Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 23F (Analysis of Functions)

For $s>\frac{1}{2}$, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

Take first $u$ in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,The Cauchy-Schwarz inequality, with weights $1+|{\xi }^{\prime }{|}^2+{\xi }_n^2$, givesAfter the substitution ${\xi }_n=(1+|{\xi }^{\prime }{|}^2{)}^{1/2}t$, the second factor iswhere finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|{\xi }^{\prime }{|}^2{)}^{s-1/2}$ and integrating in ${\xi }^{\prime }$ yieldsDensity of the Schwartz space in the Sobolev space completes the unique bounded extension.

For $1\le p<\infty$, choose nonzero $\varphi \in C_c({\mathbb{R}}^{n-1})$ and $\psi \in C_c(\mathbb{R})$ with $\psi (0)=1$. DefineThese functions are continuous and belong to the Lebesgue space $L^p({\mathbb{R}}^n)$, while their restrictions to the hyperplane are all the same:Changing variables in the normal coordinate givesIf the claimed bounded $T$ existed, thena contradiction. This scaling argument is the failure of an Lp hyperplane trace.

Compact support and the fundamental theorem of calculus give, for every $(x,y)$,whereApplying the same argument in the other coordinate gives

Multiplication and Fubini's theorem now yieldFinally, the Cauchy-Schwarz inequality givesThus the required estimate holds with $C=4$; it is the Ladyzhenskaya inequality in two dimensions.