A question about Hölder spaces. I am curious to know why we don't normally talk about Hölder spaces with Hölder exponent           $\beta$ $\ge$ $1$. I tried finding the answer and all i found out was that the function assumes constant. But i don't know why ? It would be good to get a proper explanation.  
 A: Suppose $|f(x)-f(y)|\le C|x-y|^{1+\alpha}$ for some positive constants $C, \alpha$ and for all $x$, $y$. Then we   have
$$\biggl|{ f(x)-f(y) \over x-y }\biggr|\le C|x-y|^\alpha,\quad x\ne y.$$ Since $\alpha>0$, we have, for $x$ fixed, $\lim\limits_{y\rightarrow x} |x-y|^\alpha=0$. This implies that $f$ is differentiable and that  $f'(x)$ is zero for all $x$. Thus, $f$ is a constant function.
A: For $\beta=1$, it's the set of Lipschitz continuous function, which is not uninteresting. 
But for $\beta>1$, if $f$ is such that $|f(x)-f(y)|\leq C|x-y|^{\beta}$ for an universal constant $C$, we have for a fixed $x$ that 
$$|f(x)-f(0)|=\left|\sum_{j=0}^{n-1}f\left(\frac{(j+1)x}n\right)-f\left(\frac{jx}n\right)\right|\leq C\sum_{j=0}^{n-1}\left(\frac{|x|}n\right)^{\beta}=C|x|^{\beta}\frac 1{n^{\beta-1}}$$
hence $f(x)=f(0)$ for all $x$ and $f$ is constant.
A: This is only true when the function take values in a normed vector space and domain is an convex set. If it holds Davide Giraudo answer can apply.
