How to show $\sqrt x$ is uniformly continuous at [0,1] and $[1,\infty )$
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.
lyme
Updated on August 01, 2022Comments

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 $f(x)f(y)=\frac{xy}{\sqrt x+\sqrt y}$ if we choose $L=\frac{1}{\sqrt x+\sqrt y}$ then $f(x)f(y)\leq Lxy $
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].