Research proof index

Let $1<q\le2$, let $q'$ be its Hölder conjugate, and suppose the linearized nontangential operators have the required uniform $L^2$ bound. For bounded measurable $F,G$ and measurable $f$ with $\lVert f(x)\rVert\le\mathbf 1_F(x)$, the supremum over all truncated scale intervals $-B\le s_1\le s_2\le B$ satisfies

$\int_G^+\sup_{s_1\le s_2}\lVert T_{s_1,s_2}f(x)\rVert\,dx\le C(a,q)\,\mu(G)^{1/q'}\mu(F)^{1/q}.$

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

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