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

For one coordinate, integration by parts gives ${\int }_{\Omega }({\partial }_{j}u)\phi =-{\int }_{\Omega }u{\partial }_j\phi$: the test function vanishes outside a compact subset of $\Omega$, so there is no boundary contribution. Repeat this operation once for each of the $|\alpha |$ coordinate differentiations in the multi-index $\alpha$. Each operation contributes a factor $-1$, givingAll intermediate functions are continuous on the relevant compact support, since $u\in C^k$ and $|\alpha |\le k$.

The function $v$ is the weak derivative $D^{\alpha }u$ ifThe integrals exist by local integrability and the compact support of the test function. This is the definition of a distributional derivative represented by a locally integrable function.

The fundamental lemma of the calculus of variations says that if $h\in L_{\mathrm{loc}}^1(\Omega )$ and $\int h\phi =0$ for every $\phi \in C_0^{\infty }(\Omega )$, then $h=0$ almost everywhere.

If $v_1,v_2$ both represent $D^{\alpha }u$, subtracting their defining identities gives $\int (v_1-v_2)\phi =0$ for every test function. The lemma gives $v_1=v_2$ almost everywhere, proving uniqueness. If $u\in C^1$, part (i) gives exactly the defining identity with $v={\partial }_{j}u$. Since this classical derivative is locally integrable, it is the unique weak derivative $D_{j}u$, up to equality almost everywhere.

An integrable example on the whole real line is the tent functionIt is not differentiable at $-1,0,1$. For a test function $\phi$, split the integral into $(-1,0)$ and $(0,1)$ and use integration by parts. The endpoint terms at $-1,1$ vanish because $u=0$, and those at zero cancel because $u$ is continuous. The result is $\int u{\phi }^{\prime }=-\int (Du)\phi$. Thus the corners obstruct classical differentiability but not this weak derivative.

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.