Estimate using Sobolev Embedding Theorem and Taylor Theorem Let $D \subset \mathbb{R}^{2n}$ be a smooth bounded domain.  Let $W^s(D)$ be the Sobolev space of functions with distributional derivatives up to order $s$ in $L^2(D)$ with norm $\|\cdot \|_s$.  Let $W_{0}^s(D)$ denote the $W^s$ closure of $C_0^\infty(D)$.
Let $s \geq 1$.  If $\phi \in W_0^{s+n}(D)$, then why does it follow from Taylor's theorem and the Sobolev embedding theorem that 
$$
|\phi(z)| \leq C\|\phi\|_{s+n}d(z)^s
$$
where $d(z)$ is the distance to the $\partial D$?
 A: If you fiddle with the parameters of the embedding theorem, in dimension $2n$ you should find something like $W^{s+n,2} \hookrightarrow C^s$, so $\phi$ is $s$ times classically differentiable, you can bound $\phi$ and its first $s$ derivatives  uniformly by $C\|\phi\|_{s+n}$, and $\phi \in W^{s+n,2}_0$ implies these derivatives all vanish on the boundary. 
Applying Taylor's theorem to order $s-1$ (thus with remainder estimate in terms of $\sup |D^s \phi|$, which we have controlled) at a point $y \in \partial D$ on the boundary should then give you $$|\phi(z)| \le  C\|\phi\|_{s+n} |z-y|^s,$$ so choosing $y$ to be the closest boundary point to $z$ gives you the desired estimate.
I've probably cheated a little bit here - to get $C^s$ I think you actually need $W^{s+n+\epsilon,2}$ for some $\epsilon>0$. Thus strictly speaking the embedding you actually get is $W^{s+n,2} \hookrightarrow C^{s-1,1}$; but this Lipschitz control of the first $s-1$ derivatives of $\phi$ should be enough to get the same estimate on the remainder if you're careful.
