[proofplan]
The result has two parts. For the inclusion $A_p \subset A_q$ when $1 < p \le q < \infty$, we apply Hölder's inequality with exponents tied to the dual exponents $p'$ and $q'$ to compare the $A_q$ characteristic with the $A_p$ characteristic. The cases involving $p = 1$ and $q = \infty$ reduce to direct estimates: for $A_1 \subset A_p$ we use the pointwise lower bound $w(x) \ge c \cdot \frac{1}{|Q|}\int_Q w \, d\mathcal{L}^n$ that defines $A_1$ to control $w^{1-p'}$ against an appropriate average. Throughout, the strategy is to extract the $A_p$ characteristic and absorb the remaining factor into the new average.
[/proofplan]
[step:Recall the definitions of the $A_p$ characteristics for $1 \le p \le \infty$]
Fix a weight $w$ on $\mathbb{R}^n$, that is, a non-negative locally integrable function $w: \mathbb{R}^n \to [0, \infty)$. For a cube $Q \subset \mathbb{R}^n$ with sides parallel to the coordinate axes, let
\begin{align*}
\langle w \rangle_Q := \frac{1}{|Q|} \int_Q w(x) \, d\mathcal{L}^n(x), \qquad |Q| := \mathcal{L}^n(Q).
\end{align*}
For $1 < p < \infty$, the [$A_p$ characteristic](/theorems/???) of $w$ is
\begin{align*}
[w]_{A_p} := \sup_Q \langle w \rangle_Q \, \langle w^{1-p'} \rangle_Q^{p-1}, \qquad p' := \frac{p}{p-1},
\end{align*}
where the supremum runs over all cubes $Q \subset \mathbb{R}^n$. We say $w \in A_p$ when $[w]_{A_p} < \infty$.
For $p = 1$, the $A_1$ characteristic is
\begin{align*}
[w]_{A_1} := \sup_Q \langle w \rangle_Q \, \big\| w^{-1} \big\|_{L^\infty(Q)},
\end{align*}
and equivalently $w \in A_1$ iff there exists $C > 0$ such that $M w(x) \le C w(x)$ a.e., where $M$ is the [Hardy–Littlewood maximal operator](/theorems/???); the smallest such $C$ equals $[w]_{A_1}$.
[/step]
[step:Reduce the inclusion $A_p \subset A_q$ to the case $1 < p < q < \infty$]
Three cases must be handled separately because the definition of $[w]_{A_p}$ changes form at $p = 1$ and $p = \infty$:
- $1 < p \le q < \infty$ with $p < q$: handled in the next step via Hölder.
- $p = q$: elementary since the inclusion is the identity.
- $p = 1, q > 1$: handled in the dedicated step below.
The case $q = \infty$ is the definition $A_\infty := \bigcup_{r \ge 1} A_r$, so $A_p \subset A_\infty$ holds by definition for every $p \ge 1$.
[/step]
[step:Prove $A_p \subset A_q$ for $1 < p < q < \infty$ via Hölder applied to $w^{1-q'}$]
Fix $w \in A_p$ with $1 < p < q < \infty$, and fix a cube $Q \subset \mathbb{R}^n$. The objective is to bound $\langle w \rangle_Q \langle w^{1-q'} \rangle_Q^{q-1}$ by a constant multiple of $[w]_{A_p}$, with constant independent of $Q$.
Since $1 < p < q$ implies $p' > q' > 1$, we have $q' - 1 < p' - 1$, that is $1 - q' > -(p'-1) \cdot \tfrac{q'-1}{p'-1}$, so we can write the exponent $1 - q'$ as a convex combination relevant for Hölder. Concretely, set
\begin{align*}
r := \frac{p' - 1}{q' - 1} > 1, \qquad r' = \frac{p'-1}{p'-q'},
\end{align*}
so that $1/r + 1/r' = 1$ and $r(1-q') = (1-q') \cdot \frac{p'-1}{q'-1} = -(p'-1) = 1 - p'$. By [Hölder's inequality](/theorems/???) with exponents $r, r'$ applied to the functions $w^{1-q'}$ and $\mathbb{1}_Q$:
\begin{align*}
\frac{1}{|Q|} \int_Q w^{1-q'} \, d\mathcal{L}^n &= \frac{1}{|Q|} \int_Q w^{1-q'} \cdot 1 \, d\mathcal{L}^n \\
&\le \left( \frac{1}{|Q|} \int_Q w^{r(1-q')} \, d\mathcal{L}^n \right)^{1/r} \left( \frac{1}{|Q|} \int_Q 1 \, d\mathcal{L}^n \right)^{1/r'} \\
&= \langle w^{1-p'} \rangle_Q^{1/r}.
\end{align*}
Raising both sides to the $(q-1)$-th power and using $1/r = (q'-1)/(p'-1) = (q-1)^{-1}(p-1)$ — which gives $(q-1)/r = p-1$ — we obtain
\begin{align*}
\langle w^{1-q'} \rangle_Q^{q-1} \le \langle w^{1-p'} \rangle_Q^{p-1}.
\end{align*}
Multiplying through by $\langle w \rangle_Q$ and taking the supremum over cubes:
\begin{align*}
[w]_{A_q} = \sup_Q \langle w \rangle_Q \, \langle w^{1-q'} \rangle_Q^{q-1} \le \sup_Q \langle w \rangle_Q \, \langle w^{1-p'} \rangle_Q^{p-1} = [w]_{A_p} < \infty.
\end{align*}
Hence $w \in A_q$ with $[w]_{A_q} \le [w]_{A_p}$.
[/step]
[step:Prove $A_1 \subset A_p$ for $p > 1$ via the pointwise $A_1$ inequality]
Fix $w \in A_1$ and $p > 1$. We show $w \in A_p$ with $[w]_{A_p} \le [w]_{A_1}$.
By definition of $[w]_{A_1}$, for every cube $Q$:
\begin{align*}
\langle w \rangle_Q \le [w]_{A_1} \cdot \operatorname*{ess\,inf}_{x \in Q} w(x), \qquad \text{equivalently}\qquad \big\| w^{-1} \big\|_{L^\infty(Q)} \le \frac{[w]_{A_1}}{\langle w \rangle_Q}.
\end{align*}
Since $1 - p' < 0$, the function $w^{1-p'} = (w^{-1})^{p'-1}$ is bounded above by its $L^\infty$ norm on $Q$:
\begin{align*}
w^{1-p'}(x) \le \big\| w^{-1} \big\|_{L^\infty(Q)}^{p'-1} \le \left( \frac{[w]_{A_1}}{\langle w \rangle_Q} \right)^{p'-1} \qquad \text{for a.e. } x \in Q.
\end{align*}
Averaging over $Q$:
\begin{align*}
\langle w^{1-p'} \rangle_Q \le \left( \frac{[w]_{A_1}}{\langle w \rangle_Q} \right)^{p'-1}.
\end{align*}
Raising to the $(p-1)$-th power and using $(p'-1)(p-1) = 1$ (since $p' = p/(p-1)$):
\begin{align*}
\langle w^{1-p'} \rangle_Q^{p-1} \le \frac{[w]_{A_1}^{(p'-1)(p-1)}}{\langle w \rangle_Q^{(p'-1)(p-1)}} = \frac{[w]_{A_1}}{\langle w \rangle_Q}.
\end{align*}
Multiplying both sides by $\langle w \rangle_Q$:
\begin{align*}
\langle w \rangle_Q \, \langle w^{1-p'} \rangle_Q^{p-1} \le [w]_{A_1}.
\end{align*}
Taking the supremum over cubes gives $[w]_{A_p} \le [w]_{A_1} < \infty$, so $w \in A_p$.
[/step]
[step:Conclude the chain of inclusions]
Combining the previous two steps: $A_1 \subset A_p$ for every $p > 1$, and $A_p \subset A_q$ whenever $1 < p \le q < \infty$. Together with the elementary inclusion $A_p \subset A_\infty := \bigcup_{r \ge 1} A_r$, we obtain the full nesting
\begin{align*}
A_1 \subset A_p \subset A_q \subset A_\infty, \qquad 1 < p \le q < \infty,
\end{align*}
with the quantitative bound $[w]_{A_q} \le [w]_{A_p} \le [w]_{A_1}$ at each inclusion (whenever both sides make sense). This completes the proof.
[/step]
