Whether these two functions are uniformly continuous:
1)$f$ such that $|f(x)-f(y)|\le\sqrt{|x-y|}\ \forall x\in(0,1)$
2)$f$ such that $f(x)=x^{\frac1{2}}\sin(\frac1{x^3})\forall\ x\in(0,1)$
For 1), I think there exists a counterexample because $|x-y|\le\sqrt{|x-y|}\forall x\in(0,1)$. As for 2), I think this is similar to $x\sin \frac1{x}$. Any ideas. Thanks beforehand.
1) If $f:[0,1]\to \mathbb{R}$ is such that :
$$\forall (x,y)\in[0,1]^2,\vert f(x)-f(y)\vert\le\sqrt{\vert x-y\vert}$$
Then $f$ is uniformly continuous because given any $\epsilon>0$, if we choose $\delta=\epsilon^2$, we readily see that :
$$\forall (x,y)\in[0,1]^2, \vert x-y\vert\le\delta\implies\vert f(x)-f(y)\vert\le\epsilon^2$$
2) If $f$ is defined by :
$$f:[0,1]\to\mathbb{R},x\mapsto\cases{\sqrt{x}\sin(\frac{1}{x^3})\quad\mathrm{if}\,0<x\le1\cr0\quad\mathrm{otherwise}}$$
then $f$ is continuous (because $\sin$ is bounded and therefore $\lim_{x\to0}f(x)=0=f(0)$) and hence uniformly continuous by Heine theorem