Degree-explicit polynomial trace inverse inequality on simplices

Fix $y\in F$ and let $q_y(t)=v((1-t)y+ta)$. This is a polynomial of degree at most $p$ in $t$, and $q_y(0)=v(y)$. Apply [F1] with $m=p$, then integrate the resulting inequality over $F$. In the map $(y,t)\mapsto (1-t)y+ta$, each tangent direction in $F$ is scaled by $1-t$, while the remaining component is the altitude $H_F$; hence the absolute volume Jacobian is $H_F(1-t{)}^{d-1}$, as stated in [F2]. By Fubini [F4], $\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2(F)}^2\le (p+1)(p+d){\int }_F{\int }_0^1(1-t{)}^{d-1}\mathrm{\mid }v((1-t)y+ta){\mathrm{\mid }}^{2}dtd{S}_y=\frac{(p+1)(p+d)\mathrm{\mid }F\mathrm{\mid }}{d\mathrm{\mid }K\mathrm{\mid }}\mathrm{\parallel }v{\mathrm{\parallel }}_{L^2(K)}^2$. This proves the exact reference-geometric inequality.

The cellwise normal derivative ${\mathrm{\partial }}_{n}v=n_F\cdot \mathrm{\nabla }v$ is a polynomial of degree at most $p-1$. If $p=0$ it is zero. If $p\ge 1$, repeat step 1.1 with degree $m=p-1$; the endpoint factor becomes $p(p+d-1)$.

In a shape-regular family, apply [F3] to the factor $\mathrm{\mid }F\mathrm{\mid }\mathrm{/}\mathrm{\mid }K\mathrm{\mid }$ in step 1.1 and in the same argument for ${\mathrm{\partial }}_{n}v$. This gives both $h_K^{-1}$ estimates. The only declared choice assumption is the Countable Choice required by [F3]; [F4] adds no choice hypothesis, and the linewise kernel uses finite-dimensional algebra.