The broken comparison expression is a norm

Let $z\in X_h$ and suppose $\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{♯},h}=0$. Each nonnegative term in the finite sum is zero, so $\mathrm{\nabla }(z{\mathrm{\mid }}_K)=0$ on every cell and every interior and boundary facet jump vanishes. By [F2], each restriction $z{\mathrm{\mid }}_K$ is a constant $c_K$ almost everywhere. To see that the cell-adjacency graph is connected, choose interior points in any two cells and join them by a polygonal path in $\mathrm{\Omega }$ using [F3]. Its compact image has positive distance from $\mathrm{\partial }\mathrm{\Omega }$. Subdivide its segments if necessary to obtain at least one intermediate vertex; perturb each intermediate vertex in a small open ball so all joining segments stay in $\mathrm{\Omega }$. For a fixed endpoint $a$ off a mesh face $S$ of codimension at least two, candidate points in a bounded perturbation ball whose segment from $a$ meets $S$ lie in a truncated cone $\{a+t(s-a):s\in S,1\le t\le L\}$; $L<\mathrm{\infty }$ because $a$ is a positive distance from the compact face and the candidate ball is bounded. If $S$ has dimension $k\le d-2$, cover the bounded parameter set $S\times [1,L]$ by $O({\delta }^{-(k+1)})$ cubes of side $\delta$. The cone map is Lipschitz, so their images fit in $d$-cubes of side $C\delta$ with total volume $O({\delta }^{d-k-1})\to 0$. Bounded portions of facet hyperplanes are also null by the same covering count with $k=d-1$. Thus all forbidden sets are measurable null sets, and [F6] makes each finite union null. By [F5], a perturbation ball has positive measure and so contains a point outside that union. Choose the intermediate vertices successively outside the finitely many cones from the previous vertex and facet hyperplanes; for the last vertex also avoid the cones from the fixed endpoint at the end of the path. No segment meets the codimension-at-least-two skeleton or lies in a facet hyperplane, so cell transitions occur only through relative interiors of whole facets. When $d=1$, the mesh faces are endpoints and each crossing is itself a whole facet. The crossings give a cell-adjacency chain. Zero interior jumps therefore make all constants $c_K$ equal, while a zero boundary-facet jump makes the constant zero. Every cell is connected to that boundary cell, so $z=0$ in $L^2(\mathrm{\Omega })$. Thus the kernel is trivial.

Map $z\in X_h$ to the finite direct sum $Y={⨁}_K{L}^2(K;{\mathbb{R}}^d)\oplus {⨁}_F{L}^2(F)$ by $Jz=((\mathrm{\nabla }z{\mathrm{\mid }}_K{)}_K,(h_F^{-1\mathrm{/}2}[z{]}_F{)}_F)$. The sum of the component integral pairings is an inner product on $Y$, and its induced norm satisfies $\mathrm{\parallel }Jz{\mathrm{\parallel }}_Y=\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{♯},h}$. Cauchy--Schwarz [F7] and expansion of the squared norm give the triangle inequality; step 1.1 shows $J$ is injective, proving positive definiteness on $X_h$. If $z=u\in H_0^1(\mathrm{\Omega })$, its interior and boundary jumps vanish by the local trace theorem, so the expression equals $\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_{L^2(\mathrm{\Omega })}$, equivalent to the $H^1$ norm by [F4]. Finally, on the finite-dimensional space $V_h^{CR,0}$ the restriction is an inner-product norm; in its explicit face basis its Gram matrix is positive definite, so that norm is equivalent to the Euclidean coefficient norm and is complete. Thus $V_h^{CR,0}$ is Hilbert in the comparison norm.