prove that the function $f$ is locally lipschitz

Question

Let $f:(0,1] \to \mathbb{R}$, $f(x) = \sqrt{x}$. Prove that $f$ is locally Lipschitz.

Definition: A function $f:X \to \mathbb{R}$ is locally lipschitz if $\forall x \in X$, there is a open ball $B$ with center $x$, such that $f$ restricted to $B$, $f: X \cap B \to \mathbb{R}$ is Lipschitz.

In $(0,1)$ it's ok. For every $x\in (0,1)$, there is a $r>0$ such that $(x-r,x+r) \subset (0,1)$ and $f$ is lipschitz in $(x-r,x+r)$.

In fact, if you take $y \in (x-r,x+r)$, then

$\displaystyle \frac{|f(x) - f(y)|}{|x-y|} = \frac{|\sqrt{x}- \sqrt{y}|}{|x-y|}= \frac{1}{|\sqrt{x}+\sqrt{y}|} < \frac{1}{2\sqrt{x-r}} $.

But what about the point 1???

Answer 1

Let $x_0>0$.

Then take $\delta=\frac{x_0}{2}>0$,so in the interval $I_{x_0}=[x_0-\delta,x_0+\delta]=[\frac{x_0}{2},\frac{3x_0}{3}],$ $f(x)$ has bounded derivative. Thus it is Lipschitz on $I_{x_0}$

For $x=1$ apply the same idea on the interval $[\frac{1}{2},1]$

Answer 2

Hint: Suppose $f$ is differentiable on an interval $I,$ with$f'$ bounded on $I.$ Then $f$ is Lipschitz on $I.$ Proof: MVT.