# Comments on From Vector Spaces to Periodic Functions

## Eschborn said:

This is a nice example for why the axiom of choice is controversial. The proof from this blog post only works because of it. However, the result seems counter-intuitive. Let us have a look at whether this is only our intuition being off or whether functions with these properties are just "weird".

When speaking of periodic functions the first that comes to mind is the \( \sin \) function. Can something similar to such a function be used to construct the functions \( g \) and \( h \) from the blog post? If we define them similar to \( \sin \) as periodic and continuous, then the answer is no. Every periodic and continuous function has a maximum value. The sum of two functions with a maximum value has a maximum value. However, the identity is unbounded. The sum can therefore not be the identity.

Okay, so \( \sin \)-like functions are out. What about \( \tan \)-like functions? Take \( \tan(x) \) and \( -\tan(2x) \) for example. Those are not continuous everywhere. Well, we can also rule out many functions like this. Suppose \( g \) and \( h \) are two periodic functions with period lengths \( a \) and \( b. \) If \( \frac{a}{b} \) is rational, then the sum of \( g \) and \( h \) cannot be the identity function. Suppose that \( \frac{a}{b} \) was rational, then \( ab \) is a common period of both functions, i.e., \( g(x) = g(x + ab) \) and \( h(x) = h(x + ab). \) It follows that \( (g + h)(x) = (g + h)(x + ab). \) As \( a \) and \( b \) are non-zero, it follows that \( g + h \) cannot be the identity function. Functions of the form \( a \tan(bx) \) for rational \( b \) thus do not work. Any two periodic functions where you can reason about their sum by just looking at what happens for individual periods are thus ruled out.

My conclusion is that the two periodic functions from the blog post must look "weird".

A common theme with axiom of choice proofs is also that they show the existence of something but give no hint at how this something actually looks like. The proof in this blog post is a good example of this. We know that the functions "exist" but have absolutely no clue about how they look like. It is also an open question whether describing them in any other way than "the function from this proof" is even possible.

## Phillip B said:

Very nice solution! It is quite interesting that this does not work for \( e^x. \) I wonder if there is any general way to tell for any function.