# does the uniform continuity of $f$ implies uniform continuity of $f^2$ on $\mathbb{R}$?

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?

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$f(x)=x^2$ is not uniformly continuous on $(-\infty, +\infty)$. –  Babak S. Jul 20 '12 at 17:58

No. For example, $f(x)=x$ is uniformly continuous, but $f(x)=x^2$ is not, and neither for that matter is $f(x)=x^n$ for any $n>1$.

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hey, you beat me by 12 secs or less. And your answer is more general. I'm frustrated now ;-) –  user20266 Jul 20 '12 at 17:55

No. If $f(x)= x$ then $f^2(x)$ is not unifomly continuous. A statement like that will only hold on compact sets.

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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.  $\square$

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As mentioned, this is not true. However, a related statement is true: If $f$ is uniformly continuous and bounded, then $f^2$ is uniformly continuous (and bounded). (Hint to prove this: $|f(x)^2 - f(y)^2| = |f(x)+f(y)| |f(x)-f(y)|$). Indeed, if $f,g$ are both uniformly continuous and bounded, then so is $fg$. Thus any finite product of uniformly continuous bounded functions is another such; in particular, if $f$ is, then so is $f^n$ for any $n$.

A fancy version of this statement is that the set $C_u(\mathbb{R})$ of all uniformly continuous bounded functions on $\mathbb{R}$ is a $C^*$-algebra.

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