# Are "most" continuous functions also differentiable?

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 $0$.

• You'd have to define how to compute probabilities here... Apr 16, 2014 at 17:56
• Are you asking about measure theory? Apr 16, 2014 at 17:59
• 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. Apr 16, 2014 at 18:00
• @vonbrand What if I instead asked for your credence in $f$'s differentiability? Apr 16, 2014 at 18:00
• 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. Apr 16, 2014 at 18:35

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 $1$".