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Symmetry implies that all eigenvalues of $A(x)$ lie between $\lambda$ and $\Lambda$, so $\Vert A(x){\Vert }_{\mathrm{op}}\le \Lambda$. In particular the bilinear form $B$ from part (i) is bounded and symmetric under the printed assumptions. The symmetric elliptic Dirichlet energy with a nonpositive potential isAll its terms are finite. For an admissible variation $\phi \in H_0^1$, its first variation is $B(u,\phi )+\int f\phi$. Thus stationarity is exactly the weak formulation of $Lu=f$.

Here is the direct method in the calculus of variations. Put $u=\psi +w$. Expanding the symmetric quadratic form givesThe Poincaré inequality controls $\Vert w{\Vert }_2$ by $\Vert Dw{\Vert }_2$. Therefore $J$ is bounded below on the admissible class, and a minimizing sequence is bounded in $H^1$. By weak compactness in this reflexive Banach space, a subsequence converges weakly to $u$. The affine closed subspace $\psi +H_0^1$ is weakly closed, so $u$ remains admissible.

The symmetric square root $A^{1/2}(x)$ is measurable and bounded. Since $-q\ge 0$, the two bounded linear mapstake weak $H^1$ convergence to weak $L^2$ convergence. The weak lower semicontinuity of the Hilbert norm shows that $B(u,u)$ is weakly lower semicontinuous. The term $\int fu$ is weakly continuous. Hence $J(u)\le \mathrm{liminf}J(u_k)$ and $u$ attains the minimum. Varying in either sign yields the weak equation.

The sign $q\le 0$ was used both to make the potential energy nonnegative, hence weakly lower semicontinuous by this argument, and to obtain coercivity. Finally $J(u+z)-J(u)=B(z,z)/2\ge \lambda \Vert Dz{\Vert }_2^2/2$ for $z\in H_0^1$, since the first variation at $u$ vanishes. Therefore

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

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.

Take the weak topology on the reflexive Banach space $\mathcal{U}$. A bounded linear operator is weak-to-weak continuous: $u_n\rightharpoonup u$ implies $⟨K{u}_n,v⟩\to ⟨Ku,v⟩$ for every $v\in \mathcal{V}$, by its adjoint operator. The weak lower semicontinuity of the Hilbert norm therefore makes $D(u)=\Vert Ku-f\Vert$ weakly sequentially lower semicontinuous.

This functional is finite everywhere and nonnegative. If it has coercivity, the direct method in the calculus of variations supplies a global minimizer $\bar{u}$. Squaring the residual does not change its global minimizers. Here the adjoint operator has codomain ${\mathcal{U}}^{\ast }$, since the source is a Banach space. Differentiating the squared residual along every real direction $h\in \mathcal{U}$ gives $\mathrm{Re}⟨K^{\ast }(K\bar{u}-f),h⟩=0$ in the dual pairing, hence the normal equation for a linear inverse problem $K^{\ast }(K\bar{u}-f)=0$. HenceThe converse fails. For $K:{\mathbb{R}}^2\to \mathbb{R}$, $K(x,y)=x$, and $f=1$, the datum is in the operator range, but $D(1,n)=0$ while $\Vert (1,n)\Vert \to \infty$. Thus admissible data do not imply coercivity of the residual; an unpenalized null space already prevents it.

If $x=0$, the claim is immediate. Otherwise weak convergence givesBy Cauchy-Schwarz,Taking the lower limit and dividing by $\Vert x\Vert$ proves the weak lower semicontinuity of the Hilbert norm:

The second estimate in part (a) bounds $({\omega }_m)$ in $H_{\mathrm{per}}^2$. The periodic Poisson equation $-\Delta {\Psi }_m={\omega }_m$ and the supplied curl identity then bound $({\Psi }_m)$ in $H_{\mathrm{per}}^4$ and $(u_m)$ in $H_{\mathrm{per}}^3$. After passing to a subsequence,The Rellich-Kondrachov compactness theorem also gives strong convergence in the corresponding spaces with one fewer derivative. In particular, $u_m\to u$ in $L^{\infty }$ and $\nabla {\omega }_m\to \nabla \omega$ in $L^2$, soPassing to the limit in the Galerkin equations givesThese identities have the claimed Sobolev regularity, and the first holds in $L_{\mathrm{per}}^2$. Finally, the weak lower semicontinuity of the Hilbert norm preserves the estimates