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!

  • 1
    $\begingroup$ "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. $\endgroup$ – Jonas Meyer May 7 '13 at 6:02
  • $\begingroup$ @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) $\endgroup$ – user66807 May 7 '13 at 6:09
  • $\begingroup$ 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... $\endgroup$ – Did May 7 '13 at 6:54
  • $\begingroup$ 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. $\endgroup$ – Martin Sleziak May 7 '13 at 7:50
  • $\begingroup$ You can probably find a several examples of derivatives, which are not continuous, for example here. $\endgroup$ – Martin Sleziak May 7 '13 at 7:50

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.).

  • $\begingroup$ Defined in such way $f(x)$ is continuous. $\endgroup$ – user48672 Apr 23 '16 at 12:34
  • $\begingroup$ @user48672 Neither of the functions mentioned in this post is continuous. $\endgroup$ – Martin Sleziak Apr 23 '16 at 13:20

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