TITLE: does the uniform continuity of $f$ implies uniform continuity of $f^2$ on $\mathbb{R}$?
QUESTION [10 upvotes]: my question is if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, does it implies that $f^2$ is so?and in general even or odd power of that function?

REPLY [9 votes]: By $f^2(x)$, do you mean $(f(x))^2$, or $f(f(x))$?  (The notation $f^2(x)$ can mean both of these; which is standard depends on the area of mathematics you work in.)
If you mean $(f(x))^2$, then as other answers have pointed out, this may not be uniformly continuous.
However, if you mean $f(f(x))$, then yes, this is uniformly continuous whenever f is.
Proof.  Given $\epsilon > 0$, we need some $\delta > 0$ such that whenever $|x - y| < \delta$, $|f(f(x)) - f(f(y))| < \epsilon$.
We know, by uniform continuity of $f$ applied to $\epsilon$, that there’s some $\epsilon_1$ such that whenever $|x - y| < \epsilon_1$, $|f(x) - f(y)| < \epsilon$.  By the uniform continuity of $f$ again, applied to $\epsilon_1$, there’s $\delta$ such that whenever $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon_1$.
So putting these together, if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon_1$, and so $|f(f(x)) - f(f(y))| < \epsilon$; so $\delta$ is as required.  $\square$