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Let $U\subseteq D$ be open. Regard $H_{\mathrm{supp}}=H_0^1(U)$ as functions on $D$ by zero extension of H01, and let $H_{\mathrm{harm}}$ consist of the weakly harmonic Sobolev functions on $U$ belonging to $H_0^1(D)$. For $h\in H_{\mathrm{harm}}$, the Dirichlet inner product $(h,\varphi {)}_{\nabla }$ vanishes for each test function supported in $U$. Approximate any $u\in H_{\mathrm{supp}}$ by those test functions in the gradient norm and use the Cauchy-Schwarz inequality. This gives $\boxed{(h,u{)}_{\nabla }=0}$. The proof also works with the inhomogeneous zero-boundary Sobolev space convention, and with a homogeneous completion realized as weak functions.

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.

Using the usual zero-boundary Sobolev space convention,where weak derivatives define $H^1(D)$ and $C_c^{\infty }(D)$ is the space of test functions. In the Gaussian free field convention, the same notation often denotes the Dirichlet energy space, the completion in the gradient norm alone. On bounded domains the Poincaré inequality makes the two definitions equivalent; on unbounded domains one must distinguish them. The following gradient-pairing argument applies in either setting whenever the energy completion is realized as weak functions.

Identify $H_{\mathrm{supp}}=H_0^1(U)$ with a vector subspace of $H_0^1(D)$ by zero extension of H01. Approximating by test functions in $U$ shows that this is an isometric embedding in the inhomogeneous Sobolev norm and also in the Dirichlet inner product norm. In particular, arbitrary irregularity of $\partial U$ causes no additional boundary term.

DefineThese are weakly harmonic Sobolev functions. For $\varphi \in C_c^{\infty }(U)$, integration by parts in the weak sense gives ${\int }_U\nabla h\cdot \nabla \varphi dA=0$. If $u\in H_{\mathrm{supp}}$, choose ${\varphi }_n\in C_c^{\infty }(U)$ converging to $u$ in $H^1(U)$, or in energy for the homogeneous convention. The Cauchy-Schwarz inequality givesHence the orthogonality of supported and harmonic Dirichlet functions isBoth are linear vector subspaces. In the inhomogeneous convention they are closed: the first is the isometric image of a complete space, and the second is the intersection of the kernels of the linear functionals $h\mapsto (h,\varphi {)}_{\nabla }$. No spanning assertion is needed. The PDF contains this orthogonality statement; the TeX has badly corrupted it into an assertion about openness.

In three dimensions let $v\in H_0^1(B_R)$ solve $-\Delta v=f$, where $f\in L^{6/5}(B_R)$ is supported in $B_a$ and $0<a<R$. Write $M=\int f$ and $\bar{v}(s)=(4\pi {)}^{-1}{\int }_{S^2}v(s\omega )d\omega$. Outside $B_a$, the weak solution is a harmonic function. The weak formulation against radial test functions gives $-4\pi s^2{\bar{v}}^{\prime }(s)=M$ for $a<s<R$. The zero Sobolev trace gives $\bar{v}(R)=0$; integration proves the displayed formula, without assuming the source is a radial function.

Cauchy-Schwarz inequality on spheres gives $\Vert v{\Vert }_2^2\ge M^2(4\pi {)}^{-1}{\int }_a^R(1-s/R{)}^{2}ds$. For fixed nonzero $M$ and fixed $a$, this grows linearly in $R$. Such solutions on expanding balls cannot remain bounded in the full $H^1({\mathbb{R}}^3)$ norm after zero extension of H01.