I tried to find a Counter-example for the following relation for all $x,y\in C$ where $C$is a convex set:
$$|f(x)-f(y)|\leq L|x-y|\Rightarrow |f'(x)-f'(y)|\leq \beta|x-y|$$
Is it possible to find a function which is lipschitz but its derivative is not lipschitz?
The converse is not true for example derivative of $f(x)=x^2$ is lipschitz but the function itself is not lipschitz.
We have that $x\sqrt{x}$ is Lipschitz continuous on $[0,1]$, because its derivative is bounded on $[0,1]$.
However, its derivative $\frac{3}{2}\sqrt{x}$ is not Lipschitz continuous on $[0,1]$, because the derivative of this function gets unboundedly large as we let $x\downarrow 0$.