Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 105 / 2

For an integer $s\ge 0$, the Sobolev space iswhere $D^{\beta }$ is a weak derivative. For $p<\infty$ one may use the Sobolev norm $({\sum }_{|\beta |\le s}\Vert D^{\beta }u{\Vert }_p^p{)}^{1/p}$; for $p=\infty$ use the maximum of the finitely many essential-supremum norms.

For $1\le p<d$, the Sobolev inequality is $\Vert u{\Vert }_{p^{\ast }}\le C_{d,p}\Vert \nabla u{\Vert }_p$, with Sobolev conjugate exponent $p^{\ast }=dp/(d-p)$. For $d<p<\infty$, Morrey's inequality supplies a continuous representative satisfyingAt $p=\infty$ the representative is Lipschitz continuous. At the critical exponent $p=d>1$, first-order Sobolev regularity gives every finite $L^q$ embedding for $q\ge d$, with an inhomogeneous norm, but generally no $L^{\infty }$ embedding. The one-dimensional endpoint $W^{1,1}(\mathbb{R})\hookrightarrow L^{\infty }(\mathbb{R})$ is an exception.

For the proof of Morrey's inequality, start with a smooth $u$ and write $u_{B(x,r)}$ for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables givesThe last step is the Holder inequality; integrability of the kernel to power $p^{\prime }$ is exactly $(d-1)p^{\prime }<d$. For $r=|x-y|$, translate the averaging ball along the segment from $x$ to $y$. The fundamental theorem of calculus along a line segment and the Holder inequality giveCombining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with $r=1$, together with $|u_{B(x,1)}|\le |B_1{|}^{-1/p}\Vert u{\Vert }_p$, gives the supremum estimate. Density of smooth functions in a Sobolev space then gives a uniformly convergent sequence of smooth representatives, preserving both bounds. For $p=\infty$, mollification gives the Lipschitz version.

For the decay conclusion assume $d<p<\infty$. The representative is uniformly continuous. If $|u(x_j)|\ge 2\theta >0$ along points escaping to infinity, the Hölder bound gives a radius $\rho >0$, independent of $j$, on which $|u|\ge \theta$. A subsequence has disjoint radius-$\rho$ balls, each contributing at least ${\theta }^p|B_{\rho }|$ to $\Vert u{\Vert }_p^p$, a contradiction. This is uniformly continuous integrable functions vanish at infinity.

The finite-$p$ restriction is necessary. If the printed range includes $p=\infty$, its decay assertion is false: $u\equiv 1$ belongs to $W^{1,\infty }$ but does not tend to zero.

The Arzelà-Ascoli theorem says that a uniformly bounded, equicontinuous family of continuous functions on a compact metric space is relatively compact in the uniform norm. Thus every sequence in that family has a uniformly convergent subsequence, and its limit is continuous.

On a bounded Lipschitz domain $\Omega \subset {\mathbb{R}}^d$, the Rellich-Kondrachov compactness theorem states that $W^{1,p}(\Omega )\hookrightarrow L^q(\Omega )$ is compact for $1\le p<\infty$ andwith $1/q=0$ for $q=\infty$. Thus $q<p^{\ast }$ when $p<d$, every finite $q$ is allowed when $p=d$, and the embedding into continuous functions with their uniform norm is compact when $p>d$.

For the main proof, a Sobolev extension operator puts a bounded sequence into a common compactly supported region of ${\mathbb{R}}^d$. The Sobolev fundamental theorem of calculus on lines gives the uniform translation estimate $\Vert u(\cdot +h)-u{\Vert }_p\le |h|\Vert \nabla u{\Vert }_p$. Convolution with a mollifier therefore approximates the sequence uniformly in Lp space. At any fixed smoothing scale, its derivatives and supremum are uniformly bounded, so the Arzelà-Ascoli theorem gives a convergent subsequence. A diagonal choice and the uniform approximation give strong Lp space convergence. The finite measure of $\Omega$ handles $q<p$, while Lp interpolation inequality with the bounded Sobolev embeddings upgrades this to the larger subcritical finite Lp spaces; for $p>d$, Morrey's inequality gives uniform equicontinuity directly and the Arzelà-Ascoli theorem gives uniform convergence. Boundedness of the domain is essential.

Choose continuous representatives using Morrey's inequality. On the closed unit ball they are uniformly bounded by $CM$ and satisfyThe Arzelà-Ascoli theorem therefore gives a uniformly convergent subsequence with a continuous limit. For $p=\infty$, use the uniform Lipschitz estimate instead; this part remains valid at that endpoint.

For weak lower semicontinuity of the Hilbert norm, weak convergence gives $⟨u_n-\varphi ,\varphi ⟩\to 0$, and henceTaking the lower limit proves $\boxed{\Vert \varphi \Vert \le \underset{n}{\mathrm{liminf}}\Vert u_n\Vert }$.

For the nonlinear integral, use local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Select a subsequence attaining the lower limit of the integrals. The Rellich-Kondrachov compactness theorem on successively larger balls gives a diagonal subsequence converging strongly in local L2 space to $\varphi$. After a further subsequence it converges almost everywhere. Since $1-\cos s\ge 0$, the Fatou lemma givesEach integral is finite because $0\le 1-\cos s\le s^2/2$. The argument uses local compactness rather than convexity: $1-\cos s$ is not a convex function on the whole real line.

Work with real-valued functions. The screened sine-Gordon energy is well defined: $0\le 1-\cos u\le u^2/2$, and $fu$ is integrable by the Cauchy-Schwarz inequality. It is coercive, becauseA minimizing sequence is bounded in the Hilbert space $H^1$. Weak sequential compactness of bounded sequences in a reflexive Banach space supplies a weakly convergent subsequence. The squared H1 space norm is weakly lower semicontinuous, the source pairing is weakly continuous, and the nonlinear term is covered by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. The direct method in the calculus of variations therefore gives a minimizer $\varphi$.

Taking its first variation in any $v\in H^1$ givesThus the Euler-Lagrange equation isas a weak solution, equivalently in distributions when tested against smooth compactly supported functions. The derivative of the nonlinear term is justified by $|\sin \varphi |\le |\varphi |$ and the second-order remainder bound $|\cos |\le 1$.

Now $G=f-\sin \varphi \in L^2$, so the supplied elliptic regularity estimate puts $\varphi$ in $H^2$. Applying the Sobolev inequality to $\varphi$ and each first weak derivative gives $\varphi \in W^{1,6}({\mathbb{R}}^3)$. Morrey's inequality and uniformly continuous integrable functions vanish at infinity prove that its continuous representative tends to zero.

The final printed supremum estimate is false in general. The maximum bound for a monotone reaction term involves $h(s)=s+\sin s$, which is odd and strictly increasing: $h^{\prime }=1+\cos s\ge 0$, and its zeros are isolated. If $M=\Vert f{\Vert }_{\infty }<\infty$, a positive maximum $m$ of $\varphi$ satisfies $h(m)\le f\le M$; apply the same argument to $-\varphi$. The valid general estimate isThe bound by $M$ is valid if $M\le \pi$, but $h(s)$ can be smaller than $s$ for larger positive $s$.

For an explicit failure of the source-size bound for the screened sine-Gordon equation, take $a=3\pi /2$, $L=10$, $\varphi (x)=a{e}^{-|x{|}^2/L^2}$ and $f=-\Delta \varphi +\varphi +\sin \varphi$. These are smooth functions in the required spaces. With $q=|x{|}^2/L^2$,Hence $\Vert f{\Vert }_{\infty }\le a-1+6a/L^2<a=\Vert \varphi {\Vert }_{\infty }$. Moreover $V(s)=s^2/2+1-\cos s$ has $V^{\prime \prime }=1+\cos s\ge 0$, so the energy is convex and this critical point is a minimizer. The example therefore satisfies even the minimizing and $C^2$ hypotheses of the printed claim.