Sobolev threshold for Lagrange point evaluations on a reference simplex

Suppose $p<\mathrm{\infty }$ and $\mathrm{\ell }>d\mathrm{/}p$. Then $\mathrm{\ell }p>d$, so apply [F2] with $k=\mathrm{\ell }$, $n=d$, and extension domain [F1]. It gives a continuous representative on the closed simplex and a bound for its supremum by $C\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{\mathrm{\ell },p}}$. Two continuous representatives of the same almost-everywhere class agree everywhere: if their difference were nonzero at a point of the closed simplex, continuity on the closure and the fact that the simplex is the closure of its interior would give a nonempty interior ball on which the difference stays nonzero, contradicting almost-everywhere equality. Thus every $u(x_j)$ is uniquely defined and obeys the displayed bound.

If $p=\mathrm{\infty }$, choose any finite $q>d$. Since the simplex has finite volume, [F4] gives $\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^q}\le \mathrm{\mid }{\widehat{S}}_d{\mathrm{\mid }}^{1\mathrm{/}q}\mathrm{\parallel }{D}^{\alpha }u{\mathrm{\parallel }}_{L^{\mathrm{\infty }}}$ for $\mathrm{\mid }\alpha \mathrm{\mid }\le 1$, so $W^{\mathrm{\ell },\mathrm{\infty }}\subset W^{1,q}$ continuously. The same simplex extension result [F1] and embedding [F2], now with $k=1$ and $q>d$, give a unique continuous representative and the required bound by $\mathrm{\parallel }u{\mathrm{\parallel }}_{W^{\mathrm{\ell },\mathrm{\infty }}}$. In either branch, evaluation is linear because a linear combination of continuous representatives is the unique continuous representative of the corresponding Sobolev class. On a polynomial it is the ordinary nodal value, so these bounded maps extend any prescribed nodal degrees of freedom as required by [F5], without asserting unisolvence of an arbitrary node list; for $N\ge 1$, the nodes have a common bound by taking the maximum of their constants, while for $N=0$ the assertion is vacuous and any finite constant, such as $C=1$, suffices.