How to prove $x^{n}$ is not uniformly continuous on the interval $[0, +\infty)$?
How to prove $x^{n}$ is not uniformly continuous
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1Presumably you mean for $n>1$. – 2012-08-12
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1Have you tried something? Welcome to MSE :) What about $n$? Is it an integer? Is it $\ge 0$, $\ge 1$, we don't know. – 2012-08-12
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0Maybe you can look at a related question: [Showing a function is not uniformly continuous](http://math.stackexchange.com/questions/135234/showing-a-function-is-not-uniformly-continuous) – 2012-08-12
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0Look at $(x+\delta)^n-x^n$. – 2012-08-12
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1n is integer and n > 1. I know how to prove when n = 2. proof it by contradiction when n = 2. f(x) = x^{2} `x_{1}=\sqrt{n},x_{2}=\sqrt{n+2},for \varepsilon = 1 I get |x_{1}-x_{2}|=\frac{1}{{\sqrt{n}}+{\sqrt{n+2}}} \to 0 but |f(x_{1})-f(x_{2})| = 2 > 1` but how would I chose the sequence when n > 2? – 2012-08-12
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0@MartinSleziak: Is it enough to prove, that $f_n(x)=x^n$ not uniformly continuous on $E=[0,1] \subseteq [0,\infty)$ and then conclude that, $f_n(x)=x^n$ is not uniformly continuous on $[0,\infty)$. Is it true that, if $f_n(x)$ is not uniformly continuous on $E\subseteq D $, then $f_n(x)$ is not uniformly continuous on $D$ ? Thank you! – 2012-08-12
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0@SalechAlhasov It would be sufficient to show this for a subset. But $[0,1]$ is not a good choice for this particular case. A continuous function on a compact set (compact space) is uniformly continuous, see e.g. [ProofWiki](http://www.proofwiki.org/wiki/Heine-Cantor_Theorem) or [Wikipedia](http://en.wikipedia.org/wiki/Heine%E2%80%93Cantor_theorem). – 2012-08-12
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0@MartinSleziak: Yes, forgot that *little detail*. Thanks. – 2012-08-12
2 Answers
You want to show that there exists some $\epsilon>0$ such that there is no $\delta>0$ for which $$|x-y|<\delta\implies |x^n-y^n|<\epsilon.$$ Try $\epsilon=1$. Thus you want to show that for any $\delta>0$, there is some pair $x,y\in [0,\infty)$ such that $|x-y|<\delta$ yet $|x^n-y^n|\geq 1$. Let's try letting $x=y+\delta/2$. Then we have $|x-y|=\delta/2<\delta$ and $$|x^n-y^n|=(y+\delta/2)^n-y^n\geq (y+\delta/2)y^{n-1}-y^n=\delta/2\cdot y^{n-1}.$$ Can you find some $y$ such that $\delta/2\cdot y^{n-1}$ is at least $1$?
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0Thank you, I know how to do next. – 2012-08-12
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0@Mobius, have you interested(sure you do) upvoiting the given answer? You can do it by clicking on the arrow pointing forward. And welcome to M.SE! – 2012-08-12
We note that since we have freedom to choose an $\varepsilon>0$, we let $\varepsilon=1$. Therefore, we want to show that for some $x,y$ that $|x^n-y^n|<1$ does not arise from $|x-y|<\delta$. For any $\delta>0$,
if we let $x=\delta$ and $y=\frac{1}{2\delta}$, then $$|x-y|=\delta-\frac{1}{2\delta}<\delta$$ Then, $$|f(x)-f(y)|=|x^n-y^n|=|\delta^n-\frac{1}{2^n\delta^n}|\geq 1$$ for any $\delta$ as $n\rightarrow \infty$.
Therefore, since no $\delta$ can satisfy the uniform continuity condition for this $\varepsilon$, then we conclude that the function $f(x)=x^n$ is not uniformly continuous.
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0We use $y=1/2\delta$ because that avoids the issue of $\delta=1$ satisfying the less than $\varepsilon$ requirement. We are not saying the function is not continuous, just that it is not uniformly continuous. – 2018-09-24