Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 3 / 22F (Analysis of Functions)

A weak solution is a function $u\in H_0^1(U)$ satisfyingThe zero boundary condition is encoded by membership in $H_0^1(U)$.

On the Hilbert space $H_0^1(U)$ defineThe form is bounded and coercive:The functional ${\ell }_f$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution and a boundThus $T:L^2(U)\to H_0^1(U)$ is bounded. Composing it with the compact Rellich-Kondrashov compactness theorem for H01 embeddingshows that $T:L^2(U)\to L^2(U)$ is the compact massive-Laplacian resolvent.

Extend $u_k$ and $u$ by zero outside the bounded set $U$. Weak convergence makes $(u_k)$ bounded in $L^2(U)$. For each fixed $p$,because $e^{-ip\cdot x}\in L^2(U)$.

Fix $t>0$. On the finite-measure ball $|p|\le t$, the Fourier transforms are uniformly bounded byDominated convergence therefore givesOn the complementary region,whose integral tends to zero uniformly in $k$ as $t\to \infty$ by hypothesis. Splitting first at large $t$ and then taking $k\to \infty$, the Plancherel theorem yieldsThis is the same low-frequency compactness mechanism used in the Fourier proof of Rellich compactness.

Taking the Fourier transform of the distributional equation givesbecause $2022$ is even. The multiplier therefore gives the regularity gain for one plus an even power of the Laplacian

The Sobolev embedding theorem in its derivative form says thatTo make all derivatives through order $4044$ continuous, and simultaneously make $f\in H^r$ continuous, it is enough and in general necessary to haveUnder this condition the distributional identity is a pointwise identity, so $u$ is a classical solution. ThusAt the borderline $n=2r$, the standard Sobolev embedding does not in general give continuity.