# 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 $n\in {ℕ}^{\ast }$ such that $A\subset {B}_{n}$. The sequence on functions $\left({f}_{n}\right)$ is called pointwise convergent on A if the sequence of numbers $\left({f}_{n}\left(x\right)\right)$ is convergent for all $x\in A$. In this case the function $f:A\to ℝ$ defined by

 $f\left(x\right)≔\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{f}_{n}\left(x\right)$ [6.10.1]

is called the limit function of $\left({f}_{n}\right)$. We use the symbol $f=\mathrm{lim}\phantom{\rule{0.3em}{0ex}}{f}_{n}$ or the notation

${f}_{n}\underset{pw}{\to }f$

to denote the pointwise convergence of $\left({f}_{n}\right)$ to its limit function f. If all ${B}_{n}=A$ we normally omit the adjunct "on A".

Consider:

• As each value $f\left(x\right)$ 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:

$\sum _{i=0}^{n}{a}_{i}{\left(\mathrm{X}-a\right)}^{i}\underset{pw}{\to }\sum _{i=0}^{\infty }{a}_{i}{\left(\mathrm{X}-a\right)}^{i}$

• 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

$\left({f}_{n}\right)\underset{pw}{\to }f\text{ }⇔\text{ }\left({f}_{n+k}\right)\underset{pw}{\to }f$

•

Example:

• $\frac{n}{n+1}{\mathrm{X}}^{2}\underset{pw}{\to }{\mathrm{X}}^{2}$, because:  $\frac{n}{n+1}{x}^{2}\to {x}^{2}$ for each $x\in ℝ$.

• $\left({\mathrm{X}}^{n}\right)$ is not pointwise convergent on $ℝ$ as at least one of the number sequences $\left({x}^{n}\right)$, e.g. $\left({2}^{n}\right)$, is divergent.

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

 [6.10.2]

Proof:  For $0\le x<1$ 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 ${\mathrm{X}}^{n}$, are continuous on $\left[0,1\right]$ this is no longer true
 i f is in fact discontinuous at 1 as $1-\frac{1}{n}\to 1$, but $f\left(1-\frac{1}{n}\right)=0\to 0\ne 1$.
for the limit function [6.10.2]. This may happen due to the fact that the individual convergences ${f}_{n}\left(x\right)\to f\left(x\right)$ are not coordinated: Though there is an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ with

$|{f}_{n}\left(x\right)-f\left(x\right)|<\epsilon$

for all $n\ge {n}_{0}$ 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 $\left({f}_{n}\right)$ is called uniformly convergent on A to a function $f:A\to ℝ$ if there is an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that

 $|{f}_{n}\left(x\right)-f\left(x\right)|<\epsilon$  for all $n\ge {n}_{0}$ and all $x\in A$. [6.10.3]

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

${f}_{n}\underset{uf}{\to }f$.

It is not uncommon to use ${f}_{n}⇒f$ instead.

Consider:

• A uniformly convergent sequence $\left({f}_{n}\right)$ 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:   $\left({f}_{n}\right)\underset{uf}{\to }f\text{ }⇔\text{ }\left({f}_{n+k}\right)\underset{uf}{\to }f$

•

Example:

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

 ${\mathrm{X}}^{n}\underset{uf}{\to }0|\left[0,q\right]$

Proof:   For any given $\epsilon >0$ there is an ${n}_{0}\in {ℕ}^{\ast }$ according to [5.7.2] such that the following estimate holds for all $n\ge {n}_{0}$ and for all $x\in \left[0,q\right]$

$|{x}^{n}-0|\le {q}^{n}<\epsilon$.

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 $\left({f}_{n}\right)$ 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] ${f}_{n}\in {\mathcal{C}}^{0}\left(A\right)$ for all n$\text{ }⇒\text{ }f\in {\mathcal{C}}^{0}\left(A\right)$. [6.10.6]

Proof:

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

$|f\left(x\right)-f\left(y\right)|<\epsilon$

for all $x,y\in A$ with $|x-y|<\delta$. As ${f}_{n}\underset{gm}{\to }f$ we firstly find an ${n}_{0}\in {ℕ}^{\ast }$ for $\frac{\epsilon }{3}>0$ such that

$|{f}_{{n}_{0}}\left(x\right)-f\left(x\right)|<\frac{\epsilon }{3}$  for all  $x\in A$.[1]

Now, as ${f}_{{n}_{0}}$ is uniformly continuous there is a $\delta >0$ such that

$|{f}_{{n}_{0}}\left(x\right)-{f}_{{n}_{0}}\left(y\right)|<\frac{\epsilon }{3}$[2]

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

$|f\left(x\right)-f\left(y\right)|\le |f\left(x\right)-{f}_{{n}_{0}}\left(x\right)|+|{f}_{{n}_{0}}\left(x\right)-{f}_{{n}_{0}}\left(y\right)|+|{f}_{{n}_{0}}\left(y\right)-f\left(y\right)|<\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon$

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

$|f\left(x\right)-f\left(a\right)|<\epsilon$

for all $x\in A$ with $|x-a|<\delta$. 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 $\left({f}_{n}\right)$ and any function $f:A\to ℝ$ we have:

 ${f}_{n}\underset{uf}{\to }f\text{ }⇔\text{ }$ there is a zero sequence $\left({a}_{n}\right)$ in ${ℝ}^{\ge 0}$ and a $k\in {ℕ}^{\ast }$ such that $|{f}_{n}\left(x\right)-f\left(x\right)|\le {a}_{n}$ for all $x\in A$ and all $n\ge k$. [6.10.7]

Proof:

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

$|{f}_{n}\left(x\right)-f\left(x\right)|<1$  for all $n\ge k$ and all $x\in A$.

For an n like this the set $\left\{|{f}_{n}\left(x\right)-f\left(x\right)||x\in A\right\}$ 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

${a}_{n}≔\mathrm{sup}\left\{|{f}_{n}\left(x\right)-f\left(x\right)||x\in A\right\}$

for each $n\ge k$. Setting ${a}_{1}≔\dots ≔{a}_{k-1}≔0$ additionally yields a sequence $\left({a}_{n}\right)$ in ${ℝ}^{\ge 0}$ 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 $\epsilon >0$. As ${f}_{n}\underset{uf}{\to }f$ we find an ${n}_{0}\in {ℕ}^{\ast }$, without restriction ${n}_{0}\ge k$, such that

$|{f}_{n}\left(x\right)-f\left(x\right)|<\frac{\epsilon }{2}$  for all $n\ge {n}_{0}\ge k$ and all $x\in A$.

For these n we thus have: $|{a}_{n}-0|={a}_{n}\le \frac{\epsilon }{2}<\epsilon$.

"$⇐$":  As ${a}_{n}\to 0$ there is an ${n}_{0}\in {ℕ}^{\ast }$, let's say ${n}_{0}\ge k$, for each $\epsilon >0$ such that

$|{f}_{n}\left(x\right)-f\left(x\right)|\le {a}_{n}=|{a}_{n}-0|<\epsilon$  for all $n\ge {n}_{0}$ and all $x\in A$.

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

$|\frac{nx}{1+{n}^{2}{x}^{2}}-0|\le \frac{n\cdot 1}{1+{n}^{2}{c}^{2}}$  for all $x\in \left[c,1\right]$

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

Proposition (Cauchy test):  A sequence of functions $\left({f}_{n}\right)$ converges uniformly on A if and only if there is an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that

 $|{f}_{n}\left(x\right)-{f}_{m}\left(x\right)|<\epsilon$  for all $n,m\ge {n}_{0}$ and all $x\in A$. [6.10.8]

Proof:

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

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

$f\left(x\right)≔\underset{m\to \infty }{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}{f}_{m}\left(x\right)$[3]

hence defines a function $f:A\to ℝ$. We now take an arbitrary $\epsilon >0$ to show that ${f}_{n}\underset{uf}{\to }f$. Due to the premise there is an ${n}_{0}\in {ℕ}^{\ast }$ such that

$|{f}_{n}\left(x\right)-{f}_{m}\left(x\right)|<\frac{\epsilon }{2}$  for all $n,m\ge {n}_{0}$ and all $x\in A$.[4]

For a fixed $n\ge {n}_{0}$ 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:

$|{f}_{n}\left(x\right)-f\left(x\right)|=\underset{m\to \infty }{\mathrm{lim}}|{f}_{n}\left(x\right)-{f}_{m}\left(x\right)|\le \frac{\epsilon }{2}<\epsilon$.