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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

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The given condition is Holder continuous which is not the same as uniform continuity. – JSchlather Dec 17 '12 at 6:16
Is every differentiable function uniformly continuous? – Vishal Gupta May 29 '13 at 16:31
There is a bit of vocabulary-based confusion here. These days, we tend to call this condition Holder continuity. However, in the older literature (e.g. harmonic analysis papers from the 1970s), this is often called a Lipschitz condition with an exponent. – Ray Yang May 29 '13 at 16:38

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.

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For part 3), most polynomials also work as counterexamples (all with degree 2 or higher). – Jonas Meyer Dec 17 '12 at 6:27
@JonasMeyer Very good point! Prompted me to take a closer look at the condition and make an edit, thank you! – Tom Oldfield Dec 17 '12 at 6:29

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: $\times$: $|\text{sign }x-\text{sign }y|\le|x-y|$

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I am not sure how you choose such $\delta=\dfrac{\epsilon^{1/{\alpha}}}{2K}>0$. I think $\delta=(\frac{\epsilon}{2K})^{1/\alpha}>0$ would have worked as well. – learner May 3 '14 at 11:40

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