Let $F=\{f: \mathbb{R} \to\mathbb{R}:|f(x)-f(y)|\leq K|(x-y)^\alpha|\}$ for all $x,y\in\mathbb{R}$ and for some $\alpha >0$ and some $K > 0$ Let  $F=\{f: \mathbb{R} \to\mathbb{R}:|f(x)-f(y)|\leq K|(x-y)^\alpha|\}$ for all $x,y\in\mathbb{R}$ and for some $\alpha >0$ and some $K > 0$ .
Which of the following is/are true?
1. Every $ f \in F$ is continuous
2. Every $f\in F$ is uniformly continuous
3. Every differentiable function $ f$  is in $F$
4. Every $f \in F$  is differentiable   
The given condition is Lipchitz condition. So 1 and 2 are true. but what can I say about the others
 A: The condition you give is the Holder condition, not the Lipschitz condition as you state in your question. (As Jacob Schlather points out in the comments). It is a more general condition, in that any function which satisfies the Lipschitz condition also satisfies the Holder condition, but not conversely. 
However, $1)$ and $2)$ can be proved directly fairly easily (also, $2) \implies 1)$
For part $3)$, can you think of a differentiable function which grows faster than any fixed polynomial? (this property means that it cannot be in $F$)
Edit: As Jonas Meyer says in the comments, any polynomial with degree greater than $2$ will provide a counterexample to part $3)$. In fact, from part $2)$ we can make the much more general statement that any differentiable function which is not uniformly continuous will provide a counterexample.
For part $4)$, think about $|x|$ for example.
A: 1, 2: $\surd$: For $\epsilon>0$ choose $\delta=\dfrac{\epsilon^{1/{\alpha}}}{2K}>0.$ Then $|x-y|<\delta$$\implies|f(x)-f(y)|$$\le K|x-y|^\alpha\le\epsilon.$ So 2 is true and 2 implies 1. 
3: $\times$: Choose $K=1,\alpha=1.$ Then $x^2\notin F:$ $|2^2-1^2|>|2-1|.$
4: $\times$: $|\text{sign }x-\text{sign }y|\le|x-y|$
