Uniform continuity of the function $f(x)=x^a$ 
Prove that $f:[0, \infty ) \to [0, \infty)$, $f(x)=x^a$ with $a>0$ is uniformally continous if and only if $0<a\leq 1$.

I guess that we could start with the scratch work like this:
$\forall \epsilon$>0 $\exists \delta$ s.t $\forall x,y \in [0, \infty) $ $|x-y|< \delta \implies |x^{a}-y^{a}|$
Then,
$|x^{a}-y^{a}|$= $|x-y||x^{n-1}-y^{n-2}x+...-y^{n-2}x+y^{n-1}|\leq \delta |x^{n-1}-x^{n-2}y+...-y^{n-2}x |$
the problem is I don't know how to keep it going with it, maybe bernoulli's inequality? if it is possible I would like to know if there's a sigma notation for ($a^n-b^n$).
 A: Some Hints.
We shall analyse three cases here.
a > 1
Let $f(x) = x^a$ with $x\in [0, +\infty)$. From the mean value theorem we have the existence of $c \in (p, q)$ such that
$$|f(q)- f(p)| = |f'(c)|\cdot |q-p| = a\cdot c^{a-1}\cdot |q-p|$$
Say $\epsilon = a/2$, and $\delta_{\epsilon} > 0$ (but small); we have
$$p = \dfrac{1}{\delta_{\epsilon}}^{1/(a-1)}$$
$$q = \dfrac{\delta_{\epsilon}}{2} + \dfrac{1}{\delta_{\epsilon}}^{1/(a-1)}$$
then
$$|f(q) - f(p)| = f'(c)|\cdot |q-p| = a\cdot c^{a-1} \dfrac{\delta_{\epsilon}}{2} \geq \dfrac{a}{2} = \epsilon$$
This shows the failure fo $f(x) = x^a$ to be uniformly continuous for $a > 1$.
a = 1
Here it's easy: given $\epsilon > 0$,
$$|f(q) - f(p)| = |q-p| < \epsilon \qquad \qquad \text{when} \qquad |q-p| < \delta_{\epsilon} = \epsilon$$
a < 1
A particular case this one, in which I also choose $p = 0$.
Here we have simply:
$$|f(q) - f(p)| = |f(q)| = q^a < \epsilon \qquad \qquad \text{when}\qquad |q-p| = q < \delta_{\epsilon} = \epsilon^{1/a}$$
Can you proceed with all this?
A: Theorem$\quad$ Power function $y=f(x)=x^{\alpha}(\alpha \in \mathbb{R})$,
(1) when $\alpha>1$, is not uniformly continuous on $[0,+\infty)$, but uniformly continuous on any finite interval $I\subset[0,+\infty)$;
(2) when $0<\alpha \leq 1$, is uniformly continuous on $[0,+\infty)$;
(3) when $\alpha<0$, is not uniformly continuous on $(0,+\infty)$, but uniformly continuous on $[a,+\infty)$ (where $a>0$).
Proof$\quad$ (1) Let $\varepsilon_{0}=1$. For all $\delta>0$, let $x_{1}=(n+1)^{\frac{1}{\alpha}}, x_{2}=n^{\frac{1}{\alpha}}$. Obviously, $x_{1}, x_{2} \in[0,+\infty)$. Note that $0<\frac{1}{\alpha}<1$ because $\alpha>1$. Therefore,
$$\lim _{n \rightarrow \infty}\left|x_{1}-x_{2}\right|=\lim _{n \rightarrow \infty}\left[(n+1)^{\frac{1}{\alpha}}-n^{\frac{1}{\alpha}}\right]=0.$$
So, when $n$ is sufficiently large, $\left|x_{1}-x_{2}\right|<\delta$ must be true. But
$$\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=\left((n+1)^{\frac{1}{\alpha}}\right)^{\alpha}-\left(n^{\frac{1}{\alpha}}\right)^{\alpha}=n+1-n=1=\varepsilon_{0}.$$
So, when $\alpha>1$, $y=f(x)=x^{\alpha}$ is not uniformly continuous on $[0,+\infty)$.
For any finite subinterval $I\subset[0,+\infty)$, there is always a closed interval $[0, M]$, such that $I \subset[0, M]$.
Since continuous functions on closed intervals must be uniformly continuous, $y=f(x)=x^{\alpha}(\alpha>1)$ is uniformly continuous on $[0, M]$. Therefore, $f(x)$ is uniformly continuous on $I$.
(2) When $0<\alpha \leq 1$, since $f(x)=x^{\alpha}$ is continuous on $[0,2]$, it is also uniformly continuous on $[0,2]$.
When $x \in[2,+\infty)$, for any $\varepsilon>0$, there exists $\delta=\varepsilon>0$, such that for all $x_{1}, x_{2} \in[2,+\infty)$, $\left|x_{1}-x_{2}\right|< \delta$.
Because the function $x^{\alpha}$ is differentiable on $[x_{1}, x_{2}]$, the Lagrange mean value theorem shows that $x_{1}^{\alpha}-x_{2}^{\alpha}=\alpha \xi^{\alpha-1}\left(x_{1}-x_{2}\right)$, where $\xi\in[x_{1}, x_{2}]$. Therefore,
$$\left|x_{1}^{\alpha}-x_{2}^{\alpha}\right|=\left|\alpha \xi^{\alpha-1}\left(x_{1}-x_{2}\right)\right|=\alpha \xi^{\alpha-1}\left|x_{1}-x_{2}| \leq\right| x_{1}-x_{2} \mid<\delta=\varepsilon.$$
So, $f(x)=x^{\alpha}$ is uniformly continuous on $[2,+\infty)$, and is therefore uniformly continuous on $[0,+\infty)$.
(3) When $\alpha<0$, let $\varepsilon_{0}=1$, for all $\delta>0$, let $x_{1}=(n+1)^{\frac{1}{\alpha}}, x_{2}=n^{\frac{1}{\alpha}}$. Obviously, $x_{1}, x_{2} \in(0,+\infty)$ and $-\frac{1}{\alpha}>0$. Therefore,
\begin{aligned}
& \lim _{n \rightarrow \infty}\left|x_{1}-x_{2}\right|=\lim _{n \rightarrow \infty}\left|(n+1)^{\frac{1}{\alpha}}-n^{\frac{1}{\alpha}}\right|=\lim _{n \rightarrow \infty}\left[\frac{1}{n^{-\frac{1}{\alpha}}}-\frac{1}{(n+1)^{-\frac{1}{\alpha}}}\right] \\
= & \lim _{n \rightarrow \infty} \frac{1}{n^{-\frac{1}{\alpha}}}-\lim _{n \rightarrow \infty} \frac{1}{(n+1)^{-\frac{1}{\alpha}}}=0-0=0.
\end{aligned}
So, when $n$ is sufficiently large, $\left|x_{1}-x_{2}\right|<\delta$ must be true. But
$$\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right|=\left|\left((n+1)^{\frac{1}{\alpha}}\right)^{\alpha}-\left(n^{\frac{1}{\alpha}}\right)^{\alpha}\right|=n+1-n=1=\varepsilon_{0}.$$
Therefore, $f(x)$ is not uniformly continuous on $(0,+\infty)$. Since $\lim _{n \rightarrow \infty} f(x)=\lim _{n \rightarrow \infty} x^{\alpha}=0$, $f(x)$ is uniformly continuous on $[a,+\infty )$ (where $a>0$).
A: Here is another approach using sequences:
The case $a<0$ is trivial, since in this case, $f$ is not even right-continuous at $x=0.$
Suppose $0< a\le 1.$ If $f$ is not uniformly continuous, then there is an $\epsilon>0$ and disjoint sequences $(x_n),(y_n)\subseteq [0,\infty)$ with the property that $|x_n-y_n|<\frac{1}{n}$ but $|f(x_n)-f(y_n)|>\epsilon.$ We may scale $f$ so that $\epsilon=1,$ and relabel the sequences so that $x_n<y_n.$ Now, by the mean value theorem, there is aequence $x_n<c_n<y_n$ such that $|f(x_n)-f(y_n)|=|y_n^a-x_n^a|=ac_n^{a-1}|y_n-x_n|<\frac{ac_n^{a-1}}{n}.$  If $(y_n)$ is bounded, then so is $(c_n)$ and we get an immediate contradiction. If not, then $c_n\to \infty.$ But then, $0<c_n^{a-1}\le 1$ for large enough $n$ and we get a contradiction in this case also.
If $a>1,$ let $x_n=n$ and $y_n=n+\frac{1}{n}.$ Then, $y_n-x_n=\frac{1}{n}\to 0$ as $n\to \infty.$ On the other hand, $y_n^a-x_n^a=\frac{anc_n^{a-1}}{n}=ac_n^{a-1}$ for some $n<c_n<n+\frac{1}{n}$ and since $ac_n^{a-1}\to \infty$ as $n\to \infty,\ f$ is not uniformly continuous.
