Research proof index

For a finite group of odd order and a three-term-progression-free set $s$, the normalized inner product between $\mu_s*\mu_s$ and the uniform measure on $2s$ is exactly $|s|^{-2}$. The identity records the precise normalized count forced by the absence of nontrivial three-term progressions.

For $x\in G\setminus G'$, summing the localized Carleson contribution over all tiles is exactly the same as summing the corresponding oscillatory kernel integral over the integer scales from $\sigma_1(x)$ to $\sigma_2(x)$. This is the identity that converts the discrete tile model back into the finitary operator.

Consider two tiles $p_1,p_2$ with $\mathfrak s(p_1)\le\mathfrak s(p_2)$ whose enlarged spatial balls overlap. If $x_i$ lies in the active set $E(p_i)$, then the separation of the two tile frequencies, measured at the scale of $p_1$, is controlled by the separation of the selected phases $Q(x_1)$ and $Q(x_2)$:

$1+d_{p_1}(\mathcal Q(p_1),\mathcal Q(p_2))\le C(a)\bigl(1+d_{x_1,D^{\mathfrak s(p_1)}}(Q(x_1),Q(x_2))\bigr).$

For two distinct forest tops $u_1,u_2$ with nested spatial cubes, and for a comparison cube $J$, the cutoff function $\chi_{u_1,u_2,J}$ is Lipschitz on $\mathcal I(u_1)$ at the scale of $J$:

$d\bigl(\chi(x),\chi(x')\bigr)\le C(a)\,\frac{d(x,x')}{D^{s(J)}}.$

The pairing of a tree boundary operator applied to $f$ with a test function $g$ is bounded by a sum over boundary cubes $J$. Each term integrates $|f|$ times a maximal function of $g$ over $J$, multiplied by a scale-decaying sum over grid cubes $I$ whose enlarged balls contain $J$ and whose scale is at least that of $J$.

For a tree $t(u)$ and one of its boundary cubes $L$, the measure of the part of $L\cap G$ covered by the active sets $E_p$ of tiles in the tree is at most

$C(a)\,\mathrm{dens}_1(t(u))\,\mu(L).$