 Solution

Choose a nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^3)$ with $\int \rho =1$, and put ${\rho }_{\epsilon }(x)={\epsilon }^{-3}\rho (x/\epsilon )$. Sobolev approximation by mollifiers in the Sobolev space under consideration statesIn particular, it gives density of smooth functions in a Sobolev space. We prove convergence of both the function and all its first weak derivatives.

The usual smoothing properties of convolution give $u_{\epsilon }\in C^{\infty }$. For a test function $\varphi \in C_c^{\infty }({\mathbb{R}}^3)$ and $i\in \{1,2,3\}$, Fubini's theorem and the definition of a weak derivative giveThe integrals are absolutely convergent: the compact supports of ${\rho }_{\epsilon }$ and $\varphi$ restrict $y$ to a compact set, on which $u$ and ${\partial }_{i}u$ are integrable by Holder inequality. ConsequentlyThe approximate identity theorem in the Lp space $L^6$ applies separately to $u$ and ${\partial }_{i}u$. ThusEach convolution belongs to $L^6$ by Young's convolution inequality. With the componentwise Sobolev norm, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent Sobolev norm gives the same convergence.