Originally Posted by

**Laurent** Hi,

this function is uniformly continuous.

The "$\displaystyle \epsilon-\delta$ way":

If $\displaystyle 0\leq x, y$, then $\displaystyle |\sqrt{y}-\sqrt{x}|\leq \sqrt{|y-x|}$. Indeed, if $\displaystyle 0\leq b\leq a$, $\displaystyle (a-b)^2=a^2-2ba+b^2\leq a^2-2b^2+b^2=a^2-b^2$; take $\displaystyle a=\max(\sqrt{x},\sqrt{y})$, $\displaystyle b=\min(\sqrt{x},\sqrt{y})$. Thus, if $\displaystyle 0\leq x,y$ and $\displaystyle |x-y|\leq \epsilon^2$, then $\displaystyle |\sqrt{y}-\sqrt{x}|\leq \epsilon$. And if $\displaystyle x\leq 0\leq y$ with $\displaystyle |x-y|\leq \epsilon^2$, then $\displaystyle |y-(-x)|=y+x\leq y\leq \epsilon^2$, hence $\displaystyle |\sqrt{|y|}-\sqrt{|x|}|=|\sqrt{y}-\sqrt{-x}|\leq\epsilon$ by the above argument. This settles all the cases.

The more sophisticated way :

On $\displaystyle [-1,1]$, the function is continuous and bounded, hence uniformly continuous. On $\displaystyle [1,+\infty)$, the function is differentiable and its derivative is less than $\displaystyle \frac{1}{2}$, hence it is $\displaystyle \frac{1}{2}$-Lipschitz and thus uniformly continuous. By symmetry, and gluing the parts together, this concludes.