Problem:
Suppose $f$ is continuous and has periodic length $L = 2$, thus $f$ is also uniformly continuous. Also, assume $f(0) > f(1)$. Let $g(x) = f(x^2)$. I have to prove that $g$ is not uniformly continuous.
I just can‘t come up with an approach, any help is appreciated, thank you.
Let $0<t<1$ and consider $g(\sqrt {2n})-g(\sqrt {2n+t})=f(2n)-f(2n+t)=f(0)-f(t)$. Letting $n \to \infty$ and noting that $\sqrt {2n} -\sqrt {2n+t} =\frac {-t} {\sqrt {2n} +\sqrt {2n+t}} \to 0$ as $n \to \infty$ we see that if $g$ is uniformly continuous then $f(t)=f(0)$ for all $t \in (0,1)$ This contradicts the fact that $f(0) >f(1)$ since $f$ is continuous.