$k\in (0,1)$, is $W^{k,p}(\Omega)$ the closure of $C^{0,k}(\overline{\Omega})$ with the $L^p$ norm Suppose that $\Omega\subset\mathbb{R}^N$ is a bounded smooth domain. It is common to see the definition of the Sobolev space $W^{k,p}(\Omega)$ as the completion of $C^{k}(\overline{\Omega})$ with the norm $\|u\|_{k,p}=\left(\sum_{|\alpha|\leq k}D^\alpha u\right)^{1/p}$ (here $k$ is a positive integer). 
My question is: For  $k\in (0,1)$, is $W^{k,p}(\Omega)$ the completion of $C^{0,k}(\overline{\Omega})$ with the norm: $$\|u\|_{k,p}=\left(\int_\Omega|u|^p+\int_\Omega\int_\Omega\frac{|u(x)-u(y)|^p}{|x-y|^{N+kp}}\right)^{1/p}$$
Remark: Here $C^{0,k}(\overline{\Omega})$ denotes the usual Holder space.
Remark 2: Usually $W^{k,p}(\Omega)$ for $k\in (0,1)$ is defined by $$W^{k,p}(\Omega)=\left\{u\in L^p(\Omega):\ \frac{|u(x)-u(y)|^p}{|x-y|^{\frac{N}{p}+k}}\in L^p(\Omega\times\Omega)\right\}$$
and $W^{k,p}(\Omega)$ is complete under the norm $\|\cdot\|_{k,p}$ defned above.
Thank you
 A: No, because  $C^{0,k}$ is not necessarily contained in $W^{k,p}$ when $k<1$. Consider the Hardy-Littlewood trigonometric series 
$$f(t)=\sum_{n=1}^\infty e^{i n\log n}n^{-\frac12-\alpha} e^{inx},\quad 0<\alpha<1 \tag1$$ 
It is true, though nontrivial (page 197 in Trigonometric Series, Volume I by Zygmund) that $f\in C^{0,\alpha}$ (on any bounded interval, since it's periodic). On the other hand, a trigonometric series is in $W^{\alpha,2}$ if and only if its coefficients $c_n$  satisfy $$\sum_{n\in\mathbb Z}|n|^{2\alpha}|c_n|^2<\infty\tag2$$
For (1) we get 
$$\sum_{n=1}^\infty n^{2\alpha} n^{-1-2\alpha }=\infty$$
Thus, $C^{0,\alpha}\not\subset W^{\alpha,2}$. 
But you can generalize the definition anyway, because $C^\infty$ functions are dense in function spaces with finite integrability exponents (e.g., Proposition 14.5 in First Course in Sobolev Spaces by Leoni). The definition of $W^{k,p}$ for integer $k$ could as well have $C^\infty$ there instead of $C^k$. And then for fractional $k$ it's the same thing.
