## A 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 whose sum is the identity function:

for all

He notes right off that, “Obviously the functions and can’t be continuous, since any continuous periodic function is bounded.” I’d like to explain why, in case you didn’t follow that.

If a function is periodic, that means it factors through a map to the circle, which we call . Why? Because “periodic” with period means we can take the interval and glue one end to the other to make a circle. As we walk along the real line we walk around the circle. When we come to the end of a period in the line, that’s like getting back to where we started on the circle. Really what we’re doing is specifying a function on the circle and then using that function over and over again to give us a function on the real line. And if is going to be continuous, the function had better be as well.

Now, I assert that the circle is compact. I could do a messy proof inside the circle itself (and I probably should in the long run) but for now we can just see the circle lying in the plane as the collection of points distance from the origin. Then this subspace of the plane is clearly bounded, and it’s not hard to show that it’s closed. The Heine-Borel theorem tells us that it’s compact!

And now since the circle is compact we know that its image under the continuous map must be compact as well! And since the image of is the same as the image of , it must also be a compact subspace of — a closed, *bounded* interval. Neat.