Let $f: \mathbb{R} \to \mathbb{R}$ be a function. Usually when proving a theorem where $f$ is assumed to be continuous, differentiable, $C^1$ or smooth, it is enough to draw intuition by assuming that $f$ is piecewise smooth (something that one could perhaps draw on a paper without lifting your pencil). What I'm saying is that in all these cases my mental picture is about the same. This works most of the time, but sometimes it of course doesn't.

Hence I would like to ask for examples of continuous, differentiable and $C^1$ functions, which would highlight the differences between the different classes. I'm especially interested in how nasty differentiable functions can be compared to continuously differentiable ones. Also if it is the case that the one dimensional case happens to be uninteresting, feel free to expand your answer to functions $\mathbb{R}^n \to \mathbb{R}^m$. The optimal answer would also list some general minimal 'sanity-checks' for different classes of functions, which a proof of a theorem concerning a particular class would have to take into account.

  • $\begingroup$ what about Dirac delta function? $\endgroup$
    – Dilawar
    Mar 4, 2011 at 10:48
  • 1
    $\begingroup$ @Dilawar: $\delta$ is neither continuous, differentiable, $C^1$, nor smooth. In fact, it isn't even a function. $\endgroup$ Mar 4, 2011 at 10:54
  • $\begingroup$ @Willie: It's not a Real function, it is still a function from the real numbers into the extended real line. $\endgroup$
    – Asaf Karagila
    Mar 4, 2011 at 14:51
  • 2
    $\begingroup$ @Asaf: No, the Dirac delta can't be properly understood in that way. If you think of $\delta$ as a function $\delta : \mathbb{R} \to [-\infty, +\infty]$ with $\delta(0) = +\infty$, $\delta(x) = 0$ otherwise, how can you account for the fact that $\delta \ne 2\delta$? One really does need distribution theory to obtain all the desired properties of this object. $\endgroup$ Mar 4, 2011 at 15:15
  • $\begingroup$ It really depends on which theorem you want; for example, the absolute value is sufficient to show the necessity of differentiability in the Mean Value Theorem (or Rolle's Theorem), and it's not particularly nasty... $\endgroup$ Mar 4, 2011 at 16:30

3 Answers 3


The Wikipedia article http://en.wikipedia.org/wiki/Pompeiu_derivative gives one example of how bad a non-continuous derivative can be.

One can show that any set whose complement is a dense intersection of countably many open sets is the point of discontinuities for some derivative. In particular a derivative can be discontinuous almost everywhere and on a dense set.

See the book "Differentiation of Real Functions" by Andrew Bruckner for this and much more.

  • $\begingroup$ Thank you! I think this answer is sort of what I was looking for and since the question was somewhat vague (I was just interested in seeing what can be said about the subject), I have decided to accept this one as an answer. $\endgroup$
    – J. J.
    Mar 7, 2011 at 17:45

Although this is maybe not a good example of very "nasty" functions, you could look at $f_i = x^i \sin(1/x)$ for $i=0,1,2,3$ in order to see the distinction between those classes of functions. If you set $f_i(0)=0$ for all $i$, then $f_0$ is not continuous in $0$, $f_1$ is continuous but not differentiable in $0$, $f_2$ is differentiable but not $C^1$ and $f_3$ is $C^1$.

  • $\begingroup$ This is indeed a valid example, but not particularly nasty. :) (I agree that the term 'nasty' is not well-defined.) I'm wondering for example if those points where the function's derivative fails to be continuous can have an accumulation point. EDIT: Actually that's probably true. Can those points be dense? $\endgroup$
    – J. J.
    Mar 4, 2011 at 9:55
  • $\begingroup$ derivatives have the intermediate value property (even if they are discontinuous). $\endgroup$
    – yoyo
    Mar 4, 2011 at 16:38

Another example of what can go wrong is Volterra's function.

  • It is differentiable everywhere.
  • Its derivative is bounded everywhere.
  • Its derivative is not Riemann-integrable.

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