I'm trying to prove that the function $f:[0, \infty) \to \mathbb R$ defined by $f(x)=x^r$ for $0<r<1$ is uniformly continuous. What I have so far is that for $x,y \in \mathbb [0,\infty), |x^r-y^r| \leq |x^r|+|y^r|$ but I can't figure out how to get it in the form with $|x-y|$ to bound with $\delta.$ Is there some other indirect way to prove uniform continuity on this domain?
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Hint: Yes, $|x^r-y^r| \le |x|^r + |y|^r$ will get you nowhere. But what about $ |x^r-y^r| \le |x-y|^r$?
zhw.
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As zhw. suggested, you may want to use the inequality $|x^r-y^r|\le |x-y|^r$, proved here. This leads to $\delta=\epsilon^{1/r}$ in the definition of uniform continuity.
Alternatively, observe that the function has bounded derivative on $[1,\infty)$, which implies its uniform continuity. On the interval $[0,1]$, every continuous function is uniformly continuous. You can then glue these two intervals together using the following argument:
Given $\epsilon>0$, pick $\delta_1$ that works for one interval (with $\epsilon/2$ instead of $\epsilon$), and $\delta_2$ that works for the other interval. Let $\delta=\min(\delta_1,\delta_2)$. If $|x-y|<\delta$, and these numbers are in the same interval, we are done. If, say, $x<1<y$, then use $$ |f(x)-f(1)|<\epsilon/2,\qquad |f(1)-f(y)|<\epsilon/2 $$ together with the triangle inequality.
