Let be $x=(x_1,x_2,...,x_n)$ and $0<\alpha\le1$ $$f:\Bbb R^n\to\Bbb R, f(x)=||x||^\alpha $$
I know if $\alpha=1$, the euclidean norm, the function is uniformly continuous. But when I try to prove it for $0<\alpha<1$ I don't know what to do. I only got to know that $||x||+||y||\le||x||^\alpha+||y||^\alpha$ when $x,y \in B((0,0),1)$ Maybe it's useful to prove it.
I think I have to use some inequality using the case $\alpha=1$.
First prove it for $\alpha = 1$. This is trivial by the reverse triangle inequality. Putting $h: \mathbb R^n \to \mathbb R_{\ge 0}$, $h(x) = \| x \|$ for $x \in \mathbb R^n$, we see $$\lvert h(x) - h(y) \rvert = \lvert \|x \| - \|y \| \rvert \le \|x - y\|$$ for all $x, y \in \mathbb R^n$ showing that the map is Lipschitz with constant $1$, hence uniformly continuous (indeed, in the definition of uniform continuity, you can take $\delta = \epsilon$).
Next prove the map $g: \mathbb R_{\ge 0} \to \mathbb R_{\ge 0}$ defined by $g(x) = x^{\alpha}$ is uniformly continuous for $\alpha \in (0,1]$. For this, we know that $g$ is uniformly continuous as a map from $[0,2] \to \mathbb R_{\ge 0}$ since it is a continuous function on a compact set. For a fixed $\epsilon > 0$, let $\delta_1 > 0$ be as in the definition of uniform continuity for $g$ on $[0,2]$. Next, on $[1,\infty)$, we have $$\lvert g'(x) \rvert= \lvert\alpha x^{\alpha -1}\rvert \le \alpha$$ since $\alpha - 1 \le 0$ thus by the mean value theorem, $g$ is Lipschitz on $[1,\infty)$ with constant $\alpha$, hence uniformly continuous. Let $\delta_2 > 0$ as in the definition of uniform continuity for $g$ (and this particular $\epsilon$) on $[1,\infty)$. Then taking $\delta = \min\{ \delta_1,\delta_2,1/2 \}$ proves uniform continuity of $g$ on all of $\mathbb R_{\ge 0}$ (do you see why?).
Hence $f = g \circ h$ is uniformly continuous as the composition of uniformly continuous maps.