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The continuous extension of $f(x)$ at $x=c$ makes the function continuous at that point.

Can you elaborate some more? I wasn't able to find very much on "continuous extension" throughout the web.
How can you turn a point of discontinuity into a point of continuity? How is the function being "extended" into continuity?
Thank you.

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  • $\begingroup$ Can you tell us in what context you found this comment? In particular, an optimal answer to this question would require knowing what the domain and range of $f$. (Sade's answer assumes $f$ goes from a closed interval to the reals; is that the situation you are interested in?) $\endgroup$ Nov 5, 2013 at 18:36
  • $\begingroup$ @EricStucky yes, closed interval to the reals, and open interval to the reals would be nice examples. $\endgroup$
    – Emi Matro
    Nov 5, 2013 at 18:40

4 Answers 4

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Others have already answered, but perhaps it would be useful to have at least one of the answers target the elementary calculus level.

Recall the 3-part definition of "$f(x)$ is continuous at $x=a$" from elementary calculus:

1. $\;f(a)$ is defined

2. $\;\lim\limits_{x \rightarrow a}f(x)$ exists

3. $\;\lim\limits_{x \rightarrow a}f(x) = f(a)$

Suppose the #2 holds and let's consider the possibilities.

If #1 also holds, then either #3 holds or #3 doesn't hold. If #3 holds, then the function is continuous at $x=a.$ If #3 doesn't hold, then the function is not continuous at $x=a,$ but the failure is simply due to the fact that "we picked the wrong value" in defining $f(a).$ [The "wrong value" could arise from a formula or from a piecewise definition format.] In this second case, if we simply change the function in a very trivial way by defining $f(a)$ to be whatever number $\;\lim\limits_{x \rightarrow a}f(x)$ is, then we can turn the function into a continuous function. Another way of looking at it is, thinking of a function as a certain collection of ordered pairs, if we replace the single ordered pair $(a,\,f(a))$ with the ordered pair $\left(a, \; \lim\limits_{x \rightarrow a}f(x)\,\right),$ then the result will be a continuous function. This is not what you're asking about, but it's closely related. You'll sometimes hear things like "redefine the function at a certain point so that it becomes continuous" to describe this process.

Now suppose #1 doesn't hold. Then, of course, there is no chance for #3 to hold. However, still assuming that #2 holds, then all we have to do to get all three conditions to hold is to define the function $f(x)$ so that, when $x=a,$ the value is $\lim\limits_{x \rightarrow a}f(x).$ This is an example of what is meant by a continuous extension. There are other ways a function can be a continuous extension, but probably the most basic way (and likely about the only way you'll see in elementary calculus) is that you have a function that is not defined at some point (maybe more than one point), but the limit of the function exists at that point(s), so if you simply define (like how you define a piecewise defined function) the function to be whatever its limit is at that point(s), then you'll get a continuous function, and this continuous function is called a continuous extension of the original function.

To find examples and explanations on the internet at the elementary calculus level, try googling the phrase "continuous extension" (or variations of it, such as "extension by continuity") simultaneously with the phrase "ap calculus". The reason for using "ap calculus" instead of just "calculus" is to ensure that advanced stuff is filtered out. The word "calculus" is often used for some really advanced topics that have little relation to what's in an elementary calculus course, but "ap calculus" is pretty specific to elementary calculus content.

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  • $\begingroup$ Very nice answer. $\endgroup$
    – little o
    Apr 30, 2018 at 19:33
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"Extending" a function $f$ means defining it at points where it wasn't defined before. As an example, the function $$f(x):={\sin x\over x}\quad(x\in \dot{\mathbb R})$$ is not defined at $0$. But it makes very much sense to "extend" the domain of $f$ by defining $f(0):=1$. When it matters under the given circumstances, one could even denote the the extended function by $\tilde f:\ {\mathbb R}\to{\mathbb R}$.

When some $f$ a priori is defined on a subset $A\subset{\mathbb R}$, and if $\xi\in{\mathbb R}\setminus A$ is an accumulation point of $A$, then it makes sense to check whether the $$\lim_{x\to \xi,\ x\in A} f(x)$$ exists. Assume this is the case, so that $\lim_{x\to \xi,\ x\in A} f(x)=\eta$. Then defining $$f(\xi):=\eta$$ extends $f$ to $\tilde A:=A\cup\{\xi\}$ and makes $f$ continuous at $\xi$.

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Hint Suppose $f:[a,b]\to \mathbb{R}$ is a continous function, and you want to exten this as a function from $\mathbb{R}\to\mathbb{R}$

what is the most natural way to do it?

say $g(x)$ be the extension of $f$

just define $g(x)=f(x)\forall x\in [a,b]$

and $g(x)=f(a)\forall x<a$ and $g(x)=f(b)\forall x>b$

Do you see that $g:\mathbb{R}\to\mathbb{R}$ is a continuous extension of $f(x):[a,b]\to\mathbb{R}$?

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I guess I'll work on interval case then :)

Suppose $f:(a,b)\to\mathbb R$ is continuous. If we can extend it to a closed interval, then we can use Sade's answer. There are cases when we cannot do so; all of the problem functions have the same "kind of problem" as either $f(x)=\frac1x$ or $f(x)=\sin\frac1x$ for the domain $(0,1)$.

However, we can guarantee it under some assumptions. If $f$ is monotone increasing on an interval [doesn't even need to be the whole interval $(a,b)$] with endpoint $b$, then let $$f(b)=\sup\{f(x):x\in (a,b)\}.$$ One plan for showing this is continuous is by contradiction; suppose there was an $\varepsilon$ such that for every $\delta$ there is some a $x\in(b-\delta,b]$ such that $f(x)\notin (f(b-\delta),f(b))$. If you don't see why this is a problem, draw it. You should be able to see the contradiction and it would just need to be formalized.

We can probably find a different condition, but those two counterexamples rule out lots of good tries. Lipschitz continuous, differentiable, and even smooth are insufficient. Bounded is insufficient; but bounded derivative probably works. Uniform continuity looks good.

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