How to show g is not periodic?

Question

Suppose f:R --> R is continuous and has a fundamentl period.. Show that g(x) .= f(x^2) is not periodic on R.

Answer 1

I think I answered a similar question before. Not sure if it was in English.

Dr D is great at Math, but this time I don't quite agree with his proof. In my opinion, he showed p is not a period of f, but this doesn't show g is not periodic at all.

I'll prove a stronger result, from which the non-periodicity of f is a consequence.

We know if g is continuous and periodic on R, then g is uniformly continuous on R. So, if we prove g is continuous but not uniformly continuous, then, by contraposition, we''l have automatically proved g is not periodic.

Continuity of g follows immediately from its being the composition of 2 continuous functions, f and x --> x^2. To prove it's not uniformly continuous, we'll consider that g is uniformly continuous on R if, and only if, for every sequences u_n and v_n in R such that (u_n - v_n) -> 0, we have (f(u_n) - f(v(n)) --> 0.

Since f has a fundamental period p and is continuous, it's not constant. Therefore, there exists u and v in R with f(u) <> f(v). Put u_n = sqrt(np + u) and v_n = sqrt(np + v). Since np --> oo, it follows that lim u_n - v_n = 0. On the other hand, from the definition of g it follows that, for every n, g(u_n) - g(v_n) = f(np + u) - f(np + v) = f(u) - f(v) <> 0 , because p is period of f and f(u) <> f(v). Hence, g(u_n) - g(v_n) approaches something different from 0, and the condition for uniform continuity isn't satisfied.

So, we've proved g is continuous but not uniformly continuous on R, which, in turn, implies g is not periodic.

Answer 2

Let's say f has a period p.
That means f(a) = f(a+p) = f(a+2p) etc
f(x^2) also repeats itself but does not have a constant period.
e.g. when x = √a,
f(x^2) = f(x^2 + p) = f(x^2 + 2p)
or g(x) = g(√(x^2 + p)) = g(√(x^2 + 2p))

It is easy to verify that
√(x^2 + p) - x ≠ √(x^2 + 2p) - √(x^2 + p)