# 6.9. Properties of Continuously Extendable Functions

It is quite natural that continuously extendable functions inherit a lot of their properties from the continuous ones. But there are also some results from sequences - the nesting theorem and the limit theorems to name a few - that have mirror theorems within our current subject.

As with continuity itself continuous extendability proves to be a local property. The initial propositions work out the details.

Proposition:  Let a be an accumulation point of $A\subset B$. If  $f:B\to ℝ$ is continuously extendable at a then  $f|A$ is continuously extendable at a and has the same limit:

 $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f|A\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)$ [6.9.1]

In general the reverse does not hold.

Proof:  From [6.2.8] we see that if $g:B\cup \left\{a\right\}\to ℝ$ is a continuous extension of  f at a then $g|A\cup \left\{a\right\}$ continuously extends  $f|A$ at a. And from this the identity of the limits is obvious.

The Heaviside step function H
 i
is not continuously extendable at 0 in contrast to its restriction $\mathrm{H}|{ℝ}^{>0}$.

The local character is also stressed by the fact that two functions that coincide locally don't differ in their limit behaviour (cf. [6.2.11]).

Proposition:  Let a be an accumulation point of $A\cap B$. If  $f:A\to ℝ$ and $g:B\to ℝ$ coincide locally at a the following equivalence holds:

 f continuously extendable at a$\text{ }⇔\text{ }$g continuously extendable at a [6.9.2]

If the limits exist they coincide:  $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$

Proof:  Due to the premise there are relative $\epsilon$-neighbourhoods ${A}_{a,\epsilon }$
 i ${A}_{a,\epsilon }=A\cap \right]a-\epsilon ,a+\epsilon \left[$
and ${B}_{a,\epsilon }$ such that

${A}_{a,\epsilon }={B}_{a,\epsilon }$  and  [0]

We only prove one direction, for instance "$⇒$", using the sequence criterion [6.8.4].

Let  f be continuously extendable at a. If $\left({a}_{n}\right)$ is a sequence in B with ${a}_{n}\to a$ we find a number ${n}_{0}\in {ℕ}^{\ast }$ such that ${a}_{n}\in {B}_{a,\epsilon }$ for all $n\ge {n}_{0}$. According to [0] these n now satisfy

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

which in fact is the assertion.

Consider:

• As in [6.2.11] we get for an accumulation point a of A by specialisation

f  is continuously extendable at a$\text{ }⇔\text{ }f|{A}_{a,\epsilon }$  is continuously extendable at a for one $\epsilon >0$

in which case their limits coincide: $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f|{A}_{a,\epsilon }\left(x\right)$.

Similar to [6.4.1] and [6.4.9] we can notice the typical interaction between fixed value of  f and its neighbouring values. Again we only note the result for $.

Proposition: Let  $f:A\to ℝ$ be continuously extendable at an accumulation point a of $A\subset ℝ$. Then for each $c\in ℝ$ and for every relative $\epsilon$-neighbourhood  ${A}_{a,\epsilon }$  the following holds:

 If $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)then there is a $\delta >0$ such that $f\left(x\right) for all $x\in {A}_{a,\delta }\\left\{a\right\}$. [6.9.3] If  $f\left(x\right)\le c$ for all $x\in {A}_{a,\epsilon }\phantom{\rule{0.1em}{0ex}}$ then $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)\le c$. [6.9.4]

Proof:  Let g be the continuous extension of  f at a.

1.  From $g\left(a\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right) the assertion follows directly with [6.4.1].

2.  We may employ [6.4.9] as $g\left(x\right)=f\left(x\right)\le c$ for all $x\in {A}_{a,\epsilon }\\left\{a\right\}$.

When studying limits of sequences the limit theorems proved to be a valuable tool. Fortunately they are still at hand for limits of functions.

Proposition (limit theorems):  Let  $f:A\to ℝ$ and $g:B\to ℝ$ be arbirary functions, a an accumulation point of $A\cap B$ (consequently also of A resp. B). If  f and g are continuously extendable at a the same is true for

 $f+g$  and  $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f+g\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)+\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$ [6.9.5] $f-g$  and  $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f-g\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)-\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$ [6.9.6] $f\cdot g$  and  $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\cdot g\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)\cdot \underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)$ [6.9.7] $\frac{f}{g}$  and  $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}\frac{f}{g}\left(x\right)=\frac{\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)}{\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)}$  if $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)\ne 0$ [6.9.8]

Proof:  From the premises we have continuous extensions at a for  f and for g, let's say

$r:A\cup \left\{a\right\}\to ℝ\text{ }\text{and}\text{ }s:B\cup \left\{a\right\}\to ℝ$

1.  Now $r+s:\left(A\cap B\right)\cup \left\{a\right\}\to ℝ$ continuously extends  $f+g$ at a and the limit calculates to

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

2./3.  The proof is just a copy of the above.

4.  The formula results as before again. In this case however we need to check if a is also an accumulation point of  $\left\{x\in A\cap B|g\left(x\right)\ne 0\right\}$:

As a is an accumulation point of $A\cap B$ we find a sequence $\left({a}_{n}\right)$ in $\left(A\cap B\right)\\left\{a\right\}$ converging to a, and as $s\left(a\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)\ne 0$ the continuity of g (see [6.4.2]) provides a relative $\epsilon$-neighbourhood ${B}_{a,\epsilon }$ of a such that

From a certain ${n}_{0}$ onwards all sequence members are within ${B}_{a,\epsilon }$ so that $\left({a}_{n+{n}_{0}}\right)$ is a sequence in $\left\{x\in A\cap B|g\left(x\right)\ne 0\right\}$ converging to a.

And we have a fifth limit theorem as the composition of functions is also compatible with the continuous extendability.

Proposition:  Let $g:A\to ℝ$ and  $f:B\to ℝ$ be any two functions and a an accumulation point of $\left\{x\in A|g\left(x\right)\in B\right\}$. If g is continuously extendable at a and if $\underset{y\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(y\right)$ is an accumulation point of B at which  f is continuously extendable then  $f\circ g$ is continuously extendable at a as well with the limit

 $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\circ g\left(x\right)=\underset{x\to \underset{y\to a}{\mathrm{lim}}\phantom{\rule{0.1em}{0ex}}g\left(y\right)}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)$ [6.9.9]

Proof:  We employ the sequence criterion [6.8.4]: If $\left({a}_{n}\right)$ is a sequence in $\left\{x\in A|g\left(x\right)\in B\right\}\subset A$ such that ${a}_{n}\to a$ we have in turn:

• $g\left({a}_{n}\right)\to \underset{y\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(y\right)$  because g is continuously extendable at a

• $f\circ g\left({a}_{n}\right)=f\left(g\left({a}_{n}\right)\right)\to \underset{x\to \underset{y\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(y\right)}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)$  because  f is continuously extendable at $\underset{y\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(y\right)$

Alongside with tools for calculating limits we need methods for estimating limits.

Proposition (nesting theorem):  Let a be an accumulation point of A and  $f,g,h:A\to ℝ$ be three functions such that

.

If  f and h are continuously extendable at a and if $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}h\left(x\right)$ then g is continuously extendable at a by

 $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}g\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}h\left(x\right)$ [6.9.10]

Proof:  Again we use the sequence criterion [6.8.4] and take a sequence $\left({a}_{n}\right)$ in A converging to a. From the premises we have

so that the assertion immediately results from the original nesting theorem [5.5.8].

The nesting theorem is often used after succeeding in estimating a difference by absolute value. The following proposition guarantees that this practice leads to limits.

Proposition:  Let a be an accumulation point of $A\subset ℝ$ and $b\in ℝ$. Then for any function  $f:A\to ℝ$ the following equivalence holds:

 f is continuously extendable at a with $\underset{x\to a}{\mathrm{lim}}\phantom{\rule{0.3em}{0ex}}f\left(x\right)=b$ [6.9.11] $\text{ }⇔\text{ }$ $|f-b|$ is continuously extendable at a with $\underset{x\to a}{\mathrm{lim}}|f\left(x\right)-b|=0$

Proof:  The first direction "$⇒$" is done immediately with the calculation rules [6.9.6] and [6.9.9]. For the second one "$⇐$" we employ the sequence criterion:

$\begin{array}{ll}A\ni {a}_{n}\to a\hfill & \text{ }⇒\text{ }|f\left({a}_{n}\right)-b|\to 0\hfill \\ \hfill & \text{ }⇒\text{ }f\left({a}_{n}\right)-b\to 0\hfill \\ \hfill & \text{ }⇒\text{ }f\left({a}_{n}\right)\to b\hfill \end{array}$

 6.8. 6.10.