Is $\sqrt x$ Lipschitz in the interval $[1,10]$ ? , I know it is not in $[0,\infty]$ Is $\sqrt{x}$ Lipschitz in the interval $[1,10]$? , I know it is not in $[0,\infty)$.
Here $f(x)=\sqrt{x}$, 
$\frac{df}{dx}=\frac{1}{2} x^{-1/2}$ .
Therefore
$\frac{df}{dx} \in \left[\frac{1}{2\sqrt{10}},\frac{1}{2}\right]$
Thus here the Lipschitz constant should be $\frac{1}{2\sqrt{10}}$, isn't? 
 A: No, the constant should be $1/2$.
The Lipschitz condition is that there's a constant $C$ such that:
$$|f(a)-f(b)|\le C|a-b|$$
Now if $f$ is differentiable we have that $f(a)-f(b) = (a-b)f'(\xi)$ for some $\xi\in(a,b)$, so we certainly have the condition fulfilled if $C\ge|f'(x)|$ for all $x$ in the interval of interrest.
On the other hand if $f$ is continuosly differentiable we can for every $\epsilon>0$ find an interval $(a,b)$ such that $|f'(\xi)|>\sup |f'(x)|-\epsilon$ which means that $|f(a)-f(b)| = |f'(\xi)||a-b|>\sup |f'(x)|-\epsilon$. This means that the Lipschitz constant must be at least $\sup |f'(x)|$ (ie an upper bound for the absolute derivate).
By the way this means that if $0<a<b$ we have Lipschitz continuity with the constant $1\over 2\sqrt a$ which is regardless of $b$. This means that the Lipschitz constant works on the entire interval $[a,\infty)$. 
A: If $1\le x < y,$ then
$$\sqrt {y} - \sqrt {x} = \frac{y-x}{\sqrt {y} + \sqrt {x}} \le \frac{y-x}{2}.$$
Thus $\sqrt x$ is Lipschitz on $[1,\infty)$ with Lipschitz constant no more than $1/2.$ To see the Lipschitz constant equals $1/2$ on this interval, consider $(\sqrt x - \sqrt 1)/(x-1)$ as $x\to 1^+.$
