My roommates and I have an argument you guys can help to settle (peace is at stake, don't let us down!) In undergrad calculus courses, one usually explains what it means for a function to be differentiable at a point x, and then differentiable in a domain. Then the focus is entirely on this latter notion. My question is:

Has the notion of differentiability at a point any interest?

That is, I'm looking for a theorem which is valid for a function regular at some point, but which needs significantly less regularity in a neighborhood of this point, or a good reason for which such a theorem doesn't exist.

Of course, this question is very flexible, and any insight is welcome.


Differentiability at a point is useful when you want to formulate a theorem like the fundamental theorem of calculus in the Lebesgue setting. For instance a continuous nondecreasing $f:[a,b]\to\mathbb{R}$ that satisfies some nice properties (maps sets of measure 0 to sets of measure 0 and is absolutely continuous) will be differentiable almost everywhere and satisfy the familiar $f(b) - f(a) = \int_a^b f'(x)d\mu(x)$ where $\mu$ is the Lebesgue measure.

Here of course $f'(x)$ really means any function which is equal to $f'(x)$ whenever $f'$ exists.

Often functions will be differentiable almost everywhere or, you'll only need to consider functions differentiable almost everywhere, and in this case differentiability at a point is inherent in such a definition.

Of course, sets of measure $0$ are small in some sense but such sets can still be dense so you can still easily have functions whose points of nondifferentiability are lurking in any interval about a point, although to construct such weird functions isn't trivial.


Though somewhat trivial, there is the Local Extremum Theorem: If $f$ is differentiable at $c$ and has a local extremum at $c$, then $f'(c) = 0$.


There are functions that are only differentiable at one point, here is an example:

Consider the function $d:\mathbb{R}\to\mathbb{R}$ defined by $d(x)=0$ if $x$ is rational and $d(x)=1$ if $x$ is irrational. This function is not differentiable anywhere, since it is not continuous, however, it is surely bounded. Now look at $f(x)=x^2\cdot d(x)$.

Note that for a rational $x$ we have $f(x)=0$ and while for irrational $x$ we have $f(x)=x^2$.

Next since there are rational numbers arbitrary close to any irrational $x$ our function is not continuous at irrationals. The same argument holds for rational $x$'s whenever $x\ne0$.

At $x=0$ we have, for $h$ rational and $\ne0$, $$\frac{f(x+h)-f(x)}{h}= \frac{f(0+h)-f(0)}{h}=\frac{f(h)-f(0)}{h}=\frac{0}{h}=0$$ while for irrational $h$ we get $$\frac{f(x+h)-f(x)}{h} = \frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h$$ which tends to $0$ as $h\to0$, in particular $f$ is differentiable at $0$ and $f'(0)=0$.

  • $\begingroup$ This evokes very old memories. It strikes me that even the Dirac derivative may not exist. It is easy to see that the Dirac on each rational number would be -1, but then it needs to be +1 infinitely close to rational numbers but not at rational numbers. Hmmm - no Dirac for this function! Or is it +1 for every non rational number? I would not be able to begin proving or disproving this. $\endgroup$ – asoundmove Mar 1 '11 at 21:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.