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The zero extension of H01 makes $v_m$ an element of $H^1({\mathbb{R}}^3)$. The chain rule and change of variables give the exact scaling identitiesIndeed $\nabla v_m(x)=r_m^{3/2}\nabla u_m(r_{m}x)$ inside the expanded ball, and there is no boundary jump because the Sobolev trace is zero.

First take the printed convergence assertion as an assumption, as the question requests. It would imply $\Vert \nabla v_m{\Vert }_2\to \Vert \nabla v{\Vert }_2>0$. The strict inequality follows because an $H^1({\mathbb{R}}^3)$ function with zero weak gradient is constant almost everywhere, and the only constant in $L^2({\mathbb{R}}^3)$ is zero; this contradicts the asserted positivity of $v$. If $u_m$ converged strongly in $H^1(B)$, its limit would have to be its weak limit zero from (iii), forcing $\Vert \nabla u_m{\Vert }_2\to 0$. The first scaling identity contradicts this. Thus the intended answer is

There is nevertheless a substantive printing error: the asserted strong convergence of $v_m$ in the full $H^1({\mathbb{R}}^3)$ is false for these positive sources. Here is a check that does not rely on any assertion about the limiting profile. Put $R_m=r_m^{-1}$ and $c=|E{|}^{-5/6}$. Rescaling the weak formulation givesThe spherical mean of a Poisson solution outside its source provides an exact obstruction. The total source is $M=c|E|=|E{|}^{1/6}>0$. Let $a_m(s)=(4\pi {)}^{-1}{\int }_{S^2}{v}_m(s\omega )d\omega$ be its spherical mean. On $1<s<R_m$ the function is a harmonic function. The radial flux identity, obtained by testing against radial test functions or by the divergence theorem, gives $-4\pi s^2{a}_m^{\prime }(s)=M$, while the zero boundary value gives $a_m(R_m)=0$. HenceCauchy-Schwarz inequality on each sphere, followed by spherical coordinates, now yieldsThus the printed sequence is not even bounded in the full $H^1$ norm. The result on Poisson convergence on expanding balls in a homogeneous Sobolev space gives the appropriate corrected convergence in the homogeneous Sobolev space ${\dot{H}}^1({\mathbb{R}}^3)$, which controls the gradient in $L^2$ and the function in $L^6$, without requiring it to belong to $L^2$.

For completeness, the desired nonconvergence of $u_m$ has a direct proof that needs neither the incorrect claim nor its correction. Choose a nonnegative nonzero test function $\eta \in C_c^{\infty }(E)$, which exists since $E$ is a domain containing the origin. The rescaled weak formulation and Cauchy-Schwarz inequality giveConsequently, for every $m$,Together with (iii), this rigorously proves failure of strong convergence for the actual sequence defined in the paper.

We use the homogeneous version of the Sobolev embedding theorem in three dimensions:Here homogeneous Sobolev space ${\dot{H}}^1$ is the completion of compactly supported smooth functions in the L2 norm of the gradient, identified with its $L^6$ representative. Apply this Sobolev inequality both to $\varphi$ and to its spatial derivatives.

LetThe Plancherel theorem identifies the L2 norm of the Hessian matrix with $\Vert \varphi {\Vert }_{{\dot{H}}^2}$, because ${\sum }_{i,j}{\xi }_i^2{\xi }_j^2=|\xi {|}^4$. In particular,Differentiating the defocusing semilinear wave equation gives $\square ({\partial }_i\varphi )=3{\varphi }^2{\partial }_i\varphi$. By the Holder inequality with exponents $3$ and $6$, the preceding Sobolev inequality, and conservation of the positive wave energy,Use the inhomogeneous wave energy estimate simultaneously for the three spatial derivatives. It givesThe Gronwall inequality therefore yields $B(t)\le D\exp (CEt)$. We may take the continuous, locally bounded functionThe constant is universal; the dependence on the initial data is only through $D$ and $E$. The H2 bound for the defocusing cubic wave equation holds on every existing smooth interval, without assuming the global conclusion. If $D=0$ or $E=0$, the zero solution satisfies the same estimate.

The wave map scaling symmetry is ${\varphi }_{\lambda }(t,x)=\varphi (t/\lambda ,x/\lambda )$. Its conserved wave map energy isThus energy is subcritical for $n=1$, critical for $n=2$, and supercritical for $n\ge 3$. The scaling-critical homogeneous Sobolev spaces for perturbations of a constant map are ${\dot{H}}^{n/2}\times {\dot{H}}^{n/2-1}$. These relations constitute wave map energy and criticality.

In three dimensions let $f\in L^{6/5}({\mathbb{R}}^3)$ have compact support. Let $v_R\in H_0^1(B_R)$ solve $-\Delta v_R=f$ and extend it by zero. There is a unique $v\in {\dot{H}}^1({\mathbb{R}}^3)$ with $\int \nabla v\cdot \nabla \varphi =\int f\varphi$ for all $\varphi \in {\dot{H}}^1$. Indeed Holder inequality and the homogeneous Sobolev inequality make the right side a continuous linear functional in the gradient norm, and the Riesz representation theorem applies to this Hilbert space.

The weak formulation says that $v_R$ is the orthogonal projection of $v$ onto the closed subspace consisting of zero extensions of $H_0^1(B_R)$. These subspaces increase with $R$ and their union contains every test function, which is dense by the definition of the homogeneous Sobolev space. The best-approximation property of orthogonal projection therefore gives $\Vert \nabla (v_R-v){\Vert }_2\to 0$. The homogeneous Sobolev inequality also gives $\Vert v_R-v{\Vert }_6\to 0$. This convergence does not generally extend to the inhomogeneous Sobolev space $H^1$, as the spherical mean of a Poisson solution outside its source shows for a source of nonzero total mass.

Under ${\varphi }_{\lambda }(t,x)=\varphi (t/\lambda ,x/\lambda )$, the conserved wave map energy scales as ${\lambda }^{n-2}E$. Energy is subcritical for $n=1$, critical for $n=2$, and supercritical for $n\ge 3$. The critical derivative index of the homogeneous Sobolev space is $s_c=n/2$. Small energy alone is not an appropriate general small-data regularity hypothesis in supercritical dimensions.