Weighted simplex polynomial projection has explicit p-rates

Since ${\omega }_{\alpha }>0$ in the interior, a nonzero polynomial cannot have weighted square integral zero: continuity gives a ball on which its absolute value is bounded below by a positive number. Integrability follows from ${\alpha }_j\ge 0$ on the bounded simplex. Hence the weighted inner product is positive definite on each finite-dimensional polynomial space. Orthogonal projection onto ${\mathbb{P}}_{n-1}$ therefore gives the finite-dimensional orthogonal decomposition ${\mathbb{P}}_n=E_0\oplus \cdots \oplus E_n$. The scale in the Statement is the completion of this orthogonal-sum norm, so its coefficient description and truncation are well-defined.

For a finite sum $u=\sum u_n$, orthogonality gives $\mathrm{\parallel }u-P_{p}u{\mathrm{\parallel }}_{{\mathcal{H}}_{\alpha }^{\mathrm{\ell }}}^2={\sum }_{n>p}(1+{\mu }_n{)}^{\mathrm{\ell }}\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{\alpha }^2\le (1+{\mu }_{p+1}{)}^{\mathrm{\ell }-m}{\sum }_{n>p}(1+{\mu }_n{)}^m\mathrm{\parallel }{u}_n{\mathrm{\parallel }}_{\alpha }^2$, because ${\mu }_n=n(n+d+\mathrm{\mid }\alpha \mathrm{\mid })$ is increasing for $n\ge 0$. Taking square roots proves the tail estimate for finite sums.

Truncation is contractive in every coefficient norm ${\mathcal{H}}_{\alpha }^s$, and finite sums are dense by the definition of the completion. The finite-sum estimate in step 2.1 therefore passes to every $u\in {\mathcal{H}}_{\alpha }^m$. If $q\in {\mathbb{P}}_p$, each degree layer has $n\le p$, so $(1+{\mu }_n{)}^{m-\mathrm{\ell }}\le (1+{\mu }_p{)}^{m-\mathrm{\ell }}$ term by term; summing proves the inverse estimate. Neither argument invokes a derivative norm, which is why neither implies an unweighted Sobolev estimate.