Research proof index

Apply the optional stopping theorem to get equation 4. Note that T1 Space is needed to make sure that mesh ι n has order topology.

Let $g$ have bounded finite support, let $r>0$, and suppose $d(x,x')\le r$. The truncated Calderón-Zygmund operator changes by at most a constant times the global maximal function:

$d\bigl(T_r g(x),T_r g(x')\bigr)\le C(a)\,M g(x).$

Let $f$ be $2\pi$-periodic and belong to $L^q((0,2\pi])$ for some $q>1$. Given thresholds $\delta,\varepsilon>0$, there is an index $N_0$ such that the set where the tail error $\sup_{N>N_0}\lVert f(x)-S_Nf(x)\rVert$ exceeds $\delta$ has measure at most $\varepsilon$.

For a continuous, $2\pi$-periodic function $f$ and any $\delta,\varepsilon>0$, there is an index $N_0$ such that the set of $x\in(0,2\pi]$ for which $\sup_{N>N_0}\lVert f(x)-S_Nf(x)\rVert$ exceeds $\delta$ has measure at most $\varepsilon$.

If a tile $p$ lies in the $n$th minimal layer of a set of tiles $A$, then there is a tile $p'$ in the zeroth minimal layer with $p'\le p$, and the scale of $p$ is at least the scale of $p'$ plus $n$.

There is a measurable exceptional set $G' \subseteq G$ with $2\mu(G') \le \mu(G)$ such that, for every measurable $f$ bounded by $\mathbf{1}_F$, the integral over $G \setminus G'$ of the finitary oscillatory singular integral, summed only over the scales from $\sigma_1(x)$ to $\sigma_2(x)$, is at most $C(a,q)\mu(G)^{1-1/q}\mu(F)^{1/q}$.