Enflo's Walsh-block estimates and symmetry average

Given: The objects and hypotheses in the Statement.

${\sum }_{m=0}^{2n}{z}^m{F}_m(a)=(1-z{)}^{\mathrm{\mid }a\mathrm{\mid }}(1+z{)}^{2n-\mathrm{\mid }a\mathrm{\mid }}\mathrm{.}$

If $b$ is the coordinatewise complement of $a$, then each Walsh monomial of degree $m$ changes by $(-1{)}^m$, which proves item 4. The same generating polynomial also satisfies

$z^{2n}{P}_a(1\mathrm{/}z)=(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{P}_a(z),P_a(z):=(1-z{)}^{\mathrm{\mid }a\mathrm{\mid }}(1+z{)}^{2n-\mathrm{\mid }a\mathrm{\mid }}\mathrm{.}$

Comparing reciprocal coefficients gives

$F_{2n-m}(a)=(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{F}_m(a),$

and in particular $F_{n+1}(a)=(-1{)}^{\mathrm{\mid }a\mathrm{\mid }}{F}_{n-1}(a)$. [finite product, coefficient comparison]

$F_m(a)=\frac{2^{2n}}{2\pi }{\int }_{-\pi }^{\pi }(-i{)}^r{e}^{i(n-m)\theta }{\sin }^r(\theta \mathrm{/}2){\cos }^{2n-r}(\theta \mathrm{/}2)d\theta \mathrm{.}$

The reciprocal identity in step 2.1 gives $\mathrm{\mid }{F}_{n-1}(a)\mathrm{\mid }=\mathrm{\mid }{F}_{n+1}(a)\mathrm{\mid }$. Taking absolute values in the integral gives

$\mathrm{\mid }{F}_{n\pm 1}(a)\mathrm{\mid }\le \frac{2^{2n}}{\pi }{I}_r,I_r:={\int }_0^{\pi }{\sin }^r(\theta \mathrm{/}2){\cos }^{2n-r}(\theta \mathrm{/}2)d\theta \mathrm{.}$

Here $I_2=I_{2n-2}$ by the substitution $\theta \mapsto \pi -\theta$. When $n\ge 3$ and $2<r<2n-2$, put $\lambda =(2n-2-r)\mathrm{/}(2n-4)$. The integrand defining $I_r$ is

$({\sin }^2(\theta \mathrm{/}2){\cos }^{2n-2}(\theta \mathrm{/}2){)}^{\lambda }({\sin }^{2n-2}(\theta \mathrm{/}2){\cos }^2(\theta \mathrm{/}2){)}^{1-\lambda },$

so weighted Hölder gives $I_r\le I_2^{\lambda }{I}_{2n-2}^{1-\lambda }=I_2$. For a vector $e$ of weight two, the endpoint calculation in the same Cauchy formula gives $(2^{2n}\mathrm{/}\pi )I_2=\mathrm{\mid }{F}_n(e)\mathrm{\mid }$, and direct coefficient comparison gives

$\mathrm{\mid }{F}_n(e)\mathrm{\mid }=\mid \left(\genfrac{}{}{0px}{}{2n-2}{n}\right)-2\left(\genfrac{}{}{0px}{}{2n-2}{n-1}\right)+\left(\genfrac{}{}{0px}{}{2n-2}{n-2}\right)\mid =\frac{1}{2n-1}\left(\genfrac{}{}{0px}{}{2n}{n}\right)\le \frac{1}{n}\left(\genfrac{}{}{0px}{}{2n}{n-1}\right)$

for $n\ge 2$. The weights $r=1$ and $r=2n-1$ follow from item 2 and complementation; $r=2$ and $r=2n-2$ follow from the endpoint estimate. For $n=1$, the only intermediate weight is $r=1$, already covered by item 2. This proves item 3 in every case. [step 2.1, coefficient integral, weighted Hölder, binomial arithmetic]

Write the matrix of $\tilde{T}$ in the Walsh basis. For two distinct Walsh characters $v,w$, choose a translation $U_t$ for which $U_{t}v=v(t)v$ and $U_{t}w=w(t)w$ have opposite signs. Commutation with $U_t$ forces the $(v,w)$ matrix coefficient of $\tilde{T}$ to be its own negative, hence to vanish. Thus $\tilde{T}$ is diagonal in the Walsh basis. Coordinate permutations act transitively on each $W^{n\pm 1}$, so its diagonal coefficients have constant values ${\lambda }_{-}$ on $W^{n-1}$ and ${\lambda }_{+}$ on $W^{n+1}$. Consequently

$\tilde{T}f={\lambda }_{-}\frac{F_{n-1}}{\mathrm{\parallel }{F}_{n-1}{\mathrm{\parallel }}_{\mathrm{\infty }}}-{\lambda }_{+}\frac{F_{n+1}}{\mathrm{\parallel }{F}_{n+1}{\mathrm{\parallel }}_{\mathrm{\infty }}},$

and evaluation at zero gives

$\mathrm{\mid }\stackrel{}{\mathrm{Tr}}(W^{n-1},T)-\stackrel{}{\mathrm{Tr}}(W^{n+1},T)\mathrm{\mid }\le \mathrm{\parallel }\tilde{T}f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{.}$

[L1, finite group average, Walsh orthogonality]