Poincare–Wirtinger inequality on a reference simplex

First take $u$ smooth. For $x,y\in {\widehat{S}}_d$, convexity keeps $z_t=tx+(1-t)y$ in the simplex. The segment identity and $\mathrm{\mid }x-y\mathrm{\mid }\le h:=\mathrm{diam}({\widehat{S}}_d)$ give $\mathrm{\mid }u(x)-u(y){\mathrm{\mid }}^p\le h^p{\int }_0^1\mathrm{\mid }\mathrm{\nabla }u(tx+(1-t)y){\mathrm{\mid }}^{p}dt,$ by [F2] and Jensen on the unit interval. Jensen again gives $\mathrm{\mid }u(x)-u_{{\widehat{S}}_d}{\mathrm{\mid }}^p\le \mathrm{\mid }{\widehat{S}}_d{\mathrm{\mid }}^{-1}{\int }_{{\widehat{S}}_d}\mathrm{\mid }u(x)-u(y){\mathrm{\mid }}^{p}dy\mathrm{.}$

Integrate the last bound over $x$. For $t\ge 1\mathrm{/}2$, fix $y$ and use $x\mapsto tx+(1-t)y$; its Jacobian is $t^d$, and its image is contained in the simplex, so its integral is at most $2^d{\int }_{{\widehat{S}}_d}\mathrm{\mid }\mathrm{\nabla }u{\mathrm{\mid }}^p$. For $t\le 1\mathrm{/}2$, fix $x$ and use $y\mapsto tx+(1-t)y$, whose inverse Jacobian is at most $2^d$, giving the same bound. Tonelli [F3] therefore yields $\mathrm{\parallel }u-u_{{\widehat{S}}_d}{\mathrm{\parallel }}_p^p\le 2^d{h}^p\mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_p^p\mathrm{.}$ Taking $p$th roots proves the estimate for smooth $u$.

For arbitrary $u\in W^{1,p}$ choose smooth $u_j\to u$ in $W^{1,p}$ by [F1]. Holder on the finite-measure simplex gives $\int u_j\to \int u$, so $(u_j{)}_{{\widehat{S}}_d}\to u_{{\widehat{S}}_d}$; also $\mathrm{\parallel }{u}_j-u{\mathrm{\parallel }}_p\to 0$ and $\mathrm{\parallel }\mathrm{\nabla }{u}_j{\mathrm{\parallel }}_p\to \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_p$. Passing to the limit in the smooth inequality of step 2.1 proves the statement.