6.8. Continuously Extendable Functions

With a function f continuous at a we can calculate the value

f (a)

knowing alone the values of f in a neighbourhood of a. When doing this there is actually no need for

f (a)

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 {a} \to ℝ

[6.8.1]

continuous at a such that $g | A = f$
i

thus $g (x) = f (x)$ 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 $\Leftrightarrow$ 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 $f | A \ {a}$ 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

\lim_{x \to a} f (x) ≔ g (a)

[6.8.2]

is called the limit of f at a.

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

g_{1} (x) = f (x) = g_{2} (x)

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

g_{1} (a) = \lim g_{1} (a_{n}) = \lim g_{2} (a_{n}) = g_{2} (a)

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

$\lim_{x \to a} f (x) = f (a)$

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. $q = {(X - a)}^{k} \cdot r$ with $r (a) \neq 0$ . Then the following holds:

f is continuously extendable at a

\Leftrightarrow

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 $] a - ε, a + ε [$ of a with no zeros of q other than a:

] a - ε, a + ε [\subset A \cup {a}

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

Let us now turn to the actual proof:

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

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

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

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

g is a polynomial quotient and thus continuous at a
$g (x) = \frac{s (x)}{r (x)} = \frac{{(x - a)}^{k} \cdot s (x)}{{(x - a)}^{k} \cdot r (x)} = \frac{p (x)}{q (x)} = f (x)$ for all $x \in A$

Some examples will practise the zero criterion:

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

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

will reveal the continuous extension $X + 1$ and thus the limit $\lim_{x \to 1} \frac{x^{2} - 1}{x - 1} = X + 1 (1) = 2$ .
$\frac{X + 3}{X - 2}$ is not continuously extendable at 2 because 2 is no zero for the numerator.
$\frac{X^{4} - 6 X^{3} + 10 X^{2} - 6 X + 9}{X^{3} - 2 X^{2} - 15 X + 36} = \frac{{(X - 3)}^{2} (X^{2} + 1)}{{(X - 3)}^{2} (X + 4)}$ 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 $\frac{X^{2} + 1}{X + 4}$ we get the limit

$\lim_{x \to 3} \frac{x^{4} - 6 x^{3} + 10 x^{2} - 6 x + 9}{x^{3} - 2 x^{2} - 15 x + 36} = \frac{X^{2} + 1}{X + 4} (3) = \frac{10}{7}$
$\frac{X^{2} + X}{X^{2}} = \frac{X (X + 1)}{X^{2}} = \frac{X + 1}{X}$ 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.

$\frac{X^{2} + X - 6}{2 X + 6}$ ?

−3 is a single zero for the denominator and for the numerator as well. Thus the quotient is continuously extendabel at −3 with the limit ?

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

The polynomial quotient is not continuously extendable at 0 because this is only a zero for the denominator and not for the numerator.
It is however continuously extendable at the denominator's double zero 2 as 2 is also a double zero for the numerator. The limit calculates to ?

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

1 is a double zero for the denominator but only a single one for the numerator. So there is no limit at 1.
−1 is a single zero for both denominator and numerator. Thus the quotient is continuously extendable at −1 with ?

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 $\lim_{x \to a} f (x) = b$	[6.8.4]
$\Leftrightarrow$	every sequence $(a_{n})$ in A satisfies $a_{n} \to a \Rightarrow f (a_{n}) \to b$	[6.8.4]

Proof:

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

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

g (x) ≔ {\begin{array}{l} f (x) if x \neq a \\ b if x = a \end{array}

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

We now turn to prove the continuity of g at a. Let $(a_{n})$ be any sequence in $A \cup {a}$ 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 $k \in ℕ^{*}$ such that $a_{n} = a$ for all $n \geq k$ . But these n then satisfy $g (a_{n}) = b$ so that the convergence $g (a_{n}) \to b = g (a)$ is valid.
For each $k \in ℕ^{*}$ there is an $n \geq k$ such that $a_{n} \neq a$ . In this case we modify the original sequence $(a_{n})$ : Every sequence member equal to a will be substituted by the next member different from a. The resulting sequence $(a_{n_{k}})_{k > 0}$ where

$n_{k} ≔ \min {n \geq k | a_{n} \neq a}$

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

$| a_{n_{k}} - a | < ε$

From the premise we now conclude $g (a_{n_{k}}) = f (a_{n_{k}}) \to b$ and thus get an $n_{0} \in ℕ^{*}$ for each $ε > 0$ such that

$| g (a_{n_{k}}) - b | < ε for all k \geq n_{0}$

With these k we have $| g (a_{k}) - b | = {\begin{array}{l} | g (a_{n_{k}}) - b | < ε, if k = n_{k} \\ | b - b | = 0 < ε, if k < n_{k} \end{array}$ . Thus the convergence $g (a_{k}) \to b = g (a)$ 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:

$\frac{| X |}{X}$ is not continuously extendable at 0. Take e.g. $0 \neq \frac{{(- 1)}^{n}}{n} \to 0$ and note that

$(\frac{| X |}{X} (\frac{{(- 1)}^{n}}{n})) = ({\begin{cases} 1, if n is even \\ - 1, if n is odd \end{cases})$

is divergent.

[6.8.5]

$\frac{\sin}{X}$ is continuously extentable at 0 and $\lim_{x \to 0} \frac{\sin x}{x} = 1$ .

[6.8.6]

Proof: Take $0 \neq a_{n} \to 0$ . Without restriction we may assume that $| a_{n} | \leq 1$ so that

| a_{n} |^{i} \leq | a_{n} |

holds for all

i, n \in ℕ^{*}

[1]

Consider the sine represented by $\sin = \sum_{i = 0}^{\infty} {(- 1)}^{i} \frac{X^{2 i + 1}}{(2 i + 1)!}$ according to [5.9.19]. For a fixed n and an arbitraey $k \in ℕ^{*}$ we now estimate as follows:

\begin{array}{l} | \sum_{i = 0}^{k} {(- 1)}^{i} \frac{a_{n}^{2 i + 1}}{(2 i + 1)! a_{n}} - 1 | & = | 1 + \sum_{i = 1}^{k} {(- 1)}^{i} \frac{a_{n}^{2 i}}{(2 i + 1)!} - 1 | \\ = | \sum_{i = 1}^{k} {(- 1)}^{i} \frac{a_{n}^{2 i}}{(2 i + 1)!} | \\ \leq \sum_{i = 1}^{k} \frac{| a_{n} |^{2 i}}{(2 i + 1)!} \\ \underset{[1]}{\leq} | a_{n} | \sum_{i = 1}^{k} \frac{1}{(2 i + 1)!} \\ \leq | a_{n} | \sum_{i = 0}^{\infty} \frac{1}{i!} \\ = | a_{n} | \cdot e \end{array}

As the absolute value function is continuous this estimate is valid for the limit ( $k \to \infty$ ) as well:

0 \leq | \frac{\sin a_{n}}{a_{n}} - 1 | = | \sum_{i = 0}^{\infty} {(- 1)}^{i} \frac{a_{n}^{2 i + 1}}{(2 i + 1)! a_{n}} - 1 | \leq | a_{n} | \cdot e

The nesting theorem [5.5.8] now finishes the proof:

| \frac{\sin a_{n}}{a_{n}} - 1 | \to 0 \Leftrightarrow \frac{\sin 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

f \cup g (x) ≔ {\begin{array}{l} f (x) if x \in A \\ g (x) if x \in B \end{array}

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
$\Leftrightarrow$	f and g are continuously extentable at a and $\lim_{x \to a} f (x) = \lim_{x \to a} g (x)$

[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.

" $\Rightarrow$ ": 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

\lim_{x \to a} f (x) = \lim_{x \to a} f \cup g (x) = \lim_{x \to a} g (x)

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

\begin{array}{l} r (x) = f (x) = g (x) = s (x) for all x \in A \cap B, x \neq a \\ r (a) = \lim_{x \to a} f (x) = \lim_{x \to a} g (x) = s (a) \end{array}

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 $ε > 0$ be arbitrary. As r and s are continuous at a we find $δ_{1}, δ_{2} > 0$ such that

\begin{array}{l} x \in A \land | x - a | < δ_{1} \Rightarrow | r (x) - r (a) | < ε \\ x \in B \land | x - a | < δ_{2} \Rightarrow | s (x) - s (a) | < ε \end{array}

With $δ ≔ \min {δ_{1}, δ_{2}}$ we thus have for all $x \in A \cup B$ :

| x - a | < δ \Rightarrow | r \cup s (x) - r \cup s (a) | = {\begin{array}{l} | r (x) - r (a) |, if x \in A \\ | s (x) - s (a) |, if x \in B \end{array}} < ε

6.7. 6.9.