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For a conforming triangular finite element mesh, takeThis is the quadratic conforming finite element space. Across a common edge of a polygon, the restrictions from neighboring triangles are univariate quadratic polynomials. Their agreement at the two endpoints and the midpoint makes their difference a degree-at-most-two polynomial with three distinct roots of a polynomial, so it vanishes on the entire common edge of a polygon. Thus matching the local finite element degrees of freedom produces the required continuity. Similarly, setting the three nodal values on every boundary edge carrying the Dirichlet boundary condition to zero makes its entire restriction vanish.

Continuity across element interfaces means the piecewise gradients give the global weak gradient, with no jump contribution, so $V_h\subset V$. There are finitely many local finite element degrees of freedom, hence $V_h$ is finite-dimensional. The Galerkin method seeksThe bounded bilinear form and coercive bilinear form estimates from part (b) remain valid on $V_h$. Since a finite-dimensional subspace is closed, the Lax-Milgram theorem gives exactly one $u_h$. Alternatively, the stiffness matrix plus the mass matrix is a positive-definite matrix in any basis of $V_h$. The same existence and uniqueness conclusion holds for any finite-dimensional vector subspace of this full space.

For $u,w\in X$, the forcing terms cancel, givingThe scalar function $s\mapsto s^7$ is strictly increasing on all of $\mathbb{R}$, so the second integrand is nonnegative. For the first one, use the quartic gradient monotonicity inequality. Put $d=\xi -\eta$ and $m=(\xi +\eta )/2$ for real vectors $\xi ,\eta$. Expanding givesConsequentlyThe last strict inequality uses the Poincaré inequality: if the weak gradient of $u-w\in W_0^{1,4}(\Omega )$ vanishes, then $u=w$. This proves that $A$ is a strictly monotone operator.

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.

First use weak compactness of bounded sets in the Hilbert space $L^2(\Omega )$ to pass to a subsequence with $u_k\rightharpoonup u$ globally in $L^2$. The weak lower semicontinuity of the Hilbert norm gives $\Vert u{\Vert }_{L^2(\Omega )}\le K$. This global step also applies if $\Omega$ is unbounded.

Choose a nested exhaustion $U_j⋐U_{j+1}⋐\Omega$ by bounded smooth open sets. Part (i) bounds $u_k$ in $H^1(U_j)$ for every $j$. By the Rellich-Kondrachov compactness theorem, $H^1(U_j)$ embeds compactly in $L^2(U_j)$. Weak compactness in $H^1(U_j)$ and a diagonal subsequence argument therefore give, for the same subsequence,The local limit agrees with the global weak $L^2$ limit. Using smooth exhaustion sets avoids imposing any unprinted boundary regularity on a general ${\Omega }^{\prime }$.

For a smooth compactly supported test function, the weak equation passes to the limit: the fixed bounded coefficients multiply fixed test functions, and local weak gradient convergence handles both the principal and drift terms. The zeroth-order term passes by local $L^2$ convergence. Thus $Lu=f$ weakly, $u\in H_{\mathrm{loc}}^1(\Omega )$, and the global bound already proved gives $u\in S_K$.

To improve convergence of the gradients, put $w_k=u_k-u$. Crucially the forcing is the same for all $k$, so $L{w}_k=0$. Given ${\Omega }^{\prime }⋐\Omega$, choose $j$ with $\overline{{\Omega }^{\prime }}\subset U_j$. Apply part (i) on $U_j$ to the homogeneous equation for $w_k$:This establishes strong local Sobolev compactness for a fixed elliptic equation and proves

No additional hypothesis on $F$ is needed here for scalar interior $C^{1,\beta }$ regularity. In the elliptic sense of uniform convexity, the assumptions already give constants $0<\lambda \le M<\infty$ with $\lambda I\le D^{2}F(p)\le MI$ for all $p$. The following argument gives interior gradient regularity for uniformly convex autonomous energies.

The first variation at the minimizer exists because $|DF(p)|\le C(1+|p|)$, and its weak Euler-Lagrange equation isFor a coordinate difference quotient ${\delta }_{h}u$, subtract the translated equation and the original equation. The averaged linearization of a nonlinear divergence-form equation givesThese measurable symmetric matrices satisfy $\lambda I\le A_h\le MI$ independently of $h$. The Caccioppoli inequality and the standard bound $\Vert {\delta }_{h}u{\Vert }_2\le \Vert D_{k}u{\Vert }_2$ bound $D{\delta }_{h}u$ on every smaller interior set. Quoting the Sobolev characterization by bounded difference quotients yields $u\in H_{\mathrm{loc}}^2$.

Now $DF$ is globally Lipschitz continuous, so the Sobolev chain rule and commutation of distributional derivatives allow differentiation of the weak equation. Each $v_k=D_{k}u\in H_{\mathrm{loc}}^1$ satisfiesThe De Giorgi-Nash-Moser theorem states that weak solutions of this scalar divergence-form equation with bounded measurable uniformly elliptic coefficients are locally Hölder continuous. It gives a common $\beta \in (0,1)$ depending only on $n$ and the ellipticity ratio $M/\lambda$, with $D_{k}u\in C_{\mathrm{loc}}^{0,\beta }$ for every $k$. A Sobolev function whose weak gradient is continuous has a $C^1$ representative, as follows by local mollification and integration of its gradient. HenceThe higher-regularity conclusion comes from differentiating an autonomous scalar equation; it does not rely on boundary regularity or a higher derivative such as $D^{3}F$.

There is a normalization error in the PDF. With the printed unnormalized integral, testing a constant function equal to one would give a left side $|U|(1-|U|{)}^2$ and a right side zero. Thus that formulation fails whenever $|U|\ne 1$.

Use instead the average $v_U=|U{|}^{-1}{\int }_{U}v$. The Poincare-Wirtinger inequality, also called the Neumann-Poincare inequality, isIf no $C_P$ exists, subtract the average and normalize a violating sequence to obtain $w_j\in H^1(U)$ with ${\int }_U{w}_j=0$, $\Vert w_j{\Vert }_2=1$, and $\Vert \nabla w_j{\Vert }_2\to 0$. This sequence is bounded in the Sobolev space $H^1(U)$. The Rellich-Kondrachov compactness theorem supplies a subsequence converging strongly in $L^2(U)$ to $w$.

For every compactly supported test function $\phi$, integration by parts and these convergences give $\int w{\partial }_i\phi =0$. Thus $w$ has zero weak gradient. The Sobolev function with zero weak gradient result and connectedness make $w$ constant. Strong $L^2$ convergence preserves its zero integral, so $w=0$. It also preserves its norm one, a contradiction. This proves the correctly normalized Neumann-Poincare inequality.

The constant function $v=1$ is an admissible $H^1(U)$ test for the Neumann Poisson problem. Its weak gradient is zero, so the weak identity immediately givesThis is necessary, and the preceding Hilbert-space construction proves sufficiency for $f\in L^2(U)$. It is the balance condition corresponding to zero total boundary flux.

For real vectors $\xi ,\eta$, put $m=(\xi +\eta )/2$ and $d=\xi -\eta$. Expansion givesThis proves strict operator monotonicity of the gradient of $\xi \mapsto |\xi {|}^4/4$. Integrating the inequality controls the difference of two weak gradients in the fourth power of their Lp norm, which yields uniqueness for suitable nonlinear Dirichlet boundary conditions.

Away from the origin, the radial derivative is the directional derivative along the outward radial unit vector. For a Sobolev space function it is the same scalar product with the weak gradient, defined almost everywhere. On an annulus, it has no singular direction at the origin.

On a connected open set, a Sobolev space function with zero weak gradient is constant almost everywhere. Local mollification gives constant smooth representatives on interior balls; overlaps and connectedness identify their constants. Without connectedness, the constant may differ between components. This explains the constant ambiguity in a Neumann Poisson problem.