[proofplan]
The strategy is to apply the [Reverse Hölder Inequality](/theorems/3216) to the *dual weight* $\sigma := w^{1-p'}$, which lies in $A_{p'}$ when $w \in A_p$ by the duality of the $A_p$ class. The reverse Hölder inequality gives an exponent $\delta > 0$ such that $\langle \sigma^{1+\delta} \rangle_Q^{1/(1+\delta)} \le C \langle \sigma \rangle_Q$ for all cubes $Q$. Translating back to $w$ through the algebraic identity $\sigma^{1+\delta} = w^{1-q'}$ for an appropriate $q < p$, this yields exactly the $A_q$ characterisation. Solving for $q$ in terms of $p, p', \delta$ produces the required $\varepsilon = p - q > 0$.
[/proofplan]
[step:Recall the duality of $A_p$ classes]
[claim:If $w \in A_p$ for $1 < p < \infty$, then $\sigma := w^{1-p'} \in A_{p'}$ with $[\sigma]_{A_{p'}} = [w]_{A_p}^{1/(p-1)} = [w]_{A_p}^{p'-1}$.]
[proof]
Recall $p' := p/(p-1)$, so $(p')' = p$ and $(1 - p')(1 - p) = 1$. Compute
\begin{align*}
\sigma^{1 - (p')'} = (w^{1-p'})^{1-p} = w^{(1-p')(1-p)} = w.
\end{align*}
Therefore for every cube $Q \subset \mathbb{R}^n$:
\begin{align*}
\langle \sigma \rangle_Q \, \langle \sigma^{1 - (p')'} \rangle_Q^{p' - 1} = \langle w^{1-p'} \rangle_Q \, \langle w \rangle_Q^{p' - 1} = \big( \langle w \rangle_Q^{1/(p'-1)} \langle w^{1-p'} \rangle_Q^{1/(p'-1) \cdot (p'-1)} \big)^{p'-1}.
\end{align*}
More cleanly: raise the $A_p$ inequality $\langle w \rangle_Q \langle w^{1-p'} \rangle_Q^{p-1} \le [w]_{A_p}$ to the $(p'-1)$-th power, and use $(p-1)(p'-1) = 1$:
\begin{align*}
\langle w \rangle_Q^{p'-1} \langle w^{1-p'} \rangle_Q \le [w]_{A_p}^{p'-1}.
\end{align*}
The left side is $\langle \sigma^{1-(p')'} \rangle_Q^{p'-1} \langle \sigma \rangle_Q$, the $A_{p'}$ characteristic of $\sigma$. Taking the supremum over $Q$:
\begin{align*}
[\sigma]_{A_{p'}} \le [w]_{A_p}^{p'-1} < \infty.
\end{align*}
The reverse inequality follows by symmetry, applying the same argument to $\sigma$ and using $\sigma^{1-(p')'} = w$.
[/proof]
[/claim]
[/step]
[step:Apply the Reverse Hölder Inequality to the dual weight $\sigma$]
By the previous step, $\sigma := w^{1-p'} \in A_{p'}$ with $[\sigma]_{A_{p'}} = [w]_{A_p}^{p'-1}$. By the [Reverse Hölder Inequality](/theorems/3216) applied to $\sigma$, there exist $\delta > 0$ and $C > 0$, depending only on $n$, $p'$, and $[\sigma]_{A_{p'}}$ — equivalently on $n$, $p$, $[w]_{A_p}$ — such that for every cube $Q \subset \mathbb{R}^n$:
\begin{align*}
\langle \sigma^{1+\delta} \rangle_Q^{1/(1+\delta)} \le C \, \langle \sigma \rangle_Q.
\end{align*}
Raising to the $(1+\delta)(p-1)$-th power (note $p - 1 > 0$):
\begin{align*}
\langle \sigma^{1+\delta} \rangle_Q^{p-1} \le C^{(1+\delta)(p-1)} \, \langle \sigma \rangle_Q^{(1+\delta)(p-1)}.
\end{align*}
[/step]
[step:Solve for the exponent $q < p$ that recasts $\sigma^{1+\delta}$ as $w^{1-q'}$]
We want to find $q \in (1, p)$ with
\begin{align*}
1 - q' = (1 + \delta)(1 - p').
\end{align*}
Solving:
\begin{align*}
q' = 1 - (1+\delta)(1 - p') = 1 + (1+\delta)(p' - 1) = (1 + \delta) p' - \delta.
\end{align*}
Since $p' > 1$ and $\delta > 0$, we have $q' > p' > 1$, hence $q = q'/(q' - 1) < p$. Compute $q$:
\begin{align*}
q = \frac{q'}{q' - 1} = \frac{(1+\delta) p' - \delta}{(1+\delta) p' - \delta - 1} = \frac{(1+\delta) p' - \delta}{(1+\delta)(p' - 1)} = \frac{(1+\delta) p' - \delta}{(1+\delta)(p-1)^{-1} \cdot (p-1)} \cdot \frac{1}{1+\delta}.
\end{align*}
Equivalently, using $p' - 1 = 1/(p-1)$:
\begin{align*}
q' - 1 = (1+\delta)(p' - 1) = \frac{1+\delta}{p-1}, \qquad q - 1 = \frac{1}{q' - 1} = \frac{p - 1}{1 + \delta}.
\end{align*}
Therefore $q = 1 + (p-1)/(1+\delta) < p$, and we set
\begin{align*}
\varepsilon := p - q = (p - 1)\left( 1 - \frac{1}{1+\delta} \right) = \frac{(p-1)\delta}{1+\delta} > 0.
\end{align*}
This $\varepsilon$ depends only on $n$, $p$, and $[w]_{A_p}$, since $\delta$ does.
[/step]
[step:Verify the $A_q$ characteristic of $w$ using the Reverse Hölder bound]
With $q$ chosen so that $\sigma^{1+\delta} = w^{(1+\delta)(1-p')} = w^{1-q'}$, we rewrite the bound from Step 2:
\begin{align*}
\langle w^{1-q'} \rangle_Q^{p-1} = \langle \sigma^{1+\delta} \rangle_Q^{p-1} \le C^{(1+\delta)(p-1)} \, \langle \sigma \rangle_Q^{(1+\delta)(p-1)} = C^{(1+\delta)(p-1)} \, \langle w^{1-p'} \rangle_Q^{(1+\delta)(p-1)}.
\end{align*}
However, what we want to bound is $\langle w^{1-q'} \rangle_Q^{q-1}$, with exponent $q - 1 = (p-1)/(1+\delta)$, not $p - 1$. Take the $(q-1)/(p-1) = 1/(1+\delta)$-th power of the previous inequality:
\begin{align*}
\langle w^{1-q'} \rangle_Q^{q-1} = \langle w^{1-q'} \rangle_Q^{(p-1)/(1+\delta)} \le C^{p-1} \, \langle w^{1-p'} \rangle_Q^{p-1}.
\end{align*}
Multiplying both sides by $\langle w \rangle_Q$:
\begin{align*}
\langle w \rangle_Q \, \langle w^{1-q'} \rangle_Q^{q-1} \le C^{p-1} \, \langle w \rangle_Q \, \langle w^{1-p'} \rangle_Q^{p-1} \le C^{p-1} \, [w]_{A_p}.
\end{align*}
Taking the supremum over cubes:
\begin{align*}
[w]_{A_q} \le C^{p-1} \, [w]_{A_p} < \infty,
\end{align*}
hence $w \in A_q = A_{p-\varepsilon}$.
[/step]
[step:Combine the steps to conclude $w \in A_{p - \varepsilon}$]
Setting $\varepsilon = (p-1)\delta/(1+\delta) > 0$, we have shown $[w]_{A_{p-\varepsilon}} \le C^{p-1} [w]_{A_p} < \infty$. Both $\delta$ and $C$ depend only on $n$, $p$, $[w]_{A_p}$, hence so does $\varepsilon$. This completes the proof of the openness of $A_p$.
[/step]
