Past exam of the functional analytic methods for partial differential equations course of the University of Oxford / 2023 / 1

Choose a nonnegative mollifier $\rho \in C_c^{\infty }({\mathbb{R}}^3)$ with $\int \rho =1$, and put ${\rho }_{\epsilon }(x)={\epsilon }^{-3}\rho (x/\epsilon )$. Sobolev approximation by mollifiers in the Sobolev space under consideration statesIn particular, it gives density of smooth functions in a Sobolev space. We prove convergence of both the function and all its first weak derivatives.

The usual smoothing properties of convolution give $u_{\epsilon }\in C^{\infty }$. For a test function $\varphi \in C_c^{\infty }({\mathbb{R}}^3)$ and $i\in \{1,2,3\}$, Fubini's theorem and the definition of a weak derivative giveThe integrals are absolutely convergent: the compact supports of ${\rho }_{\epsilon }$ and $\varphi$ restrict $y$ to a compact set, on which $u$ and ${\partial }_{i}u$ are integrable by Holder inequality. ConsequentlyThe approximate identity theorem in the Lp space $L^6$ applies separately to $u$ and ${\partial }_{i}u$. ThusEach convolution belongs to $L^6$ by Young's convolution inequality. With the componentwise Sobolev norm, the sixth power of the required difference is the sum of these four sixth powers, proving the boxed claim. Any equivalent Sobolev norm gives the same convergence.

Here Morrey's inequality has Hölder exponent $1-3/6=1/2$. Every $u\in W^{1,6}({\mathbb{R}}^3)$ has a unique continuous representative $u^{\ast }$ such thatWe use the Hölder norm $\Vert f{\Vert }_{C^{0,1/2}}=\Vert f{\Vert }_{\infty }+[f{]}_{C^{0,1/2}}$, where the Hölder seminorm is ${\sup }_{x\ne y}|f(x)-f(y)|/|x-y{|}^{1/2}$. The second bound is the homogeneous form of Morrey's inequality; the first includes control of the values themselves.

Let $u_j={\rho }_{1/j}\ast u$ be the mollifications from (i). They converge to $u$ in the Sobolev norm. Apply the assumed smooth version of Morrey's inequality to the differences $u_j-u_{\ell }$:Thus $(u_j)$ is a Cauchy sequence in the complete Hölder space $C^{0,1/2}({\mathbb{R}}^3)$ and converges there to a function $w$. In particular the convergence is uniform. On every bounded set $K$ of finite volume,Exhausting ${\mathbb{R}}^3$ by balls proves $w=u$ almost everywhere. Set $u^{\ast }=w$. Taking limits in the smooth Hölder norm bound gives the first boxed estimate. For any fixed $x\ne y$, taking limits ingives the second boxed estimate after taking the supremum. Finally, two continuous representatives equal almost everywhere must coincide everywhere, since a nonzero continuous difference would remain nonzero on an open ball of positive volume.

On the unbounded domain ${\mathbb{R}}^3$, translations already prevent compactness in every indicated Hölder space. Choose a bump function $\eta \in C_c^{\infty }(B_1)$ with $\eta (0)=1$, put $e_1=(1,0,0)$, and defineA change of variables in the Lebesgue integrals defining the Sobolev norm gives $\Vert u_m{\Vert }_{W^{1,6}({\mathbb{R}}^3)}=\Vert \eta {\Vert }_{W^{1,6}({\mathbb{R}}^3)}$, so the sequence is bounded. Each $u_m$ is a smooth function of compact support and belongs to every Hölder space $C^{0,\gamma }({\mathbb{R}}^3)$, $0<\gamma \le 1$: its bounded gradient gives the Lipschitz estimate for distances at most one, and its bounded values control larger distances.

For $m\ne \ell$, the supports are disjoint, and at $x=3m{e}_1$ we have $u_m(x)=1$ and $u_{\ell }(x)=0$. ThereforeNo subsequence can be a Cauchy sequence in any of these Hölder norms, hence none converges. This uses the translation mechanism for noncompactness of a Sobolev embedding by translation; concentration at the critical Hölder exponent alone would not rule out convergence at every smaller Hölder exponent.

Write $r=|x|$, $u(x)=\bar{u}(r)$, and let $g(r)$ be the radial profile of the radial derivative $\nabla u(x)\cdot x/|x|$, which may be assumed to be a radial function here. The spherical coordinate system givesSince $1\le r^2\le 4$, both $\bar{u}$ and $g$ belong to the Lp space $L^6(1,2)$. It remains to identify the weak derivative of the radial profile directly.

For $\varphi \in C_c^{\infty }(1,2)$, each component of $\varphi (r)x/r$ is a test function on $A$. Sum the three defining weak derivative identities, using the stated divergence formula:Given an arbitrary test function $\zeta \in C_c^{\infty }(1,2)$, choose $\varphi (r)=\zeta (r)/r^2$. Dividing by $4\pi$ gives $\int g\zeta =-\int \bar{u}{\zeta }^{\prime }$, exactly the definition of the weak derivative. We have provedThis radial Sobolev profile on an annulus estimate follows from the two weighted integrals above and equivalence of the usual Sobolev norms.

Use the continuous representative of $\bar{u}$ furnished by the one-dimensional Sobolev embedding theorem, and define $u^{\ast }(x)=\bar{u}(|x|)$ on $A$. The one-dimensional Sobolev representative is absolutely continuous; Holder inequality applied to its fundamental theorem of calculus givesAlso $\Vert \bar{u}{\Vert }_{\infty }\le \Vert \bar{u}{\Vert }_6+\Vert {\bar{u}}^{\prime }{\Vert }_6$: select a point where $|\bar{u}|$ is no larger than its $L^6$ norm on the unit-length interval and apply the same fundamental theorem of calculus. Thus the one-dimensional Hölder norm is controlled by $\Vert \bar{u}{\Vert }_{W^{1,6}(1,2)}$.

The two Hölder seminorms must be compared using the correct distances. By the reverse triangle inequality,Conversely, take $x=re$ and $y=se$ for any fixed unit vector $e$. Then $|x-y|=|r-s|$, so taking suprema yields equality of the two Hölder seminorms. Their supremum norms are equal as well. Part (i) now givesThe representative agrees with $u$ almost everywhere, and the assignment is linear and injective on the radial Sobolev space. This proves the requested continuous embedding. The inner radius being positive makes the radial profile Sobolev norm comparable to its weighted radial version.