$\int_{\ell(P_{i})^{2}}^{\ell(\hat{P}_{i})^{2}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\hat{P}_{i}}|T{\mathcal{Q}}_{t}^{(N)}(f)|^{q_{0}}\,\mathrm{d}\mu\right)^{\frac{1}{q_{0}}}\,\frac{\mathrm{d}t}{t}\leq\int_{0}^{\ell(\hat{P}_{i})^{2}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\hat{P}_{i}}|T{\mathcal{Q}}_{t}^{(N)}(f)|^{q_{0}}\,\mathrm{d}\mu\right)^{\frac{1}{q_{0}}}\,\frac{\mathrm{d}t}{t}.$

The difference is that the absence of the
$\mathbbm{1}_{(\lambda\hat{P}_{i})^{c}}$ term forces the sum, and in turn the
supremum, to start from $k_{0}=0$:

$\int_{\ell(P_{i})^{2}}^{\ell(\hat{P}_{i})^{2}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\hat{P}_{i}}|T{\mathcal{Q}}_{t}^{(N)}(f)|^{q_{0}}\,\mathrm{d}\mu\right)^{\frac{1}{q_{0}}}\,\frac{\mathrm{d}t}{t}\lesssim\sup_{k\geq
0}\lVert
f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(2^{k+1}\hat{P}_{i})}.$

But now, $\hat{P}_{i}\subseteq 2^{k+1}\hat{P}_{i}$ is true for all $k\geq 0$,
and therefore through 2.3 again one gets

$\sup_{k\geq 0}\lVert
f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(2^{k+1}\hat{P}_{i})}\lesssim\inf_{y\oldin\nolinebreak[4]\hat{P}_{i}}{\mathcal{M}}_{p_{0}}f(y).$

Hence, (4.9) along with the above inequalities give us that

(4.15)
$\lVert{\mathcal{T}}_{2}(P_{i})\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}}(P_{i})}\lesssim\eta\lVert
f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{0})}.$

Combining (4.14) and (4.15) with (4.12), we finally have that

$\lVert(T-T\Lambda_{P_{i}})f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}}(P_{i})}\lesssim\eta\lVert
f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{0})}.$

And substituting this back to (4.11), we get

(4.11’) $\sum_{i}\int_{P_{i}}|T(f-\Lambda_{P_{i}}f)\cdot
g|\,\mathrm{d}\mu\lesssim\eta\lVert
f\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{0})}\sum_{i}\mu(P_{i})\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{P_{i}}|g|^{q_{0}^{\prime}}\,\mathrm{d}\mu\Big{)}^{\frac{1}{q_{0}^{\prime}}}.$

For $q_{0}^{\prime}=1$ (or $q_{0}=\infty$),

$\sum_{i}\mu(P_{i})\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{P_{i}}|g|^{q_{0}^{\prime}}\,\mathrm{d}\mu\Big{)}^{\frac{1}{q_{0}^{\prime}}}=\sum_{i}\int_{P_{i}}|g|\,\mathrm{d}\mu=\int_{\cup_{i}P_{i}}|g|\,\mathrm{d}\mu\lesssim\mu(Q_{0})\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\lambda
Q_{0}}|g|\,\mathrm{d}\mu,$

since the $P_{i}$’s are disjoint and ${\cup_{i}P_{i}}\subseteq Q_{0}$.
Inputting this back to (4.11’), we get exactly (4.5).

For $q_{0}^{\prime}\oldin\nolinebreak[4](1,2)$ (or
$q_{0}\oldin\nolinebreak[4](2,\infty)$), first observe that
$\big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|g|^{q_{0}^{\prime}}\,\mathrm{d}\mu\big{)}^{\frac{1}{q_{0}^{\prime}}}\leq\inf_{y\oldin\nolinebreak[4]Q}{\mathcal{M}}_{q_{0}^{\prime}}g(y)$
for any $Q\oldin\nolinebreak[4]\mathscr{D}$. Then, since the $P_{i}$’s are
disjoint, we have that

$\displaystyle\sum_{i}\mu(P_{i})\Big{(}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{P_{i}}|g|^{q_{0}^{\prime}}\,\mathrm{d}\mu\Big{)}^{\frac{1}{q_{0}^{\prime}}}$
$\displaystyle\leq\sum_{i}\mu(P_{i})\inf_{y\oldin\nolinebreak[4]P_{i}}{\mathcal{M}}_{q_{0}^{\prime}}g(y)\leq\sum_{i}\int_{P_{i}}{\mathcal{M}}_{q_{0}^{\prime}}g(y)\,\mathrm{d}\mu(y)$
$\displaystyle=\int_{\cup_{i}P_{i}}{\mathcal{M}}_{q_{0}^{\prime}}g(y)\,\mathrm{d}\mu(y)\lesssim\mu(\cup_{i}P_{i})^{1-\frac{1}{q_{0}^{\prime}}}\lVert
g^{q_{0}^{\prime}}\rVert_{L^{1}(\cup_{i}P_{i})}^{\frac{1}{q_{0}^{\prime}}}$
$\displaystyle\leq\mu(Q_{0})\lVert
g\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(Q_{0})}\lesssim\mu(Q_{0})\lVert
g\rVert_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
Q_{0})}.$

For the inequality in the second line, we have used Kolmogorov’s inequality
and the boundedness of ${\mathcal{M}}_{q_{0}^{\prime}}$; in the third line, we
used the fact that $\sum_{i}\mu(P_{i})\leq\mu(Q_{0})$ and 2.3. This combined
with (4.11’) again gives us exactly (4.5).

Therefore, since both eqs. 4.4 and 4.5 are true, the assumptions of 4.3 are
fulfilled. Consequently, there exists a dyadic family $\\{Q_{1}^{j}\\}_{j}$ of
pairwise disjoint sets in $\mathscr{D}(Q_{0})$ with
$\sum_{j}\mu(Q_{1}^{j})\leq\frac{dK}{\eta}\mu(E)$ and such that

$\int_{Q_{0}}|\mathbf{T}(\Lambda_{Q_{0}}f)(x)\cdot g(x)|\,\mathrm{d}\mu(x)\leq
Cd^{3/2}|\llangle
f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{0};{\mathbb{C}}^{d})}\cdot\llangle
g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
Q_{0};{\mathbb{C}}^{d})}|+\sum_{j}\int_{Q_{1}^{j}}|\mathbf{T}(\Lambda_{Q_{1}^{j}}f)(x)\cdot
g(x)|\,\mathrm{d}\mu(x)$

for all $f\oldin\nolinebreak[4]L^{p}(M,\mu;{\mathbb{C}}^{d})$ and
$g\oldin\nolinebreak[4]L^{p^{\prime}}(M,\mu,{\mathbb{C}}^{d})$ whose
coordinates are supported on $\lambda Q_{0}$.

Now, we need to construct our sparse family. This can be done by following
Step 2 in the proof of Bernicot et al. almost without any changes; the fact
that here we are dealing with vector-valued functions does not affect their
argument. The conclusion is similar to [Hyt24, Corollary 5.4].

Here, we explain the iterative procedure with which one constructs the sparse
family. Initially, set $\mathscr{S}:=\\{Q_{0}\\}$. Then, we can rewrite the
above inequality in the following form

$\displaystyle\int_{Q_{0}}|\mathbf{T}(\Lambda_{Q_{0}}f)(x)\cdot
g(x)|\,\mathrm{d}\mu(x)$ $\displaystyle\leq Cd^{3/2}|\llangle
f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{0};{\mathbb{C}}^{d})}\cdot\llangle
g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
Q_{0};{\mathbb{C}}^{d})}|$
$\displaystyle+\sum_{j}\int_{Q_{1}^{j}}|\mathbf{T}(\Lambda_{Q_{1}^{j}}f^{j})(x)\cdot
g^{j}(x)|\,\mathrm{d}\mu(x),$

where (for all $j$) $f^{j}=\mathbbm{1}_{\lambda Q^{j}_{1}}f$, and
$g^{j}=\mathbbm{1}_{\lambda Q^{j}_{1}}g$; $f^{j}$ and $g^{j}$ are supported on
$\lambda Q_{1}^{j}$ (by the definition of $\Lambda_{P}$ and since the
integration is over $Q_{1}^{j}$).

Next, we can add the collection of all such $Q_{1}^{j}$ into $\mathscr{S}$ by
redefining $\mathscr{S}:=\mathscr{S}\cup\bigcup_{j}\\{Q_{1}^{j}\\}$, and then
iterate on each $Q_{1}^{j}$ to get

$\displaystyle\int_{Q_{1}^{j}}|\mathbf{T}(\Lambda_{Q_{1}^{j}}f^{j})(x)\cdot
g^{j}(x)|\,\mathrm{d}\mu(x)$ $\displaystyle\leq Cd^{3/2}|\llangle
f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q_{1}^{j};{\mathbb{C}}^{d})}\cdot\llangle
g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
Q_{1}^{j};{\mathbb{C}}^{d})}|$
$\displaystyle+\sum_{k}\int_{Q_{2}^{j,k}}|\mathbf{T}(\Lambda_{Q_{2}^{j,k}}f^{j,k})(x)\cdot
g^{j,k}(x)|\,\mathrm{d}\mu(x),$

and so on.

Eventually, as we show below, the remainder terms go to $0$ and we conclude
that

$\int_{Q_{0}}|\mathbf{T}f\cdot g|\,\mathrm{d}\mu\leq
Cd^{3/2}\eta\sum_{Q\oldin\nolinebreak[4]\mathscr{S}}\textmu(Q)\Big{|}\llangle
f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
Q;{\mathbb{C}}^{d})}\cdot\llangle
g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
Q;{\mathbb{C}}^{d})}\Big{|}.$

Following the argument in [BFP16] one needs to pick an $\eta$ large enough so
that the series $\sum_{\alpha\geq
1}\big{(}d\frac{K}{\eta}\big{)}^{\alpha}\leq\frac{dK}{\eta-dK}$ converges. For
such large $\eta$, the dyadic collection
$\mathscr{S}\subseteq\mathscr{D}(Q_{0})$ is $\varepsilon$-sparse with
$\varepsilon=1-\frac{dK}{\eta-dK}$.

All that is left to do is to check that the remainder term indeed goes to $0$.

Let ${\mathcal{G}}_{n}$ be the family of all cubes added at the initial sparse
collection $\\{Q_{0}\\}$ at the $n$-th step of our construction. The remainder
term at the $n$-th step is clearly

$\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|\mathbf{T}(\Lambda_{P}f)(x)\cdot
g(x)|\,\mathrm{d}\mu(x).$

By construction, $\lim_{n\rightarrow\infty}\mu(F_{n})=0$, where
$F_{n}:=\bigcup_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}P$.

Write now $f=(f_{1},\ldots,f_{d})$ and $g=(g_{1},\ldots,g_{d})$. Then,
similarly to the proof of 4.3 we have

$\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|\mathbf{T}(\Lambda_{P}f)(x)\cdot
g(x)|\,\mathrm{d}\mu(x)\leq\sum_{i=1}^{d}\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|T(\Lambda_{P}f_{i})(x)g_{i}(x)|\,\mathrm{d}\mu(x).$

Fix $i\oldin\nolinebreak[4]\\{1,\ldots,d\\}$. We show that

$\lim_{n\rightarrow\infty}\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|T(\Lambda_{P}f_{i})(x)g_{i}(x)|\,\mathrm{d}\mu(x)=0.$

Although this follows by a standard argument, we include the details for the
reader’s convenience.

Apply first Hölder’s inequality:

$\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|T(\Lambda_{P}f_{i})(x)g_{i}(x)|\,\mathrm{d}\mu(x)\leq\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\left(\int_{P}|T(\Lambda_{P}f_{i})(x)|^{p}\,\mathrm{d}\mu(x)\right)^{1/p}\left(\int_{P}|g_{i}(x)|^{p^{\prime}}\,\mathrm{d}\mu(x)\right)^{1/p^{\prime}}.$

By 2.6 (iii) and Equation 2.9, we deduce that the operators $\Lambda_{P}$,
$P\oldin\nolinebreak[4]\mathscr{D}(Q_{0})$ are $L^{p}(M,\mu)$-bounded,
_uniformly in the cube $P$_. Thus, since $T$ is also by assumption
$L^{p}(M,\mu)$ bounded we obtain

$\displaystyle\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\left(\int_{P}|T(\Lambda_{P}f_{i})(x)|^{p}\,\mathrm{d}\mu(x)\right)^{1/p}\left(\int_{P}|g_{i}(x)|^{p^{\prime}}\,\mathrm{d}\mu(x)\right)^{1/p^{\prime}}$
$\displaystyle\lesssim\lVert
T\rVert_{L^{p}(M,\mu)}\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\left(\int_{P}|f_{i}(x)|^{p}\,\mathrm{d}\mu(x)\right)^{1/p}\left(\int_{P}|g_{i}(x)|^{p^{\prime}}\,\mathrm{d}\mu(x)\right)^{1/p^{\prime}}.$

Applying Hölder’s inequality (for sums) and taking into account the
disjointness of the sets $P\oldin\nolinebreak[4]{\mathcal{G}}_{n}$, we get

$\displaystyle\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\left(\int_{P}|f_{i}(x)|^{p}\,\mathrm{d}\mu(x)\right)^{1/p}\left(\int_{P}|g_{i}(x)|^{p^{\prime}}\,\mathrm{d}\mu(x)\right)^{1/p^{\prime}}$
$\displaystyle\leq\left(\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|f_{i}(x)|^{p}\,\mathrm{d}\mu(x)\right)^{1/p}\left(\sum_{P\oldin\nolinebreak[4]{\mathcal{G}}_{n}}\int_{P}|g_{i}(x)|^{p^{\prime}}\,\mathrm{d}\mu(x)\right)^{1/p^{\prime}}$
$\displaystyle=\left(\int_{F_{n}}|f_{i}|^{p}\,\mathrm{d}\mu\right)^{1/p}\left(\int_{F_{n}}|g_{i}|^{p^{\prime}}\,\mathrm{d}\mu\right)^{1/p^{\prime}}.$

Since the functions $|f_{i}|^{p}$ and $|g_{i}|^{p^{\prime}}$ are
$\mu$-integrable and $\lim_{n\rightarrow\infty}\mu(F_{n})=0$, we deduce the
desired result. ∎

##  5\. Matrix weighted bounds for bilinear convex body sparse forms

In this section we obtain matrix weighted bounds for bilinear convex body
sparse forms of the type appearing in 1.2. Our main result is the following.

######  1.3.

Let $1\leq p_{0}<q_{0}\leq\infty$ and $p\oldin\nolinebreak[4](p_{0},q_{0})$.
Let $\mathscr{D}$ be a dyadic system in $M$ as in Subsection 2.2 and let
$\lambda\geq 2$. Let $0<\varepsilon<1$ and let
$\mathscr{S}\subseteq\mathscr{D}$ be an $\varepsilon$-sparse collection. Set

$t:=\frac{p}{p_{0}},\quad s:=\frac{q_{0}}{p},\quad
a:=\frac{t^{\prime}p}{t},\quad b:=s^{\prime}p.$

Let $W$ be a $d\times d$ matrix weight on $M$ such that
$[W^{s^{\prime}}]_{A_{a^{\prime},b}}<\infty$. Then, there exists a constant
$C=(\mu,d,p_{0},q_{0},p)$ such that for all
$f\oldin\nolinebreak[4]L^{p}_{W}(M,\mu)$ and all
$g\oldin\nolinebreak[4]L^{p^{\prime}}_{W^{\prime}}(M,\mu)$ the following
estimate holds:

$\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)|\llangle
f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
P)}\cdot\llangle
g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
P)}|\leq\frac{C}{\varepsilon}([W^{s^{\prime}}]_{A_{a^{\prime},b}})^{\alpha}\lVert
f\rVert_{L^{p}_{W}(M,\mu)}\lVert
g\rVert_{L^{p^{\prime}}_{W^{\prime}}(M,\mu)},$

where $W^{\prime}:=W^{-1/(p-1)}$ and

$\alpha:=\frac{1}{s^{\prime}(p-p_{0})}+\frac{1}{p^{\prime}}.$

After reviewing some notation and facts on the so-called _reducing operators_
in Subsection 5.1, which furnish a major tool for manipulating matrix weights
in estimates, we introduce the so-called _Muckenhoupt characteristics_ for
matrix weights in Subsection 5.2. Finally, we prove 1.3 in Subsection 5.3.

Further in the paper, we implement the following elementary facts without
special mention:

  1. (i)

For all $x_{1},\ldots,x_{k}\geq 0$, for all $a>0$, and for all positive
integers $k$, we have

(5.1) $(x_{1}+\ldots+x_{k})^{a}\simeq_{a,k}x_{1}^{a}+\ldots+x_{k}^{a}.$

In the rest of the text, we mostly apply this for $k=1,2,3,4$ or $d$ non-
negative numbers. Moreover, the place of $a$ will be taken by the exponent or
exponents at hand.

  2. (ii)

For a matrix $A\oldin\nolinebreak[4]M_{d}({\mathbb{C}})$, we denote by $|A|$
the _spectral norm_ of $A$, that is, the largest singular value of $A$ (i.e.
the square root of the largest eigenvalue of the matrix $A^{*}A$). Then, it is
true that

(5.2) $|A|\simeq_{d}\sum_{i=1}^{d}|Ae_{i}|,$

where $(e_{1},\ldots,e_{d})$ is the standard orthonormal basis of
${\mathbb{C}}^{d}$.

###  5.1. Reducing operators

Integral averages of scalar weights play a major role in the investigation of
scalar weighted estimates. In the context of matrix weighted estimates, it is
often the case that plain integral averages of positive definite matrix valued
functions are poorly suited for the problems at hand. The concept of the so-
called _reducing operators_ offers in many cases a possible remedy.

Let $(X,m)$ be a measure space and let $E$ be a measurable subset of $X$ with
$0<m(E)<\infty$. Recall that an $m$-measurable function $f:E\to
M_{d}({\mathbb{C}})$ (for some positive integer $d$) is said to be _$m$
-integrable over $E$_ if

$\int_{E}|f(x)|\,\mathrm{d}m(x)<\infty,$

and that in this case we denote

$\langle
f\rangle_{E}:=\frac{1}{m(E)}\int_{E}f(x)\,\mathrm{d}m(x)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}f(x)\,\mathrm{d}m(x).$

######  Definition 5.1.

  1. (1)

Let $1\leq p<\infty$, and $W:E\to\mathrm{M}_{d}({\mathbb{C}})$ be an
$m$-integrable function such that $W(x)$ is a positive definite matrix for
$m$-a.e. $x\oldin\nolinebreak[4]E$. Then, one can consider the norm

$e\mapsto\rho_{W,E,p}(e):=\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/p}e|^{p}\,\mathrm{d}m(x)\right)^{1/p}$

acting on vectors $e\oldin\nolinebreak[4]{\mathbb{C}}^{d}$. Let
$K\subseteq{\mathbb{C}}^{d}$ be the closed unit ball of $\rho_{W,E,p}$. In the
terminology of Section 3.3, $K$ is a convex body in ${\mathbb{C}}^{d}$ with
$0\oldin\nolinebreak[4]\mathrm{Int}(K)$. Let ${\mathcal{E}}$ be the John
ellipsoid of $K$. Subsequently, as explained in Section 3.3, there exists a
positive definite matrix $A\oldin\nolinebreak[4]\mathrm{M}_{d}({\mathbb{C}})$
with $A\overline{\mathbf{B}}={\mathcal{E}}$, where $\overline{\mathbf{B}}$ is
the standard (Euclidean) closed unit ball in ${\mathbb{C}}^{d}$. Following
[Gol03], we call the matrix $A^{-1}$ _the reducing operator of $W$ over $E$
with respect to the exponent $p$_ and denote ${\mathcal{W}}_{E,p}:=A^{-1}$.

(We omit the measure $m$ from the notation of the reducing operator.) Since
${\mathcal{E}}\subseteq K\subseteq d^{1/2}{\mathcal{E}}$ (see [Gol03,
Naz+17]), one easily deduces that

$\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/p}e|^{p}\,\mathrm{d}m(x)\right)^{1/p}\leq|{\mathcal{W}}_{E,p}e|\leq
d^{1/2}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/p}e|^{p}\,\mathrm{d}m(x)\right)^{1/p}\quad\forall
e\oldin\nolinebreak[4]{\mathbb{C}}^{d}.$

Notice that if $p=2$ or $d=1$, we simply have
${\mathcal{W}}_{E,p}=\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}W(x)\,\mathrm{d}m(x)\right)^{1/p}$.

  2. (2)

If $1<p<\infty$ and $W:E\to\mathrm{M}_{d}({\mathbb{C}})$ is a $m$-measurable
function with $W(x)\oldin\nolinebreak[4]\mathrm{P}_{d}({\mathbb{C}})$ for
$m$-a.e. $x\oldin\nolinebreak[4]E$ such that the function $V:=W^{-1/(p-1)}$ is
$m$-integrable over $E$, then we denote
${\mathcal{W}}^{\prime}_{E,p}:={\mathcal{V}}_{E,p^{\prime}}$.

Notice that, if $1<s<\infty$, and $W:E\to\mathrm{M}_{d}({\mathbb{C}})$ is a
$m$-integrable function with
$W(x)\oldin\nolinebreak[4]\mathrm{P}_{d}({\mathbb{C}})$ for $m$-a.e.
$x\oldin\nolinebreak[4]E$, then, by Jensen’s inequality, we have

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/s}|\,\mathrm{d}m(x)=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)|^{1/s}\,\mathrm{d}m(x)\leq\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)|\,\mathrm{d}m(x)\right)^{1/s}<\infty.$

Therefore, the function $W^{1/s}$ is also $m$-integrable over $E$. However,
this might not be the case for $W^{s}$.

The following standard lemma, whose proof we omit, helps to avoid ambiguous
terminology below.

######  Lemma 5.2.

Let $(X,m)$ be a measure space and let $E$ be a measurable subset of $X$ with
$0<m(E)<\infty$. Let $1\leq a,b<\infty$, and
$W,V:E\to\mathrm{M}_{d}({\mathbb{C}})$ be $m$-measurable functions with
$W(x),V(x)\oldin\nolinebreak[4]\mathrm{P}_{d}({\mathbb{C}})$ for $m$-a.e.
$x\oldin\nolinebreak[4]E$. Then, we have

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/b}V(y)^{1/a}|^{a}\,\mathrm{d}m(y)\right)^{b/a}\,\mathrm{d}m(x)<\infty\iff
W,\leavevmode\nobreak\ V\text{ are $m$-integrable over }E.$

Moreover, in this case, the following holds:

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{E}|W(x)^{1/b}V(y)^{1/a}|^{a}\,\mathrm{d}m(y)\right)^{b/a}\,\mathrm{d}m(x)\simeq_{d,a,b}|{\mathcal{W}}_{E,b}{\mathcal{V}}_{E,a}|^{b}.$

###  5.2. Matrix Muckenhoupt weights

The weight $W$ in 1.3 was assumed to satisfy the condition
$[W^{s^{\prime}}]_{A_{a^{\prime},b}}<\infty$, which is a so-called _matrix
Muckenhoupt condition_. We explain the terminology in this subsection.

######  Definition 5.3.

Let $1<p\leq q<\infty$, and $W:M\to\mathrm{M}_{d}({\mathbb{C}})$ be a
measurable function with values $\mu$-a.e. in the set of positive-definite
$d\times d$ complex-valued matrices. We define

(5.3)
$[W]_{A_{p,q}}:=\sup_{B}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B}|W(x)^{\frac{1}{q}}W(y)^{-\frac{1}{q}}|^{p^{\prime}}\,\mathrm{d}\mu(y)\right)^{\frac{q}{p^{\prime}}}\,\mathrm{d}\mu(x),$

where the supremum ranges over all open balls $B\subseteq M$. If
$[W]_{A_{p,q}}<\infty$, then we say that $W$ is a _$d\times d$ matrix
$A_{p,q}$ weight_, and we write $W\oldin\nolinebreak[4]A_{p,q}$. If $p=q$,
then we denote $A_{p,p}:=A_{p}$.

Variants of the matrix $A_{p,q}$ weights were already introduced and studied
in [NT96, LT07]. The particular form of condition (5.3) considered here was
first introduced and studied in [IM19] and further investigated in [CI22].

Below, we present several properties of the matrix $A_{p,q}$ weights that are
important for our purposes. These were all established in [IM19] in the case
when the ambient space $M$ is simply ${\mathbb{R}}^{n}$ equipped with the
Euclidean metric and the $n$-dimensional Lebesgue measure, but they still hold
true in arbitrary doubling measure metric spaces simply by replacing cubes
with balls. See [IM19, Section 3] for the proofs.

######  Lemma 5.4.

Let $1<p\leq q<\infty$, and let $W$ be a $d\times d$ matrix $A_{p,q}$ weight
on $M$. Set $r:=1+\frac{q}{p^{\prime}}$. Then, the following are true:

$\displaystyle[W^{-p^{\prime}/q}]_{A_{q^{\prime},p^{\prime}}}$
$\displaystyle\simeq_{d,p,q}[W]_{A_{p,q}}^{1/(r-1)};$
$\displaystyle\sup_{e\oldin\nolinebreak[4]{\mathbb{C}}^{d}}[|W^{1/q}e|^{q}]_{A_{r}}$
$\displaystyle\lesssim_{d,p,q}[W]_{A_{p,q}};$
$\displaystyle\sup_{e\oldin\nolinebreak[4]{\mathbb{C}}^{d}}[|W^{-1/q}e|^{p^{\prime}}]_{A_{r^{\prime}}}$
$\displaystyle\lesssim_{d,p,q}[W^{-p^{\prime}/q}]_{A_{q^{\prime},p^{\prime}}}.$

We also need the so-called “weak” reverse Hölder property of scalar
Muckenhoupt weights, in general, doubling measure metric spaces, as obtained
from [HPR12, Theorem 1.1] coupled with (a trivial generalization of) [Naz+17,
Lemma 4.1].

######  Lemma 5.5.

Let $w$ be a scalar weight on $M$ with $[w]_{A_{p}}<\infty$ for some
$1<p<\infty$. Then, there exist constants $a_{1},a_{2}>0$ depending only on
the doubling measure $\mu$ and $p$ such that for all
$0<\varepsilon\leq\frac{1}{1+a_{1}[w]_{A_{p}}}$ we have

$\langle w^{1+\varepsilon}\rangle_{B}^{1/(1+\varepsilon)}\leq a_{2}\langle
w\rangle_{2B},$

for all open balls $B\subseteq M$, where $2B$ denotes the ball co-centric with
$B$ and twice its radius.

In particular, 5.5 can be used to show that scalar $A_{p}$ weights for
$1<p<\infty$ satisfy an openness property, see, for instance, [HPR12] for a
precise statement. However, counterexamples due to [Bow01] show that matrix
$A_{p}$ weights on ${\mathbb{R}}^{n}$ fail to satisfy an analogous openness
property for any $1<p<\infty$.

###  5.3. Proving the matrix-weighted bound

We devote this subsection to a proof of 1.3.

First, we need some preparation. Fix $1\leq p_{0}<p<q_{0}\leq\infty$. Next,
set

$t:=\frac{p}{p_{0}},\quad
s:=\frac{q_{0}}{p},\quad\tilde{s}:=\frac{p^{\prime}}{q_{0}^{\prime}},$

and

$a:=\frac{t^{\prime}p}{t},\quad b:=s^{\prime}p.$

Observe that $t,s,\tilde{s},a,b>1$, and $t,\tilde{s},a,b<\infty$. Direct
computation gives

$a=\frac{p}{t-1}=t^{\prime}p_{0},\quad
b=\frac{p^{\prime}}{\tilde{s}-1}=\tilde{s}^{\prime}q_{0}^{\prime},$

and

$\frac{1}{a}+\frac{1}{b}=\frac{1}{p_{0}}-\frac{1}{q_{0}}\leq 1.$

In particular, we have $b\geq a^{\prime}$. Finally, define

$r:=s^{\prime}(t-1)+1.$

Observe that $r\oldin\nolinebreak[4](1,\infty)$ and

$\frac{b}{a}=r-1=\frac{s^{\prime}t}{t^{\prime}}.$

Let $W$ be a $d\times d$ matrix weight on $M$. Set $U:=W^{-1/(t-1)}$ and
$V:=W^{s^{\prime}}$. Notice that

$U=V^{-1/(r-1)}\quad\text{and}\quad V^{1/b}=W^{1/p}=U^{-1/a}.$

Assume now that

$[V]_{A_{a^{\prime},b}}<\infty.$

Then, by 5.2 we deduce the following:

(5.4)
$|{\mathcal{U}}_{B,a}{\mathcal{V}}_{B,b}|\lesssim_{d,a,b}[V]_{A_{a^{\prime},b}}^{1/b},$

for all open balls $B\subseteq M$. Also, 5.4 for $p=a^{\prime}$ and $q=b$
gives us that

(5.5)
$\displaystyle[|V^{1/b}e|^{b}]_{A_{r}}\lesssim_{d,a,b}[V]_{A_{a^{\prime},b}}\quad\forall
e\oldin\nolinebreak[4]{\mathbb{C}}^{d}\setminus\\{0\\},$ (5.6)
$\displaystyle[|U^{1/a}e|^{a}]_{A_{r^{\prime}}}\lesssim_{d,a,b}[U]_{A_{b^{\prime},a}}\quad\forall
e\oldin\nolinebreak[4]{\mathbb{C}}^{d}\setminus\\{0\\},$ and (5.7)
$\displaystyle[U]_{A_{b^{\prime},a}}\simeq_{d,a,b}[V]_{A_{a^{\prime},b}}^{1/(r-1)}.$

Now, we are finally ready to prove 1.3

###### Proof (of 1.3).

We adapt the arguments from [Naz+17, Section 5]. We keep the setup and
notation from the previous subsection. Throughout the proof, we ignore all
implied constants depending only on $\mu,d,p_{0},q_{0}$ and $p$. First of all,
it suffices to prove that

$\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)|\llangle
W^{-1/p}f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
P)}\cdot\llangle
W^{1/p}g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
P)}|\lesssim\frac{1}{\varepsilon}([W^{s^{\prime}}]_{A_{a^{\prime},b}})^{\alpha}\lVert
f\rVert_{L^{p}(M,\mu;{\mathbb{C}}^{d})}\lVert
g\rVert_{L^{p^{\prime}}(M,\mu;{\mathbb{C}}^{d})},$

for all $f\oldin\nolinebreak[4]L^{p}(M,\mu;{\mathbb{C}}^{d})$ and all
$g\oldin\nolinebreak[4]L^{p^{\prime}}(M,\mu;{\mathbb{C}}^{d})$. In what
follows, we fix such $f$ and $g$. From 2.3 we have that for all
$P\oldin\nolinebreak[4]\mathscr{S}$, there exists an open ball $B(P)\subseteq
M$ such that $\lambda P\subseteq B(P)$ and $\mu(B(P))\lesssim\mu(\lambda P)$.
Thus, we begin by estimating

$\displaystyle A$
$\displaystyle:=\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)|\llangle
W^{-1/p}f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
P)}\cdot\llangle
W^{1/p}g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
P)}|$
$\displaystyle=\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)|(\mathcal{U}_{2B(P),a}^{-1}\llangle
W^{-1/p}f\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{p_{0}}(\lambda
P)})\cdot(\mathcal{U}_{2B(P),a}\llangle
W^{1/p}g\rrangle_{\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-2.59586pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-1.91917pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.35042pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.06459pt}}\\!L^{q_{0}^{\prime}}(\lambda
P)})|$
$\displaystyle\leq\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}W^{-1/p}f|^{p_{0}}\right\rangle_{\lambda
P}^{1/p_{0}}\left\langle|\mathcal{U}_{2B(P),a}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{\lambda
P}^{1/q_{0}^{\prime}}$
$\displaystyle\lesssim\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}W^{-1/p}f|^{p_{0}}\right\rangle_{B(P)}^{1/p_{0}}\left\langle|\mathcal{U}_{2B(P),a}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}^{1/q_{0}^{\prime}}.$

We apply Hölder’s inequality for the exponents $p$ and $p^{\prime}$:

$A\leq\left(\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}W^{-1/p}f|^{p_{0}}\right\rangle_{B(P)}^{p/p_{0}}\right)^{1/p}\left(\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}^{p^{\prime}/q_{0}^{\prime}}\right)^{1/p^{\prime}}=:A_{1}\cdot
A_{2}.$

Now we estimate $A_{1}$ and $A_{2}$ from above separately. We start with
$A_{1}$:

$\displaystyle A_{1}^{p}$
$\displaystyle=\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}W^{-1/p}f|^{p_{0}}\right\rangle_{B(P)}^{p/p_{0}}=\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}U^{1/a}f|^{a/t^{\prime}}\right\rangle_{B(P)}^{t}$
$\displaystyle\leq\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}U^{1/a}|^{a/t^{\prime}}\cdot|f|^{a/t^{\prime}}\right\rangle_{B(P)}^{t}.$

Set $\theta:=t^{\prime}(1+\delta)$, for some $\delta>0$ to be chosen later.
Then, applying Hölder’s inequality for the exponents $\theta$ and
$\theta^{\prime}$, we get

$\displaystyle A_{1}^{p}$
$\displaystyle\leq\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}U^{1/a}|^{a\theta/t^{\prime}}\right\rangle_{B(P)}^{t/\theta}\cdot{\left\langle|f|^{a\theta^{\prime}/t^{\prime}}\right\rangle}_{B(P)}^{t/\theta^{\prime}}$
$\displaystyle=\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|\mathcal{U}_{2B(P),a}^{-1}U^{1/a}|^{a(1+\delta)}\right\rangle_{B(P)}^{(t-1)/(1+\delta)}\cdot{\left\langle|f|^{a\theta^{\prime}/t^{\prime}}\right\rangle}_{B(P)}^{t/\theta^{\prime}}.$

For every $P\oldin\nolinebreak[4]\mathscr{S}$, consider the scalar weight

$\omega_{P}:=|\mathcal{U}_{2B(P),a}^{-1}U^{1/a}|^{a}=|U^{1/a}\mathcal{U}_{2B(P),a}^{-1}|^{a}$

on $M$. Then, by (5.6) coupled with (a tiny modification of) [Dom+24, Lemma
3.4], we deduce

$\sup_{P\oldin\nolinebreak[4]\mathscr{S}}[\omega_{P}]_{A_{r^{\prime}}}\lesssim[U]_{A_{b^{\prime},a}}.$

Thus, by the weak Reverse Hölder property for scalar Muckenhoupt weights, 5.5,
there exists a constant $a_{1}>0$ depending only on $\mu,d,p_{0},q_{0}$ and
$p$ such that for the choice

$\delta:=\frac{1}{1+a_{1}[U]_{A_{b^{\prime},a}}}$

we have, for every $P\oldin\nolinebreak[4]\mathscr{S}$,

$\left\langle\omega_{P}^{1+\delta}\right\rangle_{B}^{1/(1+\delta)}\lesssim\left\langle\omega_{P}\right\rangle_{2B},$

for every open ball $B\subseteq M$. Thus, for every
$P\oldin\nolinebreak[4]\mathscr{S}$, we obtain

$\left\langle\omega_{P}^{1+\delta}\right\rangle_{B(P)}^{1/(1+\delta)}\lesssim\left\langle\omega_{P}\right\rangle_{2B(P)}=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{2B(P)}|U(x)^{1/a}\mathcal{U}_{2B(P),a}^{-1}|^{a}\,\mathrm{d}\mu(x)\simeq|{\mathcal{U}}_{2B(P),a}{\mathcal{U}}_{2B(P),a}^{-1}|^{a}=1.$

Thus, we have

$\displaystyle A_{1}^{p}$
$\displaystyle\lesssim\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|f|^{a\theta^{\prime}/t^{\prime}}\right\rangle_{B(P)}^{t/\theta^{\prime}}\leq\frac{1}{\varepsilon}\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(F_{P})\left\langle|f|^{a\theta^{\prime}/t^{\prime}}\right\rangle_{B(P)}^{t/\theta^{\prime}}$
$\displaystyle\leq\frac{1}{\varepsilon}\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\int_{F_{P}}(\mathcal{M}(|f|^{a\theta^{\prime}/t^{\prime}})(x))^{t/\theta^{\prime}}\,\mathrm{d}\mu(x)\leq\frac{1}{\varepsilon}\int_{M}(\mathcal{M}(|f|^{a\theta^{\prime}/t^{\prime}})(x))^{t/\theta^{\prime}}\,\mathrm{d}\mu(x),$

since the sets $F_{P}$, $P\oldin\nolinebreak[4]\mathscr{S}$ are pairwise
disjoint. Observe that $\theta>t^{\prime}$, therefore $\theta^{\prime}<t$,
that is $q:=t/\theta^{\prime}>1$. Thus, applying (2.6) we obtain

$A_{1}^{p}\lesssim\frac{1}{\varepsilon}(q^{\prime})^{q}\int_{M}(|f(x)|^{a\theta^{\prime}/t^{\prime}})^{q}\,\mathrm{d}\mu(x)=(q^{\prime})^{q}\frac{1}{\varepsilon}\int_{M}|f(x)|^{p}\,\mathrm{d}\mu(x).$

5.6 below shows that $(q^{\prime})^{q}\lesssim\delta^{-1}$, therefore we
finally obtain

$A_{1}\lesssim\frac{1}{\varepsilon^{1/p}}[U]_{A_{b^{\prime},a}}^{1/p}\lVert
f\rVert_{L^{p}(M,\mu;{\mathbb{C}}^{d})}.$

To estimate $A_{2}$, we first observe that for all
$P\oldin\nolinebreak[4]\mathscr{S}$, by (5.4) we get

$\displaystyle\left\langle|{\mathcal{U}}_{2B(P),a}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}$
$\displaystyle\leq\left\langle|{\mathcal{U}}_{2B(P),a}{\mathcal{V}}_{2B(P),b}{\mathcal{V}}_{2B(P),b}^{-1}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}$
$\displaystyle\leq|{\mathcal{U}}_{2B(P),a}{\mathcal{V}}_{2B(P),b}|^{q_{0}^{\prime}}\cdot\left\langle|{\mathcal{V}}_{2B(P),b}^{-1}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}$
$\displaystyle\lesssim[V]_{A_{a^{\prime},b}}^{q_{0}^{\prime}/b}\cdot\left\langle|{\mathcal{V}}_{2B(P),b}^{-1}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}.$

Similarly to the above and using (5.5) instead of (5.6), we obtain

$\left(\sum_{P\oldin\nolinebreak[4]\mathscr{S}}\mu(P)\left\langle|{\mathcal{V}}_{2B(P),b}^{-1}W^{1/p}g|^{q_{0}^{\prime}}\right\rangle_{B(P)}^{p^{\prime}/q_{0}^{\prime}}\right)^{1/p^{\prime}}\lesssim\frac{1}{\varepsilon^{1/p^{\prime}}}[V]_{A_{a^{\prime},b}}^{1/p^{\prime}}\lVert
g\rVert_{L^{p^{\prime}}(M,\mu;{\mathbb{C}}^{d})}.$

Putting all estimates together and using (5.7), we finally deduce

$A\lesssim\frac{1}{\varepsilon}([V]_{A_{a^{\prime},b}})^{\alpha}\lVert
f\rVert_{L^{p}(M,\mu;{\mathbb{C}}^{d})}\lVert
g\rVert_{L^{p^{\prime}}(M,\mu;{\mathbb{C}}^{d})},$

where

$\alpha:=\frac{1}{p(r-1)}+\frac{1}{b}+\frac{1}{p^{\prime}}=\frac{1}{s^{\prime}p(t-1)}+\frac{1}{s^{\prime}p}+\frac{1}{p^{\prime}}=\frac{t}{s^{\prime}p(t-1)}+\frac{1}{p^{\prime}}=\frac{1}{s^{\prime}(p-p_{0})}+\frac{1}{p^{\prime}},$

concluding the proof. ∎

######  Lemma 5.6.

Let $0<\delta<1$. Set $\theta:=t^{\prime}(1+\delta)$ and
$q:=\frac{t}{\theta^{\prime}}$. Then, we have
$(q^{\prime})^{q}\lesssim\delta^{-1}$.

###### Proof.

Set $y:=q-1>0$, then

$y=\frac{t}{\theta^{\prime}}-1=\frac{t}{\frac{\theta}{\theta-1}}-1=\frac{t(\theta-1)-\theta}{\theta}.$

Since $\theta=t^{\prime}(1+\delta)$, we rewrite

$y=\frac{t(t^{\prime}(1+\delta)-1)-t^{\prime}(1+\delta)}{t^{\prime}(1+\delta)},$

which gives

(5.8)
$y=\frac{t\left(\frac{t}{t-1}(1+\delta)-1\right)-\frac{t}{t-1}(1+\delta)}{\frac{t}{t-1}(1+\delta)}=\frac{(t(1+\delta)-t+1)-(1+\delta)}{(1+\delta)}=\frac{\delta}{1+\delta}(t-1).$

Now, we observe that

$(q^{\prime})^{q}=((y+1)^{\prime})^{y+1}=\left(1+\frac{1}{y}\right)^{y+1}\leq
e\left(1+\frac{1}{y}\right).$

Thus, using (5.8), we get

$(q^{\prime})^{q}\leq
e\left(1+\frac{1+\delta}{\delta(t-1)}\right)=e\left(\frac{\delta
t+1}{\delta(t-1)}\right)<e\frac{t+1}{\delta(t-1)}\lesssim\delta^{-1}$

concluding the proof. ∎

###  5.4. An open question on sharpness of estimates

Although we expect our estimates to be both qualitatively and quantitatively
sharp, especially in the view of [Dom+24a], we have not been able to prove
that as of the moment of writing.

##  6\. A hierarchy of two-exponent matrix Muckenhoupt conditions

In this section we want to understand the relationship of the various two-
exponent matrix Muckenhoupt characteristics $[W]_{A_{p,q}}$ for $1<p\leq
q<\infty$ with $\frac{q}{p^{\prime}}+1=r$ for some _fixed_ $r$. In the scalar
case, all these characteristics coincide. However, in the matrix case, they
turn out to be quite different from each other, and only one-sided inclusions
remain in general true. We provide explicit, fully constructive
counterexamples demonstrating that. As a result, we deduce that the matrix
weighted bound of 1.3 improves the one obtained earlier in [Lau23]. Our
methods are very flexible and allow us to consider even several endpoint cases
that play an important role in the limited range extrapolation in Section 7
later.

Throughout this section, for the sake of clarity and simplicity we focus only
on ${\mathbb{R}}^{n}$ equipped with the Euclidean metric and the
$n$-dimensional Lebesgue measure. In particular, we simply write $\mathrm{d}x$
for integration under the Lebesgue measure, and $|E|$ for the $n$-dimensional
Lebesgue measure of any Lebesgue measurable subset $E$ of ${\mathbb{R}}^{n}$.

###  6.1. The members of the hierarchy

Let us first recall the two-exponent matrix Muckenhoupt conditions from 5.3.
In the case of ${\mathbb{R}}^{n}$, one usually takes the supremum in 5.3 to
range over cubes $Q\subseteq{\mathbb{R}}^{n}$ (with faces parallel to the
coordinate hyperplanes) instead of open balls $B\subseteq{\mathbb{R}}^{n}$.
Then, the resulting Muckenhoupt characteristics are equivalent to their “ball”
versions up to constants depending only on $n$ and the exponents in question.
The resulting equivalence classes of (scalar and matrix) weights coincide with
their “ball” versions.

######  Definition 6.1.

Let $1<p\leq q<\infty$, and
$W:{\mathbb{R}}^{n}\to\mathrm{M}_{d}({\mathbb{C}})$ be a measurable function
with values a.e. in the set of positive-definite $d\times d$ complex-valued
matrices. We define

$[W]_{A_{p,q}}:=\sup_{Q}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{\frac{1}{q}}W(y)^{-\frac{1}{q}}|^{p^{\prime}}\,\mathrm{d}y\right)^{\frac{q}{p^{\prime}}}\,\mathrm{d}x<\infty,$

where the supremum ranges over all cubes $Q\subseteq{\mathbb{R}}^{n}$ (with
faces parallel to the coordinate hyperplanes). If $[W]_{A_{p,q}}<\infty$, then
we say that $W$ is a _$d\times d$ matrix $A_{p,q}$ weight_, and we write
$W\oldin\nolinebreak[4]A_{p,q}$. If $p=q$, then we denote $A_{p,p}:=A_{p}$.

Furthermore, we are interested in the following endpoint cases.

######  Definition 6.2.

Let $W:{\mathbb{R}}^{n}\to\mathrm{M}_{d}({\mathbb{C}})$ be a measurable
function with values a.e. in the set of positive-definite $d\times d$ complex-
valued matrices, and $1\leq t<\infty$. We define

$[W]_{A_{t,\infty}}:=\sup_{B}\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]B}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{B}|W(x)^{1/t}W(y)^{-1/t}|^{t}\,\mathrm{d}y,$

where the supremum ranges over all cubes $Q\subseteq{\mathbb{R}}^{n}$ (with
faces parallel to the coordinate hyperplanes). If $[W]_{A_{t,\infty}}<\infty$,
then we say that $W$ is a _$d\times d$ matrix $A_{t,\infty}$ weight_, and we
write $W\oldin\nolinebreak[4]A_{t,\infty}$. Moreover, we denote
$A_{1,\infty}:=A_{\infty}$.

Our definition of $A_{\infty}$ follows [BC22]. However, our definitions of
$A_{t,\infty}$ and $A_{\infty}$ deviate from the ones appearing in other
places in the literature, even in the scalar case. See, for example, [NT96,
Vol97, HPR12, Naz+17, CIM18, ID24]. We also refer to [DMO16] for a detailed
discussion of the relationship between the various definitions in the scalar
case.

It is worth noting that if $W:{\mathbb{R}}^{n}\to\mathrm{P}_{d}({\mathbb{C}})$
is a measurable function with $W\oldin\nolinebreak[4]A_{t,\infty}$ for some
$1\leq t<\infty$, then the function $|W|$ is locally bounded and the function
$W^{-1}$ is locally integrable.

We turn our attention to reverse Hölder classes.

######  Definition 6.3.

If $w$ is a scalar weight on ${\mathbb{R}}^{n}$ and $1<s<\infty$, then we
define

$[w]_{RH_{s}}:=\sup_{Q}\frac{\langle w^{s}\rangle_{Q}^{1/s}}{\langle
w\rangle_{Q}},$

where the supremum ranges over all cubes $Q$ in ${\mathbb{R}}^{n}$ (with faces
parallel to the coordinate hyperplanes). We say that $w$ belongs to the
_Reverse Hölder class of order $s$_, whenever $[w]_{RH_{s}}<\infty$, and we
write $w\oldin\nolinebreak[4]RH_{s}$.

Moreover, following [AM07], for a scalar weight $w$ on ${\mathbb{R}}^{n}$ we
define

$[w]_{RH_{\infty}}:=\sup_{Q}\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]Q}\frac{w(x)}{\langle
w\rangle_{Q}},$

where the supremum ranges over all cubes $Q$ in ${\mathbb{R}}^{n}$ (with faces
parallel to the coordinate hyperplanes). We say that $w$ belongs to the
_Reverse Hölder class of order $\infty$_ whenever $[w]_{RH_{\infty}}<\infty$,
and we write $w\oldin\nolinebreak[4]RH_{\infty}$.

######  Definition 6.4.

If $W$ is a $d\times d$ matrix weight on ${\mathbb{R}}^{n}$, $1<t<\infty$, and
$1<s\leq\infty$, then following [Lau23] we define

$[W]_{RH_{t,s}}:=\sup_{e\oldin\nolinebreak[4]{\mathbb{C}}^{d}\setminus\\{0\\}}[|W^{1/t}e|^{t}]_{RH_{s}}<\infty.$

We say that $W$ belongs to the _Reverse Hölder class of exponent $t$ and order
$s$_ and write $W\oldin\nolinebreak[4]RH_{t,s}$ when $[W]_{RH_{t,s}}<\infty$.

###  6.2. Situation in the scalar case

The various condition introduced in Section 6.1 are related to each other in
the case of scalar weights in a very simple way. We briefly review how below.

Let $w$ be a scalar weight on ${\mathbb{R}}^{n}$. Then, for all $1\leq p\leq
q<\infty$, we clearly have

(6.1) $[w]_{A_{p,q}}=[w]_{A_{r}},\quad\text{where }r:=1+\frac{q}{p^{\prime}}.$

It is also obvious that

(6.2) $[w]_{A_{t,\infty}}=[w]_{A_{\infty}},$

for all $1\leq t<\infty$.

Finally, let $1<s,t<\infty$ and set $r:=s(t-1)+1$. It is a well-known fact
that

(6.3) $w\oldin\nolinebreak[4]A_{t}\cap RH_{s}\iff
w^{s}\oldin\nolinebreak[4]A_{r},$

see, for example, [JN91, AM07, BFP16].

###  6.3. Commonalities with the scalar case

In this subsection we show that at least parts of (6.1) and (6.2), as well as
an appropriate form of (6.3) survive in the matrix case.

We begin with the correct analog of (6.3).

######  Lemma 6.5.

Let $W$ be a $d\times d$ matrix weight on ${\mathbb{R}}^{n}$, and
$1<t,s<\infty$. Then, the following hold:

  1. (1)

$\max([W]_{A_{t}},[W]_{RH_{t,s}})^{s}\lesssim_{d,t,s}[W^{s}]_{A_{t,st}}\lesssim_{d,t,s}([W]_{A_{t}}[W]_{RH_{t,s}})^{s}.$

In particular, $W\oldin\nolinebreak[4]A_{t}\cap RH_{t,s}$ if and only if
$W^{s}\oldin\nolinebreak[4]A_{t,st}$.

  2. (2)

$\max([W]_{A_{t}},[W]_{RH_{t,\infty}})^{t^{\prime}/t}\lesssim_{d,t}[W^{t^{\prime}/t}]_{A_{t^{\prime},\infty}}\lesssim_{d,t}([W]_{A_{t}}[W]_{RH_{t,\infty}})^{t^{\prime}/t}.$

In particular, $W\oldin\nolinebreak[4]A_{t}\cap RH_{t,\infty}$ if and only if
$W^{t^{\prime}/t}\oldin\nolinebreak[4]A_{t^{\prime},\infty}$.

###### Proof.

For (1), set $V:=W^{s}$. Then, $V^{1/(st)}=W^{1/t}$.

First, assume that $[W^{s}]_{A_{t,st}}<\infty$. For all cubes
$Q\subseteq{\mathbb{R}}^{n}$, since $s\geq 1$, Jensen’s inequality yields

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}W(y)^{-1/t}|^{t^{\prime}}\,\mathrm{d}y\right)^{t/t^{\prime}}\,\mathrm{d}x\leq\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}W(y)^{-1/t}|^{t^{\prime}}\,\mathrm{d}y\right)^{st/t^{\prime}}\,\mathrm{d}x\right)^{1/s},$

which directly implies that $[W]_{A_{t}}\leq[V]_{A_{t,st}}^{1/s}$.

Let us now show that
$[W]_{RH_{t,s}}\lesssim_{d,t,s}[V]_{A_{t,s^{\prime}t}}^{1/s}$. Let
$Q\subseteq{\mathbb{R}}^{n}$ be any cube and let
$e\oldin\nolinebreak[4]{\mathbb{C}}^{d}\setminus\\{0\\}$. Since
$[W]_{A_{t}}<\infty$, $W^{-1/(t-1)}$ is integrable over $Q$. Therefore,
repeatedly applying (5.2) and (5.1), we can estimate

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}{\mathcal{W}}_{Q,t}^{\prime}|^{st}\,\mathrm{d}x\simeq_{d,t,s}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}W(y)^{-1/t}|^{t^{\prime}}\,\mathrm{d}y\right)^{st/t^{\prime}}\mathrm{d}x\leq[V]_{A_{t,st}}.$

It follows that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}e|^{st}\,\mathrm{d}x=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}{\mathcal{W}}_{Q,t}^{\prime}({\mathcal{W}}_{Q,t}^{\prime})^{-1}e|^{st}\,\mathrm{d}x\lesssim_{d,t,s}[V]_{A_{t,st}}|({\mathcal{W}}_{Q,t}^{\prime})^{-1}e|^{st}.$

It is well-known (see e.g. [NT96, LT07, Dom+24]) that
$|({\mathcal{W}}_{Q,t}^{\prime})^{-1}e|\leq|{\mathcal{W}}_{Q,t}e|$. From this,
we get that

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}e|^{st}\,\mathrm{d}x\lesssim_{d,t,s}[V]_{A_{t,st}}|{\mathcal{W}}_{Q,t}e|^{st}\simeq_{d,t,s}[V]_{A_{t,st}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}e|^{t}\,\mathrm{d}x\right)^{s}.$

Consequently, $[W]_{RH_{t,s}}\lesssim_{d,t,s}[V]_{A_{t,s^{\prime}t}}^{1/s}$.

Conversely, assume that $W\oldin\nolinebreak[4]A_{t}\cap RH_{t,st}$. Then,
repeatedly applying (5.2) and (5.1), we can estimate

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}W(y)^{-1/t}|^{t^{\prime}}\,\mathrm{d}y\right)^{st/t^{\prime}}\,\mathrm{d}x$
$\displaystyle\simeq_{d,t,s}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}{\mathcal{W}}_{Q,t}^{\prime}|^{st}\,\mathrm{d}x$
$\displaystyle\lesssim_{d,t,s}[W]_{RH_{t,s}}^{s}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/t}{\mathcal{W}}_{Q,t}^{\prime}|^{t}\,\mathrm{d}x\right)^{s}$
$\displaystyle\simeq_{d,t,s}[W]_{RH_{t,s}}^{s}|{\mathcal{W}}_{Q,t}{\mathcal{W}}_{Q,t}^{\prime}|^{ts}$
$\displaystyle\lesssim_{d,t,s}[W]_{RH_{t,s}}^{s}[W]_{A_{t}}^{s},$

for all cubes $Q\subseteq{\mathbb{R}}^{n}$. It follows that
$[V]_{A_{t,st}}\lesssim_{d,t,s}[W]_{RH_{t,s}}^{s}[W]_{A_{t}}^{s}$.

We omit the proof for (2) as it is very similar to the proof for part (1). ∎

Next, we treat the parts of (6.1), (6.2) that survive in the scalar case.

######  Lemma 6.6.

Let $W$ be a $d\times d$ matrix weight on ${\mathbb{R}}^{n}$.

  1. (1)

Let $1<p_{1},q_{1},p_{2},q_{2}<\infty$ with $p_{i}\leq q_{i}$ (for $i=1,2$),
$p_{1}\leq p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$. Then,

$[W]_{A_{p_{1},q_{1}}}\leq[W]_{A_{p_{2},q_{2}}}.$

  2. (2)

Let $1\leq s\leq t<\infty$. Then,

$[W]_{A_{t,\infty}}\leq[W]_{A_{s,\infty}}.$

In the proof, we apply the following inequality due to Cordes [Cor87].

######  Lemma 6.7 (Cordes Inequality).

Let $A,B\oldin\nolinebreak[4]\mathrm{P}_{d}({\mathbb{C}})$ and $0\leq a\leq
1$. Then, it holds that

$|A^{a}B^{a}|\leq|AB|^{a}.$

###### Proof (of 6.6).

For (1), first observe that

$0<\frac{q_{2}}{q_{1}}=\frac{p_{2}^{\prime}}{p_{1}^{\prime}}\leq 1.$

Therefore, by Cordes inequality, we estimate

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\right)^{q_{1}/p_{1}^{\prime}}\,\mathrm{d}x$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|(W(x)^{1/q_{2}})^{q_{2}/q_{1}}(W(y)^{-1/q_{2}})^{q_{2}/q_{1}}|^{q_{1}^{\prime}}\,\mathrm{d}y\right)^{q_{1}/p_{1}^{\prime}}\,\mathrm{d}x$
$\displaystyle\leq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/q_{2}}W(y)^{-1/q_{2}}|^{q_{2}p_{1}^{\prime}/q_{1}}\,\mathrm{d}y\right)^{q_{1}/p_{1}^{\prime}}\,\mathrm{d}x$
$\displaystyle=\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{Q}|W(x)^{1/q_{2}}W(y)^{1/q_{2}}|^{p_{2}^{\prime}}\,\mathrm{d}y\right)^{q_{2}/p_{2}^{\prime}}\,\mathrm{d}x,$

for all cubes $Q\subseteq{\mathbb{R}}^{n}$. The desired result follows.

The proof for (2) is analogous to the proof of part (1). ∎

###  6.4. Differences from the scalar case

Despite the positive results in Subsection 6.3, (6.1) and (6.2) can actually
fail in the matrix case. The following two propositions demonstrate that. Let

${\mathcal{D}}:=\\{[m2^{k},(m+1)2^{k}):\leavevmode\nobreak\
m,k\oldin\nolinebreak[4]{\mathbb{Z}}\\}$

be the family of dyadic intervals in ${\mathbb{R}}$. For $1<p\leq q<\infty$,
we denote

$[W]_{A_{p,q},{\mathcal{D}}}:=\sup_{I\oldin\nolinebreak[4]{\mathcal{D}}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|W(x)^{1/q}W(y)^{-1/q}|^{p^{\prime}}\,\mathrm{d}y\right)^{q/p^{\prime}}\,\mathrm{d}x.$

Similarly to 6.2 (with the cubes replaced by the dyadic intervals of
${\mathbb{R}}$), we define $[W]_{A_{t,\infty},{\mathcal{D}}}$ for $1\leq
t<\infty$.

######  Proposition 6.8.

  1. (1)

Let $1<p_{1}\leq q_{1}<\infty$. Then, there exists a $2\times 2$ matrix weight
$W$ on ${\mathbb{R}}$ with $[W]_{A_{p_{1},q_{1}}}<\infty$ such that for all
$1<p_{2}\leq q_{2}<\infty$ with $p_{1}<p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$ we have
$[W]_{A_{p_{2},q_{2}},{\mathcal{D}}}=\infty$.

  2. (2)

Let $1<t<\infty$. Then, there exists a $2\times 2$ matrix weight $W$ on
${\mathbb{R}}$ with $[W]_{A_{t,\infty}}<\infty$ such that
$[W]_{A_{s,\infty},{\mathcal{D}}}=\infty$ for all $1\leq s<t$.

######  Proposition 6.9.

  1. (1)

Let $1<p_{2}\leq q_{2}<\infty$. Then, there exists a $2\times 2$ matrix weight
$W$ on ${\mathbb{R}}$ with $[W]_{A_{p_{2},q_{2}},{\mathcal{D}}}=\infty$ such
that $[W]_{A_{p_{1},q_{1}}}<\infty$, for all $1<p_{1}\leq q_{1}<\infty$ with
$\frac{q_{1}^{\prime}}{p_{1}}=\frac{q_{2}^{\prime}}{p_{2}}$ and $p_{1}<p_{2}$.

  2. (2)

Let $1\leq s<\infty$. Then, there exists a $2\times 2$ matrix weight $W$ on
${\mathbb{R}}$ with $[W]_{A_{s,\infty},{\mathcal{D}}}=\infty$ such that
$[W]_{A_{t,\infty}}<\infty$, for all $s<t<\infty$.

Before proving 6.8 and 6.9, let us briefly examine their implications for the
relationship of the matrix weighted bounds of 1.3 to those obtained earlier by
[Lau23]. We first recall the notation of 1.3: we consider $1\leq
p_{0}<q_{0}\leq\infty$, $p\oldin\nolinebreak[4](p_{0},q_{0})$ and let

$t:=\frac{p}{p_{0}},\quad s:=\frac{q_{0}}{p},\quad a:=\frac{p}{t-1},\quad
b:=s^{\prime}p.$

In this notation, the matrix-weighted bounds in [Lau23] are valid for matrix
weights $W\oldin\nolinebreak[4]A_{t}\cap RH_{t,s^{\prime}}$. By 6.5, this
condition is equivalent to
$W^{s^{\prime}}\oldin\nolinebreak[4]A_{t,s^{\prime}t}$. Observe that
$t=\frac{p}{p_{0}}\leq p$. Therefore

$a=\frac{p}{t-1}\geq\frac{t}{t-1}=t^{\prime},$

and consequently $a^{\prime}\leq t$. Since
$\frac{b}{a}=\frac{s^{\prime}t}{t^{\prime}}$, by 6.6 we deduce that the
condition $W^{s^{\prime}}\oldin\nolinebreak[4]A_{a^{\prime},b}$ featuring in
1.3 is weaker than the condition $W\oldin\nolinebreak[4]A_{t}\cap
RH_{t,s^{\prime}}$. Observe also that $a^{\prime}=t$ if and only if $p_{0}=1$.
Therefore, if $p_{0}>1$ and $d\geq 2$, then using 6.8 one can find a $d\times
d$ matrix weight $W$ on ${\mathbb{R}}^{n}$ with
$W^{s^{\prime}}\oldin\nolinebreak[4]A_{a^{\prime},b}$ but $W\notin A_{t}\cap
RH_{t,s^{\prime}}$.

Let us now prove 6.8 and 6.9. The core of both proofs boils down to
constructing appropriate families of counterexamples to the Cordes inequality.
One such family is given in the lemma below.

######  Lemma 6.10.

Set

$C_{n}:=\begin{bmatrix}1&0\\\ 0&n\end{bmatrix},\quad
D_{n}:=\begin{bmatrix}1&\frac{1}{2n^{1/2}}\\\
\frac{1}{2n^{1/2}}&\frac{1}{n}\end{bmatrix},\quad n=1,2,\ldots.$

Then, $C_{n},D_{n}\oldin\nolinebreak[4]\mathrm{P}_{2}({\mathbb{C}})$ with

$\operatorname{tr}(C_{n}D_{n}),\leavevmode\nobreak\
\operatorname{tr}(C_{n}^{-1}D_{n}^{-1})\simeq 1,$

for all $n=1,2,\ldots$. Moreover, for all $a>1$,

$\lim_{n\to\infty}\operatorname{tr}(C_{n}^{a}D_{n}^{a})=\infty.$

We apply 6.10 to prove 6.11. For completeness, we give a proof of 6.10 in
Section 8, together with a second family of counterexamples to the converse of
the Cordes inequality. The latter family allows one to prove in a similar way
6.9. Thus, we concentrate on proving 6.8. We only prove part (1) of 6.8; it is
clear that part (2) can be proved similarly.

Going forward, by “interval” we always mean a left-closed, right-open, non-
trivial bounded interval in ${\mathbb{R}}$. Before proving part (1) of 6.8, we
construct an analogous counterexample on the unit interval $I_{0}:=[0,1)$. A
_weight_ $W$ on $I_{0}$ is an integrable function
$W:I_{0}\to\mathrm{M}_{d}({\mathbb{C}})$ with
$W(x)\oldin\nolinebreak[4]\mathrm{P}_{d}({\mathbb{C}})$ for a.e.
$x\oldin\nolinebreak[4]I_{0}$. Moreover, for $1<p\leq q<\infty$ we denote

$[W]_{A_{p,q}(I_{0})}:=\sup_{I\subseteq
I_{0}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|W(x)^{1/q}W(y)^{-1/q}|^{p^{\prime}}\,\mathrm{d}y\right)^{q/p^{\prime}}\,\mathrm{d}x,$

and we define $[W]_{A_{p,q},{\mathcal{D}}(I_{0})}$ analogously.

######  Lemma 6.11.

Let $1<p_{1}\leq q_{1}<\infty$. Then, there exists a $2\times 2$ matrix weight
$W$ on $I_{0}$ with $[W]_{A_{p_{1},q_{1}}(I_{0})}<\infty$ such that for all
$1<p_{2}\leq q_{2}<\infty$ with $p_{1}<p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$, we have
$[W]_{A_{p_{2},q_{2}},{\mathcal{D}}(I_{0})}=\infty$.

###### Proof.

Throughout the proof, we suppress from the notation dependence of implied
constants on $p_{1}$ and $q_{1}$.

For all positive integers $n$, consider the matrices $C_{n},D_{n}$ from 6.10
and set

$A_{n}:=C_{n}^{-q_{1}/2},\quad B_{n}:=D_{n}^{q_{1}/2}.$

Notice that for all $1<p_{2}\leq q_{2}<\infty$ with $p_{1}<p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$ we have

$|A_{n}^{-1/q_{2}}B_{n}^{1/q_{2}}|=|C_{n}^{q_{1}/(2q_{2})}D_{n}^{q_{1}/(2q_{2})}|\simeq\operatorname{tr}(C_{n}^{q_{1}/q_{2}}D_{n}^{q_{1}/q_{2}})^{1/2},$

for all $n=1,2,\ldots$, therefore since
$\frac{q_{1}}{q_{2}}=\frac{p_{1}^{\prime}}{p_{2}^{\prime}}>1$, 6.10 implies
$\lim_{n\to\infty}|A_{n}^{-1/q_{2}}B_{n}^{1/q_{2}}|=\infty$.

Set now

$W:=\sum_{n=0}^{\infty}(\mathbbm{1}_{(J_{n})_{-}}A_{n+1}+\mathbbm{1}_{(J_{n})_{+}}B_{n+1}),$

where

$J_{n}:=\left[\frac{1}{2^{n+1}},\frac{1}{2^{n}}\right),\quad n=0,1,2,\ldots,$

and for all intervals $I=[a,b)\subseteq{\mathbb{R}}$ we denote

$I_{-}:=\left[a,\frac{a+b}{2}\right),\quad
I_{+}:=\left[\frac{a+b}{2},b\right).$

Notice that for all $1<p_{2}\leq q_{2}<\infty$ with $p_{1}<p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$ we have

$\displaystyle[W]_{A_{p_{2},q_{2}}(I_{0})}$
$\displaystyle\geq\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{J_{n}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{J_{n}}|W(x)^{1/q_{2}}W(y)^{-1/q_{2}}|^{p_{2}^{\prime}}\,\mathrm{d}y\right)^{q_{2}/p_{2}^{\prime}}\,\mathrm{d}x$
$\displaystyle\gtrsim_{q_{2},p_{2}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{(J_{n})_{+}}\left(\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{(J_{n})_{-}}|W(x)^{1/q_{2}}W(y)^{-1/q_{2}}|^{p_{2}^{\prime}}\,\mathrm{d}y\right)^{q_{2}/p_{2}^{\prime}}\,\mathrm{d}x$
$\displaystyle=|B_{n+1}^{1/q_{2}}A_{n+1}^{-1/q_{2}}|^{q_{2}}=|A_{n+1}^{-1/q_{2}}B_{n+1}^{1/q_{2}}|^{q_{2}},$

for all $n=0,1,2,\ldots$, therefore
$[W]_{A_{p_{2},q_{2}},{\mathcal{D}}(I_{0})}=\infty$.

It remains now to show that $[W]_{A_{p_{1},q_{1}}(I_{0})}<\infty$. We show a
slightly stronger property, that is:

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,$

for all subintervals $I$ of $I_{0}$.

Let us first show that for all $k=0,1,2,\ldots$ we have

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,$

where

$I_{k}:=\left[0,\frac{1}{2^{k}}\right),\quad k=0,1,2\ldots.$

Let $k$ be any nonnegative integer. Observe that
$I_{k}=\bigcup_{i=k}^{\infty}J_{k}$. Thus, it suffices to prove that

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]J_{i}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,\quad\forall i=k,k+1,\ldots.$

We show the following more general estimate, which is later useful for the
proof of 6.8:

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]J_{i}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim\max\left(1,\frac{k+1}{i+1}\right)^{p_{1}^{\prime}/2},\quad\forall
i=0,1,2,\ldots.$

Let $i$ be any nonnegative integer. For all
$x\oldin\nolinebreak[4](J_{i})_{-}$, we compute

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\sum_{j=k}^{\infty}2^{k-j}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{J_{j}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle\begin{split}&\simeq\sum_{j=k}^{\infty}2^{k-j}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{(J_{j})_{-}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\\\
&\qquad\qquad+\sum_{j=k}^{\infty}2^{k-j}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{(J_{j})_{+}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\end{split}$
$\displaystyle=\sum_{j=k}^{\infty}2^{k-j}|A_{i+1}^{1/q_{1}}A_{j+1}^{-1/q_{1}}|^{p_{1}^{\prime}}+\sum_{j=k}^{\infty}2^{k-j}|A_{i+1}^{1/q_{1}}B_{j+1}^{-1/q_{1}}|^{p_{1}^{\prime}}$
$\displaystyle\simeq\sum_{j=k}^{\infty}2^{k-j}\operatorname{tr}(C_{i+1}^{-1}C_{j+1})^{p_{1}^{\prime}/2}+\sum_{j=k}^{\infty}2^{k-j}\operatorname{tr}(C_{i+1}^{-1}D_{j+1}^{-1})^{p_{1}^{\prime}/2}.$

A similar computation yields

$\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\simeq\sum_{j=k}^{\infty}2^{k-j}\operatorname{tr}(D_{i+1}C_{j+1})^{p_{1}^{\prime}/2}+\sum_{j=k}^{\infty}2^{k-j}\operatorname{tr}(D_{i+1}D_{j+1}^{-1})^{p_{1}^{\prime}/2},$

for all $x\oldin\nolinebreak[4](J_{i})_{+}$. Direct computation gives

$\displaystyle\operatorname{tr}(C_{m}^{-1}C_{n})\simeq\max\left(1,\frac{n}{m}\right),\quad\operatorname{tr}(D_{m}C_{n})\simeq\max\left(1,\frac{n}{m}\right),$
$\displaystyle\operatorname{tr}(C_{m}^{-1}D_{n}^{-1})\simeq\max\left(1,\frac{n}{m}\right),\quad\operatorname{tr}(D_{m}D_{n}^{-1})\simeq\max\left(1,\frac{n}{m}\right)$

for all $m,n=1,2,\ldots$. We distinguish two cases:

  1. Case 1.

$i\geq k$. Then, for all $x\oldin\nolinebreak[4]J_{i}$, we have

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle\simeq\sum_{j=k}^{i}2^{k-j}+\sum_{j=i+1}^{\infty}2^{k-j}\left(\frac{j+1}{i+1}\right)^{p_{1}^{\prime}/2}$
$\displaystyle\lesssim
1+\sum_{j=k}^{\infty}2^{k-j}\left(\frac{j+1}{k+1}\right)^{p_{1}^{\prime}/2}=1+\sum_{j=k}^{\infty}2^{k-j}\left(\frac{j-k}{k+1}+1\right)^{p_{1}^{\prime}/2}$
$\displaystyle\lesssim
1+\sum_{j=k}^{\infty}2^{k-j}((j-k)^{p_{1}^{\prime}/2}+1)\lesssim 1.$

  2. Case 2.

$i<k$. Then, for all $x\oldin\nolinebreak[4]J_{i}$, we have

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle\simeq\sum_{j=k}^{\infty}2^{k-j}\left(\frac{j+1}{i+1}\right)^{p_{1}^{\prime}/2}$
$\displaystyle\lesssim\frac{1}{(i+1)^{p_{1}^{\prime}/2}}\sum_{j=k}^{\infty}2^{k-j}((j-k)^{p_{1}^{\prime}/2}+(k+1)^{p_{1}^{\prime}/2})$
$\displaystyle\lesssim\left(\frac{k+1}{i+1}\right)^{p_{1}^{\prime}/2}.$

Let now $I$ by any subinterval of $I_{0}$. Let $k$ be the largest nonnegative
integer with $I\subseteq I_{k}$. If $|I|\geq\frac{|I_{k}|}{512}$, then clearly

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

Assume now that $|I|<\frac{|I_{k}|}{512}$. Then, at least one of the two must
hold:

  * •

$I\subseteq J_{k}=(J_{k})_{+}\cup(J_{k})_{-}$

  * •

$I\subseteq(J_{k+1})_{+}\cup(J_{k})_{-}$.

And since

$\operatorname{tr}(C_{i}D_{j}),\leavevmode\nobreak\
\operatorname{tr}(C_{i}^{-1}D_{j}^{-1}),\leavevmode\nobreak\
\operatorname{tr}(C_{i}C_{j}^{-1}),\leavevmode\nobreak\
\operatorname{tr}(D_{i}D_{j}^{-1})\simeq 1$

for all $i,j\oldin\nolinebreak[4]\\{k+1,k+2\\}$, it easily follows that

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,$

concluding the proof. ∎

To extend the counterexample of 6.11 to the whole real line, we use a standard
trick that was also used in [Kak19, Lemma 2.2].

###### Proof (of part (1) of 6.8).

Throughout the proof, we suppress from the notation dependence of implied
constants on $p_{1}$ and $q_{1}$.

Let $W$ be the $2\times 2$ matrix weight on $I_{0}$ from the proof of 6.11.
Consider the function
$\widetilde{W}:{\mathbb{R}}\to\mathrm{M}_{2}({\mathbb{R}})$ given by

$\widetilde{W}(x):=w(x-k),\quad x\oldin\nolinebreak[4](k,k+1),\quad\text{if
}k\text{ is even},$ $\widetilde{W}(x):=w(k+1-x),\quad
x\oldin\nolinebreak[4](k,k+1),\quad\text{if }k\text{ is odd},$
$\widetilde{W}(k):=I\oldin\nolinebreak[4]M_{2}({\mathbb{R}}),\quad
k\oldin\nolinebreak[4]{\mathbb{Z}}.$

It is obvious that $[\widetilde{W}]_{A_{p_{2},q_{2}},{\mathcal{D}}}=\infty$,
for all $1<p_{2}\leq q_{2}<\infty$ with $p_{1}<p_{2}$ and
$\frac{q_{2}}{p_{2}^{\prime}}=\frac{q_{1}}{p_{1}^{\prime}}$. It remains to
show that $[\widetilde{W}]_{A_{p_{1},q_{1}}}<\infty$. We prove a slightly
stronger fact, namely:

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,$

for all intervals $I\subseteq{\mathbb{R}}$.

First of all, let $m,n\oldin\nolinebreak[4]{\mathbb{Z}}$ with $m<n$ be
arbitrary. We check that

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4][m,n)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[m,n)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

Since $[m,n)=\bigcup_{k=m}^{n-1}[k,k+1)$, we may compute

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4][m,n)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[m,n)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{0}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[m,n)}|W(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{0}}\sum_{k=m}^{n-1}\frac{1}{n-m}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[k,k+1)}|W(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{0}}\sum_{k=m}^{n-1}\frac{1}{n-m}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{0}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{0}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{0}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

Moreover, let $k,\ell$ be arbitrary non-negative integers. We check that

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

First of all, we have

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},0\right)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}$
$\displaystyle|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle\begin{split}&\leq\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},0\right)}\frac{2^{-k}}{2^{-k}+2^{-\ell}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},0\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\\\
&\qquad\qquad+\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},0\right)}\frac{2^{-\ell}}{2^{-k}+2^{-\ell}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[0,\frac{1}{2^{\ell}}\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\end{split}$
$\displaystyle\begin{split}&=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\frac{2^{\ell}}{2^{k}+2^{\ell}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\\\
&\qquad\qquad+\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\frac{2^{k}}{2^{k}+2^{\ell}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{\ell}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y.\end{split}$

Since $I_{k}=\bigcup_{i=k}^{\infty}J_{i}$, by the proof of 6.11 we deduce

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{k}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1$

and

$\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I_{k}}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I_{\ell}}|W(x)^{1/q_{1}}W(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim\max\left(1,\frac{\ell+1}{k+1}\right)^{p_{1}^{\prime}/2}.$

It follows that

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},0\right)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1+\frac{2^{k}}{2^{k}+2^{\ell}}\max\left(1,\frac{\ell+1}{k+1}\right)^{p_{1}^{\prime}/2}.$

Analogously, we obtain

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[0,\frac{1}{2^{\ell}}\right)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1+\frac{2^{\ell}}{2^{k}+2^{\ell}}\max\left(1,\frac{k+1}{\ell+1}\right)^{p_{1}^{\prime}/2}.$

If $k<\ell$, then

$\displaystyle\frac{2^{k}}{2^{k}+2^{\ell}}\left(\frac{\ell+1}{k+1}\right)^{p_{1}^{\prime}/2}\leq
2^{k-\ell}\left(\frac{\ell-k}{k+1}+1\right)^{p_{1}^{\prime}/2}\lesssim
2^{k-\ell}(\ell-k)^{p_{1}^{\prime}/2}\lesssim 1.$

The case $\ell<k$ is symmetric, thus we deduce the desired estimate.

Let now $I$ be any interval in ${\mathbb{R}}$. We distinguish the following
cases:

  1. Case 1.

There exists $m\oldin\nolinebreak[4]{\mathbb{Z}}$ with $I\subseteq[m,m+1)$. It
is then immediate by 6.11 that

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

  2. Case 2.

There does not exist $m\oldin\nolinebreak[4]{\mathbb{Z}}$ with
$I\subseteq[m,m+1)$, and $|I|\geq\frac{1}{2}$. Then, there exist
$m,n\oldin\nolinebreak[4]{\mathbb{Z}}$ with $m+1<n$ such that
$I\subseteq[m,n)$ and $I\nsubseteq[m+1,n)$, $I\nsubseteq[m,n-1)$. It is clear
that $|I|\simeq|[m,n)|$, thus

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4][m,n)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{[m,n)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

  3. Case 3.

There does not exist $m\oldin\nolinebreak[4]{\mathbb{Z}}$ with
$I\subseteq[m,m+1)$, and $|I|<\frac{1}{2}$. Then, there exists
$m\oldin\nolinebreak[4]{\mathbb{Z}}$ with $I\subseteq[m,m+2)$ and
$m+1\oldin\nolinebreak[4]\mathrm{Int}(I)$.

If $m+1$ is odd, that is $m$ is even, then it is clear that

$\displaystyle\widetilde{W}(x)\oldin\nolinebreak[4]\\{A_{1},B_{1}\\},\quad\text{for
a.e. }x\oldin\nolinebreak[4]I,$

therefore we trivially deduce

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1.$

If $m+1$ is even, that is $m$ is odd, then it is easy to see that there exist
positive integers $k,\ell$ with

$I\subseteq J:=\left[m+1-\frac{1}{2^{k}},m+1+\frac{1}{2^{\ell}}\right)$

and $|I|\geq\frac{1}{2}|J|$, therefore

$\displaystyle\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]I}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{I}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle\lesssim\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]J}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{J}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y$
$\displaystyle=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int_{\left[-\frac{1}{2^{k}},\frac{1}{2^{\ell}}\right)}|\widetilde{W}(x)^{1/q_{1}}\widetilde{W}(y)^{-1/q_{1}}|^{p_{1}^{\prime}}\,\mathrm{d}y\lesssim
1,$

concluding the proof.

∎

##  7\. Limited range extrapolation for matrix weights

In this section, we establish a limited-range extrapolation theorem for matrix
weights. Although the proof is a straightforward adaptation of [BC22, Theorem
9.1] guided by [AM07, Theorem 4.9], we carefully track all details. We
restrict ourselves to ${\mathbb{R}}^{n}$ equipped with the Euclidean metric
and $n$-dimensional Lebesgue measure as underlying space. In particular, we
simply write $\mathrm{d}x$ for integration under the Lebesgue measure, and
$|E|$ for the $n$-dimensional Lebesgue measure of any Lebesgue-measurable
subset $E$ of ${\mathbb{R}}^{n}$.

###  7.1. Preliminaries

Here we briefly recall some important definitions and facts from [BC22], where
we refer for details and proofs.

####  7.1.1. Integration of convex set valued maps

We consider a complete $\sigma$-finite measure space
$(\Omega,{\mathcal{A}},\mu)$.

######  Definition 7.1.

Let $F:\Omega\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$ be an
${\mathcal{A}}$-measurable map. We say that this map is _$\mu$ -integrably
bounded_, if there exists a $\mu$-integrable function $k:\Omega\to[0,\infty)$
with $F(\omega)\subseteq k(\omega)\overline{\mathbf{B}}$, for all
$\omega\oldin\nolinebreak[4]\Omega$.

Observe that an ${\mathcal{A}}$-measurable map
$F:\Omega\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$ is
$\mu$-integrably bounded if and only if the function $|F|:\Omega\to[0,\infty)$
is $\mu$-integrable.

######  Definition 7.2.

Let $F:\Omega\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$ be a
$\mu$-integrably bounded map. Following [BC22, Definition 3.11], we define the
_Aumann $\mu$-integral of $F$ over $\Omega$_,
$\int_{\Omega}F(\omega)\,\mathrm{d}\mu(\omega)$, as the following subset of
${\mathbb{C}}^{d}$:

$\int_{\Omega}F(\omega)\,\mathrm{d}\mu(\omega):=\left\\{\int_{\Omega}f(\omega)\,\mathrm{d}\mu(\omega)\,:\,f:\Omega\to{\mathbb{C}}^{d}\;\mu\text{-integrable
selection function for }F\right\\}.$

Let $F:\Omega\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$ be a
$\mu$-integrably bounded map. In [BC22, Section 3] it is proved that the
Aumann integral $\int_{X}F(\omega)\,\mathrm{d}\mu(\omega)$ is a convex body in
${\mathbb{C}}^{d}$. For all $A\oldin\nolinebreak[4]{\mathcal{A}}$, we denote

$\int_{A}F(\omega)\,\mathrm{d}\mu(\omega):=\int_{X}\mathbbm{1}_{A}(\omega)F(\omega)\,\mathrm{d}\mu(\omega)=\int_{A}F(\omega)\,\mathrm{d}\mu|_{A}(\omega).$

######  Remark 7.3.

Let $F:\Omega\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$ be a
$\mu$-integrably bounded map. It is proved in [BC22, Lemma 3.19] that
$\int_{X}F(\omega)\,\mathrm{d}\mu(\omega)=\\{0\\}$ if and only if
$F(\omega)=\\{0\\}$ for $\mu$-a.e. $\omega\oldin\nolinebreak[4]\Omega$.

####  7.1.2. Convex body valued maximal function

######  Definition 7.4.

  1. (1)

For any measurable function
$F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$ and for
any $d\times d$ matrix weight $W$ on ${\mathbb{R}}^{n}$, we denote

$\lVert
F\rVert_{L^{p}_{{\mathcal{K}}}(W)}:=\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/p}F(x)|^{p}\,\mathrm{d}x\right)^{1/p},\quad
0<p<\infty,$

and we also denote

$\lVert
F\rVert_{L^{\infty}_{{\mathcal{K}}}(W)}:=\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]{\mathbb{R}}^{n}}|W(x)F(x)|.$

Then, for any $0<p\leq\infty$, we denote by $L^{p}_{{\mathcal{K}}}(W)$ the
family of all (Lebesgue-a.e. equivalence classes of) measurable functions
$F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$ with
$\lVert F\rVert_{L^{p}_{{\mathcal{K}}}(W)}<\infty$. Observe that if
$F\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$ for some $0<p\leq\infty$,
then the subset $F(x)$ of ${\mathbb{C}}^{d}$ is bounded, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

  2. (2)

If $1\leq p,q\leq\infty$ and $W,V$ are $d\times d$ matrix weights on
${\mathbb{R}}^{n}$, then a map $T$ is said to be
$L^{p}_{{\mathcal{K}}}(W)$-$L^{q}_{{\mathcal{K}}}(V)$ bounded, if it is a map
$T:L^{p}_{{\mathcal{K}}}(W)\to L^{q}_{{\mathcal{K}}}(V)$ and there exists
$0<C<\infty$ such that

$\lVert Tf\rVert_{L^{q}_{{\mathcal{K}}}(V)}\leq C\lVert
f\rVert_{L^{p}_{{\mathcal{K}}}(W)},$

for all $f\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$; in this case, we
denote by $\|T\|_{L^{p}_{{\mathcal{K}}}(W)\to L^{q}_{{\mathcal{K}}}(V)}$ the
best such constant. Moreover, we denote
$\|T\|_{L^{p}_{{\mathcal{K}}}(W)}:=\|T\|_{L^{p}_{{\mathcal{K}}}(W)\to
L^{p}_{{\mathcal{K}}}(W)}$.

  3. (3)

A measurable map
$F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$ is said
to be _locally integrable_ if the function $|F|:{\mathbb{R}}^{n}\to[0,\infty]$
is locally integrable.

######  Remark 7.5.

If $F\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$ for some $1<p<\infty$ and
some $d\times d$ matrix weight $W$ such that $W^{1/(p-1)}$ is also a $d\times
d$ matrix weight, then it is easy to see that $F$ is locally integrable on
${\mathbb{R}}^{n}$.

######  Definition 7.6.

  1. (1)

A map $T$ acting from a space of
${\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$-valued functions on
${\mathbb{R}}^{n}$ that is closed under the (pointwise) Minkowski sum and
(pointwise) multiplication with complex numbers into a space of
${\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$-valued functions on
${\mathbb{R}}^{n}$ is said to be _sublinear_ , if for all $F,G$ in the domain
of definition of $T$ and for all $a\oldin\nolinebreak[4]{\mathbb{C}}$ we have

$T(F+G)(x)\subseteq TF(x)+TG(x)$

and

$T(aF)(x)=aTF(x)=|a|TF(x),$

for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

  2. (2)

A map $T$ acting from a space of
${\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$-valued functions on
${\mathbb{R}}^{n}$ into a space of
${\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$-valued functions on
${\mathbb{R}}^{n}$ is said to be _monotone_ if $TF(x)\subseteq TG(x)$ for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$, whenever $F,G$ are in the domain of
definition of $T$ such that $F(x)\subseteq G(x)$ for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

######  Definition 7.7.

If $F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$
is a locally integrable function, then following [BC22] we define its _convex-
set-valued maximal function_ by

${\mathcal{M}}^{{\mathcal{K}}}F(x):=\mathrm{clos}\bigg{(}\mathrm{conv}\bigg{(}\bigcup_{\begin{subarray}{c}Q\text{
cube in }{\mathbb{R}}^{n}\\\ x\oldin\nolinebreak[4]Q\end{subarray}}\langle
F\rangle_{Q}\bigg{)}\bigg{)}$

where $\langle F\rangle_{Q}:=\frac{1}{|Q|}\int_{Q}F(x)\,\mathrm{d}x$. Observe
that
${\mathcal{M}}^{{\mathcal{K}}}F(x)\oldin\nolinebreak[4]{\mathcal{K}}_{\mathrm{cs}}({\mathbb{C}}^{d})$
for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. It is proved in [BC22,
Section 5] that this gives rise to an operator ${\mathcal{M}}^{{\mathcal{K}}}$
that is sublinear and monotone moreover satisfies
$F(x)\subseteq{\mathcal{M}}^{{\mathcal{K}}}F(x)$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

######  Remark 7.8.

Let $F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$
be a locally integrable function that is not a.e. equal to $\\{0\\}$. Then,
for all $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$, there exists a cube
$Q\subseteq{\mathbb{R}}^{n}$ with $x\oldin\nolinebreak[4]Q$ such that $F$ is
not a.e. equal to $\\{0\\}$ on $Q$, therefore by 7.3 we deduce $\langle
F\rangle_{Q}\neq\\{0\\}$. It follows that
${\mathcal{M}}^{{\mathcal{K}}}F(x)\neq\\{0\\}$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

We need the following weighted bound for the maximal function
${\mathcal{M}}^{{\mathcal{K}}}$ proved in [BC22].

######  Theorem E ([BC22, Theorem 6.9]).

Let $1<p<\infty$ and let $W$ be a $d\times d$ matrix weight on
${\mathbb{R}}^{n}$. Then, ${\mathcal{M}}^{{\mathcal{K}}}$ acts boundedly on
$L^{p}_{{\mathcal{K}}}(W)$,
${\mathcal{M}}^{{\mathcal{K}}}:L^{p}_{{\mathcal{K}}}(W)\to
L^{p}_{{\mathcal{K}}}(W)$ with

$\lVert{\mathcal{M}}^{{\mathcal{K}}}\rVert_{L^{p}_{{\mathcal{K}}}(W)}\lesssim_{d,n,p}[W]_{A_{p}}^{1/(p-1)}.$

####  7.1.3. Rubio de Francia algorithm for convex body valued functions

######  Definition 7.9.

Let $W:{\mathbb{R}}^{n}\to\mathrm{M}_{d}({\mathbb{C}})$ be a measurable
function with values a.e. in the set of positive-definite $d\times d$ complex-
valued matrices. Following Bownik and Cruz-Uribe [BC22], we define

$[W]_{A_{1}}:=\sup_{Q}\operatorname*{ess\,sup}_{x\oldin\nolinebreak[4]Q}\mathchoice{{\vbox{\hbox{$\textstyle-$}}\kern-4.86108pt}}{{\vbox{\hbox{$\scriptstyle-$}}\kern-3.25pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-2.29166pt}}{{\vbox{\hbox{$\scriptscriptstyle-$}}\kern-1.875pt}}\\!\int|W^{-1}(x)W(y)|\,\mathrm{d}y,$

where the supremum ranges over all cubes $Q\subseteq{\mathbb{R}}^{n}$. If
$[W]_{A_{1}}<\infty$, then we say that $W$ is a $d\times d$ matrix $A_{1}$
weight, in symbols $W\oldin\nolinebreak[4]A_{1}$. Observe that
$[W]_{A_{1}}=[W^{-1}]_{A_{\infty}}$.

######  Definition 7.10.

Following [BC22], we say that a locally integrable function
$F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$
belongs to the $A_{1}^{{\mathcal{K}}}$ class, in symbols
$F\oldin\nolinebreak[4]A^{{\mathcal{K}}}_{1}$, if there exists $0<C<\infty$
such that

${\mathcal{M}}^{{\mathcal{K}}}F(x)\subseteq F(x),$

for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$; in this case, we denote the
infimum of all such constants $C$ by $[F]_{A_{1}^{{\mathcal{K}}}}$.

In what follows, we denote by $\overline{\mathbf{B}}$ the closed unit ball in
${\mathbb{C}}^{d}$. Bownik and Cruz-Uribe [BC22, Corollary 7.4] describe
precisely the relation between the $A_{1}^{{\mathcal{K}}}$ class and matrix
$A_{1}$ weights.

######  Lemma F ([BC22, Corollary 7.4]).

If $W$ is a $d\times d$ matrix weight on ${\mathbb{R}}^{n}$, then
$W\overline{\mathbf{B}}\oldin\nolinebreak[4]A_{1}^{{\mathcal{K}}}$ if and only
if $W\oldin\nolinebreak[4]A_{1}$.

Bownik and Cruz-Uribe [BC22, Theorem 7.6] establish a Rubio de Francia-type
algorithm to prove their extrapolation algorithm. This algorithm is of
fundamental importance also for us, so we recall it below. We refer to [BC22,
Theorem 7.6] for a detailed proof.

######  Theorem G ([BC22, Theorem 7.6]).

Let $1\leq p\leq\infty$ and let $W$ be a $d\times d$ matrix weight on
${\mathbb{R}}^{n}$. Let $T:L^{p}_{{\mathcal{K}}}(W)\to
L^{p}_{{\mathcal{K}}}(W)$ be a bounded sublinear and monotone map. Given
$G\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$, define $SG$ by

$SG:=\sum_{k=0}^{\infty}\frac{1}{2^{k}\lVert
T\rVert_{L^{p}_{{\mathcal{W}}}(K)}^{k}}T^{k}G,$

where $T^{0}G:=G$ and $T^{k}$ denotes the $k$-fold composition of $T$ with
itself. Then, this series converges in $L^{p}_{{\mathcal{K}}}(W)$. Moreover,
it has the following properties:

  1. (1)

$G(x)\subseteq SG(x)$, for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$

  2. (2)

$\lVert SG\rVert_{L^{p}_{{\mathcal{K}}}(W)}\leq 2\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(W)}$

  3. (3)

$T(SG)(x)\subseteq 2\lVert T\rVert_{L^{p}_{{\mathcal{K}}}(W)}SG(x)$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

######  Definition 7.11.

If $F:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\textnormal{bcs}}({\mathbb{C}}^{d})$
is a measurable function and $W:{\mathbb{R}}^{n}\to M_{d}({\mathbb{C}})$ is a
measurable function, then following [BC22] we way that $F$ is _ellipsoid-
valued with respect to $W$_ is there exists a function
$r:{\mathbb{R}}^{n}\to{\mathbb{R}}$ with $F(x)=r(x)W(x)$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

Cruz-Uribe and Bownik [BC22, Lemma 8.5] show that their Rubio de Francia
algorithm in G behaves well for input functions that are ellipsoid-valued with
respect to a given matrix function.

######  Lemma H ([BC22, Lemma 8.5]).

Let $V:{\mathbb{R}}^{n}\to M_{d}({\mathbb{C}})$ be a measurable function and
$T$ be an operator as in G such that whenever
$G\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$ is ellipsoid-valued with
respect to $V$, then $T(G)$ is also ellipsoid-valued with respect to $V$.
Then, $SG$ is also ellipsoid-valued with respect to $V$, for all
$G\oldin\nolinebreak[4]L^{p}_{{\mathcal{K}}}(W)$ that are ellipsoid-valued
with respect to $V$.

Finally, we need the following reverse Jones factorization type lemma for
matrix weights [BC22, Corollary 8.9], which is itself a special case of a much
more general result [BC22, Proposition 8.8].

######  Theorem I ([BC22, Corollary 8.9]).

Let $1<p<\infty$, and let $W$ be a $d\times d$ matrix weight on
${\mathbb{R}}^{n}$.

  1. (1)

Let $s:{\mathbb{R}}^{n}\to[0,\infty)$ be a measurable function such that
$W_{1}:=sW^{1/p}\oldin\nolinebreak[4]A_{\infty}$. Then, for all
$p_{0}\oldin\nolinebreak[4](p,\infty)$, we have

$\overline{W}:=s^{p_{0}-p}W^{p_{0}/p}\oldin\nolinebreak[4]A_{p_{0}},$

with

$[\overline{W}]_{A_{p_{0}}}\lesssim_{d,n,p,p_{0}}[W]_{A_{p}}[W_{1}]_{A_{\infty}}^{p_{0}-p}.$

  2. (2)

Let $r:{\mathbb{R}}^{n}\to[0,\infty)$ be a measurable function such that
$W_{0}:=rW^{1/p}\oldin\nolinebreak[4]A_{1}$. Then, for all
$p_{0}\oldin\nolinebreak[4](1,p)$, we have

$\overline{W}:=r^{p_{0}-p^{\prime}(p_{0}-1)}W^{p_{0}/p}\oldin\nolinebreak[4]A_{p_{0}},$

with

$[\overline{W}]_{A_{p_{0}}}\lesssim_{d,n,p,p_{0}}[W_{0}]_{A_{1}}^{p_{0}-p^{\prime}(p_{0}-1)}[W]_{A_{p}}^{(p_{0}-1)/(p-1)}.$

###  7.2. Statement of main result

In the rest of this section, ${\mathcal{F}}$ denotes a non-empty family of
pairs $(F,G)$ of measurable functions
$F,G:{\mathbb{R}}^{n}\to{\mathcal{K}}_{\textrm{bcs}}({\mathbb{C}}^{d})$ such
that neither $F$ nor $G$ is a.e. equal to $\\{0\\}$. Whenever we write an
inequality of the form

$\lVert F\rVert_{L^{p}_{{\mathcal{K}}}(W)}\leq C\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(W)},\quad(F,G)\oldin\nolinebreak[4]{\mathcal{F}}$

we understand that the inequality holds for all pairs
$(F,G)\oldin\nolinebreak[4]{\mathcal{F}}$ with $\lVert
F\rVert_{L^{p}_{{\mathcal{K}}}(W)}<\infty$. The constant $C$ depends on some
concrete Muckenhoupt characteristic of some concrete power of the weight $W$,
but not on the particular weight $W$ itself.

We need some notation. Let $0<p_{0}<q_{0}\leq\infty$. For all
$p\oldin\nolinebreak[4][p_{0},q_{0})$, we set

$t(p):=\frac{p}{p_{0}}\geq 1,\quad s(p):=\frac{q_{0}}{p}\geq 1,\quad
r(p):=s(p)^{\prime}(t(p)-1)+1\geq 1,$

where the usual convention $\infty^{\prime}=1$ is used; notice that
$1<s(p)\leq\infty$, so $1\leq(s(p))^{\prime}<\infty$. Observe that for all
$p,q\oldin\nolinebreak[4][p_{0},q_{0})$ with $p<q$, we have $r(p)<r(q)$.

######  1.6.

Let $0<p_{0}<q_{0}\leq\infty$. Assume that for some fixed
$p\oldin\nolinebreak[4][p_{0},q_{0}]$, there exists an increasing function
$K_{p}$ such that one of the following holds:

  1. (1)

If $p<q_{0}$, then

$\lVert F\rVert_{L^{p}_{{\mathcal{K}}}(W)}\leq
K_{p}([W^{s(p)^{\prime}}]_{A_{r(p)}})\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(W)},\quad(F,G)\oldin\nolinebreak[4]{\mathcal{F}},$

for all $d\times d$ matrix weights $W$ on ${\mathbb{R}}^{n}$ with
$[W^{s(p)^{\prime}}]_{A_{r(p)}}<\infty$.

  2. (2)

If $p=q_{0}<\infty$, then

$\lVert F\rVert_{L^{p}_{{\mathcal{K}}}(W)}\leq
K_{p}([W^{t(p)^{\prime}/t(p)}]_{A_{\infty}})\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(W)},\quad(F,G)\oldin\nolinebreak[4]{\mathcal{F}},$

for all $d\times d$ matrix weights $W$ on ${\mathbb{R}}^{n}$ with
$[W^{t(p)^{\prime}/t(p)}]_{A_{\infty}}<\infty$.

  3. (3)

If $p=q_{0}=\infty$, then

$\lVert F\rVert_{L^{p}_{{\mathcal{K}}}(W)}\leq
K_{p}([W^{1/p_{0}}]_{A_{\infty}})\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(W)},\quad(F,G)\oldin\nolinebreak[4]{\mathcal{F}},$

for all $d\times d$ matrix weights $W$ on ${\mathbb{R}}^{n}$ with
$[W^{1/p_{0}}]_{A_{\infty}}<\infty$.

Then, for all $q\oldin\nolinebreak[4](p_{0},q_{0})$,

$\lVert F\rVert_{L^{q}_{{\mathcal{K}}}(W)}\leq
K_{q}([W^{s(q)^{\prime}}]_{A_{r(q)}})\lVert
G\rVert_{L^{q}_{{\mathcal{K}}}(W)},\quad(F,G)\oldin\nolinebreak[4]{\mathcal{F}}$

holds, for all $d\times d$ matrix weights $W$ on ${\mathbb{R}}^{n}$ with
$[W^{s(q)^{\prime}}]_{A_{r(q)}}<\infty$, where

$K_{q}([W^{s(q)^{\prime}}]_{A_{r(q)}})=c_{1}K_{p}(c_{2}[W^{s(q)^{\prime}}]_{A_{r(q)}}^{\alpha(p,q)}),$

for some constants $c_{1},c_{2}>0$ depending only on $n,d,p_{0},q_{0},p$ and
$q$, and the exponent $\alpha(p,q)$ is given by

(7.1) $\alpha(p,q):=\begin{cases}\frac{r(p)-1}{r(q)-1},\text{ if }q<p<q_{0}\\\
\\\ \frac{1}{r(q)-1},\text{ if }q<p=q_{0}\\\ \\\ 1,\text{ if }p\leq
q\end{cases}.$

######  Remark 7.12.

Given the comparison results proved in Section 6, the conditions on the
weights considered in the above theorem are reasonable. In particular, in the
case $d=1$ we recover Theorem C. Moreover, in the case $p_{0}=1$ and
$q_{0}=\infty$ we recover [BC22, Theorem 9.1].

###  7.3. Proof of main result

Here we prove 1.6 in detail. We first need the following technical lemma.

######  Lemma 7.13.

Let $p,q\oldin\nolinebreak[4][p_{0},q_{0})$. The following two identities are
true:

(7.2) $s(p)^{\prime}s(q)^{\prime}(p-q)=ps(p)^{\prime}-qs(q)^{\prime}$

and

(7.3) $\frac{r(p)}{r(q)}=\frac{ps(p)^{\prime}}{qs(q)^{\prime}}.$

###### Proof.

If $q_{0}=\infty$, then $s(p)^{\prime}=s(q)^{\prime}=1$, so the first identity
holds trivially, and the second says $\frac{t(p)}{t(q)}=\frac{p}{q}$, which
also holds trivially because $t(p)=\frac{p}{p_{0}}$ and
$t(q)=\frac{q}{p_{0}}$.

Now we prove the identities under the assumption that $q_{0}<\infty$. For the
first identity we have

$s(p)^{\prime}=(q_{0}/p)^{\prime}=\frac{q_{0}}{q_{0}-p},\quad
s(q)^{\prime}=\frac{q_{0}}{q_{0}-q},$

so what we want to see becomes

$\frac{q_{0}^{2}(p-q)}{(q_{0}-p)(q_{0}-q)}=\frac{pq_{0}}{q_{0}-p}-\frac{qq_{0}}{q_{0}-q}.$

Computing

$pq_{0}(q_{0}-q)-qq_{0}(q_{0}-p)=q_{0}^{2}p-pqq_{0}-qq_{0}^{2}+pqq_{0}=q_{0}^{2}(p-q),$

we obtain the first identity.

The second identity can be equivalently rewritten as

$\displaystyle qs(q)^{\prime}r(p)=ps(p)^{\prime}r(q),$

that is

$qs(q)^{\prime}[s(p)^{\prime}(t(p)-1)+1]=ps(p)^{\prime}[s(q)^{\prime}(t(q)-1)+1],$

that is after expanding and rearranging

$s(p)^{\prime}s(q)^{\prime}(qt(p)-q-pt(q)+p)=ps(p)^{\prime}-qs(q)^{\prime}.$

We compute

$qt(p)-q-pt(q)+p=\frac{qp}{p_{0}}-q-\frac{pq}{p_{0}}+p=p-q.$

Thus, the second identity is equivalent to the first one, which concludes the
proof. ∎

###### Proof (of 1.6).

We combine and adapt the proofs of [AM07, Theorem 4.9] and [BC22, Theorem
9.1].

Let $q\oldin\nolinebreak[4](p_{0},q_{0})$, $W$ be a $d\times d$ matrix weight
with $[W^{s(q)^{\prime}}]_{A_{r(q)}}<\infty$, and
$(F,G)\oldin\nolinebreak[4]{\mathcal{F}}$ with $0<\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}<\infty$. We may assume that $q\neq p$. We
may also assume that $0<\lVert F\rVert_{L^{q}_{{\mathcal{K}}}(W)}<\infty$,
otherwise the inequality to show is trivial. Throughout the proof, we suppress
the notation dependence of implied constants on $n,d,p_{0},q_{0},p$, and $q$.
Moreover, $c,c_{1},c_{2}$ are positive finite constants depending only on
these parameters which might change from line to line.

We now distinguish several cases.

$q<p<q_{0}$. We first need to define an associated Rubio de Francia operator.
Thus, consider the following auxiliary operator $N_{W}$ acting on functions
$H\oldin\nolinebreak[4]L^{r(q)}_{{\mathcal{K}}}(V)$:

$N_{W}H(x):=|V(x)^{1/r(q)}H(x)|W(x)^{-1/r(q)}\overline{\mathbf{B}},\quad\text{for
a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

where $V:=W^{s(q)^{\prime}}$. For this operator we may compute

$\displaystyle\lVert N_{W}H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}=\lVert
H\rVert_{L^{r(q)}_{{\mathcal{K}}}(V)}.$

It is easy to see that $N_{W}$ is sublinear and monotone.

Furthermore, we claim that

$H(x)\subseteq\Sigma(x)N_{W}H(x),\quad\text{for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

where $\Sigma:=W^{1/r(q)}\cdot V^{-1/r(q)}=W^{(1-s(q)^{\prime})/r(q)}$.
Indeed, this inclusion is equivalent to

$H(x)\subseteq|V(x)^{1/r(q)}H(x)|V(x)^{-1/r(q)}\overline{\mathbf{B}},\quad\text{for
a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

that is

$V(x)^{1/r(q)}H(x)\subseteq|V(x)^{1/r(q)}H(x)|\overline{\mathbf{B}},\quad\text{for
a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

which is obvious.

Now we define the operator $P_{W}$ acting on functions
$H\oldin\nolinebreak[4]L^{r(q)}_{{\mathcal{K}}}(W)$ by

$P_{W}H(x):=N_{W}({\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H))(x)=|V(x)^{1/r(q)}{\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H)(x)|W(x)^{-1/r(q)}\overline{\mathbf{B}},\quad\text{for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Then, using E we can compute

$\displaystyle\lVert
P_{W}H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}=\lVert{\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H)\rVert_{L^{r(q)}_{{\mathcal{K}}}(V)}\lesssim[V]_{A_{r(q)}}^{1/(r(q)-1)}\lVert\Sigma
H\rVert_{L^{r(q)}_{{\mathcal{K}}}(V)}=[V]_{A_{r(q)}}^{1/(r(q)-1)}\lVert
H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}.$

Since ${\mathcal{M}}^{{\mathcal{K}}}$ is itself sublinear and monotone, it is
easy to see that $P_{W}$ is sublinear and monotone. (Although we do not need
it, we note that it is easy to see that $H(x)\subseteq P_{W}H(x)$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.) Note also that if $H$ is not a.e.
equal to $\\{0\\}$, then $P_{W}H(x)\neq\\{0\\}$, for a.e.
$x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

All assumptions of G are fulfilled, thus we can define the following Rubio de
Francia type operator acting on functions
$H\oldin\nolinebreak[4]L^{r(q)}_{{\mathcal{K}}}(W)$:

${\mathcal{R}}_{W}H:=\sum_{k=0}^{\infty}\frac{1}{2^{k}\lVert
P_{W}\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}^{k}}(P_{W})^{k}H.$

We record the following properties of ${\mathcal{R}}_{W}$ that we obtain from
G:

$H(x)\subseteq{\mathcal{R}}_{W}H(x),$

for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$,

$\lVert{\mathcal{R}}_{W}H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}\leq 2\lVert
H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}$

and

$P_{W}({\mathcal{R}}_{W}H)(x)\subseteq\lVert
P_{W}\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}{\mathcal{R}}_{W}H(x)\subseteq
c[V]_{A_{r(q)}}^{1/(r(q)-1)}{\mathcal{R}}_{W}H(x),$

for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. In particular, we have

${\mathcal{M}}^{{\mathcal{K}}}(\Sigma{\mathcal{R}}_{W}H)(x)\subseteq\Sigma(x)P_{W}({\mathcal{R}}_{W}H)(x)\subseteq
c[V]_{A_{r(q)}}^{1/(r(q)-1)}\Sigma(x){\mathcal{R}}_{W}H(x),$

for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. This means that if $H$ is
not a.e. equal to $\\{0\\}$, then the convex body valued function
$\Sigma{\mathcal{R}}_{W}H$ belongs to the $A^{1}_{{\mathcal{K}}}$ class with
$[\Sigma{\mathcal{R}}_{W}H]_{A^{1}_{{\mathcal{K}}}}\leq
c[V]_{A_{r(q)}}^{1/(r(q)-1)}$.

Finally, notice that $P_{W}H$ is always ellipsoid-valued with respect to the
matrix valued function $W^{-1/r(q)}$, therefore, by H, we deduce that if $H$
happens to be ellipsoid-valued with respect to $W^{-1/r(q)}$, then
${\mathcal{R}}_{W}H$ is also ellipsoid-valued with respect to $W^{-1/r(q)}$.

We now define the scalar-valued function

$\bar{h}(x):=\left(\frac{|W(x)^{1/q}F(x)|}{\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}}+\frac{|W(x)^{1/q}G(x)|}{\lVert
G\rVert_{L^{q}_{{\mathcal{K}}}(W)}}\right)^{q/r(q)},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

It is obvious that

$\lVert\bar{h}\rVert_{L^{r(q)}}\lesssim 1.$

We consider the convex body valued function

$\bar{H}(x):=\bar{h}(x)W(x)^{-1/r(q)}\overline{\mathbf{B}},\quad\text{ for
a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

It is clear that $\lVert H\rVert_{L^{r(q)}_{{\mathcal{K}}}(W)}\leq 2$. Since
$\bar{H}$ is ellipsoid-valued with respect to $W^{-1/r(q)}$, we deduce that
${\mathcal{R}}_{W}\bar{H}$ is also ellipsoid-valued with respect to
$W^{-1/r(q)}$, therefore we can write
${\mathcal{R}}_{W}\bar{H}(x)={\mathcal{R}}_{W}\bar{h}(x)W(x)^{-1/r(q)}\overline{\mathbf{B}}$
for some nonnegative scalar-valued function ${\mathcal{R}}_{W}\bar{h}$. Then,
we have

$\bar{h}(x)W(x)^{-1/r(q)}\overline{\mathbf{B}}\subseteq{\mathcal{R}}_{W}\bar{h}(x)W(x)^{-1/r(q)}\overline{\mathbf{B}},$

which implies immediately that $\bar{h}(x)\leq{\mathcal{R}}_{W}\bar{h}(x)$,
for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. We now define the
nonnegative scalar-valued function

$h:=({\mathcal{R}}_{W}\bar{h})^{r(q)/q}.$

Since $\bar{H}$ is not a.e. equal to $\\{0\\}$, we deduce that $h(x)\neq 0$,
for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. We compute

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}=\left(\int_{{\mathbb{R}}^{n}}|h(x)^{-1/(p/q)^{\prime}}W(x)^{1/q}F(x)|^{q}\cdot
h(x)^{q/(p/q)^{\prime}}\,\mathrm{d}x\right)^{1/q}$
$\displaystyle\leq\left(\int_{{\mathbb{R}}^{n}}|h(x)^{-1/(p/q)^{\prime}}W(x)^{1/q}F(x)|^{p}\,\mathrm{d}x\right)^{1/p}\left(\int_{{\mathbb{R}}^{n}}h(x)^{q}\,\mathrm{d}x\right)^{1/(q(p/q)^{\prime})}.$

It is of course the case that

$\int_{{\mathbb{R}}^{n}}h(x)^{q}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}({\mathcal{R}}_{W}\bar{h}(x))^{r(q)}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}|W(x)^{1/r(q)}\bar{H}(x)|^{r(q)}\,\mathrm{d}x\lesssim
1.$

Now let us deal with the first integral. Set

$T(x):=h(x)^{-p/(p/q)^{\prime}}W(x)^{p/q},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

What we want to check is that

$T^{s(p)^{\prime}}\oldin\nolinebreak[4]A_{r(p)}.$

By above we have

$\displaystyle\Sigma{\mathcal{R}}_{W}\bar{H}\oldin\nolinebreak[4]A^{1}_{{\mathcal{K}}},$

that means concretely

$({\mathcal{R}}_{W}\bar{h})V^{-1/r(q)}\overline{\mathbf{B}}\oldin\nolinebreak[4]A^{1}_{{\mathcal{K}}}.$

Thus, by F we have
${\mathcal{R}}_{W}\bar{h}V^{-1/r(q)}\oldin\nolinebreak[4]A_{1}$, therefore,

$({\mathcal{R}}_{W}\bar{h})^{-1}V^{1/r(q)}\oldin\nolinebreak[4]A_{\infty},$

that is

$h^{-q/r(q)}W^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{\infty}.$

Set

$m(x):=h(x)^{-q/r(q)},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Then

$mW^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{\infty}\quad\text{and}\quad
W^{s(q)^{\prime}}\oldin\nolinebreak[4]A_{r(q)}.$

The fact that $p>q$ easily implies $r(p)>r(q)$. Thus, by I (1), we deduce

$m^{r(p)-r(q)}(W^{s(q)^{\prime}})^{r(p)/r(q)}\oldin\nolinebreak[4]A_{r(p)}.$

Our goal now is to check that

$m^{r(p)-r(q)}(W^{s(q)^{\prime}})^{r(p)/r(q)}=T^{s(p)^{\prime}},$

that is

$h^{-q(r(p)-r(q))/r(q)}W^{s(q)^{\prime}r(p)/r(q)}=h^{-s(p)^{\prime}p/(p/q)^{\prime}}W^{s(p)^{\prime}p/q}.$

We first check that the exponents of $h$ agree:

$\displaystyle-\frac{q(r(p)-r(q))}{r(q)}=-\frac{qr(p)}{r(q)}+q\overset{\eqref{second
identity}}{=}-\frac{qps(p)^{\prime}}{qs(q)^{\prime}}+q=-\frac{ps(p)^{\prime}}{s(q)^{\prime}}+q=-\frac{ps(p)^{\prime}-qs(q)^{\prime}}{s(q)^{\prime}}$
$\displaystyle\overset{\eqref{first
identity}}{=}-s(p)^{\prime}(p-q)=-\frac{s(p)^{\prime}p}{\frac{p}{p-q}}=-\frac{s(p)^{\prime}p}{(p/q)^{\prime}}.$

We also check that the exponents of $W$ agree:

$\displaystyle\frac{s(q)^{\prime}r(p)}{r(q)}\overset{\eqref{second
identity}}{=}\frac{ps(p)^{\prime}}{q}=\frac{s(p)^{\prime}p}{q}.$

Thus, we can estimate

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}\leq
c\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\right)^{1/p}.$

We now observe that

$h(x)=({\mathcal{R}}_{W}\bar{h}(x))^{r/q}\geq\bar{h}(x)^{r/q}\geq\frac{|W(x)^{1/q}F(x)|}{\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

thus

$\displaystyle\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}f(x)|^{p}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}h(x)^{-p/(p/q)^{\prime}}|W(x)^{1/q}F(x)|^{p}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}h(x)^{q-p}|W(x)^{1/q}F(x)|^{p}\,\mathrm{d}x$
$\displaystyle\leq\frac{1}{\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}^{q-p}}\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x=\|F\|_{L^{q}_{{\mathcal{K}}}(W)}^{p}<\infty.$

Therefore, we are entitled to apply the extrapolation hypothesis, obtaining

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}$
$\displaystyle\leq
c_{1}\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\right)^{1/p}$
$\displaystyle\leq
c_{1}K_{p}([T^{s(p)^{\prime}}]_{A_{r(p)}})\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}G(x)|^{p}\right)^{1/p}$
$\displaystyle\leq
c_{1}K_{p}(c_{2}[V]_{A_{r(q)}}^{\alpha(p,q)})\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}G(x)|^{p}\right)^{1/p},$

since

$[T^{s(p)^{\prime}}]_{A_{r(p)}}\leq
c[W^{s(q)^{\prime}}]_{A_{r(q)}}[s(x)W^{s(q)^{\prime}/r(q)}]_{A_{\infty}}^{r(p)-r(q)}\leq
c[V]_{A_{r(q)}}[V]_{A_{r(q)}}^{(r(p)-r(q))/(r(q)-1)}=c[V]_{A_{r(q)}}^{\alpha(p,q)}.$

Similarly to above, we obtain

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}G(x)|^{p}\right)^{1/p}\leq\lVert
G\rVert_{L_{{\mathcal{K}}}^{q}(W)}.$

Thus

$\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}\leq
c_{1}K_{p}(c_{2}[V]_{A_{r(q)}}^{\alpha(p,q)})\lVert
G\rVert_{L_{{\mathcal{K}}}^{q}(W)},$

concluding the proof in this case.

$q<p=q_{0}<\infty$. Similarly to Case 1, we construct a nonnegative scalar-
valued function $h$ with the following properties:

$h(x)\geq\frac{|W(x)^{1/q}F(x)|}{\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}}+\frac{|W(x)^{1/q}G(x)|}{\lVert
G\rVert_{L^{q}_{{\mathcal{K}}}(W)}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$ $h(x)>0,\quad\text{for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$
$\int_{{\mathbb{R}}^{n}}h(x)^{q}\,\mathrm{d}x\lesssim 1,$

and

$h^{q/r(q)}W^{-s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{1}.$

As in case 1 we have

$\lVert F\rVert_{L^{q}_{{\mathcal{K}}}(W)}\lesssim\lVert
F\rVert_{L^{p}_{{\mathcal{K}}}(T)},\quad\lVert
F\rVert_{L^{p}_{{\mathcal{K}}}(T)}\leq\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}\quad\text{ and }\quad\lVert
G\rVert_{L^{p}_{{\mathcal{K}}}(T)}\leq\lVert
G\rVert_{L^{q}_{{\mathcal{K}}}(W)},$

where

$T(x):=h(x)^{-p/(p/q)^{\prime}}W(x)^{p/q},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Notice that

$B:=h^{-q/r(q)}W^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{\infty}.$

We check now that

$T=B^{s(p_{0})-1},$

that is

$h^{-p/(p/q)^{\prime}}W^{p/q}=h^{-q(s(p_{0})-1)/r(q)}W^{s(q)^{\prime}(s(p_{0})-1)/r(q)}.$

We first check that the exponents of $h$ agree. We need to show that

$-q_{0}/(q_{0}/q)^{\prime}=-\frac{q}{r(q)}\left(\frac{q_{0}}{p_{0}}-1\right),$

that is that

$q_{0}-q=\frac{q(q_{0}-p_{0})}{p_{0}r(q)}.$

We compute

$\displaystyle
p_{0}r(q)=s(q)^{\prime}(q-p_{0})+p_{0}=\frac{q_{0}(q-p_{0})}{q_{0}-q}+p_{0}=\frac{q_{0}(q-p_{0})+p_{0}(q_{0}-q)}{q_{0}-q}=\frac{q(q_{0}-p_{0})}{q_{0}-q},$

yielding the desired equality. Moreover, we check that the exponents of $W$
agree. We have

$\displaystyle\frac{s(q)^{\prime}(s(p_{0})-1)}{r(q)}=\frac{q_{0}(q_{0}-p_{0})}{p_{0}r(q)}=\frac{q_{0}}{q}=\frac{p}{q}.$

Thus

$T^{t(q_{0})^{\prime}/t(q_{0})}=T^{s(p_{0})^{\prime}-1}=B\oldin\nolinebreak[4]A_{\infty}.$

The rest of the proof proceeds exactly as in case 1.

$q<p=q_{0}=\infty$. Similarly to Case 1, we construct a nonnegative scalar-
valued function $h$ with the following properties:

$h(x)\geq\frac{|W(x)^{1/q}F(x)|}{\lVert
F\rVert_{L^{q}_{{\mathcal{K}}}(W)}}+\frac{|W(x)^{1/q}G(x)|}{\lVert
G\rVert_{L^{q}_{{\mathcal{K}}}(W)}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$ $h(x)>0,\quad\text{for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$
$\int_{{\mathbb{R}}^{n}}h(x)^{q}\,\mathrm{d}x\lesssim 1,$

and

$h^{q/r(q)}W^{-s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{1}.$

We compute

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}=\left(\int_{{\mathbb{R}}^{n}}|h(x)^{-1/(p/q)^{\prime}}W(x)^{1/q}F(x)|^{q}\cdot
h(x)^{q/(p/q)^{\prime}}\,\mathrm{d}x\right)^{1/q}$ $\displaystyle\leq\lVert
F\rVert_{L^{\infty}_{{\mathcal{K}}}(T)}\left(\int_{{\mathbb{R}}^{n}}h(x)^{q}\,\mathrm{d}x\right)^{1/(q(p/q)^{\prime})}\lesssim\lVert
F\rVert_{L^{\infty}_{{\mathcal{K}}}(T)},$

where

$T(x):=h(x)^{-1/(p/q)^{\prime}}W(x)^{1/q},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Notice that

$B:=h^{-q/r(q)}W^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{\infty}.$

So, we only have to check that $T^{1/p_{0}}=B$. We check that the exponents of
$h$ agree:

$-q/(p_{0}r(q))=-q/(p_{0}t(q))=-q/q=-1,\quad-1/(p/q)^{\prime}=-1/1=-1.$

We also check that the exponents of $W$ agree:

$s(q)^{\prime}/r(q)=1/t(q)=p/q_{0}.$

Thus $T^{1/p_{0}}=B\oldin\nolinebreak[4]A_{\infty}$. The rest of the proof
proceeds exactly as in case 1.

$p_{0}\leq p<q$. We first need to define an associated Rubio de Francia
operator. So consider first the following auxiliary operator acting on convex-
body valued functions $H\oldin\nolinebreak[4]L^{r(q)^{\prime}}(U)$:

$N_{W}H(x):=|U(x)^{1/r(q)^{\prime}}H(x)|W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}},\quad\text{
for a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

where $U:=W^{-1/(t(q)-1)}$. For this operator, we may compute

$\displaystyle\lVert
N_{W}H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}=\lVert
H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(U)}.$

It is easy to see that this map is sublinear and monotone and

$H(x)\subseteq\Sigma(x)N_{W}H(x),\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

where $\Sigma:=W^{1/r(q)^{\prime}}\cdot
U^{-1/r(q)^{\prime}}=W^{t(q)^{\prime}/r(q)^{\prime}}$.

Now we define the operator acting on functions
$H\oldin\nolinebreak[4]L^{r(q)^{\prime}}(W)$

$P_{W}H(x):=N_{W}({\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H))(x)=|U(x)^{1/r(q)^{\prime}}{\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H)(x)|W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Since ${\mathcal{M}}^{{\mathcal{K}}}$ is sublinear and monotone, one can
easily observe that $P_{W}$ is sublinear and monotone. Moreover, using E we
can compute

$\displaystyle\lVert
P_{W}H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}=\lVert{\mathcal{M}}^{{\mathcal{K}}}(\Sigma
H)\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(U)}\lesssim[U]_{A_{r(q)^{\prime}}}^{1/(r(q)^{\prime}-1)}\lVert\Sigma
H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(U)}=[U]_{A_{r(q)^{\prime}}}^{1/(r(q)^{\prime}-1)}\lVert
H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}.$

(Although we do not need it, we note that it is easy to see that
$H(x)\subseteq P_{W}H(x)$, for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.)
Note also that if $H$ is not a.e. equal to $\\{0\\}$, then
$P_{W}H(x)\neq\\{0\\}$, for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$.

Thus, all assumptions of G are fulfilled, therefore we can define the
following Rubio de Francia type operator acting on functions
$H\oldin\nolinebreak[4]L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)$:

${\mathcal{R}}_{W}H:=\sum_{k=0}^{\infty}\frac{1}{2^{k}\lVert
P_{W}\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}^{k}}(P_{W})^{k}H.$

We record the following properties of ${\mathcal{R}}_{W}$ that we obtain from
G:

$H(x)\subseteq{\mathcal{R}}_{W}H(x),\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$
$\lVert{\mathcal{R}}_{W}H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}\leq
2\lVert H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}$

and

$P_{W}({\mathcal{R}}_{W}H)(x)\subseteq\lVert
P_{W}\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}{\mathcal{R}}_{W}H(x)\subseteq
c[U]_{A_{r(q)^{\prime}}}^{1/(r(q)^{\prime}-1)}{\mathcal{R}}_{W}H(x),\quad\text{
for a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

In particular, we have

${\mathcal{M}}^{{\mathcal{K}}}(\Sigma{\mathcal{R}}_{W}H)(x)\subseteq\Sigma(x)P_{W}({\mathcal{R}}_{W}H)(x)\subseteq
c[U]_{A_{r^{\prime}}}^{1/(r^{\prime}-1)}\Sigma(x){\mathcal{R}}_{W}H(x),\quad\text{
for a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

This means that if $H$ is not a.e. equal to $\\{0\\}$, then the convex body
valued function $\Sigma{\mathcal{R}}_{W}H$ belongs to the
$A^{1}_{{\mathcal{K}}}$ class with
$[\Sigma{\mathcal{R}}_{W}H]_{A^{1}_{{\mathcal{K}}}}\leq
c[U]_{A_{r(q)^{\prime}}}^{1/(r(q)^{\prime}-1)}$.

Finally, notice that $P_{W}H$ is always ellipsoid-valued with respect to
$W^{-1/r(q)^{\prime}}$, therefore by H we deduce that if $H$ happens to be
ellipsoid-valued with respect to $W^{-1/r(q)^{\prime}}$, then
${\mathcal{R}}_{W}H$ is also ellipsoid-valued with respect to
$W^{-1/r(q)^{\prime}}$.

We now observe that the Riesz representation theorem yields that there exists
a nonnegative scalar-valued function
$h_{1}\oldin\nolinebreak[4]L^{(q/p)^{\prime}}$ with $\lVert
h_{1}\rVert_{L^{(q/p)^{\prime}}}=1$ such that

$\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}(|W(x)^{1/q}F(x)|^{p})^{q/p}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{p}h_{1}(x)\,\mathrm{d}x.$

Now we define the nonnegative scalar-valued function

$\bar{h}(x):=h_{1}(x)^{(q/p)^{\prime}/r(q)^{\prime}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n},$

and the convex body valued function

$\bar{H}(x):=\bar{h}(x)W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}},\quad\text{
for a.e. }x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

It is clear that $\lVert H\rVert_{L^{r(q)^{\prime}}_{{\mathcal{K}}}(W)}=1$.
Since $\bar{H}$ is ellipsoid-valued with respect to $W^{-1/r(q)^{\prime}}$, we
deduce that ${\mathcal{R}}_{W}\bar{H}$ is also ellipsoid-valued with respect
to $W^{-1/r(q)^{\prime}}$, therefore we can write
${\mathcal{R}}_{W}\bar{H}(x)={\mathcal{R}}_{W}\bar{h}(x)W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}}$
for some nonnegative scalar-valued function ${\mathcal{R}}_{W}\bar{h}$. Then,
we have

$\bar{h}(x)W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}}\subseteq{\mathcal{R}}_{W}\bar{h}(x)W(x)^{-1/r(q)^{\prime}}\overline{\mathbf{B}},$

which implies immediately that $\bar{h}(x)\leq{\mathcal{R}}_{W}\bar{h}(x)$,
for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. We now define the
nonnegative scalar-valued function

$h:=({\mathcal{R}}_{W}\bar{h})^{r(q)^{\prime}/(q/p)^{\prime}}.$

Since $\bar{H}$ is not a.e. equal to $\\{0\\}$, we deduce that $h(x)\neq 0$,
for a.e. $x\oldin\nolinebreak[4]{\mathbb{R}}^{n}$. Observe that $h\geq h_{1}$.
Thus, we have

$\displaystyle\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\leq\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{p}h(x)\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}|h(x)^{1/p}W(x)^{1/q}F(x)|^{p}\,\mathrm{d}x.$

Set

$T(x):=h(x)W(x)^{p/q},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Our goal is to show that

$T^{s(p)^{\prime}}\oldin\nolinebreak[4]A_{r(p)}.$

By above, we have

$\displaystyle\Sigma{\mathcal{R}}_{W}\bar{H}\oldin\nolinebreak[4]A^{1}_{{\mathcal{K}}},$

that means concretely

$({\mathcal{R}}_{W}\bar{h})U^{-1/r(q)^{\prime}}\overline{\mathbf{B}}\oldin\nolinebreak[4]A^{1}_{{\mathcal{K}}},$

that is

$({\mathcal{R}}_{W}\bar{h})W^{s(q)^{\prime}/r(q)}\overline{\mathbf{B}}\oldin\nolinebreak[4]A^{1}_{{\mathcal{K}}}.$

Thus, by F we have
$({\mathcal{R}}_{W}\bar{h})V^{1/r(q)}\oldin\nolinebreak[4]A_{1}$, that is

$h^{(q/p)^{\prime}/r(q)^{\prime}}W^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{1}.$

Set

$m(x):=h(x)^{(q/p)^{\prime}/r(q)^{\prime}},\quad\text{ for a.e.
}x\oldin\nolinebreak[4]{\mathbb{R}}^{n}.$

Then

$mW^{s(q)^{\prime}/r(q)}\oldin\nolinebreak[4]A_{1}\quad\text{and}\quad
W^{s(q)^{\prime}}\oldin\nolinebreak[4]A_{r(q)}.$

The fact that $p<q$ easily implies $r(p)<r(q)$. Thus, by I (2), we deduce

$m^{r(p)-r(q)^{\prime}(r(p)-1)}(W^{s(q)^{\prime}})^{r(p)/r(q)}\oldin\nolinebreak[4]A_{r(p)},$

both for $p>p_{0}$ and $p=p_{0}$.

Our goal now is to check that

$m^{r(p)-r(q)^{\prime}(r(p)-1)}(W^{s(q)^{\prime}})^{r(p)/r(q)}=T^{s(p)^{\prime}},$

that is

$h^{(q/p)^{\prime}[r(p)-r(q)^{\prime}(r(p)-1)]/r(q)^{\prime}}W^{s(q)^{\prime}r(p)/r(q)}=h^{s(p)^{\prime}}W^{s(p)^{\prime}p/q}.$

We first check that the exponents of $h$ agree:

$\displaystyle(q/p)^{\prime}\frac{r(p)-r(q)^{\prime}(r(p)-1)}{r(q)^{\prime}}=(q/p)^{\prime}\left[\frac{r(p)}{r(q)^{\prime}}-r(p)+1\right]=(q/p)^{\prime}\left[r(p)\left(\frac{1}{r(q)^{\prime}}-1\right)+1\right]$
$\displaystyle=(q/p)^{\prime}\left(-\frac{r(p)}{r(q)}-1\right)\overset{\eqref{second
identity}}{=}(q/p)^{\prime}\left(-\frac{ps(p)^{\prime}}{qs(q)^{\prime}}-1\right)=(q/p)^{\prime}\frac{qs(q)^{\prime}-ps(p)^{\prime}}{qs(q)^{\prime}}$
$\displaystyle\overset{\eqref{second
identity}}{=}(q/p)^{\prime}\frac{s(p)^{\prime}s(q)^{\prime}(q-p)}{qs(q)^{\prime}}=\frac{q}{q-p}\cdot\frac{s(p)^{\prime}q-p}{q}=s(p)^{\prime}.$

Moreover, we check that the exponents of $W$ agree:

$\displaystyle s(p)^{\prime}\frac{r(p)}{r(q)}\overset{\eqref{second
identity}}{=}s(p)^{\prime}\frac{p}{q}.$

Therefore, we have

$\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\leq\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\,\mathrm{d}x.$

We also observe that

$\displaystyle\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{p}h(x)\,\mathrm{d}x$
$\displaystyle\leq\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{p/q}\left(\int_{{\mathbb{R}}^{n}}h(x)^{(q/p)^{\prime}}\,\mathrm{d}x\right)^{1/(q/p)^{\prime}}$
$\displaystyle=\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{p/q}\left(\int_{{\mathbb{R}}^{n}}{\mathcal{R}}_{W}\bar{h}(x)^{r(q)^{\prime}}\,\mathrm{d}x\right)^{1/(q/p)^{\prime}}.$

Notice that

$\displaystyle\int_{{\mathbb{R}}^{n}}{\mathcal{R}}_{W}\bar{h}(x)^{r(q)^{\prime}}\,\mathrm{d}x=\int_{{\mathbb{R}}^{n}}|W(x)^{1/r(q)^{\prime}}{\mathcal{R}}_{W}\bar{H}(x)|^{r(q)^{\prime}}\,\mathrm{d}x\leq
c.$

It follows that

$\displaystyle\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\,\mathrm{d}x\leq
c\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{p/q}<\infty.$

Thus, we are entitled to apply the extrapolation hypothesis, obtaining

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}\leq
c\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}F(x)|^{p}\,\mathrm{d}x\right)^{1/p}\leq
cK_{p}([T^{s(p)^{\prime}}]_{A_{r(p)}})$ $\displaystyle\leq
c_{1}K_{p}(c_{2}[V]_{A_{r(p)}}^{\alpha(p,q)})\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}G(x)|^{p}\,\mathrm{d}x\right)^{1/p},$

because

$\displaystyle[T^{s(p)^{\prime}}]_{A_{r(p)}}\leq
c[W^{s(q)^{\prime}}]_{A_{r(q)}}^{(r(p)-1)/(r(q)-1)}[r(x)W(x)^{s(q)^{\prime}/r(q)}]_{A_{1}}^{r(p)-r(q)^{\prime}(r(p)-1)}$
$\displaystyle\leq