Muckenhoupt weights

Remarks

By Hölder's inequality $[w]_{A_p} \ge 1$, and $A_1 \subseteq A_p \subseteq A_q$ for $1 < p < q$. The union $A_\infty = \bigcup_{p \ge 1} A_p$ has several equivalent descriptions, for example through a reverse Hölder inequality. For $p = 2$ the condition reads $\sup_Q \left( \frac{1}{\abs{Q}}\int_Q w \right) \left( \frac{1}{\abs{Q}}\int_Q w^{-1} \right) < \infty$.

Example: $n = 1$, $p = 2$, $w(x) = \abs{x}^a$ with $-1 < a < 1$. On $Q = [0, L]$ the two averages are $L^a/(1 + a)$ and $L^{-a}/(1 - a)$, whose product is $1/(1 - a^2)$ for every $L$. Intervals symmetric about $0$ give the same averages. On intervals far from $0$ compared with their length, $w$ is almost constant, so the product is close to $1$. The remaining intervals lie in a symmetric interval of comparable length, and this case analysis shows $w \in A_2$. More generally $\abs{x}^a \in A_p(\R^n)$ exactly when $-n < a < n(p - 1)$.

Non-example: $n = 1$, $p = 2$, $w(x) = \abs{x}$. It is a weight, but on $Q = [0, 1]$ the average of $w^{-1} = 1/x$ is infinite, so $w \notin A_2$.