Discrete compactness and uniform Poincare inequality for Nedelec fields

Put $B=\mathrm{\nabla }{H}_0^1$, closed by [F5]. For a graph-bounded sequence $v_h\in X_h$, choose the decompositions $v_h=z_h+\mathrm{\nabla }{q}_h$ from [F2], using AC on this countable family of nonempty decomposition sets. The sequence $z_h$ is bounded in $H_0^1$. By [F4], extract a subsequence on which all three components converge in $L^2$, using three successive subsequences. On this subsequence the lifts $w_h=(I-P_B)v_h=(I-P_B)z_h$ converge in $L^2$, since $I-P_B$ is a contraction.

The continuous kernel $K$ is closed in $L^2$: if $k_j\in K$ converges in $L^2$, it is Cauchy in the curl graph norm, its graph limit lies in $H_0(\mathrm{curl})$ by the closure definition, and its curl remains zero. Let $P_K$ be its orthogonal projection. A graph-bounded sequence $u_j\in Y:=H_0(\mathrm{curl})\cap K^{\perp }$ satisfies $u_j=(I-P_K)z_j$ by [F2], since $\mathrm{\nabla }{q}_j\in K$. Therefore the same componentwise Rellich argument proves compactness of $Y\hookrightarrow L^2$. If no continuous curl Poincare constant existed on $Y$, choose $u_j\in Y$ with $\mathrm{\parallel }{u}_j{\mathrm{\parallel }}_2=1$ and $\mathrm{\parallel }\mathrm{curl}{u}_j{\mathrm{\parallel }}_2<1\mathrm{/}j$. Compactness gives $u_j\to u$ in $L^2$ along a subsequence; their curls tend to zero, so graph closure puts $u\in K$, while the closed orthogonality conditions put $u\in K^{\perp }$ and $\mathrm{\parallel }u{\mathrm{\parallel }}_2=1$. This contradiction proves $\mathrm{\parallel }u{\mathrm{\parallel }}_2\le C_{\mathrm{\Omega }}\mathrm{\parallel }\mathrm{curl}u{\mathrm{\parallel }}_2$ on $Y$.

The gap estimate [F5], using the relative degree-zero commutation in [F1], gives $\mathrm{\parallel }{v}_h-w_h{\mathrm{\parallel }}_2\le \mathrm{\parallel }{\mathrm{\Pi }}_h^1{w}_h-w_h{\mathrm{\parallel }}_2\to 0$. For clarity, this last limit uses both strong convergence on each fixed field and the uniform $L^2$ operator bound; it does not follow from pointwise convergence alone on a moving sequence. The strong limit of $w_h$ need only lie in $L^2$: the bounded extensions converge on all $L^2$ by density, as proved in [F5]. Hence $v_h$ converges strongly along the same subsequence. This proves discrete compactness without any topological exactness assumption.

For $v_h\in Z_h$, set $w=(I-P_K)v_h\in Y$. Its curl equals that of $v_h$. This curl belongs to the next relative graph space: approximate $v_h$ by compactly supported smooth fields in the curl graph norm; their curls converge in $L^2$, have zero divergence, and are compactly supported, hence converge also in the divergence graph norm. Since the finite element spaces form the relative subcomplex, $\mathrm{curl}{v}_h$ belongs to its discrete degree-two target. Commutation and identity on that discrete next-degree space therefore imply $e=v_h-{\mathrm{\Pi }}_h^{1}w\in K_h$. Hence $(v_h,e)=0$, and $\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_2^2=(v_h,{\mathrm{\Pi }}_h^{1}w)\le M\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_2\mathrm{\parallel }w{\mathrm{\parallel }}_2\le M{C}_{\mathrm{\Omega }}\mathrm{\parallel }{v}_h{\mathrm{\parallel }}_2\mathrm{\parallel }\mathrm{curl}{v}_h{\mathrm{\parallel }}_2$. Division when $v_h\mathrm{\ne }0$ proves the mesh-uniform inequality on $Z_h$; the zero case is immediate. No limit in $h$ or finite-coarse-mesh exception is required for this inequality.

On the original contractible domain, [F3] implies $K_h=V_h\cap B$. If $v_h=\mathrm{\nabla }p\in K_h$ with $p\in H_0^1$, then on every tetrahedron $v_h=a+b\times x$ has curl $2b=0$, so it is constant. The scalar function $p-a\cdot x$ has zero weak gradient on the connected open tetrahedron; Zero weak gradient gives componentwise constants, whose AC hypothesis is included here, makes it constant almost everywhere there, so $p$ is affine on that cell. We establish the face values directly, without using a trace characterization. On an interior face $F$, integration by parts on its two adjacent cells shows that the distributional gradient of the piecewise-affine representative is its piecewise constant gradient plus the jump $[p]n_F$ times surface measure on $F$. A nonzero jump cannot be represented by an $L^2$ function: choose a smooth tangential bump $\eta$ supported in the relative interior of $F$ with ${\int }_F[p]\eta \mathrm{\ne }0$, choose a coordinate $i$ with $(n_F{)}_i\mathrm{\ne }0$, and test that component with $\eta (x^{\mathrm{\prime }})\chi (s\mathrm{/}\delta )e_i$, where $s$ is signed normal distance and $\chi$ is smooth with $\chi (0)=1$. The face pairing is a fixed nonzero number, while the pairing with any $L^2$ function tends to zero as $O(\sqrt{\delta })$ because the test support has volume $O(\delta )$. Since $p\in H^1$, its distributional gradient is $L^2$, so $[p]=0$ on $F$. For a boundary face, the closure definition of $H_0^1$ gives compactly supported smooth $p_j\to p$ in $H^1(\mathrm{\Omega })$; their zero extensions are Cauchy in $H^1({\mathbb{R}}^3)$ and, by the completeness of $H^1({\mathbb{R}}^3)$ under AC (Integer-order Sobolev spaces are Banach), converge to a zero extension $P$ of $p$. Apply the same face argument to $P$, whose exterior value is zero, to conclude $p{\mathrm{\mid }}_F=0$. These traces are affine polynomials, so their almost-everywhere equalities hold everywhere, including face vertices. At each mesh vertex, choose a sufficiently small neighborhood meeting only cells incident to that vertex. The matching tetrahedra triangulate this neighborhood in the closure of a Lipschitz three-manifold, whose local model is a ball at an interior vertex and a half-ball at a boundary vertex. Each incident tetrahedron is the cone from the vertex over its opposite face, so the cross-section of the star by a sufficiently small sphere is its simplicial link. The local ball or half-ball model makes that link a connected triangulated sphere or disk. Join interior points of any two link triangles by a path in that link, perturbing the path to avoid the finitely many link vertices and to cross edges transversely. Each successive pair of triangles shares a link edge, hence the corresponding tetrahedra share a face containing the original vertex. This establishes the required face chain. Equality of the affine values therefore propagates to every tetrahedron incident at the vertex. The common nodal values assemble a continuous piecewise affine function on $\bar{\mathrm{\Omega }}$, equal to $p$ almost everywhere, hence in $W_h^0$. The zero boundary-face values just proved imply all boundary nodal values vanish, so the assembled function belongs to $S_h$. Conversely $\mathrm{\nabla }{S}_h\subset K_h$. Consequently $K_h=B_h$, $Z_h=X_h$, and step 2.2 is the asserted inequality on $X_h$.

In general topology $B\subset K$ and $B_h\subset K_h$. Closedness of $B$ gives $K=B\oplus \mathcal{H}$, while finite-dimensional orthogonal decomposition gives $K_h=B_h\oplus {\mathcal{H}}_h$ and $Z_h=X_h\cap {\mathcal{H}}_h^{\perp }$. The unit ball of $\mathcal{H}$ is graph bounded and lies in $B^{\perp }$; [F2] and [F4] make it relatively compact by the argument in step 1.1. A Hilbert space with compact unit ball is finite dimensional: otherwise choose successive unit vectors orthogonal to their predecessors, obtaining pairwise distance $\sqrt{2}$ and contradicting compactness. The discrete harmonic space is finite dimensional as a subspace of $V_h$. Nonzero $a_h\in {\mathcal{H}}_h$ has $\mathrm{\parallel }{a}_h{\mathrm{\parallel }}_2>0$ but curl zero, disproving a curl Poincare bound on all $X_h$ whenever such a mode exists. Steps 1.1 and 2.1 still prove compactness of $X_h$ and step 2.2 proves the quotient inequality. The proof uses AC for selections and imported analytic suppliers, CC for the projection/Poincare interfaces, and CC/DC through the exact Rellich supplier; it asserts no choice-free version.