The broken gradient is a norm on the homogeneous Crouzeix–Raviart space

Suppose $\mathrm{\parallel }{\mathrm{\nabla }}_h{v}_h{\mathrm{\parallel }}_{L^2}=0$. Then the affine restriction $v_K$ has zero gradient on every cell and is constant $c_K$. If two cells share an interior facet, [F2] gives $\mathrm{\mid }F\mathrm{\mid }{c}_K=\mathrm{\mid }F\mathrm{\mid }{c}_{K^{\mathrm{\prime }}}$, hence $c_K=c_{K^{\mathrm{\prime }}}$.

The cell-adjacency graph is connected. Join interior points of any two cells by a polygonal path in $\mathrm{\Omega }$ using [F1]. Its compact image has positive distance from $\mathrm{\partial }\mathrm{\Omega }$. Subdivide the path if necessary to have intermediate vertices and perturb them in small open balls so every segment remains in $\mathrm{\Omega }$. For a fixed endpoint $a$ and a mesh face $S$ of codimension at least two, every endpoint $b$ for which $[a,b]$ meets $S$ lies in the proper affine subspace $\mathrm{aff}(\{a\}\cup S)$, of dimension at most $d-1$. Also avoid each facet hyperplane so no perturbed segment lies in one. A finite union of proper affine subspaces cannot cover an open ball: each has empty interior, so successively choose a smaller nonempty open ball disjoint from each one. Choose intermediate vertices successively outside the finitely many forbidden subspaces from the previous vertex; for the last vertex also avoid those from the fixed endpoint at the end of the path. Thus no segment meets the codimension-at-least-two skeleton or lies in a facet hyperplane, so every cell transition is through a relative interior of a whole facet. When $d=1$ mesh faces are endpoints and each crossing is itself a whole facet. This gives an adjacency chain, so all cell constants in step 1.1 agree. Since the bounded mesh has a boundary facet and its face mean is zero by [F2], the common constant is zero.

The broken gradient is linear, and its $L^2$ norm is nonnegative and absolutely homogeneous. For the triangle inequality, expand $\mathrm{\parallel }{\mathrm{\nabla }}_h(v_h+w_h){\mathrm{\parallel }}_2^2$ using the $L^2$ inner product; [F3] bounds the cross term by $\mathrm{\parallel }{\mathrm{\nabla }}_h{v}_h{\mathrm{\parallel }}_2\mathrm{\parallel }{\mathrm{\nabla }}_h{w}_h{\mathrm{\parallel }}_2$, so the square is at most $(\mathrm{\parallel }{\mathrm{\nabla }}_h{v}_h{\mathrm{\parallel }}_2+\mathrm{\parallel }{\mathrm{\nabla }}_h{w}_h{\mathrm{\parallel }}_2{)}^2$. Taking nonnegative square roots gives the triangle inequality. Step 2.1 proves that its kernel is $\{0\}$; hence it is a norm.