Muckenhoupt weights
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In mathematics, the class of Muckenhoupt weights Ap are those weights ω for which the Hardy-Littlewood maximal operator is bounded on Lp(dω). Specifically, we consider functions f on
and there associated maximal functions M(f) defined as
where Br is a ball in
with radius r and centre x. We wish to characterise the functions
for which we have a bound
where C depends only on
and ω. This was first done by Benjamin Muckenhoupt[1].
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[edit] Definition
For a fixed
, we say that a weight
belongs to Ap if ω is locally integrable and there is a constant C such that, for all balls B in
, we have
where 1 / p + 1 / p' = 1 and | B | is the Lebesgue measure of B. We say
belongs to A1 if there exists some C such that
for all
and all balls B.[2]
[edit] Equivalent characterisations
This following result is a fundamental result in the study of Muckenhoupt weights. A weight ω is in Ap if and only if any one of the following hold.[2]
(a) The Hardy-Littlewood maximal function is bounded on Lp(ω(x)dx), that is
for some C which only depends on p and the constant A in the above definition.
(b) There is a constant c such that for any locally integrable function f on 
for all balls B. Here
is the average of f over B and
[edit] Reverse Hölder inequalities
The main tool in the proof of the above equivalence is the following result.[2] The following statements are equivalent
(a) ω belongs to Ap for some 
(b) There exists an r > 1 and a c (both depending on ω such that
for all balls Br
(c) There exists
so that for all balls B and subsets 
We call the inequality in (b) a reverse Hölder inequality as the reverse inequality follows for any non-negative function directly from Hölder's inequality. If any of the three equivalent conditions above hold we say ω belongs to
.
[edit] Boundedness of singular integrals
It is not only the Hardy-Littlewood maximal operator that is bounded on these weighted Lp spaces. In fact, any Calderón-Zygmund singular integral operator is also bounded on these spaces.[3] Let us describe a simpler version of this here.[2] Suppose we have an operator T which is bounded on L2(dx), so we have
for all smooth and compactly supported f. Suppose also that we can realise T as convolution against a kernel K in the sense that, whenever f and g are smooth and have disjoint support
Finally we assume a size and smoothness condition on the kernel K:
for all
and multi-indices
. Then, for each
and
, we have that T is a bounded operator on
. That is, we have the estimate
for all f for which the right-hand side is finite.
[edit] A converse result
If, in addition to the three conditions above, we assume the non-degeneracy condition on the kernel K: For a fixed unit vector u0
whenever
with
, then we have a converse. If we know
for some fixed
and some ω, then
.[2]


![\frac{1}{|B|} \int_B \omega(x) \, dx [ \frac{1}{|B|} \int_B \omega(x)^\frac{-p}{p'} \, dx ]^\frac{p}{p'} \leq A < \infty,](../../../../math/d/2/6/d2629bb9a021a45ee54f89285a0342c1.png)













