6.10. Sequences of Continuous Functions

Sometimes it it difficult to access properties of a function f directly. An alternative idea is to approximate f by simpler functions which however requires that we only test properties which cooperate with the approximation method used. In this part we develop such a concept for continuity.

Definition: Let A be a non empty subset of

ℝ

and let

f_{n} : B_{n} \to ℝ

be functions numbered by

n \in ℕ^{*}

such that

A \subset B_{n}

. The sequence on functions

(f_{n})

is called pointwise convergent on A if the sequence of numbers

(f_{n} (x))

is convergent for all

x \in A

. In this case the function

f : A \to ℝ

defined by

f (x) ≔ \lim f_{n} (x)

[6.10.1]

is called the limit function of $(f_{n})$ . We use the symbol $f = \lim f_{n}$ or the notation

f_{n} \underset{p w}{\to} f

to denote the pointwise convergence of $(f_{n})$ to its limit function f. If all $B_{n} = A$ we normally omit the adjunct "on A".

Consider:

As each value $f (x)$ is the limit of an ordinary sequence it is unique. Thus any sequence of functions has at most one limit function.
The limit function of a convergent power series (c.f. [5.11.9]) is a special example for the new notion:

$\sum_{i = 0}^{n} a_{i} {(X - a)}^{i} \underset{p w}{\to} \sum_{i = 0}^{\infty} a_{i} {(X - a)}^{i}$
As known from common sequences pointwise convergence also does not depend on the first sequence members, which means: For each $k \in ℕ$ we have

$(f_{n}) \underset{p w}{\to} f \Leftrightarrow (f_{n + k}) \underset{p w}{\to} f$

Example:

$\frac{n}{n + 1} X^{2} \underset{p w}{\to} X^{2}$ , because: $\frac{n}{n + 1} x^{2} \to x^{2}$ for each $x \in ℝ$ .
$(X^{n})$ is not pointwise convergent on $ℝ$ as at least one of the number sequences $(x^{n})$ , e.g. $(2^{n})$ , is divergent.

$(X^{n})$ however has a pointwise limit on $[0, 1]$ namely the function $f : [0,1] \to ℝ$ defined by

f (x) ≔ {\begin{array}{l} 0,  if 0 \leq x < 1 \\ 1,  if x = 1 \end{array}

[6.10.2]

Proof: For $0 \leq x < 1$ we know from [5.7.2] that $x^{n} \to 0$ , and $1^{n} \to 1$ is obvious.

The last example displays a certain shortcoming in pointwise convergence: Although all sequence members, i.e. the powers $X^{n}$ , are continuous on $[0,1]$ this is no longer true
i

f is in fact discontinuous at 1 as $1 - \frac{1}{n} \to 1$ , but

$f (1 - \frac{1}{n}) = 0 \to 0 \neq 1$ .

for the limit function [6.10.2]. This may happen due to the fact that the individual convergences $f_{n} (x) \to f (x)$ are not coordinated: Though there is an $n_{0} \in ℕ^{*}$ for each $ε > 0$ with

| f_{n} (x) - f (x) | < ε

for all $n \geq n_{0}$ this $n_{0}$ however may vary from x to x. Thus we need a stronger notion of convergence to get it "continuity-proof".

Definition: A sequence of functions $(f_{n})$ is called uniformly convergent on A to a function $f : A \to ℝ$ if there is an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that

| f_{n} (x) - f (x) | < ε

for all

n \geq n_{0}

and all

x \in A

[6.10.3]

Whereas f is again called the limit function the convergence now is denoted by

f_{n} \underset{u f}{\to} f

It is not uncommon to use $f_{n} \Rightarrow f$ instead.

Consider:

A uniformly convergent sequence $(f_{n})$ is pointwise convergent as well having the same unique limit function.
The reverse is not true as the example [6.10.2] and the property [6.10.5] show.
For each $k \in ℕ$ we have: $(f_{n}) \underset{u f}{\to} f \Leftrightarrow (f_{n + k}) \underset{u f}{\to} f$

Example:

For every positive $q < 1$ the powers $X^{n}$ are uniformly convergent on $[0, q]$ :

X^{n} \underset{u f}{\to} 0 | [0, q]

Proof: For any given $ε > 0$ there is an $n_{0} \in ℕ^{*}$ according to [5.7.2] such that the following estimate holds for all $n \geq n_{0}$ and for all $x \in [0, q]$

| x^{n} - 0 | \leq q^{n} < ε

The technique used in the proof above will be generalised to a test in [6.10.7]. But first we will show that uniform convergence respects continuity.

Proposition: Let $(f_{n})$ be uniformly convergent on A and take any $a \in A$ . Then we have:

If each $f_{n}$ is uniformly continuous on A then f is uniformly continuous on A.	[6.10.4]
If each $f_{n}$ is continuous at a then f is continuous at a.	[6.10.5]
$f_{n} \in C^{0} (A)$ for all n $\Rightarrow f \in C^{0} (A)$ .	[6.10.6]

Proof:

1. ► For any $ε > 0$ we need to find a $δ > 0$ (see [6.5.2]) such that

| f (x) - f (y) | < ε

for all $x, y \in A$ with $| x - y | < δ$ . As $f_{n} \underset{g m}{\to} f$ we firstly find an $n_{0} \in ℕ^{*}$ for $\frac{ε}{3} > 0$ such that

| f_{n_{0}} (x) - f (x) | < \frac{ε}{3}

for all

x \in A

.[1]

Now, as $f_{n_{0}}$ is uniformly continuous there is a $δ > 0$ such that

| f_{n_{0}} (x) - f_{n_{0}} (y) | < \frac{ε}{3}

[2]

for all $x, y \in A$ with $| x - y | < δ$ . Due to [1] and [2] these x now satisfy

| f (x) - f (y) | \leq | f (x) - f_{n_{0}} (x) | + | f_{n_{0}} (x) - f_{n_{0}} (y) | + | f_{n_{0}} (y) - f (y) | < \frac{ε}{3} + \frac{ε}{3} + \frac{ε}{3} = ε

2. ► Now we need to find a $δ > 0$ for each $ε > 0$ (see [6.5.1]) such that

| f (x) - f (a) | < ε

for all $x \in A$ with $| x - a | < δ$ . But this is actually the proof of 1 if we replace y by a. The argument for [2] now is the continuity of $f_{n_{0}}$ at a.

3. ► is an immediate consequence of 2.

It is often difficult to see if a sequence of functions is uniformly convergent or not so that we would benefit from appropriate tests. We prove two of them, a kind of comparison test and the well-known Cauchy-test. The technical advantage of Cauchy's test is that we don't need to know the limit function, the disadvantage on the other hand that we won't get it.

Proposition (comparison test): For any sequence of functions $(f_{n})$ and any function $f : A \to ℝ$ we have:

$f_{n} \underset{u f}{\to} f \Leftrightarrow$

there is a zero sequence $(a_{n})$ in $ℝ^{\geq 0}$ and a $k \in ℕ^{*}$ such that

| f_{n} (x) - f (x) | \leq a_{n}

for all $x \in A$ and all $n \geq k$ .

[6.10.7]

Proof:

" $\Rightarrow$ ": At first we employ [6.10.3] to find a $k \in ℕ^{*}$ for $ε = 1$ such that

| f_{n} (x) - f (x) | < 1

for all

n \geq k

and all

x \in A

For an n like this the set ${| f_{n} (x) - f (x) | | x \in A}$ is non-empty and bounded. Thus the completeness axiom
i

Every non-empty, bounded subset of $ℝ$ has a greatest lower bound, its infimum, and a least upper bound, its supremum.

guarantees the existence of the number

a_{n} ≔ \sup {| f_{n} (x) - f (x) | | x \in A}

for each $n \geq k$ . Setting $a_{1} ≔ \dots ≔ a_{k - 1} ≔ 0$ additionally yields a sequence $(a_{n})$ in $ℝ^{\geq 0}$ which already satisfies [6.10.7]. To complete the proof we only need to show that $a_{n} \to 0$ . To that end we take an arbitrary $ε > 0$ . As $f_{n} \underset{u f}{\to} f$ we find an $n_{0} \in ℕ^{*}$ , without restriction $n_{0} \geq k$ , such that

| f_{n} (x) - f (x) | < \frac{ε}{2}

for all

n \geq n_{0} \geq k

and all

x \in A

For these n we thus have: $| a_{n} - 0 | = a_{n} \leq \frac{ε}{2} < ε$ .

" $\Leftarrow$ ": As $a_{n} \to 0$ there is an $n_{0} \in ℕ^{*}$ , let's say $n_{0} \geq k$ , for each $ε > 0$ such that

| f_{n} (x) - f (x) | \leq a_{n} = | a_{n} - 0 | < ε

for all

n \geq n_{0}

and all

x \in A

As an example we consider the sequence $(\frac{n X}{1 + n^{2} X^{2}})$ . For any $0 < c < 1$ this sequence converges uniformly on $[c, 1]$ to 0: The estimate

| \frac{n x}{1 + n^{2} x^{2}} - 0 | \leq \frac{n \cdot 1}{1 + n^{2} c^{2}}

for all

x \in [c, 1]

proves $(\frac{n}{1 + c^{2} n^{2}})$ to be suitable for [6.10.7].

Proposition (Cauchy test): A sequence of functions $(f_{n})$ converges uniformly on A if and only if there is an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that

| f_{n} (x) - f_{m} (x) | < ε

for all

n, m \geq n_{0}

and all

x \in A

[6.10.8]

Proof:

" $\Rightarrow$ ": The proof for this direction is just a duplicate of the corresponding sentence [5.5.7] for common sequences.

" $\Leftarrow$ ": For each $x \in A$ $(f_{m} (x))$ is a real Cauchy sequence and thus convergent due to [5.8.8]. Setting

f (x) ≔ \lim_{m \to \infty} f_{m} (x)

[3]

hence defines a function $f : A \to ℝ$ . We now take an arbitrary $ε > 0$ to show that $f_{n} \underset{u f}{\to} f$ . Due to the premise there is an $n_{0} \in ℕ^{*}$ such that

| f_{n} (x) - f_{m} (x) | < \frac{ε}{2}

for all

n, m \geq n_{0}

and all

x \in A

.[4]

For a fixed $n \geq n_{0}$ and any $x \in A$ we use the second limit theorem, the continuity of the absolute value function and the property [5.5.3] to get from [3] and [4] the following estimate:

| f_{n} (x) - f (x) | = \lim_{m \to \infty} | f_{n} (x) - f_{m} (x) | \leq \frac{ε}{2} < ε

6.9.