How to show $\sqrt x$ is uniformly continuous at [0,1] and $[1,\infty )$

2,104

Your proof of uniform continuity works fine, even on $[0,\infty)$ (and I like it).

The square root function is not Lipschitz on $[0,1]$. Indeed, your Lipschitz “constant” is not constant, so that proof does not work. Indeed, you can see that it goes to infinity as both $x$ and $y$ approach zero. But on $[1,\infty)$, you can use $L=1/2$, as follows immediately from your estimate.

Share:
2,104
lyme
Author by

lyme

Updated on August 01, 2022

Comments

  • lyme
    lyme over 1 year

    From the definition if we choose $\delta=\epsilon^2$ $|\sqrt x−\sqrt y|^2≤|\sqrt x−\sqrt y||\sqrt x+\sqrt y|=|x−y|<ϵ^2⟹|\sqrt x−\sqrt y|<ϵ.$

    does this suffice both interval [0,1] and $[1,\infty )$?

    And from Lipschitz $|f(x)-f(y)|=\frac{|x-y|}{|\sqrt x+\sqrt y|}$ if we choose $L=\frac{1}{|\sqrt x+\sqrt y|}$ then $|f(x)-f(y)|\leq L|x-y| $

    again does this suffice both interval [0,1] and $[1,\infty )$? or do we have to say [0,1] is compact from heine–borel and since $\sqrt x $ is continuous , $\sqrt x $ uniformly continuous at [0,1].