Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 105 / 2

For $u,g\in L_{\mathrm{loc}}^1(U)$, the function $g$ is the $i$th weak derivative $D_{i}u$ whenfor every test function $\phi \in C_c^{\infty }(U)$. This is the integration by parts identity with no boundary term and agrees with the ordinary derivative whenever $u$ is classically differentiable.

For $1\le p\le \infty$, the first-order Sobolev space iswith norm, for example,Functions equal almost everywhere represent the same Sobolev element.

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Let $g=Du\in L^p(0,1)$ be the weak derivative and defineThe fundamental theorem of calculus for Lebesgue integration makes $F$ an absolutely continuous function, differentiable almost everywhere, with $F^{\prime }=g$ almost everywhere. The distributional derivative of $u-F$ is zero. A locally integrable function with zero distributional derivative on a connected interval is equal almost everywhere to a constant $C$. Consequentlyis an absolutely continuous representative of $u$, and ${\tilde{u}}^{\prime }=g\in L^p(0,1)$ almost everywhere.

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Write points of $U$ as $(x,y)$ with $x>0$. For $h>0$, translate $u$ into the domain byContinuity of translations in $L^p$ applied to $u,D_{x}u,D_{y}u$ givesas $h\downarrow 0$. Choose a standard mollifier ${\rho }_{\epsilon }$ supported in a ball of radius $\epsilon <h/2$. For $(x,y)\in U$, the convolutiononly samples points with first coordinate greater than $-h$, so it is well-defined and smooth throughout $U$. The approximation-to-the-identity theorem, applied also to each weak derivative, allows $\epsilon (h)<h/2$ to be chosen so thatTaking $h=1/j$ and using the triangle inequality proves the density of smooth functions in a Sobolev space.

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Use the reflection extension from a half-spaceIt is plainly linear and restricts to $u$ on $U$. For a smooth $u$, the chain rule givesA change of variables therefore givesfor $p<\infty$, with the evident equality of essential suprema for $p=\infty$. Approximate a general $u$ by the smooth functions from part c. The estimate makes their reflections Cauchy in $W^{1,p}({\mathbb{R}}^2)$, and their limit defines a bounded Sobolev extension operator with $\Vert E\Vert \le 2^{1/p}$.

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For a smooth $u$ and fixed $y$, the fundamental theorem of calculus gives, for $0<x<1$,Average this inequality over $x\in (0,1)$, use Holder inequality on that unit interval, raise to the power $p$, and integrate in $y$. This proves the estimate behind the W1p trace theorem on a half-space:Use the Sobolev extension operator from part d, approximate $Eu$ in $W^{1,p}({\mathbb{R}}^2)$ by smooth functions, and define $Tu$ as the $L^p(\mathbb{R})$ limit of their restrictions to $x=0$. The trace inequality makes this limit independent of the approximation and proves thatis linear and bounded. For a smooth function that extends continuously to the boundary, $Tu=u{|}_{\partial U}$, so this is the trace operator required.

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The Rellich-Kondrachov compactness theorem says that if $U\subset {\mathbb{R}}^n$ is a bounded Lipschitz domain, thenfor $1\le q<p^{\ast }=np/(n-p)$ when $p<n$. When $p=n$, the embedding is compact into every finite $L^q$, and when $p>n$ it is compact into $C^0(\overline{U})$, hence into every $L^q$.

The boundedness of the domain is essential. Choose a nonzero $\varphi \in C_c^{\infty }(0,1)$ and setTranslation invariance gives $\Vert u_j{\Vert }_{W^{1,1}({\mathbb{R}}_{+})}=\Vert \varphi {\Vert }_{W^{1,1}}$, so after a fixed rescaling these functions lie in the unit ball. Their supports are pairwise disjoint andNo subsequence is Cauchy in $L^1$, so the unit ball is not compact. This is the standard failure of Rellich compactness on an unbounded domain.

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