# The Unapologetic Mathematician

## 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 $f,g:\mathbb{R}\rightarrow\mathbb{R}$ whose sum is the identity function:

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

He notes right off that, “Obviously the functions $f$ and $g$ 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 $f$ is periodic, that means it factors through a map to the circle, which we call $S^1$. Why? Because “periodic” with period $p$ means we can take the interval $\left[0,p\right)$ 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 $f$ is going to be continuous, the function $\bar{f}:S^1\rightarrow\mathbb{R}$ 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 $\mathbb{R}^2$ as the collection of points distance $1$ 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 $\bar{f}$ must be compact as well! And since the image of $f$ is the same as the image of $\bar{f}$, it must also be a compact subspace of $\mathbb{R}$ — a closed, bounded interval. Neat.

January 23, 2008 Posted by | Analysis, Calculus | 13 Comments