Local inverse trace bound for a polynomial normal flux

Apply [F1] to each component ${\mathrm{\partial }}_j{v}_h$, which has degree at most $k-1$. Summing the $d$ squared inequalities and using [F3] gives $\mathrm{\parallel }{\mathrm{\partial }}_{n_{K,F}}{v}_h{\mathrm{\parallel }}_{L^2(F)}^2\le \mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2(F)}^2\le C_0^2{h}_K^{-1}\mathrm{\parallel }\mathrm{\nabla }{v}_h{\mathrm{\parallel }}_{L^2(K)}^2$ for a finite $C_0$ depending only on $d,k$ and the shape bound.

Multiply step 1.1 by $h_F$ and use $h_F\mathrm{/}{h}_K\le C$ from [F2]. Choose $C_{\mathrm{t}\mathrm{r}}:=\max \{1,C_0\sqrt{C}\}$; this is positive, finite, uniform over the family, and proves the required bound.

If the family contains a cell, fix one of its facets and test the inequality with the affine polynomial $p(x):=n_{K,F}\cdot x$, which belongs to ${\mathbb{P}}_k(K)$ because $k\ge 1$. Then $\mathrm{\nabla }p=n_{K,F}$ and ${\mathrm{\partial }}_{n_{K,F}}p=1$, so $h_F\mathrm{\parallel }{\mathrm{\partial }}_{n_{K,F}}p{\mathrm{\parallel }}_F^2=(\mathrm{\mid }K\mathrm{\mid }\mathrm{/}\mathrm{\mid }F\mathrm{\mid })\mathrm{\mid }F\mathrm{\mid }=\mathrm{\mid }K\mathrm{\mid }=\mathrm{\parallel }\mathrm{\nabla }p{\mathrm{\parallel }}_K^2>0$. Thus, for a family containing a cell, every admissible positive constant satisfies $C_{\mathrm{t}\mathrm{r}}^2\ge 1$. For an empty family the bound is vacuous; in either case step 2.1 selects $C_{\mathrm{t}\mathrm{r}}\ge 1$.