Periodic function - how can we prove f is uniformly continuous?

Question

This is rather intuitive, but I'd like help to give a formal proof.

Suppose f:R --> R is periodic and continuous on R. Then, f is uniformly continuous on R

Thank you

Answer 1

I think the following will be sufficient to prove this:

Theorem: Let f be a continuous mapping of a compact metric space X into a metric space Y. Then f is uniformly continuous on X.

You can find a proof of this in Rudin's Principles of mathematical analysis, p.91

Now for the argument!

Let p be the fundamental period of f. Because f is continuous on R, it is continuous on [0, p]. Notice that the range of f is f[0, p], since every y in R is in [mp, (m+1)p] for some integer m; f(y) = f(y-mp), where y-mp is in [0, p].

Observe that the restriction g : [mp, (m+1)p]-->f[0, p] of f is continuous. Because [mp, (m+1)p] is compact, it follows that g is uniformly continuous on [mp, (m+1)p]. Hence f is uniformly continuous on every closed interval of length p. Thus, f is uniformly continuous.

Answer 2

Let p >0 be any period of f. Then, f is continuous on [0, 2p] and, since this set is compact, f is uniformly continuous on [0, 2p]. So, for every eps >0 there exists d >0 , which we can suppose to satisfy also d < p, such that |f(v) - f(u)| < eps for every u and v in [0, 2p] with |u - v | < d.

Let x and y be non negative real numbers satisfying |y - x | < d. Suppose, without loss of generality, that x <= y. There exists an integer m >=0 and 0 <= r < p such that x = mp + r. Then, r is in [0, p) contained in [0, 2p] and f(x) = f(mp + r) = f(r). Let s = r + y -x. Then 0 <= s = r + |y - x| < r + d < r + p < 2p, so that s is in [0, 2p]. In addition, y = s -r + x = s -r + mp + r = mp + s, so that f(y) = f(s) and, therefore, |f(y) - f(x)| = |f(s) - f(r)|.

Since r and s are in [0, 2p] and |s- r| = |y - x| < d, then, |f(y) - f(x|)| < eps whenever |y - x|

Answer 3

Let T be a period of f, and consider f restricted to [0,T]. This is a compact set, so every continuous function is uniformly continuous. Now for points outside this interval, translate them into this interval (using the fact f is periodic) to show the uniform continuity every where.