## Periodic Functions Problem

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Here’s a neat little problem that I learned about during a party in my first year of graduate school. I don’t know where it’s from originally, but I got it from Joe Miller:

Show that there exist two periodic functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ whose sum is the identity function:

$f(x)+g(x)=x$ for all $x\in\mathbb{R}$.

Here periodic has the standard definition: a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is periodic if there exists a constant $c>0$ such that $f(x+c)=f(x)$ for $x\in\mathbb{R}$.

Obviously the functions $f$ and $g$ can’t be continuous, since any continuous periodic function is bounded. Indeed, it is possible to show that $f$ and $g$ can’t both be measurable.

In case you’re interested, this paper gives a complete criterion for determining whether a function can be written as a sum of periodic functions with specified periods, and this paper investigates the question of whether such functions can be measurable.

### 10 Responses to “Periodic Functions Problem”

1. Omar Says:

Am I missing something or is this pretty easy?

Just take a basis for the reals over the rationals, and define f and g on the basis so that their sum is the identity for ever basis element. Be sure to set f equal to zero on at least one basis element a, and g equal to zero on some other basis element b. Extend f and g by linearity (over the rationals). Now, f+g is the identity, f has period a and g has period b.

Please don’t ask me for a solution that doesn’t use the axiom of choice…

2. Jim Belk Says:

Yeah that’s about it. I’m not sure, but I doubt that there’s a solution that doesn’t involve the axiom of choice. It’s not that hard of a problem, but I’ve always found the statement really surprising . . . two periodic functions summing to give the identity.

Also, the solution gives a neat way of thinking about periodic functions. Note that you can choose for f and g to each be zero for uncountably many different basis elements, in which case both functions have uncountably many different periods.

• Peter Says:

Jim: as you suspect, any solution must involve the axiom of choice (or some similar principle beyond just ZF set theory). As pointed out in the post, any solution here involves a non-measurable function, while Solovay (iirc) showed that in ZF, it is consistent that all functions from R to R are measurable.

• Peter Luthy Says:

Yeah, Solovay is probably the right guy. I know he’s responsible for showing that one needs more than just ZF to construct a non-measurable set.

3. John Armstrong Says:

Not just different periods, incommensurable periods. That is, no multiple of one period of the function is a multiple of another period. Weird. Eerie.

4. A note on the Periodic Functions Problem « The Unapologetic Mathematician Says:

[...] note on the Periodic Functions Problem Over at The Everything Seminar, Jim Belk mentions an interesting little problem. Show that there exist two periodic functions [...]

Some extensions: (1) show that every polynomial function is a finite sum of periodic functions, and (2) show that the exponential function is not a finite sum of periodic functions.

6. Jim Belk Says:

Interesting . . . for the polynomial, you can start by expressing x as the sum of n different periodic functions:

x = f_1 + f_2 + … + f_n

You have to arrange it so that any n-1 of these functions have a common period. (This is easily accomplished: for each choice of n-1 functions, choose a basis element for which they are all zero.)

Since any n-1 of these functions have a common period, any monomial in f_1,…,f_n of degree strictly less than n will be periodic. It follows that any polynomial in x of degree less than n is the sum of periodic functions. In fact, if you collect terms with the same period, you can arrange for any polynomial of degree n-1 to be the sum of at most n periodic functions.

7. Ewan Delanoy Says:

A related question : what if we demand that f and g be continuous but relax the range and only ask for f and g to be defined on a finite interval [0,M] ? (the peridocity condition is then redefined in the obvious way, f(x+T)=f(x) whenever x and x+T are both in the range). To fix ideas, say we want f to be 1-periodic and g to be T-periodic, where 0<T<1. When T is rational there will be a largest M such that a solution exists on [0,M]. When T is irrational I think that there should be solutions for any M.