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 whose sum is the identity function:

for all .

Here periodic has the standard definition: a function is *periodic* if there exists a constant such that for .

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

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January 22, 2008 at 11:15 pm |

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…

January 23, 2008 at 4:12 am |

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

uncountablymany different periods.April 26, 2009 at 12:01 am |

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.

April 26, 2009 at 6:15 pm

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.

January 23, 2008 at 1:26 pm |

Not just different periods,

incommensurableperiods. That is, no multiple of one period of the function is a multiple of another period. Weird. Eerie.January 23, 2008 at 1:50 pm |

[...] 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 [...]

January 24, 2008 at 12:30 am |

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.

January 24, 2008 at 1:15 am |

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.

January 28, 2008 at 4:22 am |

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.

August 14, 2010 at 10:26 am |

To define a periodic function is not difficult. Could you give sample how to define a periodic polynomial function!

Thanks.