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This is a general question to which I need help finding a concrete example so that I may understand the concept/strategy better, and any help will be greatly appreciated.

If given a function $F$ that is not continuous, how can I show that the given function satisfies the intermediate value property? A hint that was given by the Professor was to find an auxiliary function $f$ such that $f'=F$.

I know that all continuous functions have the intermediate value property (Darboux's property), and from reading around I know that all derivatives have the Darboux property, even the derivatives that are not continuous.

Here is what I could make sense of the Professor's hint:

If I could find a suitable function $f$ which was differentiable, and $f'=F$, the derivative would have the intermediate value theorem (since all derivatives have the intermediate value property), and thus the original discontinuous function $F$ would also have the intermediate value theorem.

Can anyone please tell me if my reasoning is correct and/or please provide me with a discontinuous function that I can practice on (or perhaps direct me to another question with such a function that I may have overlooked)?

Thanks a lot in advance!

user66807
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    "If given a function F that is not continuous, how can I show that the given function satisfies the intermediate value property?" You can't, because that's often false. From later context, perhaps you mean to ask, "How can I find an example of a function $F$ that is discontinuous but satisfies the intermediate value property, and how can I show that it has these properties?" Unless you were given the explicit function $F$, in which case, please share. – Jonas Meyer May 07 '13 at 06:02
  • @JonasMeyer Sorry, I realized where the confusion came about, and, no, I wasn't given a specific function. I think this perhaps goes along with what I was thinking better: If I am given a discontinuous function $F$ and am told to show that it has the IVP (which is a clear indicator that it has the property and my task is to prove it) how can I go about doing this? (Preferably using the Professor's hint. If you have a function in mind like this I'd be glad to hear/see it) – user66807 May 07 '13 at 06:09
  • In a nutshell you want an example of $f$ differentiable with $f'$ not continuous. Can you provide that? Such an example might be mentioned earlier on in your notes... – Did May 07 '13 at 06:54
  • There are already several threads concerning Darboux functions, which are not continuous; so they might be somewhat interesting for you. For example here or here; you can also find other links in answers and comments there. – Martin Sleziak May 07 '13 at 07:50
  • You can probably find a several examples of derivatives, which are not continuous, for example here. – Martin Sleziak May 07 '13 at 07:50
  • @MartinSleziak okay, I looked at the links you provided and think I understood things a little better. So, say I was given a function $F$ and $F=- \cos(1/x)+2x \sin(1/x)$ and was told to show that it had the IVP. This is what I would do(?): Observe that the function $F$ is not continuous at $x=0$. Let $f(x)=x^2 \sin(1/x)$ when $x \neq 0$ and $f(0)=0$, the derivative of this function is $f'(x)=- \cos(1/x)+2x \sin(1/x)$ and because all derivatives have the IVP, the original function, $F(x)$, must also contain the IVP. – user66807 May 07 '13 at 14:21
  • @user66807 I did not want to clutter this page with too many comments, so I tried to reply in chat instead. – Martin Sleziak May 07 '13 at 14:35
  • @Did is the function I mentioned in the comments the kind of example you are talking about? – user66807 May 07 '13 at 15:05
  • @user66807 I mentioned no "kind of example" hence I guess the answer to your query is "no"... – Did May 07 '13 at 15:16
  • @Did you said "In a nutshell you want an example of $f$ differentiable with $f′$ not continuous. Can you provide that?" my $f$ is $f(x)=x^2 \sin(1/x)$ and its derivative is not continuous at $x=0$. – user66807 May 07 '13 at 15:30
  • It seems @MartinSleziak is willing to help you in chat, so, if you will excuse me... (Simply, when you are clear about a solution, do not forget to post it as an answer.) – Did May 07 '13 at 15:35
  • @Did sorry if I was bothering you, I entered the chat but he was not there. If I get a solution I will be sure to post it. – user66807 May 07 '13 at 15:37

2 Answers2

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If I understood the OP correctly, he wants some simple examples of functions, which are not continuous and they have Darboux property. (He wants to practice showing that a function has intermediate value property on some concrete examples.)

I've given a few examples. I have made this post CW, so feel free to add further examples.

Functions which are not continuous, but are derivatives:

$f(x)= \begin{cases} \sin\frac1x, & x\ne 0, \\ 0 & \text{otherwise}. \end{cases} $

$g(x)= \begin{cases} 2x\cos\frac1x+\sin\frac1x, & x\ne 0, \\ 0 & \text{otherwise}. \end{cases} $

$h(x)= \begin{cases} 2x\sin\frac1{x^2}-2\frac1x\cos\frac1{x^2}, & x\ne 0, \\ 0 & \text{otherwise}. \end{cases} $

The functions $f(x)$, $g(x)$ are $h(x)$ are from the book Van Rooij-Schikhof: A Second Course in Real Analysis (in the Introduction.).

Functions which are not continuous, but have Darboux property (intermediate value property):

$f_2(x)= \begin{cases} \sin\frac1x, & x\ne 0, \\ 1 & \text{otherwise}. \end{cases} $

Again from the book Van Rooij-Schikhof: A Second Course in Real Analysis (in the Introduction.).

Martin Sleziak
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  • Defined in such way $f(x)$ is continuous. – user48672 Apr 23 '16 at 12:34
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    @user48672 Neither of the functions mentioned in this post is continuous. – Martin Sleziak Apr 23 '16 at 13:20
  • All of these functions have sin/cos of $\frac1x$. Are there any functions that can be "drawn" by pen, or otherwise simpler? – SOFe Nov 08 '19 at 06:11
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    If you want function to be discontinuous at some point $x$, oscillation at this point has to be non-zero. If we also want IVP, this implies that all values from some interval will be attained arbitrarily close to $x$. (And, consequently, those values will be attained infinitely many times.) So I do not expect functions with these properties that are much "easier to sketch" than this. (If needed, we can discuss this a bit more in the calculus chatroom some time.) – Martin Sleziak Nov 08 '19 at 07:16
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Here's a function that's nowhere continuous but satisfies the Darboux property.

$$\begin{align}f:\mathbb{R}&\to[0,1]\\\\ \overline{a_1a_2\cdots a_r.a_{r+1}a_{r+2}\cdots}_\textbf{2}&\mapsto\limsup_{n\to\infty}\frac{a_1+...+a_n}{n}\end{align}$$

Here $\overline{a_1a_2\cdots a_r.a_{r+1}a_{r+2}\cdots}_\textbf{2}$ is the unique binary non-degenerate representation of an arbitrary real number. (By non-degenerate I mean that the sequence $(a_n)_{n=1}^\infty$ does not end in an infinite array of ones.)

The Darboux property for $f$ is derived from the fact that $f(I)=[0,1]$ for every open interval $I\subseteq\mathbb{R}$; and $f$ is discontinuous everywhere. You may care to check both properties.

Luca T. Castrillón
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