# 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

• I want to clarify what exactly $k$ is. What do you mean by "smooth in $u$"? Can you give me an example of $k$? – timur Sep 17 '13 at 14:16
• I have edited the post to include the fact that $k$ is bounded away from zero, and I have also changed the domain to be a subset of $\mathbb{R}^{2}$ as this is the case I am interested in. The paper does not specify what is meant by smooth, but I am interpreting this as the fact that $k \in C^{\infty}( H^{1+\alpha} \cup H^{1}_{0} \times \Omega )$. No specific example is given, as the paper describes a general result – Keeran Brabazon Sep 17 '13 at 14:26
• A sufficient condition is boundedness of $k_u(u,x)$ and square integrability of $k_{x_i}(u,x)$. It is ok in 2 dimensions because $H^{1+\alpha}$ is in $L^\infty$. – timur Sep 17 '13 at 14:29
• OK, I see your point from a heuristic point of view, as this would mean that the right hand side of (1) above is bounded. Is this as simple as stating that $||k(u(x),x)_{x_{i}}|| \leq ||k_{u}(u,x)||\cdot||u_{x_{i}}|| + ||k_{x_{i}}(u,x)||$ where all we have to do is define the appropriate norms? Or is this an oversimplification? – Keeran Brabazon Sep 17 '13 at 14:49

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.