A continuous traction field balanced on every piecewise $C^1$ region is given by a continuous stress matrix, and $t(-n)=-t(n)$

Given: The open set $O$, the continuous traction field $t$, the bounded continuous field $g$, and the balance hypothesis (B).

A bounded piecewise $C^1$ domain is a nonempty bounded open set with compact faces covering its boundary, a compact edge set of surface measure zero on each face, and, off the edges, a local $C^1$ graph description with the domain on one side; the boundary integral is the sum of the face surface integrals with the actual outward normals (Specified finite piecewise C1 boundary presentations, Surface integration on compact C1 hypersurfaces). Boxes and tetrahedra with closures in $O$ and vertices in general position are examples: their faces are compact planar triangles or rectangles, their edges form a compact set meeting each face in a segment of zero surface measure, and off the edges the boundary is locally a plane. The cube and tetrahedron constructions below therefore fall under (B).

For a continuous $h$ on a compact set $K$ and a compact surface piece $F$ contained in $K$ with finite area $S(F)>0$, the face average satisfies

$\mid \frac{1}{S(F)}{\int }_{F}hdS-h(y_0)\mid \le {\sup }_{y\in F}\mathrm{\parallel }h(y)-h(y_0){\mathrm{\parallel }}_2\mathrm{.}$

Indeed the definition of the surface integral and the order rules for the nonnegative Lebesgue integral bound ${\int }_F\mathrm{\parallel }h-h(y_0){\mathrm{\parallel }}_{2}dS$ by the supremum times $S(F)$, and the signed integral obeys the triangle inequality (Surface integration on compact C1 hypersurfaces, Monotonicity and nonnegative homogeneity of the nonnegative integral, The Lebesgue integral is linear on $L^1(\mu )$). If moreover all points of $F$ are within distance $\rho$ of $y_0$, then by uniform continuity of $h$ on a compact neighbourhood of $y_0$ the right-hand side tends to $0$ as $\rho \to 0$ (Heine-Cantor: a continuous map from a compact metric space to any metric space is uniformly continuous, Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

(Geometry of the two test bodies.) Fix an orthonormal basis $(n,p,q)$ of ${\mathbb{R}}^3$ and $h>0$.

1. The thin box $D_h:=y_0+\{an+bp+cq:\mathrm{\mid }a\mathrm{\mid }<h^2,\mathrm{\mid }b\mathrm{\mid }<h,\mathrm{\mid }c\mathrm{\mid }<h\}$ has two faces of area $4h^2$ with outward normals $\pm n$ and four lateral faces of area $4h^3$ with outward normals $\pm p,\pm q$; its volume is $8h^4$. These areas are the products of the corresponding side lengths $2h^2,2h,2h$.
2. Let $n\in S^2$ have no zero component, define $s_i:=1$ when $n_i>0$ and $s_i:=-1$ when $n_i<0$, and put $P_i:=y_0+(h\mathrm{/}{n}_i)e_i$, and let $T_h$ be the open tetrahedron with vertices $y_0,P_1,P_2,P_3$. Equivalently, in coordinates $v=y-y_0$ it is given by $s_i{v}_i>0$ and $n\cdot v<h$. Its coordinate face $F_i$ has outward normal $-s_i{e}_i$ and area $A\mathrm{\mid }{n}_i\mathrm{\mid }$, where $A:=h^2\mathrm{/}(2\mathrm{\mid }{n}_1{n}_2{n}_3\mathrm{\mid })$; its oblique face $F_0$ has outward normal $n$ and area $A$. Indeed the coordinate-face normal points out of $s_i{v}_i>0$, and the oblique-face normal points out of $n\cdot v<h$. The volume is $Ah\mathrm{/}3$. The reflection $v_i\mapsto s_i{v}_i$ is an isometry taking this tetrahedron to the positive-coordinate tetrahedron with intercepts $h\mathrm{/}\mathrm{\mid }{n}_i\mathrm{\mid }$, so its face areas are unchanged. Alternatively the cross-product formula for triangle area gives $\frac{1}{2}\mathrm{\parallel }(P_2-P_1)\times (P_3-P_1){\mathrm{\parallel }}_2=h^2\mathrm{/}(2\mathrm{\mid }{n}_1{n}_2{n}_3\mathrm{\mid })$, since ${\sum }_i{n}_i^2=1$ (The cross product is bilinear, alternating, and orthogonal to both factors, The squared cross-product norm is the Gram determinant of two vectors). The face opposite $P_i$ is a right triangle with legs $h\mathrm{/}\mathrm{\mid }{n}_j\mathrm{\mid },h\mathrm{/}\mathrm{\mid }{n}_k\mathrm{\mid }$, and the volume is one third of the oblique-face area times its distance $h$ from $y_0$.

The columns of a matrix $\sigma \in M_3(\mathbb{R})$ determine it, and $(\sigma n{)}_k={\sum }_i{\sigma }_{ki}{n}_i$. A map into $M_3(\mathbb{R})$ is continuous if and only if its three columns are continuous.