6.9. Properties of Continuously Extendable Functions

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

In general the reverse does not hold.

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

The Heaviside step function H

i

${\displaystyle \mathrm{H}(x)=\{\begin{array}{l}1\text{, if }x\ge 0\hfill \\ 0\text{, if }x<0\hfill \end{array}}$

is not continuously extendable at 0 in contrast to its restriction ${\displaystyle \mathrm{H}|{ℝ}^{>0}}$.

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

If the limits exist they coincide:  ${\displaystyle \underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }g(x)}$

Proof:  Due to the premise there are relative ${\displaystyle \epsilon }$-neighbourhoods ${\displaystyle A_{a,\epsilon }}$

i

${\displaystyle A_{a,\epsilon }=A\cap ]a-\epsilon ,a+\epsilon [}$

and ${\displaystyle B_{a,\epsilon }}$ such that

We only prove one direction, for instance "${\displaystyle \Rightarrow }$", using the sequence criterion [6.8.4].

Let  f be continuously extendable at a. If $(a_n)$ is a sequence in B with ${\displaystyle a_n\to a}$ we find a number ${\displaystyle n_0\in {\mathbb{N}}^{\ast }}$ such that ${\displaystyle a_n\in B_{a,\epsilon }}$ for all ${\displaystyle n\ge n_0}$. According to [0] these n now satisfy

which in fact is the assertion.

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

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

1. ►  From ${\displaystyle g(a)=\underset{x\to a}{\lim }f(x)<c}$ the assertion follows directly with [6.4.1].

2. ►  We may employ [6.4.9] as ${\displaystyle g(x)=f(x)\le c}$ for all ${\displaystyle x\in A_{a,\epsilon }\backslash \{a\}}$.

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

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

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

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  ${\displaystyle \{x\in A\cap B|g(x)\ne 0\}}$:

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

From a certain ${\displaystyle n_0}$ onwards all sequence members are within ${\displaystyle B_{a,\epsilon }}$ so that ${\displaystyle (a_{n+n_0})}$ is a sequence in ${\displaystyle \{x\in A\cap B|g(x)\ne 0\}}$ converging to a.

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

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

• ${\displaystyle g(a_n)\to \underset{y\to a}{\lim }g(y)}$  because g is continuously extendable at a

• ${\displaystyle f\circ g(a_n)=f(g(a_n))\to \underset{x\to \underset{y\to a}{\lim }g(y)}{\lim }f(x)}$  because  f is continuously extendable at ${\displaystyle \underset{y\to a}{\lim }g(y)}$

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

If  f and h are continuously extendable at a and if ${\displaystyle \underset{x\to a}{\lim }f(x)=\underset{x\to a}{\lim }h(x)}$ then g is continuously extendable at a by

Proof:  Again we use the sequence criterion [6.8.4] and take a sequence $(a_n)$ in A converging to a. From the premises we have

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

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

Proof:  The first direction "${\displaystyle \Rightarrow }$" is done immediately with the calculation rules [6.9.6] and [6.9.9]. For the second one "${\displaystyle \Leftarrow }$" we employ the sequence criterion: