Uniform continuity of $\sqrt{x}, x\in (0,1)$

Question

In the lecture, lecturer said that $\sqrt{x} ,\,x \in (0,1)$ is uniformly continuous. But, I think it is not uniformly continuous because as $x$ approaches to 0, we require to have smaller $\delta$ for a given $\varepsilon$.

I have not learnt proper proof of uniform continuity, yet. If i am wrong, please explain to me..

Thanks,

Answer 1

The definition of uniform continuity is, to recall that a function $f \colon [0,1] \to \def\R{\mathbf R}\R$ is uniformly contiuous iff $$ \forall \epsilon > 0\; \exists \delta > 0\; \forall x,y \in [0,1]:\; |x-y| < \delta \Rightarrow |f(x) - f(y)| < \epsilon $$ So let $\epsilon > 0$ be given, recall that for $x,y \in [0,1]$ we have \begin{align*} |\sqrt x - \sqrt y|^2 &\le |\sqrt x - \sqrt y|\cdot |\sqrt x + \sqrt y|\\ & = |x-y|\\ \iff |\sqrt x - \sqrt y| &\le |x-y|^{1/2} \end{align*} That is, we can choose $\delta := \epsilon^2$ and everything works fine.

Another possibility is to use the compactness of $[0,1]$ and recall that every continuous function is uniformly continuous on compact sets.

Answer 2

It's uniformly continuous because it is continuous on $[0,1]$. Given an $\epsilon>0$, you choose a $\delta$ for continuity at $0$! and then that $\delta$ works for the uniform continuity definition, too.