Density of smooth functions in fractional Sobolev space I am reading a paper on the analysis of numerical methods, and am confused about a statement made. I am working in fractional Hilbert spaces, but I don't think that this has much bearing on the answer to the question. Here are the details:
We are given a function $k(u,x)$ for $u \in H^{1+\alpha}(\Omega) \cap H_{0}^{1}(\Omega)$ and $x \in \Omega \subset \mathbb{R}^{2}$ such that $k$ is smooth in $u$ and $k$ on $\overline{\Omega}$. In addition we are given that $k$ is bounded away from zero (i.e. $||k|| \geq C$). $\alpha$ is in $(0,1/2)$.
Assuming that partial differentiation (in the distributional sense) of $k$ is possible with respect to $x_{i}$ we have that
${\partial k(u(x),x)\over\partial x_{i}} = k_{u}(u,x)u_{x_{i}} + k_{x_{i}}(u,x). \tag{1}$
Now, it is stated that (1) is clearly defined for smooth $u$, which is obvious to me, and it is then stated that the fact that (1) holds "... easily follows for general $u \in H^{1+\alpha}(\Omega) \cap H_{0}^{1}(\Omega)$ from the density of smooth functions in $H^{1+\alpha}(\Omega) \cap H_{0}^{1}(\Omega)$".
This is where I am stuck. First of all I can't see why the smooth functions should be dense in $H^{1+\alpha} \cap H_{0}^{1}$, and even if I accept this fact I can't see how this helps to define the distributional derivative of $k$.
Any help would be appreciated on this, including good textbooks in which to learn about fractional Sobolev spaces
 A: The density result is completely standard and generally known as the Meyers-Serrin theorem. An accessible reference for fractional Sobolev spaces is MacLean's book on strongly elliptic systems.
There is a result by Friedrichs that says that if $f$ is an $L^2$ function, and $f_n$ is a sequence of smooth functions converging to $f$ in $L^2$, and in addition, if $\partial_k\,f_n$ converges to some $w$ in $L^2$, then the weak derivative $\partial_k\,f$ exists and is equal to $w$. Conversely, if the weak derivative $\partial_k\,f$ exists and is in $L^2$, and $f_n$ is a sequence of smooth functions with $f_n\to f$ in $L^2$, then $\partial_k\,f_n\to\partial_k\,f$ in $L^2$.
So if we want to justify (1) for $u\in H^{1+\alpha}$, it would be sufficient to show that for smooth functions $u_n$ that tends to $u$ in $H^{1+\alpha}$,
$$
k_u(u_n,\cdot)\partial_i u_n + k_{x_i}(u_n,\cdot) 
\to 
k_u(u,\cdot)\partial_i u + k_{x_i}(u,\cdot),
$$
with the convergence in $L^2$.
We can look at the two terms separately. For the first term, we have
$$
\begin{split}
\|k_u(u_n,\cdot)\partial_i u_n-k_u(u,\cdot)\partial_i u\|_{L^2}
&\leq 
\|k_u(u_n,\cdot)\partial_i u_n-k_u(u_n,\cdot)\partial_i u\|_{L^2}
+ \|k_u(u_n,\cdot)\partial_i u-k_u(u,\cdot)\partial_i u\|_{L^2} \\
&\leq \|k_u(u_n,\cdot)\|_{L^\infty}\|\partial_i u_n-\partial_i u\|_{L^2}
+ \|k_u(u_n,\cdot)-k_u(u,\cdot)\|_{L^\infty}\|\partial_i u\|_{L^2},
\end{split}
$$
and taking into account that $\|k_u(u_n,\cdot)\|_{L^\infty}$ is uniformly bounded because $\|u_n\|_{L^\infty}\leq c\|u_n\|_{H^{1+\alpha}}\leq M$, that $\partial_i u_n\to\partial_i u$ in $L^2$, and that $k_u(u_n,\cdot)\to k_u(u,\cdot)$ in $L^\infty$ because $u_n\to u$ in $L^\infty$, we conclude that 
$$
k_u(u_n,\cdot)\partial_i u_n
\to 
k_u(u,\cdot)\partial_i u
\qquad \textrm{in}\quad L^2.
$$
Let me leave how to treat the second term as an exercise.
