Sobolev lower bound for continuous parameter selection

Define $p(t)=(4t(1-t){)}^{n+1}$ for $0\le t\le 1$ and $p(t)=0$ otherwise, and put $\psi (x)={\prod }_{r=1}^{d}p(x_r)$. The one-sided derivatives of $p$ of orders $0,\dots ,n$ vanish at each endpoint because the polynomial has a zero of multiplicity $n+1$ there. Hence the zero extension is $C^n(\mathbb{R})$, and $\psi \in C^n({\mathbb{R}}^d)$ is supported in $[0,1{]}^d$, is one at its center, and lies between zero and one. Set $B={\max }_{\mathrm{\mid }\alpha \mathrm{\mid }\le n}\mathrm{\parallel }{D}^{\alpha }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\ge 1.$ This is finite since the derivatives are continuous polynomials on the compact cube and zero outside it. Their classical derivatives are weak derivatives: integrate by parts successively in each coordinate against a compactly supported smooth test function; endpoint terms vanish because the derivatives below order $n$ match their zero extensions. This also verifies the weak derivative identity after every scaling and finite sum used below.

Set $N=W+1$ and $k=\lceil N^{1\mathrm{/}d}\rceil$. Partition $Q$ into $k^d$ closed coordinate cubes of side $1\mathrm{/}k$, and take the first $N$ in lexicographic order. Write $v_j$ for their lower vertices and put ${\varphi }_j(x)=B^{-1}{k}^{-n}\psi (k(x-v_j)),j=1,\dots ,N\mathrm{.}$ Distinct supports have disjoint interiors; all derivatives of orders at most $n$ vanish on their boundaries. Thus, for $u=(u_1,\dots ,u_N)\in [-1,1{]}^N$, the sum $T(u)={\sum }_j{u}_j{\varphi }_j$ satisfies $\mathrm{\parallel }{D}^{\alpha }T(u){\mathrm{\parallel }}_{\mathrm{\infty }}\le B^{-1}{k}^{\mathrm{\mid }\alpha \mathrm{\mid }-n}\mathrm{\parallel }{D}^{\alpha }\psi {\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. The weak derivative verification in step 1.1 proves $T(u)\in F_{d,n}$. Disjoint interiors, and evaluation at the centers, give the exact identity $\mathrm{\parallel }T(u){\mathrm{\parallel }}_{\mathrm{\infty }}=B^{-1}{k}^{-n}{\max }_j\mathrm{\mid }{u}_j\mathrm{\mid }\mathrm{.}$ In particular $T$ is linear, odd and continuous into $C(Q)$ with its uniform norm.

For $z\in S^{N-1}=S^W$, put $R(z)=z\mathrm{/}\mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}$. Its denominator is nonzero, and $R$ is continuous and odd with $\mathrm{\parallel }R(z){\mathrm{\parallel }}_{\mathrm{\infty }}=1$. Therefore $z\mapsto M(T(R(z)))$ is continuous into ${\mathbb{R}}^W$. By [F2], some antipodal pair has equal images. Setting $f=T(R(z))$, we obtain $M(f)=M(-f)$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=B^{-1}{k}^{-n}$.

Let $q=\eta (M(f))=\eta (M(-f))$. The triangle inequality yields $2B^{-1}{k}^{-n}=\mathrm{\parallel }f-(-f){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }f-q{\mathrm{\parallel }}_{\mathrm{\infty }}+\mathrm{\parallel }-f-q{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$ At least one of the two errors is at least $B^{-1}{k}^{-n}$. Because $k\le 2N^{1\mathrm{/}d}$, this is at least $c_{d,n}{N}^{-n\mathrm{/}d}$ with $c_{d,n}=2^{-n}{B}^{-1}>0$. This proves the asserted supremum bound without any regularity of $\eta$.

If every error is at most $\epsilon$, step 4.1 gives $W+1\ge (c_{d,n}\mathrm{/}\epsilon {)}^{d\mathrm{/}n}$. Since $W\ge 1$ implies $W+1\le 2W$, we obtain $W\ge \frac{1}{2}{c}_{d,n}^{d\mathrm{/}n}{\epsilon }^{-d\mathrm{/}n}$. The construction works at $W=1$, $d=1$, $n=1$, on cubes touching the boundary and when $N=k^d$, with exactly the same vanishing-derivative argument. AC is inherited from [F2] and subsumes the countable choice in [F1]; choosing finitely many indexed cubes or derivatives introduces no new choice principle.