f ∣$ on $A$ , usually by studying the derivative, and then computes the limit of the numerical sequence $M_{n}$ .

Sup criterion via the derivative

$f_{n} (x) = x e^{- n x}$ on $[0, + \infty)$ . For each fixed $x \geq 0$ , $e^{- n x} \to 0$ if $x > 0$ and $f_{n} (0) = 0$ , so the pointwise limit is $f = 0$ . Since $f_{n} \geq 0$ , we study $f_{n}$ itself:

f_{n}^{'} (x) = e^{- n x} (1 - n x),

so $f_{n}$ increases on $[0, \frac{1}{n}]$ and decreases afterwards, and the maximum is

M_{n} = f_{n} (\frac{1}{n}) = \frac{1}{n e} n \to + \infty 0.

Therefore the convergence is uniform on $[0, + \infty)$ .

y = x e^{-x}

y = x e^{-3 x}

y = x e^{-8 x}

y = x e^{- n x}

for

n = 1, 3, 8

: the maximum is at

x = \frac{1}{n}

, equals

\frac{1}{n e}

and tends to

0

Warning

The same sequence can converge uniformly on one set and not on another. $f_{n} (x) = x^{n}$ converges pointwise to $0$ on $[0, 1)$ , but $sup_{x \in [0, 1)} x^{n} = 1$ for every $n$ , so the convergence is not uniform on $[0, 1)$ . On $[0, a]$ with $0 < a < 1$ we get $M_{n} = a^{n} \to 0$ , so the convergence is uniform on every interval $[0, a]$ . The set one works on is part of the answer.

y = x^{8}

y = x^{8}

with

a = \frac{4}{5}

: on

[0, a]

the maximum is

a^{8} \approx 0.17

, while on

[0, 1)

the values approach

1

x \to 1^{-}

Remark

The study of a sequence of functions follows a two-step scheme. First one computes the pointwise limit $f$ , point by point, distinguishing cases when the expression requires it. Then one studies $M_{n} = sup ∣ f_{n} - f ∣$ on the given set and on any subsets on which the convergence improves, as in the examples $[0, a] \subset [0, 1)$ or $[δ, + \infty) \subset (0, + \infty)$ .

The theory continues with 5 more sections (continuity of the uniform limit, passing to the limit under the integral sign, differentiation for sequences…), plus all worked examples, the solutions and the PDF.

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