Periodic function - how can we prove f has a fundamental period?

Question

Let f: R--> R be continuous, periodic and non constant on R. Show that f has a fundamental period p >0. If f is constant, then it's trivial that f has no fundamental period. But I'm having trouble to prove the converse.

Thank you

Answer 1

First we show there is a lower bound for the fundamental period.

We assume this is not the case and we can find fundamental periods as small as we want.

The function is continuous, so in particular it is continuous at x = 0. And we can assume f(0) = 0.

In particular, given d (think delta) there exists an e (think epsilon) such that on [-d, d], then |f(x)| < epsilon.

Now by assumption, the fundamental period is less than d. This implies |f| < epsilon on the entire real line. Since epsilon is arbitrary, it shows f is 0 and therefore constant. (if we did not assume f(0) = 0, we would have shown f is always equal to f(0).)

So there is a lower bound to the fundamental period. Let it be p.

Now we show f(p+x) = f(x) so it is periodic with period p and p is the fundamental period.

For a given x, p is the lower bound of the set of qs such that f is periodic with period q. So there exists a sequence q_n such that
f(x+q_n) = f(x)
and q_n goes to p

Thus f(x+p) = lim f(x+q_n) because f is continuous
and
f(x+q_n) = f(x) because f is periodic with period q_n

f(x+p) = lim f(x+q_n) = f(x)
or
f(x+p) = f(x)

Since x is arbitrary, f is periodic with period p. And p is the smallest such number such that this is true.

QED

Answer 2

By the definition, p is the fundamental period of a periodic function if p >0 and p = infimum {q | q is a period of f}. We know that, if f is continuous, then p is a period of f.

If f is constant, then it's really trivial that infimum {q | q is a period of f} =0 and, therefore, f has no fundamental period.

To prove the assertion, let's use contraposition. So, let's suppose f is periodic and continuous and has no fundamental period, that is infimum {q | q is a period of f} = 0. We'll show f must be constant.

Let x be any real number. Since f is continuous, for every eps > 0 we can find d >0, also satisfying d < x, such that |f(u) - f(x)| < eps for every u in (x -d, x +d). Since f doesn't have a fundamental period, it follows from the definition of fundamental period that there exists a period q < infimum {x - d, d}. So, there exists a positive integer m such that mq is in (x -d, x +d), which implies |f(mq) - f(x)| < eps.
Since m is integer and q is a period of f, f(m q) = f(0). Therefore, |f(0) - f(x) < eps. Since this is true for every eps >0, we must have f(x) = f(0), for every real x. So, f is constant on [0, oo). By a entirely similar reasoning, we concluse f is constant on (-oo, 0), so that f is constant on R..

By contraposition, we conclude that, if f is continuous, periodic and non constant on R, then f has a fundamental period p >0.