The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

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