Sunday Function

In our tour of the zoo of functions we’ve been spending time in the snake pit. These are the pathological functions of pure math, and are generally but not always useless in physics and pretty much everything else. But they’re very cool to look at! We’ll eventually get back to the useful domesticated farm animals of the functions but we’re in no hurry. Here’s the next snake and its picture:

It looks like a double-valued function, a straight line under a parabola. but it’s not. If x is rational, f(x) is just the square of x. Otherwise, it’s 0. So f(π) = 0, but f(3) = 9.

It’s loosely of the same species as the Dirichlet function. It’s chopped up in an infinite number of places, and it’s discontinuous everywhere.

Almost everywhere. In fact, this function is continuous at exactly one point – the origin. Here’s why: a function is continuous at x = C if for any tiny positive number ε, there is a positive number δ such that every x within δ of C, |f(C) – f(x)| < ε. This function fits that definition. But only at that one point x = 0. Everywhere else, the fact that the function is infinitely chopped up prevents a the definition of continuity from being satisfied. Continuous at exactly one point. Exceptionally weird. There are weirder. If this function is a snake, Thomae’s function is a basilisk. That one’s a much more difficult story for another day.

[Bonus exercise for the mathematically fluent: is f differentiable at the origin? If we replaced x2 with x in the definition of the function, what would its status be with regard to continuity and differentiability at the origin?]

1. #1 Alan D. McIntire
September 21, 2008

I found this post very entertaining. I could never have concieved of a function being continuous at exactly one point!

2. #2 Steven N. Severinghaus
September 21, 2008

Intuitively, it seems that f should be differentiable at the origin, since the tangents of both “pieces” of the function match at the origin. By the same argument, if you replaced the x^2 piece with x, I would expect it to no longer be differentiable (though still continuous).

I imagine you could construct a function like f that when graphed as above appeared arbitrarily multivalued. I’m not quite sure how you would construct the disjoint domains for the different pieces of the function. But for example, for a (seemingly) three-valued function, you could use the rationals, the irrationals minus the trancendentals, and the trancendentals as your three domains, and assign three different functions to those pieces.

3. #3 Odysseus
September 21, 2008

Steven appears to be right. For any x, the quotient (f(x)-f(0))/(x-0)<=x, so it converges as x approaches zero. If you replace x**2 with x, the function is still continuous at x=0, but you find (f(x)-f(0))/(x-0) to be zero for x in R\Q and one for x in Q. Pretty much in accordance with Steven's intuition. I remember creating another seemingly multivalued function which filled the whole screen of my TI, namely y=rand(). Not very interesting to analyze, though.

4. #4 Odysseus
September 21, 2008

For any x, the quotient (f(x)-f(0))/(x-0)

Should have used the preview; some goblin stole the right half of my inequality. It was meant to be (f(x)-f(0))/(x-0) \leq x, of course.