8
$\begingroup$

Let $A$ be a nonempty open subset of $\mathbb{R}$.

Consider a function $f : A \rightarrow \mathbb{R}$. Given that $f$ is continuous, what is the probability that it is differentiable? I suspect it is $0$.

$\endgroup$
  • 5
    $\begingroup$ You'd have to define how to compute probabilities here... $\endgroup$ – vonbrand Apr 16 '14 at 17:56
  • 2
    $\begingroup$ Are you asking about measure theory? $\endgroup$ – user5402 Apr 16 '14 at 17:59
  • $\begingroup$ If it can help : the subset of nowhere differentiable functions contains an intersection of open dense subsets. Or if your function is a Brownian motion, then the probability of being differentiable is 0. $\endgroup$ – user10676 Apr 16 '14 at 18:00
  • $\begingroup$ @vonbrand What if I instead asked for your credence in $f$'s differentiability? $\endgroup$ – Zubin Mukerjee Apr 16 '14 at 18:00
  • 1
    $\begingroup$ That question is in a finite dimensional setting, and the question plays nicely with countable operations. Probability on infinite dimensional spaces is extremely technical. For what it's worth, I would be shocked if you could find a reasonable interpretation in which the answer was not zero. $\endgroup$ – Chris Janjigian Apr 16 '14 at 18:35
9
$\begingroup$

There is a very precise sense in which the answer to your question is "$0$".

Let us denote by $ND$ the set of all continuous nowhere differentiable functions on, say, the interval $[0,1]$, and by $\mathcal C([0,1]$ the Banach space of all continuous functions on $[0,1]$. Then, it can be shown that $SD:=\mathcal C([0,1])\setminus ND$ (the set of all functions which are somewhere differentiable) is a "Haar-null" set, which means that there exists a Borel probability measure $\mu$ on $\mathcal C([0,1])$ such that $$\forall f\in\mathcal C([0,1])\;:\; \mu(f+SD)=0\, .$$

Here, "$\mu(E)=0$" for a possibly non-Borel set $E$ means that $E$ is contained in a Borel set $\widetilde E$ such that $\mu(\widetilde E)=0$. (This remark is needed because $SD$ is not Borel in $\mathcal C([0,1])$).

Thus, in this sense, a continuous function is nowhere differentiable "with probability $1$".

For a proof of this result, this the original paper by Hunt: http://www.ams.org/journals/proc/1994-122-03/S0002-9939-1994-1260170-X/S0002-9939-1994-1260170-X.pdf

| cite | improve this answer | |
$\endgroup$

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.