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For what value of $n \in \Bbb N$ the function $f(x)=x^n$ is uniformly continuous in all of the set $\Bbb R$

I've been trying to work on this problem but I don't really know what to do Can you help me understand this problem?

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  • $\begingroup$ "For a function to be uniformly continuous in $\Bbb R$ then its derivative should be bounded in $\Bbb R$." That's false actually. $\endgroup$ – zhw. Apr 18 '18 at 20:57
  • $\begingroup$ Oh yes, I guess I got that wrong. $\endgroup$ – J.Dane Apr 18 '18 at 21:01
  • $\begingroup$ Possible duplicate of math.stackexchange.com/a/181582. $\endgroup$ – Chris Custer Apr 18 '18 at 21:25
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Hint: Suppose $n>1.$ Let $\delta > 0.$ Then by the mean value theorem,

$$(x+\delta)^n-x^n = nc^{n-1}\delta,$$

where $c\in (x,x+\delta).$

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I am not sure about your claim about the boundedness of $f'$ is equivalent to the uniform continuity of $f$. However to prove that $x \mapsto x^n$ is not uniformly continuous for $n > 1$, you can use $$\left(p^2+\frac1p\right)^n - p^{2n}= \sum_{k=1}^n \binom{n}{k} \frac1{p^k} p^{2(n-k)} \ge_{k=1} np^{2n-3} \to_{p\to \infty} \infty $$ For $n = 0$ it is constant so uniform continuous and as @Ian say in the comment it is uniformly continuous for $n= 1$.


In the above reasoning, I am using that $$ f \text{ is uniformly continuous on $\mathbb R$ $\Leftrightarrow$ for every sequences $(x_n),(y_n)$, $\lim_{n\to\infty} x_n - y_n = 0 \Rightarrow \lim_{n\to\infty} f(x_n) - f(y_n) = 0$} $$

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  • $\begingroup$ $x \mapsto x$ is obviously uniformly continuous, so your bound must have some error. $\endgroup$ – Ian Apr 18 '18 at 21:06
  • $\begingroup$ Yes that's correct ! I edit my answer. $\endgroup$ – Youem Apr 18 '18 at 21:09
  • $\begingroup$ My point is that your bound itself is not saying what you want it to say. It's comparing $(x^2+\epsilon)^n$ with $x^{2n}$. You want to compare $(x+\epsilon)^n$ with $x^n$ instead. $\endgroup$ – Ian Apr 18 '18 at 21:10
  • $\begingroup$ It is the same thing. no ? $\endgroup$ – Youem Apr 18 '18 at 21:11
  • $\begingroup$ No, it's not, because you've used a non-uniformly continuous change of variable in switching from $x$ to $x^2$. $\endgroup$ – Ian Apr 18 '18 at 21:12

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