lipschitz function and its properties

Question

$f\colon\mathbb{R}\to\mathbb{R}$ satisfies $|f(x)-f(y)|\le |x-y|^{\beta}$. Which of the following are correct statements?

1. $\beta=1$ $f$ is differentiable;

2. $\beta>0$ $f$ is uniformly continuous;

3. $\beta>1$ $f$ is constant function;

4. $f$ must be a polynomial.

$1\rightarrow |x|$ is an counter example, $2\rightarrow$ is true we can choose $\delta=\epsilon^{1\over\beta}$, $3\rightarrow$ true as$-(x-y) \le {f(x)-f(y)\over x-y}\le (x-y)$, taking limit $x\to y$ and then by sandwich theorem we see $f'(x)=0$ so $f$ is constant, $4$ is not true in general $f(x)=\sin x$ satsfies same relation.

Answer 1

Everything is correct, except a misprint in the answer to question 3. It should be $$-|x-y|^{\beta-1}\leqslant \frac{f(x)-f(y)}{x-y} \leqslant |x-y|^{\beta-1}.$$

If we want to avoid the mean value theorem, we can use a chaining argument:

$$|f(x)-f(0)|\leqslant \sum_{j=0}^{n-1}|f((j+1)n^{-1}x)-f(jn^{-1}x)|\leqslant \sum_{j=0}^{n-1}(n^{-1}|x|)^{\beta}=n^{1-\beta}|x|^\beta.$$