Directional Poincaré bound for zero trace

If $\mathrm{\Omega }=\mathrm{\varnothing }$, [F3] makes each scalar Sobolev space the zero space, so its finite vector product and the closure of the sole compactly supported smooth field are $\{0\}$. Thus the closure convention in the Statement gives $H_0^1(\mathrm{\varnothing };{\mathbb{R}}^d)=\{0\}$, and all three displayed norms vanish. Henceforth assume $\mathrm{\Omega }\mathrm{\ne }\mathrm{\varnothing }$, so [F2] applies. For $v\in C_c^{\mathrm{\infty }}(\mathrm{\Omega };{\mathbb{R}}^d)$, extend $v$ by zero to the cube. For each component and fixed $x^{\mathrm{\prime }}=(x_2,\dots ,x_d)$, the fundamental theorem of calculus gives $v(x_1,x^{\mathrm{\prime }})={\int }_{-R}^{x_1}{\mathrm{\partial }}_{1}v(t,x^{\mathrm{\prime }})dt$. Cauchy–Schwarz bounds its squared norm by $2R{\int }_{-R}^R\mathrm{\mid }{\mathrm{\partial }}_{1}v(t,x^{\mathrm{\prime }}){\mathrm{\mid }}^{2}dt$.

Integrating the preceding bound over $x_1\in (-R,R)$ and $x^{\mathrm{\prime }}$ yields $\mathrm{\parallel }v{\mathrm{\parallel }}_2^2\le (2R{)}^2\mathrm{\parallel }{\mathrm{\partial }}_{1}v{\mathrm{\parallel }}_2^2$. The pointwise matrix norm inequality $\mathrm{\mid }{\mathrm{\partial }}_{1}v\mathrm{\mid }\le \mathrm{\mid }\mathrm{\nabla }v\mathrm{\mid }$ gives the second inequality.

By the closure in [F2], choose compactly supported smooth $v_n\to v$ in $H^1$. Then both $v_n\to v$ in $L^2$ and ${\mathrm{\partial }}_1{v}_n\to {\mathrm{\partial }}_{1}v$ in $L^2$, so the inequalities pass to the limit.