Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
$f(x)=x^2$ is not uniformly continuous on $(-\infty, +\infty)$. –  B. S. Jul 20 '12 at 17:58
add comment

4 Answers 4

up vote 16 down vote accepted

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 $n>1$.

share|improve this answer
5  
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
add comment

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

share|improve this answer
add comment

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.  $\square$

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.