A function f: R (reals) ? R (reals) is periodic on R (reals) with period p if th

Question

A function f: R (reals) ? R (reals) is periodic on R (reals) with period p if there exists a p > 0 such that f(x + p) = f(x) for all x is an element of R (reals) . Prove that a continuous periodic function on R (reals) is bounded and uniformly continuous on R (reals) .

Explanation / Answer

Here is a sketch: Assume f:R->R is continuous and periodic with fundamental period p. Boundedness: Let x0 in R be arbitrary. The interval [x0, x0+p] is closed. As f is continuous on [x0,x0+p], the Extreme Value Theorem ensures that f attains a maximum value M and minimum value m on [x0,x0+p]. Hence |f(x)| 0 (some delta that can depend on x), such that for all y such that |x-y|

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