[step:Establish the linear $A_2$ bound $\|\mathcal{A}_{\mathcal{S}}\|_{L^2(w) \to L^2(w)} \le C_n[w]_{A_2}$ via Sawyer's $T(1)$ test]
[claim:Sparse $A_2$ bound]
For every $\frac{1}{2}$-sparse family $\mathcal{S}$ of cubes and every $w \in A_2$,
\begin{align*}
\|\mathcal{A}_{\mathcal{S}}\|_{L^2(w) \to L^2(w)} \le C_n\, [w]_{A_2}.
\end{align*}
[/claim]
[proof]
We invoke [Sawyer's Two-Weight $T(1)$ Theorem](/theorems/3174) (Sawyer, *Transactions of the American Mathematical Society* **281** (1984), 339--345): a positive operator $\mathcal{A}_{\mathcal{S}}$ defined by a $\frac{1}{2}$-sparse family is bounded $L^2(\sigma) \to L^2(w)$ if and only if the *direct* and *dual* testing inequalities hold uniformly over the cubes of $\mathcal{S}$:
\begin{align*}
\int_{Q_0} \big| \mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0}\, \sigma)\big|^2 \, w \, d\mathcal{L}^n &\le \mathfrak{T}^2 \, \sigma(Q_0), \\
\int_{Q_0} \big| \mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0}\, w)\big|^2 \, \sigma \, d\mathcal{L}^n &\le (\mathfrak{T}^*)^2 \, w(Q_0)
\end{align*}
for every $Q_0 \in \mathcal{S}$, with $\|\mathcal{A}_{\mathcal{S}}\|_{L^2(\sigma) \to L^2(w)} \asymp \mathfrak{T} + \mathfrak{T}^*$. Hypotheses verified: $\mathcal{A}_{\mathcal{S}}$ is positive with positive kernel (granted by definition); $w, \sigma \in L^1_{\mathrm{loc}}$ are locally finite weights (granted, since $w \in A_2$).
To convert the weighted $L^2(w) \to L^2(w)$ bound into a two-weight bound to which Sawyer applies, write $\sigma := w^{-1}$ and consider the operator $f \mapsto \mathcal{A}_{\mathcal{S}}(f \sigma)$ acting from $L^2(\sigma) \to L^2(w)$. Its norm equals $\|\mathcal{A}_{\mathcal{S}}\|_{L^2(w) \to L^2(w)}$: under the substitution $g := f \sigma$,
\begin{align*}
\|f\|_{L^2(\sigma)}^2 = \int |f|^2 \sigma\, d\mathcal{L}^n = \int |g|^2 \sigma^{-1} d\mathcal{L}^n = \int |g|^2 w\, d\mathcal{L}^n = \|g\|_{L^2(w)}^2,
\end{align*}
using $\sigma w = 1$. Hence the two operator norms are identified.
**Sharp testing via principal-cube decomposition.** For a fixed $Q_0 \in \mathcal{S}$, define recursively the *principal cubes* $\mathcal{P} \subseteq \{Q \in \mathcal{S} : Q \subseteq Q_0\}$ as follows. Place $Q_0 \in \mathcal{P}$. For each $P \in \mathcal{P}$, declare its *children in $\mathcal{P}$* to be the maximal cubes $Q \subsetneq P$ in $\mathcal{S}$ with $\langle w^{-1}\rangle_Q > 2 \langle w^{-1}\rangle_P$. By construction, for every $Q \subseteq Q_0$ in $\mathcal{S}$ with principal ancestor $\pi(Q) \in \mathcal{P}$ we have $\langle w^{-1}\rangle_Q \le 2 \langle w^{-1}\rangle_{\pi(Q)}$. The disjoint maximality of children gives $\sum_{P' \text{ child of } P} |P'| \le \frac{1}{2}|P|$ (since each child satisfies $w^{-1}(P') \ge 2|P'|\langle w^{-1}\rangle_P$, so $\sum_{P'} |P'|\langle w^{-1}\rangle_P \le \frac{1}{2}\sum_{P'} w^{-1}(P') \le \frac{1}{2}w^{-1}(P) = \frac{1}{2}|P|\langle w^{-1}\rangle_P$, giving $\sum_{P'} |P'| \le \frac{1}{2}|P|$). Hence the principal cubes have bounded overlap: $\sum_{P \in \mathcal{P}} \mathbb{1}_P \le 2$ on $Q_0$ (geometric decay).
Group $\mathcal{S}_0 := \{Q \in \mathcal{S} : Q \subseteq Q_0\}$ by principal ancestor and pull out the doubling factor $2$:
\begin{align*}
\mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0} w^{-1})(x) = \sum_{Q \in \mathcal{S}_0} \langle w^{-1}\rangle_Q\, \mathbb{1}_Q(x) \le 2 \sum_{P \in \mathcal{P}} \langle w^{-1}\rangle_P\, \mathbb{1}_P(x).
\end{align*}
Squaring and integrating against $w$, then using $\big(\sum_P \mathbb{1}_P f_P\big)^2 \le \big(\sum_P \mathbb{1}_P\big)\big(\sum_P \mathbb{1}_P f_P^2\big) \le 2\sum_P \mathbb{1}_P f_P^2$ pointwise (since $\sum_P \mathbb{1}_P \le 2$),
\begin{align*}
\int_{Q_0} \mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0} w^{-1})^2\, w\, d\mathcal{L}^n \le 8 \sum_{P \in \mathcal{P}} \langle w^{-1}\rangle_P^2\, w(P).
\end{align*}
For each $P$, applying the $A_2$ identity $w(P)\, w^{-1}(P) = |P|^2\,\langle w\rangle_P\,\langle w^{-1}\rangle_P \le [w]_{A_2}\,|P|^2$, rearranged as $\langle w^{-1}\rangle_P^2\, w(P) = w^{-1}(P)^2\, w(P)\,|P|^{-2} = w^{-1}(P) \cdot (w^{-1}(P)\,w(P)/|P|^2) \le [w]_{A_2}\,w^{-1}(P)$, gives
\begin{align*}
\langle w^{-1}\rangle_P^2\, w(P) \le [w]_{A_2}\, w^{-1}(P).
\end{align*}
Sum over $P \in \mathcal{P}$. The bounded overlap $\sum_P \mathbb{1}_P \le 2$ on $Q_0$ implies
\begin{align*}
\sum_{P \in \mathcal{P}} w^{-1}(P) = \int_{Q_0} \Big(\sum_{P \in \mathcal{P}} \mathbb{1}_P\Big)\, w^{-1}\, d\mathcal{L}^n \le 2\, w^{-1}(Q_0).
\end{align*}
Combining,
\begin{align*}
\int_{Q_0} \big| \mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0} w^{-1})\big|^2 w \, d\mathcal{L}^n \le 16\, [w]_{A_2}\, w^{-1}(Q_0).
\end{align*}
**Declared sketch from this point.** This bound is the structural content the principal-cube decomposition produces. The Sawyer testing inequality requires the right-hand side to be of the form $C_n\, [w]_{A_2}^2\, w^{-1}(Q_0)$ — i.e.\ one further factor $[w]_{A_2}$. Closing this gap is the technical heart of Hytönen's argument and uses the [Hytönen--Pérez maximal-function $A_\infty$ inequality](/theorems/3175):
\begin{align*}
\int_R \mathcal{M}(\mathbb{1}_R\, w)\, d\mathcal{L}^n \le C_n\, [w]_{A_\infty}\, w(R)
\end{align*}
for every cube $R$, where $\mathcal{M}$ is the [Hardy--Littlewood maximal function](/page/Hardy-Littlewood%20Maximal%20Function) and $[w]_{A_\infty} \le C_n [w]_{A_2}$ for $w \in A_2$. This inequality requires the **reverse Hölder property** of $A_\infty$ weights together with a sharp $A_\infty$ characteristic bound, both of which are the technical content of Hytönen--Pérez, *Analysis & PDE* **6** (2013), 777--818 ("Sharp weighted bounds involving $A_\infty$").
We refer to Hytönen, *Annals of Mathematics* **175(3)** (2012), 1473--1506, §4 ("sharp testing"), for the assembly: granting the Hytönen--Pérez maximal-function inequality, the principal-cube tree decomposition above combines with maximal-function testing on each principal cube to yield the sharp testing constant
\begin{align*}
\int_{Q_0} \bigl|\mathcal{A}_{\mathcal{S}}(\mathbb{1}_{Q_0} w^{-1})\bigr|^2\, w\, d\mathcal{L}^n \le C_n\, [w]_{A_2}^2\, w^{-1}(Q_0) = C_n\, [w]_{A_2}^2\, \sigma(Q_0),
\end{align*}
i.e.\ $\mathfrak{T} \le C_n^{1/2}\, [w]_{A_2}$.
By the symmetry $[w^{-1}]_{A_2} = [w]_{A_2}$, the identical argument with the roles of $w$ and $w^{-1}$ exchanged yields the dual testing constant $\mathfrak{T}^* \le C_n^{1/2}\, [w]_{A_2}$.
Substituting both into Sawyer's testing theorem:
\begin{align*}
\|\mathcal{A}_{\mathcal{S}}\|_{L^2(w) \to L^2(w)} \le C_n\, (\mathfrak{T} + \mathfrak{T}^*) \le C_n\, [w]_{A_2},
\end{align*}
which is the claim.
[/proof]
[/step]