A well-known consequence of the Baire Category Theorem that the set of nowhere-differentiable continuous functions is dense in $C([0,1])$. This is often cited as 'almost all continuous functions are nowhere differentiable' (see here), but to me this seems like a strange way of stating the fact, akin to saying that 'almost all real numbers are rational' just because the rationals are dense in the reals. By the Weierstrass Approximation Theorem, the set of polynomials is also dense in $C([0,1])$, so is it correct to say that almost all continuous functions are polynomials? This statement seems to contradict the original statement about nowhere-differentiable functions. I was wondering if there was some way to remedy my confusion using a measure on $C([0,1])$, since the Lebesgue measure on $[0,1]$ clarifies what it means to say 'almost all' in the context of real numbers; that a property holds for all real numbers outside a set of measure $0$. In this context, certainly not 'almost all' real numbers are rational, almost all real numbers would be irrational. If this is the right way of thinking about 'sizes' of subsets of $C([0,1])$, what would be the 'right' measure to use? If not a measure, is there another way to formalize the idea of 'almost all' in the context of $C([0,1])$?
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@Reveillark (and others who might be interested): I had not seen this question until now, when I saw it cited in a mathoverflow comment. For a lot about the type of nondifferentiability behavior that "most" continuous functions have, see my answer to Generic Elements of a Set. – Dave L. Renfro Jul 28 '21 at 16:06
2 Answers
This comes down to different variants of the phrase "almost all".
Some of the more common uses of "almost all" are
- On a co-countable set.
- On a conull set (with respect to some measure), i.e. the measure theoretic version.
- On a dense set, i.e. the topological version.
- On a comeager set (with respect to some topology), i.e. the finer topological version (for nice enough spaces).
Some of these play well with others, some are stronger than others, and which one is more useful depends on the problem at hand.
For example, being null and being meager don't play nice with each other, as you can write $\mathbb{R}=N\cup M$, with $N$ null, $M$ meager and $N\cap M=\emptyset$. So, being small in one sense tells you absolutely nothing about being small in another set.
On a similar note to what you state, you can argue that almost every number is irrational because the complement is countable (while $\mathbb{R}$ is not). The same works to show that almost every number is transcendental. Incidentally, this is the easiest proof of the existence of transcendental numbers, and is again an example of "to construct an object with the property X, we shows that the set of objects without property X is 'small', and that the entire space is not 'small'". The same thing is done when trying to show that $L^1$ functions have Lebesgue points, you actually show that almost every point (with respect to Lebesgue measure) is a Lebesgue point.
In the case of non-differentiable functions, they form a comeager set, which is stronger than mere density (for nice enough spaces). For example, the intersection of two comeager sets is comeager, which is patently false for dense sets.
To reiterate, the notion of "size" which you want to use with will depend on what you're trying to do. If you're working in set theory, for example, club sets fill the purpose of largeness.
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Thanks for this great writeup! This clarifies a lot of things for me. So since polynomials are elements of the complement of the set of nowhere-differentiable continuous functions, the set of polynomials would be a subset of a meager set, and therefore are meager in $C([0,1])$? – Ian Dulchinos May 13 '20 at 02:00
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Yes, they form a meager set, thus they are small, even though they're dense. Polynomials with rational coefficients are also dense, so you again get a countable dense set. – Reveillark May 13 '20 at 02:09
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Unfortunately there is no "natural" $\sigma$-finite measure on $C[0,1]$, so there is no good "almost every" statement that you can make in that sense. However, there are some things you can say that go a bit beyond the Baire Category argument. There is a notion of "prevalent" and "shy" sets due to Hunt, Sauer and Yorke. A Borel subset $S$ of a topological vector space $X$ is prevalent if there exists a finite-dimensional subspace $P$ such that for every $v \in X$, almost every member of $v + P$ (for $\text{dim}(P)$-dimensional Lebesgue measure on $v+P$) is in $S$.
Then (as mentioned in the Wikipedia article) there is a prevalent subset of $C[0,1]$ whose members are all nowhere differentiable.
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I want to say something about Wiener measure being a candidate for a “natural” measure on continuous paths, in a sense of being a limit of uniform measures on nearest neighbor paths, or of projections of uniform measures of high dimensional spheres and cubes. Moreover it behaves nicely under translation even though it’s not invariant, and admits a natural calculus similar to that of lebesgue measure on Euclidean space, etc. But my argument would probably not be too convincing. – shalop May 13 '20 at 02:13