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Choose continuous representatives using Morrey's inequality. On the closed unit ball they are uniformly bounded by $CM$ and satisfyThe Arzela-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