# 6.8. Continuously Extendable Functions

With a function  f continuous at a we can calculate the value $f\left(a\right)$ knowing alone the values of  f in a neighbourhood of a. When doing this there is actually no need for $f\left(a\right)$ to exist. The mere calculation works even if a does not belong to the domain of  f.

This observation will enable us to assign additional values to certain functions!

Definition:  Let a be an accumulation point of $A\subset ℝ$. A function  $f:A\to ℝ$ is said to be continuously extendable at a if there is another function

 $g:A\cup \left\{a\right\}\to ℝ$ [6.8.1]

continuous at a such that $g|A=f$
 i thus $g\left(x\right)=f\left(x\right)$ for all $x\in A$.
. Any function g like this is called a continuous extension of  f at a.

Consider:

• $a\notin A$ will be the standard situation. This actually corresponds to our original intention to create additional values. Nevertheless the case $a\in A$ is not excluded, an easy case however to deal with:

f is continuously extendable at a $\text{ }⇔\text{ }$ f is continuous at a

And now that $g=g|A=f$f is its own continuous extension.

• For the case $a\in A$ it thus might be better to modify our concept: A function  $f:A\to ℝ$ is called continuously replaceable at $a\in A$ if  $f|A\\left\{a\right\}$ is continuously extentable at a.

• We only consider accumulation points in our discussion. In fact if a would be no accumulation point of A then every function  $f:A\to ℝ$ would be continuously extendable at a, even in any arbitrary way.

As there are sufficient many values of  f in any neighbourhood of a we can prove the uniqueness of a continuous extension.

Proposition and Notation:  For each  $f:A\to ℝ$ there is at most one continuous extension at an accumulation point a of A. If  f is continuously extendable at a the unique number

 $\underset{x\to a}{\mathrm{lim}}$ [6.8.2]

is called the limit of  f at a.

Proof:  Suppose ${g}_{1},{g}_{2}:A\cup \left\{a\right\}\to ℝ$ are two different continuous extensions of  f at a. As a start we then have

${g}_{1}\left(x\right)=f\left(x\right)={g}_{2}\left(x\right)$

for all $x\in A$. As a is an accumulation point of A there is at least one sequence $\left({a}_{n}\right)$ in $A\\left\{a\right\}$ converging to a, and as our extensions are both continuous at a we get:

${g}_{1}\left(a\right)=\mathrm{lim}\phantom{\rule{0.2em}{0ex}}{g}_{1}\left({a}_{n}\right)=\mathrm{lim}\phantom{\rule{0.2em}{0ex}}{g}_{2}\left({a}_{n}\right)={g}_{2}\left(a\right)$

Thus we have  ${g}_{1}={g}_{2}$, a contradiction.

Consider:

• If  $f:A\to ℝ$ is continuous at a, which especially means $a\in A$, we have

$\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=f\left(a\right)$

because  f is its own continuous extension as explained above.

In a first series of examples we consider polynomial quotients. The crucial points are the denominator's zeros. The following criterion is a valuable tool when studying these kind of functions.

Proposition (zero criterion):  Let  $f=\frac{p}{q}$ be a polynomial quotient and a a zero of order k for the denominator q, i.e. $q={\left(\mathrm{X}-a\right)}^{k}\cdot r$  with $r\left(a\right)\ne 0$. Then the following holds:

 f is continuously extendable at a$\text{ }⇔\text{ }$ a is a zero of at least order k for the numerator p [6.8.3]

In this case we can calculate the continuous extension by polynomial division and subsequent cancelling.

Proof:  Let A be the domain of  f. At first we need to show that a is an accumulation point of A:

As r is continuous we know that r, and with it q as well, is non-zero in whole neighbourhood of a. Thus q is not the zero polynomial which guarantees that there are only finitely many zeros, for instance $a,{x}_{1},\dots ,{x}_{n}$. Therefor we find a neighbourhood $\right]a-\epsilon ,a+\epsilon \left[$ of a with no zeros of q other than a:

$\right]a-\epsilon ,a+\epsilon \left[\subset A\cup \left\{a\right\}$.

$\left(a+\frac{\epsilon }{n+1}\right)$ for instance is thus a sequence in $A\\left\{a\right\}$ converging to a.

Let us now turn to the actual proof:

"$⇒$":  If g denotes the continuous extension of  f at a we have for all $x\in A\cup \left\{a\right\}$

$p\left(x\right)=q\left(x\right)\cdot g\left(x\right)={\left(x-a\right)}^{k}\cdot r\left(x\right)\cdot g\left(x\right)$

which proves a to be a zero of at least order k for p.

"$⇐$":  If  $p={\left(\mathrm{X}-a\right)}^{k}\cdot s$ the function $g≔\frac{s}{r}:A\cup \left\{a\right\}\to ℝ$ is a continuous extension of  f at a:

• g is a polynomial quotient and thus continuous at a

• $g\left(x\right)=\frac{s\left(x\right)}{r\left(x\right)}=\frac{{\left(x-a\right)}^{k}\cdot s\left(x\right)}{{\left(x-a\right)}^{k}\cdot r\left(x\right)}=\frac{p\left(x\right)}{q\left(x\right)}=f\left(x\right)$  for all $x\in A$

Some examples will practise the zero criterion:

• $\frac{{\mathrm{X}}^{2}-1}{\mathrm{X}-1}$ is continuously extendabel at the denominator's (single) zero 1 because 1 is also a zero for the numerator. Cancelling the factorized representation

$\frac{{\mathrm{X}}^{2}-1}{\mathrm{X}-1}=\frac{\left(\mathrm{X}-1\right)\left(\mathrm{X}+1\right)}{\mathrm{X}-1}$

will reveal the continuous extension $\mathrm{X}+1$ and thus the limit   $\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{2}-1}{x-1}=\mathrm{X}+1\left(1\right)=2$.

• $\frac{\mathrm{X}+3}{\mathrm{X}-2}$ is not continuously extendable at 2 because 2 is no zero for the numerator.

• $\frac{{\mathrm{X}}^{4}-6{\mathrm{X}}^{3}+10{\mathrm{X}}^{2}-6\mathrm{X}+9}{{\mathrm{X}}^{3}-2{\mathrm{X}}^{2}-15\mathrm{X}+36}=\frac{{\left(\mathrm{X}-3\right)}^{2}\left({\mathrm{X}}^{2}+1\right)}{{\left(\mathrm{X}-3\right)}^{2}\left(\mathrm{X}+4\right)}$ is continuously extendable at the denominator's double zero 3 because 3 is a double zero for the numerator as well. And from the continuous extension $\frac{{\mathrm{X}}^{2}+1}{\mathrm{X}+4}$ we get the limit

$\underset{x\to 3}{\mathrm{lim}}\frac{{x}^{4}-6{x}^{3}+10{x}^{2}-6x+9}{{x}^{3}-2{x}^{2}-15x+36}=\frac{{\mathrm{X}}^{2}+1}{\mathrm{X}+4}\left(3\right)=\frac{10}{7}$

• $\frac{{\mathrm{X}}^{2}+\mathrm{X}}{{\mathrm{X}}^{2}}=\frac{\mathrm{X}\left(\mathrm{X}+1\right)}{{\mathrm{X}}^{2}}=\frac{\mathrm{X}+1}{\mathrm{X}}$ is not continuously extendable at 0 because 0 is a double zero for the denominator but only a single one for the numerator.

Task:  Check if the following polynomial quotients are continuously extendable at the denominator's zeros. Compute the limits in the positive cases.

 $\frac{{\mathrm{X}}^{2}+\mathrm{X}-6}{2\mathrm{X}+6}$ ? $\frac{{\mathrm{X}}^{3}-3{\mathrm{X}}^{2}+4}{{\mathrm{X}}^{3}-4{\mathrm{X}}^{2}+4\mathrm{X}}$ ? $\frac{{\mathrm{X}}^{2}-1}{{\mathrm{X}}^{3}-{\mathrm{X}}^{2}-\mathrm{X}+1}$ ?

With more general functions it is often rather difficult to check their limit behaviour. The following criterion at least offers a technical approach for an investigation.

Proposition (sequence criterion):  Let  $f:A\to ℝ$ be arbitrary, a an accumulation point of $A\subset ℝ$ and $b\in ℝ$. The following statements are equivalent:

 f is continuously extendable at a and $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=b$ [6.8.4] $\text{ }⇔\text{ }$ every sequence $\left({a}_{n}\right)$ in A satisfies  ${a}_{n}\to a\text{ }⇒\text{ }f\left({a}_{n}\right)\to b$

Proof:

"$⇒$":  Let g be the continuous extension of  f at a. For any sequence $\left({a}_{n}\right)$ in A with ${a}_{n}\to a$ we then have:  $f\left({a}_{n}\right)=g\left({a}_{n}\right)\to g\left(a\right)=b$.

"$⇐$":  We need to find a continuous extension of  f at a. For $x\in A\cup \left\{a\right\}$ we set

If $a\notin A$ the identity $g|A=f$ is obvious. If $a\in A$ we only need to show that $f\left(a\right)=b$, but this succeeds immediately when choosing the constant sequence $\left(a\right)$ in [6.8.4].

We now turn to prove the continuity of g at a. Let $\left({a}_{n}\right)$ be any sequence in $A\cup \left\{a\right\}$ converging to a. As it is not for sure that all sequence members differ from a we cannot employ the premise directly. Therefor we need to consider two cases:

• There is a number $k\in {ℕ}^{\ast }$ such that ${a}_{n}=a$ for all $n\ge k$. But these n then satisfy $g\left({a}_{n}\right)=b$ so that the convergence  $g\left({a}_{n}\right)\to b=g\left(a\right)$ is valid.

• For each $k\in {ℕ}^{\ast }$ there is an $n\ge k$ such that ${a}_{n}\ne a$. In this case we modify the original sequence $\left({a}_{n}\right)$: Every sequence member equal to a will be substituted by the next member different from a. The resulting sequence $\left({a}_{{n}_{k}}{\right)}_{k>0}$ where

${n}_{k}≔\mathrm{min}\left\{n\ge k|{a}_{n}\ne a\right\}$

is now a sequence in A converging to a as well: As ${a}_{n}\to a$ there is an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that $|{a}_{n}-a|<\epsilon$ for all $n\ge {n}_{0}$. Due to our construction we have ${n}_{k}\ge k$ so that more than ever all $k\ge {n}_{0}$ satisfy:

$|{a}_{{n}_{k}}-a|<\epsilon$

From the premise we now conclude $g\left({a}_{{n}_{k}}\right)=f\left({a}_{{n}_{k}}\right)\to b$ and thus get an ${n}_{0}\in {ℕ}^{\ast }$ for each $\epsilon >0$ such that

With these k we have . Thus the convergence $g\left({a}_{k}\right)\to b=g\left(a\right)$ is valid and consequently g is continuous at a.

The following examples illustrate the sequence criterion. The first one also proves that the zero criterion [6.8.3] may fail with arbitrary quotients.

Example:

 $\frac{|\mathrm{X}|}{\mathrm{X}}$ is not continuously extendable at 0. Take e.g. $0\ne \frac{{\left(-1\right)}^{n}}{n}\to 0$ and note that is divergent. [6.8.5]
 $\frac{\mathrm{sin}}{\mathrm{X}}$ is continuously extentable at 0 and  $\underset{x\to 0}{\mathrm{lim}}\frac{\mathrm{sin}x}{x}=1$. [6.8.6]
 Proof:  Take $0\ne {a}_{n}\to 0$. Without restriction we may assume that $|{a}_{n}|\le 1$ so that $|{a}_{n}{|}^{i}\le |{a}_{n}|$  holds for all $i,n\in {ℕ}^{\ast }$[1] Consider the sine represented by $\mathrm{sin}=\sum _{i=0}^{\infty }{\left(-1\right)}^{i}\frac{{\mathrm{X}}^{2i+1}}{\left(2i+1\right)!}$ according to [5.9.19]. For a fixed n and an arbitraey $k\in {ℕ}^{\ast }$ we now estimate as follows: $\begin{array}{ll}|\sum _{i=0}^{k}{\left(-1\right)}^{i}\frac{{a}_{n}^{2i+1}}{\left(2i+1\right)!{a}_{n}}-1|\hfill & =|1+\sum _{i=1}^{k}{\left(-1\right)}^{i}\frac{{a}_{n}^{2i}}{\left(2i+1\right)!}-1|\hfill \\ \hfill & =|\sum _{i=1}^{k}{\left(-1\right)}^{i}\frac{{a}_{n}^{2i}}{\left(2i+1\right)!}|\hfill \\ \hfill & \le \sum _{i=1}^{k}\frac{|{a}_{n}{|}^{2i}}{\left(2i+1\right)!}\hfill \\ \hfill & \underset{{\left[1\right]}}{\le }|{a}_{n}|\sum _{i=1}^{k}\frac{1}{\left(2i+1\right)!}\hfill \\ \hfill & \le |{a}_{n}|\sum _{i=0}^{\infty }\frac{1}{i!}\hfill \\ \hfill & =|{a}_{n}|\cdot e\hfill \end{array}$ As the absolute value function is continuous this estimate is valid for the limit ($k\to \infty$) as well: $0\le |\frac{\mathrm{sin}\phantom{\rule{0.15em}{0ex}}{a}_{n}}{{a}_{n}}-1|=|\sum _{i=0}^{\infty }{\left(-1\right)}^{i}\frac{{a}_{n}^{2i+1}}{\left(2i+1\right)!{a}_{n}}-1|\le |{a}_{n}|\cdot e$ The nesting theorem [5.5.8] now finishes the proof: $|\frac{\mathrm{sin}\phantom{\rule{0.15em}{0ex}}{a}_{n}}{{a}_{n}}-1|\to 0\text{ }⇔\text{ }\frac{\mathrm{sin}\phantom{\rule{0.15em}{0ex}}{a}_{n}}{{a}_{n}}-1\to 0$

To extend a function continuously is a concept primarily designed to assign additional values to certain functions. But it also contributes to a further problem: Which are the conditions that allow to continuously concatenate two functions at a given point?

Definition:  Let  $f:A\to ℝ$ and $g:B\to ℝ$ be two functions such that  $f|A\cap B=g|A\cap B$ and $a\in ℝ$.

 The function  $f\cup g:A\cup B\to ℝ$ given by is called the concatenation of  f and g. [6.8.7] We say that  f and g are continuously concatenable at a if  $f\cup g$ is continuously extendable at a. [6.8.8]

Consider:

• As  f and g coincide on $A\cap B$ the concatenation in [6.8.7] is well defined.

The ability of two functions to be continuously concatenable is solely controlled by their limit behaviour at a.

Proposition:  Let  $f:A\to ℝ$ and $g:B\to ℝ$ be two functions such that  $f|A\cap B=g|A\cap B$, a an accumulation point of A and of B. Then the following holds:

 f and g are continuously concatenable at a $⇔\text{ }$ f and g are continuously extentable at a and $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$
[6.8.9]

Proof:  First note that the accumulation point a of A (resp. B) is an accumulation point of $A\cup B$ as well.

"$⇒$":  Let  $f\cup g$ be continuously extendable at a. According to [6.9.1] this is also true for $f=f\cup g|A$ and for $g=f\cup g|B$  with their limits satisfying

$\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\cup g\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$.

"$⇐$":  Let $r:A\cup \left\{a\right\}\to ℝ$ be the continuous extension of  f and $s:B\cup \left\{a\right\}\to ℝ$ be the one of g at a. The identities

show that the concatenation $r\cup s$ is well defined and that $r\cup s|A\cup B=f\cup g$ holds. To prove the continuity of $r\cup s$ at a let $\epsilon >0$ be arbitrary. As r and s are continuous at a we find ${\delta }_{1},{\delta }_{2}>0$ such that

$\begin{array}{l}x\in A\text{ }\wedge \text{ }|x-a|<{\delta }_{1}\text{ }⇒\text{ }|r\left(x\right)-r\left(a\right)|<\epsilon \hfill \\ x\in B\text{ }\wedge \text{ }|x-a|<{\delta }_{2}\text{ }⇒\text{ }|s\left(x\right)-s\left(a\right)|<\epsilon \hfill \end{array}$

With  $\delta ≔\mathrm{min}\left\{{\delta }_{1},{\delta }_{2}\right\}$  we thus have for all $x\in A\cup B$:

 6.7. 6.9.