# What does “the average continuous function is nowhere monotonic” mean?

I plan on asking my professor what he meant by "average continuous function," but as it is possible that this is a concept as vague as the statement, I was hoping to get some interesting answers/interpretations from stack exchange first.

How do you think of the average of some infinite group of things?

Or does this just mean that the real line is so dense/big that it is somehow likely that a function would bounce around everywhere except on some countable number of points?

I'm sorry this is vague, I will be sure to post his response if I get a good one.

I would also appreciate any resources or reading; googling around hasn't been fruitful.

• This paper seems to express a similar statement. It seems that they mean that the set of nowhere monotone functions is thick in the set of continuous functions, which they define in the paper. What this notion of "thickness" has to do with a set being "average" or "generic" I don't know. – Jack M May 13 '16 at 23:45
• Hmm I always thought that such statements meant that, given any reasonable probability distribution over all continuous functions, any random function drawn from this distribution is nowhere monotonic with probability $1$. The problem is that I don't know how exactly to define such a probability distribution, and none of the current answers do so. Is this possible? I'm sure that my vague notion can be made precise... – user21820 May 14 '16 at 9:46
• I would have thought "average" here meant something like "typical". in the kind of sense that a typical natural number is composite or a typical real number is transcendental – Henry May 14 '16 at 12:01
• @user21820 Kind of like saying the average real number is irrational? – steven gregory May 14 '16 at 12:08
• @StevenGregory: Exactly. Those are easy because the set of rationals or algebraic reals are countable and hence is a zero-measure subset of $\def\rr{\mathbb{R}}$$\rr and so any continuous random variable from \rr is algebraic with probability 0. So how to define a measure on the space of continuous functions? – user21820 May 14 '16 at 12:12 ## 3 Answers He probably meant that either: 1. The space of continuous functions which are nowhere monotonic has probability one using Wiener measure. I.e. a continuous function is "almost surely" nowhere monotonic using the standard probability measure for that space. 2. That set of nowhere monotonic continuous functions is of the first category (Baire category theorem) with respect to the sup topology/ topology of uniform convergence. Also see this question for a discussion of the definition of "nowhere monotonic". Nowhere monotonic continuous function Here is one possible interpretation of what he meant: every continuous, real-valued function on an interval can be approximated as well as you like by continuous, nowhere monotonic functions. In other words, these functions are dense. In more detail, consider the space C([0,1]) (for example) of continuous functions [0,1]\to\mathbb R. To talk about the nowhere monotonic functions being dense, the space C([0,1]) needs to have some sort of topology, or better yet, a metric. Given f,g\in C([0,1]), say that the distance d(f,g) between f and g is$$ d(f,g) = \sup_{t\in[0,1]} \big|f(t) - g(t)\big|.$$(This is a metric!) Theorem: Under this metric, the space of continuous, nowhere monotonic functions is a dense subspace of$C([0,1])$. In other words, I can approximate continuous functions on the interval by nowhere monotonic functions: if you give me some random continuous function$f:[0,1]\to\mathbb R$, and some tolerance$\varepsilon>0$, I can find a nowhere monotonic function$g:[0,1]\to\mathbb R$such that$|f(t)-g(t)|<\varepsilon$for all$t\in[0,1]\$.

This theorem has an easy proof using the Baire Category Theorem. Your professor can probably give it, or one of us at MSE can if you would like.

Denseness is only one notion of a behavior being "average", and it's not always the best one. For instance, the rationals are a dense subset of the reals even though the "average" real number is irrational. Several other answers give alternative notions of average.

• Your last paragraph shows that “dense” does not at all imply “typical”. – Carsten S May 14 '16 at 9:07

One way to understand this is that nowhere monotonic functions form a "prevalent" set in the space of all continuous functions on, say, an interval, in the sense of Hunt, Sauer and Yorke.