Past exam of the methods of applied functional analysis course of the University of Oxford / 2010 / A / 2 / b

Choose a standard mollifier $\rho \in C_0^{\infty }({\mathbb{R}}^n)$ with $\rho \ge 0$, $\int \rho =1$ and support in the unit ball, and set ${\rho }_{\epsilon }(z)={\epsilon }^{-n}\rho (z/\epsilon )$. Define the convolutionThe smoothing properties of a mollifier give $u_{\epsilon }\in C^{\infty }$ and ${\partial }_j{u}_{\epsilon }={\rho }_{\epsilon }\ast D_{j}u$. Apply the chain rule to $t\mapsto u_{\epsilon }(y+t(x-y))$ and integrate its derivative by the fundamental theorem of calculus:

Write $g=(g_1,\dots ,g_n)$ for the continuous representatives of the weak derivatives. Integrating the preceding identity over $B=B(0,1)$ givesThe first term converges to ${\int }_{B}u$ because $u_{\epsilon }\to u$ in $L^1$. For fixed $x$, all segments in the second term lie in a compact set. Continuity of $g$ implies ${\rho }_{\epsilon }\ast g\to g$ uniformly on that set. Hence every $x$ has the finite limitThe same argument is uniform for $x$ in any fixed compact set: all the segments then lie in a common compact set. Thus $u_{\epsilon }\to U$ locally uniformly, so $U$ is continuous; the local $L^1$ convergence also gives $U=u$ almost everywhere.

Passing to the limit in the segment identity, and taking its endpoints to be $x$ and $x+h$, givesTherefore $U$ is differentiable with continuous gradient $g$: $\boxed{U\in C^1,U=u\text{ almost everywhere},\nabla U=g}$. This is the continuous weak gradient gives a C1 representative criterion.