Can you please help me with proving Lipschitz condition?
Let: $f\colon\mathbb{R}\longrightarrow \mathbb{R}$ satisfy $\lvert f(x)-f(y) \rvert\leq C\lvert x-y\rvert^\alpha$ for some constants $C\gt 0$, $0\lt\alpha\lt\infty$, and for any $x,y\in\mathbb{R}$.
(1) Prove that if $0\lt\alpha\le1$, then $f$ is uniformly continuous.
(2) Prove that if $1\lt\alpha\lt\infty$, then $f$ is a constant.
Use the definition of uniform continuity. For any fixed $\varepsilon > 0$, you want to find $\delta=\delta(\varepsilon) > 0$ such that, for any $x,y\in\mathbb{R}$, $\lvert x-y\rvert \leq \delta$ implies $\lvert f(x)-f(y)\rvert \leq \varepsilon$. Using your hypothesis, why is it sufficient to choose $\delta$ such that $C\delta^\alpha \leq \varepsilon$? (And can you conclude from there?)
Prove that $f$ is then differentiable everywhere, by fixing an arbitrary $x$, taking $y=x+h$ and letting $h\to 0$. This will in particular show you that $f^\prime(x)$ not only exists, but equals 0. You will need the assumption $\alpha > 1$ for this, since then $\lvert h\rvert^\alpha = \lvert h\rvert^{1+\beta}$ with $\beta>0$.