If a function is bounded and continuous does it imply it is Lipshitz function

Question

Let f be a real valued function. If f is bounded and continuous on an Interval does it imply it is Lipshitz function on that interval? Also does there exist an unbounded function on (0,1) which is Lipshitz function?

I couldn'nt think of any counter example and intuitively it feels true to me but I was not able to prove it.

Answer 1

Consider $f(x)=\sqrt{x}$ on $(0,1)$, it is bounded and continuous on $(0,1)$ But it is not Lipschitz...

Suppose it is Lipschitz,

then there $\exists$ $L$, such that $\forall x,y \in (0,1)$ $|\sqrt{x}-\sqrt{y}| \leq L|x-y|$,

Let $x=0$, then $\sqrt{y} \leq L y,$

$\implies \frac{1}{\sqrt{y}}\leq L, \forall y \in (0,1)$

But, given $L \in \mathbb{R}, \exists n: n>L$,

Now choose $y=\frac{1}{n^2}$ , Then $\frac{1}{\sqrt{y}}\leq L \implies n \leq L$

but that's a contradiction , because we choose $n$, such that $n>L.$

Hence $f(x)=\sqrt{x}$ is not Lipschitz on $(0,1)$, but it is bounded and continuous on $(0,1) $

For the second question refer "https://math.stackexchange.com/a/790975/1077989"