Weighted endpoints are not obtained by setting p equal to one

The strong weighted estimates of this page, which are stated for $1<p<\mathrm{\infty }$, are not the instance $p=1$ of an $A_p$ theory, in three distinct senses.

First, the $A_p$ condition itself degenerates at $p=1$: the factor $w^{-1\mathrm{/}(p-1)}$ in $[w{]}_{A_p}$ is undefined, and the class $A_1$ is defined instead by the pointwise bound $M^{\ast }w\le Cw$ almost everywhere (Muckenhoupt A_p and A_1 weights). The correct replacement for the maximal function is a weak-type estimate, $w(\{Mf>\lambda \})\le 5^n[w{]}_{A_1}{\lambda }^{-1}\int \mathrm{\mid }f\mathrm{\mid }wd\lambda$ (Weighted weak (1,1) bound for the maximal function under A_1), and the maximal characterisation of $A_p$ is likewise a statement about the strict range $1<p<\mathrm{\infty }$ (The Hardy-Littlewood maximal operator characterises A_p).

Second, the weighted Calderón–Zygmund theorem attaches the weak $(1,1)$ bound to the endpoint $w\in A_1$, not a strong $L^1(w)$ bound (Weighted L-p bounds for standard Calderon-Zygmund maximal truncations); its constant depends on the $A_1$ characteristic and is not universal, exactly as in the maximal case.

Third, the obstruction is not an artefact of the weights: already for the Lebesgue weight $w=1$, which lies in $A_1$, the Hardy–Littlewood maximal operator is not of strong type $(1,1)$ (The Hardy-Littlewood maximal operator is not strong type $(1,1)$). Hence no strong $L^1(w)$ endpoint can be expected for all $w\in A_1$, and the strict range $1<p<\mathrm{\infty }$ in the strong weighted theorems is essential.