Rubio de Francia extrapolation theorem

Remarks

In the typical use, the pairs are $(\abs{Tf}, \abs{f})$ for an operator $T$. Assume $T$ is bounded on $L^{p_0}(w)$ for all $w \in A_{p_0}$, with a bound depending only on $[w]_{A_{p_0}}$. Then $T$ is bounded on $L^p(w)$ for all $1 < p < \infty$ and all $w \in A_p$. In particular, taking $w = 1$, one exponent $p_0$ gives unweighted bounds for all $1 < p < \infty$.

The proof uses the Rubio de Francia algorithm. Let $v$ be a weight and $q > 1$ such that the maximal operator $M$ is bounded on $L^q(v)$, with norm $\norm{M}$. For $h \ge 0$ in $L^q(v)$, set $Rh = \sum_{k \ge 0} \frac{M^k h}{(2\norm{M})^k}$, where $M^k$ is the $k$-th iterate. Then $h \le Rh$, $\norm{Rh}_{L^q(v)} \le 2 \norm{h}_{L^q(v)}$, and $M(Rh) \le 2 \norm{M} \, Rh$, so $Rh$ is an $A_1$ weight. These $A_1$ weights are combined into an $A_{p_0}$ weight to which the hypothesis applies, and Muckenhoupt's theorem controls the constants.

Going deeper

For a student

The theorem is a main reason to prove weighted inequalities. A single weighted $L^2$ bound for a Calderón–Zygmund operator, a commutator or a square function yields all the $L^p$ bounds, weighted and unweighted. It also gives vector-valued inequalities almost for free, since a sum $\left( \sum_j \abs{T f_j}^q \right)^{1/q}$ can be treated by applying the pair formulation to suitable functions. It relies on the theory of $A_p$ weights, in particular on Muckenhoupt's theorem and on the factorization of $A_p$ weights as $w = w_1 w_2^{1 - p}$ with $w_1, w_2 \in A_1$.