I know that the function $\sqrt x$ is not Lipschitz in $[0,1]$ and, thinking about this fact, a couple of question has arisen to me:
- Let us call $Lip^+([0,1])$ the space of Lipschitz functions $f:[0,1]\to \mathbb{R^+}$ with the norm $\|f\|_\infty$. Is the functional $$F:Lip^+([0,1])\to L^\infty,\; F[f](t)=\sqrt{f(t)}\space \text{Lipschitz?}$$, that is, we are wondered if $$\max_{[0,1]}|\sqrt{f(t)}-\sqrt{g(t)}|\leq C\max_{[0,1]}|{f(t)}-{g(t)}|$$ Clearly the answer is negative and this can be shown by taking $f_n(t)=\frac{t}{n},\;g_n(t)=\frac{t}{2n}$.
- Does the same conclusion hold if we replace $Lip^+([0,1])$ with $BiLip_M^+([0,1])$?, where $BiLip^+([0,1])$ is the space of functions $f:[0,1]\to \mathbb{R^+}$ such that $$\frac{1}{M}|t-s|\leq |f(t)-f(s)|\leq M|t-s|$$ I have not found any counterexample for this case.
Could someone give me proof of a counterexample for the second question? Any help will be really welcome. Thanks in advance