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

Question

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?

Answer 1

Hint: Suppose $n>1.$ Let $\delta > 0.$ Then by the mean value theorem,

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

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

Answer 2

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 $n= 1$.

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