Enflo's Walsh-block assembly

Given: The objects and hypotheses in the Statement.

$n_m=\lfloor {\alpha }^m\rfloor ,t_m=\left(\genfrac{}{}{0px}{}{2n_m}{n_m-1}\right),k_m=\lfloor t_m^{\gamma }\rfloor \mathrm{.}$

After deleting finitely many initial indices, all are positive and strictly increasing. Let $K_m$ be the disjoint union of $k_m$ copies $K_{m,j}$ of ${\mathbb{Z}}_2^{2n_m}$, and let $B_1=({⨁}_{m}C(K_m){)}_2$. [explicit parameters]

$\mathrm{\mid }{N}_{m,i}\cap N_{m,j}\mathrm{\mid }\le \frac{2t_{m+1}}{n_{m+1}}(i\mathrm{\ne }j),$

$\mathrm{\mid }{N}_{m,j}\cap M_{m,i}\mathrm{\mid }\le \min \{\frac{t_{m+1}}{n_{m+1}},\frac{t_m}{n_m}\},$

and, with $\sigma (e)=\mathrm{\mid }\{j:e\in N_{m,j}\}\mathrm{\mid }$,

${\sum }_{e\in M_m}\mid \frac{1}{k_m{t}_m}-\frac{\sigma (e)}{k_{m+1}{t}_{m+1}}\mid \le \frac{1}{n_{m+1}}\mathrm{.}$

These are Enflo's conditions 4--6. Let $B$ be the closed span in $B_1$ of ${\bigcup }_m{M}_m$. The unique lowest nonzero block proves independence. To check [L4]'s property A, fix a finite combination $x={\sum }_e{a}_{e}e$, a participating generator $e_0$, and one component block $K_{r,i}$. If $e_0$ vanishes there, the componentwise estimate is trivial. Otherwise the nonzero restrictions of the participating generators are distinct Walsh characters: property (iii) handles characters from the same $M_m$, and the two possible adjacent layers have different degrees. Walsh orthogonality makes the normalized $L^2$ norm of $x{\mathrm{\mid }}_{K_{r,i}}$ at least $\mathrm{\mid }{a}_{e_0}\mathrm{\mid }$, so its supremum norm is at least $\mathrm{\mid }{a}_{e_0}\mathrm{\mid }$ as well. Taking the maximum over $i$ for each $C(K_r)$ coordinate and then the Hilbertian sum over $r$ gives $\mathrm{\parallel }x\mathrm{\parallel }\ge \mathrm{\mid }{a}_{e_0}\mathrm{\mid }\mathrm{\parallel }{e}_0\mathrm{\parallel }$. Thus property A holds with the full generator norm, including both adjacent nonzero coordinates. Countably many generators give separability, and [A1] is used through [L3] to make the closed subspace $B$ reflexive. [A1, L3, L4, steps 2.1, 3.1]

$\mid \stackrel{}{\mathrm{Tr}}(M_m,T)-\frac{1}{k_{m+1}}{\sum }_{j=1}^{k_{m+1}}\stackrel{}{\mathrm{Tr}}(N_{m,j},T)\mid \le \frac{\mathrm{\parallel }T\mathrm{\parallel }}{n_{m+1}}\mathrm{.}$

Indeed the left side is the weighted sum of the diagonal coefficients $a(e)$, each bounded by $\mathrm{\parallel }T\mathrm{\parallel }$ via property A, and condition 6 is exactly the total weight error. [L4, step 4.1]

$\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel ={\max }_{p\in K_{m+1,j}}\mathrm{\mid }x(p)\mathrm{\mid }\mathrm{.}$

The vector supplied by [L2] therefore satisfies

$\mid \stackrel{}{\mathrm{Tr}}(N_{m,j},T)-\stackrel{}{\mathrm{Tr}}(M_{m+1,j},T)\mid \le \frac{2}{n_{m+1}}\frac{\parallel \text{\hspace{0.17em}⁣}\mid T^{\mathrm{\prime }}x\mid \text{\hspace{0.17em}⁣}\parallel }{\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel }\mathrm{.}$

Its values on every other $K_{m+1,i}$ have modulus at most $\mathrm{\mid }{N}_{m,i}\cap N_{m,j}\mathrm{\mid }\mathrm{/}{t}_{m+1}\le 2\mathrm{/}{n}_{m+1}$ by condition 4. Its values on $K_{m,i}$ and $K_{m+2,i}$ have modulus at most $1\mathrm{/}{n}_{m+1}$ by the two parts of condition 5, while [L2] gives $\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel \ge 2\mathrm{/}{n}_{m+1}$. Thus the middle block has sup norm $\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel$, each adjacent outer block has sup norm at most half of that, and all other blocks vanish. Since the ambient sum is Hilbertian, $\mathrm{\parallel }x\mathrm{\parallel }\le \sqrt{3\mathrm{/}2}\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel <2\parallel \text{\hspace{0.17em}⁣}\mid x\mid \text{\hspace{0.17em}⁣}\parallel$. Also $\parallel \text{\hspace{0.17em}⁣}\mid T^{\mathrm{\prime }}x\mid \text{\hspace{0.17em}⁣}\parallel \le \mathrm{\parallel }Tx\mathrm{\parallel }\le \mathrm{\parallel }T\mathrm{\parallel }\mathrm{\parallel }x\mathrm{\parallel }$. Hence

$\mathrm{\mid }\stackrel{}{\mathrm{Tr}}(N_{m,j},T)-\stackrel{}{\mathrm{Tr}}(M_{m+1,j},T)\mathrm{\mid }\le \frac{4\mathrm{\parallel }T\mathrm{\parallel }}{n_{m+1}}\mathrm{.}$

Averaging in $j$ and combining with step 5.1 yields

$\mathrm{\mid }\stackrel{}{\mathrm{Tr}}(M_{m+1},T)-\stackrel{}{\mathrm{Tr}}(M_m,T)\mathrm{\mid }\le \frac{5\mathrm{\parallel }T\mathrm{\parallel }}{n_{m+1}}\mathrm{.}$

[L2, steps 3.1, 4.1, 5.1]

$L_m=\lfloor \frac{k_m}{t_{m+1}}\rfloor ,{\nu }_m=\lfloor \frac{k_{m+1}}{L_m{t}_m}\rfloor ,$

and identify each Walsh layer with a cyclic group of the corresponding cardinality. Inside $\{1,\dots ,L_m{t}_{m+1}\}\times {\mathbb{Z}}_{t_m}$, enumerate

$(j,j\rho +k),\rho =0,1,2,\dots ,k=0,\dots ,t_m-1,j=1,\dots ,L_m{t}_{m+1},$

first by increasing $\rho$, then $k$, then $j$. Take the first $k_{m+1}$ successive blocks of $t_{m+1}$ points as $N_{m,1},\dots ,N_{m,k_{m+1}}$. Each such block meets every $M_{m,i}=\{i\}\times {\mathbb{Z}}_{t_m}$ in at most one point, so condition 5 holds eventually. [explicit lexicographic construction]

${\sum }_{e\in M_m}\mid \frac{1}{k_m{t}_m}-\frac{\sigma (e)}{k_{m+1}{t}_{m+1}}\mid \le \frac{2t_{m+1}}{k_m}+\frac{k_m{t}_m}{k_{m+1}{t}_{m+1}}\mathrm{.}$

Both terms are $o(1\mathrm{/}{n}_{m+1})$: their exponential orders are respectively $t_m^{\alpha -\gamma +o(1)}$ and $t_m^{(\gamma +1)(1-\alpha )+o(1)}$. Thus condition 6 holds after discarding finitely many indices. [step 1.1, balanced incidence count]

$t_m\mid \mu ({\rho }_1-{\rho }_2),\mu \ge \frac{t_m}{\mathrm{\mid }{\rho }_1-{\rho }_2\mathrm{\mid }}\ge \frac{t_m}{{\nu }_m}\mathrm{.}$

Moreover

$\frac{t_m}{{\nu }_m}\sim \frac{L_m{t}_m^2}{k_{m+1}}\sim \frac{k_m{t}_m^2}{t_{m+1}{k}_{m+1}}=t_m^{\gamma +2-\alpha (\gamma +1)+o(1)}\mathrm{.}$

The exponent is positive precisely because $\alpha <(2+\gamma )\mathrm{/}(1+\gamma )$, so eventually $t_m\mathrm{/}{\nu }_m>n_{m+1}$. The common first coordinates lie in an interval of length less than $t_{m+1}$ and are more than $n_{m+1}$ apart. There are therefore at most $1+t_{m+1}\mathrm{/}{n}_{m+1}\le 2t_{m+1}\mathrm{/}{n}_{m+1}$ of them. This proves condition 4. Together with steps 7.1--8.1, all three incidence conditions hold after a finite reindexing. [step 1.1, finite arithmetic count, Stirling estimate]