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Let's consider the function $f:\mathbb R\to\mathbb R$ such that $|f(x)-f(y)|\leq|x-y|^\sqrt2$ for all $x$ and $y$ in $\mathbb R$. Then how do I prove if the function is differentiable or not?

I firstly replaced $x-y$ by $h$ and put the limit $h\to 0$ so it becomes $\lim\limits_{h\to 0} |f(y+h)-f(y)|≤h^\sqrt2$. I don't know what to do next.

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    You would need to divide by $h$ before taking the limit to get something useful, but also you have a typo I think: you want $|f(y+h)-f(y)|$, right? – 2017-02-10

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The following proof may be repeated with $x$ substituted for $y+h$ if it makes the reader feel more comfortable. Either is valid.

If a function $f:\mathbb R \to \mathbb R$ satisfies the condition $|f(x)-f(y)|\leq |x-y|^{\sqrt 2}$ for all $x$ and $y$

then for $x\neq y$ it satisfies $$\dfrac{|f(x)-f(y)|}{|x-y|} \leq |x-y|^{\sqrt{2}-1}$$ Now $f$ is differentiable at $x$ if and only if the left hand side of the equation approaches a limit whenever $y$ approaches $x$. The right hand side implies this limit is $0$. Thus, $f'$ exists and is $0$ everywhere. $f$ must be constant.

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Hint: Divide your last expression by $h$ and take the limit $h\rightarrow 0$.

Hint 2: Very few functions verify this property.

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    Alright. And this would be right hand derivative. The same again had to be done to get left hand derivative by replacing x-y by -h. If both are equal it's differentiable. Otherwise not. Right? – 2017-02-10
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    You could just replace h by |h| on the RHS. Then you don't have to treat positive and negative h differently. – 2017-02-10
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    I think that the first hint may have been referring to $|h|$. – 2017-02-11