Proving uniform continuity for $x^{a}$ in $[0,\infty)$, $0

Question

I can prove it for $x^{\frac{1}{2}}$ and easy rationals like that by just algebraic manipulation.

But how should I approach proving the uniform continuity for all $0<a<1$. At $a=1$ it is just lipschitz. But how to use the lipschitz criteria for any $a>0$?.

Another argument I tried is the sequential criteria.

If I have $x_{n}$ and $y_{n}$ are two sequences such that $|x_{n}-y_{n}|\to 0$. Then $|(x_{n})^{a}-(y_{n})^{a}|\to 0$. So if it be possible that $|(x_{n})^{a}-(y_{n})^{a}|$ does not go to $0$. Then I can find subsequence $|(x_{n_{k}})^{a}-(y_{n_{k}})^{a}|\geq \epsilon$. But then again I run into the problem of algebraically equating $|(x_{n_{k}})^{a}-(y_{n_{k}})^{a}|$ to $|x_{n_{k}}-y_{n_{k}}|$.

Any help is appreciated.

Answer 1

Uniform continuity follows from the fact that $$|x^a -y^a| \leq |x-y|^a $$ for $x,y\geq 0, 1\geq a>0.$

To prove this fact observe that $$ (x+y)^a =\frac{x+y}{(x+y)^{1-a} } \leq \frac{x}{x^{1-a} } +\frac{y}{y^{1-a}} = x^a +y^a .$$ Substiuting $x=u, y=v-u $ where $v\geq u $ we obtain $$v^a \leq u^a +(v-u)^a $$ and cosequently $$|u^a - v^a|\leq |u-v|^a$$