# Weights theory basics, pt. II

This is the 2nd part of a post on basic weights theory. In the previous part I included definitions and fundamental facts, namely the nestedness of the ${A_p}$ classes, the weighted inequalities for ${M}$ and an extrapolation result for general operators. Refer to part I for notation and those results. This part includes instead a fundamental result on Calderón-Zygmund operators, together with inequalities relevant to the proof, and an elegant proof of Marcinkiewicz’s multiplier theorem as an application.

All of the following is taken from the excellent [Duo].

4. Calderón-Zygmund operators

This section is about the following result: if ${T}$ is now a Calderón-Zygmund operator, then ${T}$ is bounded on any ${L^p (A_p)}$ for all ${1; moreover, if ${p=1}$ then ${T}$ is weakly bounded on ${L^1 (A_1)}$. This is exactly what one would’ve hoped for, and suggests that the weights theory is indeed one worth pursuing.

Recall a CZ-operator is an ${L^2}$ bounded operator given by integration against a CZ-kernel ${K}$, defined on ${\mathcal{S}(\mathbb{R}^d)}$ at least, i.e.

$\displaystyle Tf(x) = \int{K(x,y)f(y)}\,dy,$

with the properties (for some ${\delta > 0}$)

1. ${|K(x,y)|\lesssim \frac{1}{|x-y|^n}}$;
2. ${|K(x,y)- K(x,y')|\lesssim \frac{|y-y'|^\delta}{(|x-y|+|x-y'|)^{d+\delta}}\qquad}$ for ${|y-y'| \leq \frac{1}{2}\max\{|x-y|,|x-y'|\}}$;
3. ${|K(x,y)- K(x',y)|\lesssim \frac{|x-x'|^\delta}{(|x-y|+|x'-y|)^{d+\delta}}\qquad}$ for ${|x-x'| \leq \frac{1}{2}\max\{|x-y|,|x'-y|\}}$.

Then, let’s state formally

Theorem 6 Let ${T}$ be a Calderón-Zygmund operator. Then ${T}$ is bounded on all ${L^p (A_p)}$ for ${1:

$\displaystyle \int{|Tf|^p w} \lesssim \int{|f|^p w}\qquad \forall w \in A_p.$

Moreover,

$\displaystyle w\left(\{|Tf|>\lambda\}\right) \lesssim \frac{1}{\lambda}\int{|f|w}\qquad \forall w \in A_1.$

The proof of the second part is a close adaptation of the standard proof of weak ${(1,1)}$ boundedness of a CZ-operator, using CZ-decomposition. The first part relies instead on some inequalities relating maximal functions to the sharp maximal function ${M_\#}$, the maximal function that controls oscillation around the average:

$\displaystyle M_\# f(x) := \sup_{Q\ni x}{\int_{Q}{|f-f_Q|}},$

where ${f_Q = \frac{1}{|Q|}\int_{Q}{f}}$. In particular, one needs to prove the two lemmas

Lemma 7 For ${s>1}$

$\displaystyle M_\# (Tf) \lesssim_s M(|f|^s)^{1/s}.$

Lemma 8 Suppose ${M_{\mathrm{dyad}}f \in L^{p_0}(w)}$ for some ${p_0> 1}$ and ${w \in A_p}$ for ${p\geq p_0}$. Then

$\displaystyle \int{|M_{\mathrm{dyad}}f|^p w}\lesssim \int{|M_\# f|^p w}.$

Then the first part of theorem 6 follows at once if you can prove ${M_{\mathrm{dyad}}T f \in L^p (w)}$:

$\displaystyle \int{|Tf|^p w}\leq \int{|M_{\mathrm{dyad}} (T f)|^p w} \stackrel{\text{Lemma 8}}{\lesssim} \int{|M_\# (Tf)|^p w}$

$\displaystyle \stackrel{\text{Lemma 7}}{\lesssim}\int{|M(|f|^s)|^{p/s} w} \lesssim \int{|f|^p w},$

where the last inequality holds by the strong version of theorem 2 provided ${s}$ is close enough to ${1}$ for ${w}$ to belong in ${A_{p/s}}$ (again, by reverse Hölder inequality). Again, the reason we resort to maximal functions of the form ${M(|f|^s)^{1/s}}$ is because equality (7) gives us some room to stretch our exponents. As for proving ${M_{\mathrm{dyad}}Tf \in L^p(w)}$, it is enough to prove ${Tf \in L^p (w)}$, and it is enough to prove the result for ${f}$ bounded and compactly supported since these functions are dense in every ${L^p (w)}$. Therefore, if ${\mathrm{Supp}(f) \subset B(0,R)}$, the mass of ${Tf}$ around the support of ${f}$ is bounded by

$\displaystyle \int_{B(0,2R)}{|Tf|^p w} \leq \left(\int{|Tf|^{p(1+\varepsilon)'}}\right)^{1/(1+\varepsilon)'}\left(\int_{B(0,2R)}{w^{1+\varepsilon}}\right)^{1/(1+\varepsilon)},$

which is finite since ${Tf \in L^q}$ for every ${q}$ and by reverse Hölder inequality the second factor is finite as long as ${\varepsilon}$ is sufficiently small. As for the mass of ${Tf}$ away from the support of ${f}$, we can estimate pointwise, for ${x \not\in B(0,2R)}$

$\displaystyle |Tf(x)| \lesssim \int_{y\in B(0,R)}{\frac{|f(y)|}{|x-y|^d}}\,dy\lesssim \frac{\|f\|_{L^\infty}}{|x|^n},$

and therefore

$\displaystyle \int_{B(0,2R)^c}{|Tf|^p w} \lesssim \|f\|_{L^\infty}\sum_{k \in \mathbb{N}}{\int_{\{|x|\sim R 2^k\}}{\frac{w(x)}{|x|^{dp}}}\,dx}$

$\displaystyle \sim \|f\|_{L^\infty}R^{-dp} \sum_{k \in \mathbb{N}}{2^{-kdp} w\left(\{|x|\sim R 2^k\}\right)};$

once again we have to use the reverse Hölder inequality: ${w \in A_q}$ for some ${q, therefore

$\displaystyle w\left(\{|x|\lesssim R 2^k\}\right) \lesssim \left(\frac{|B(0,CR2^k)|}{|B(0,CR)|}\right)^q w(B(0,R))= 2^{kdq} C_R,$

and since ${q the above series is convergent (we’ve used (5)).

It remains to prove the lemmas, and to prove the weak boundedness part of theorem 6. We prove the latter here and move the proofs of the lemmas to an appendix to ease the presentation.

So, we want to prove that for ${w\in A_1}$ and ${T}$ with CZ-kernel, ${L^2}$ bounded,

$\displaystyle w\left(\{|Tf|>\lambda\}\right)\lesssim \frac{1}{\lambda}\int{|f| w}.$

Do a CZ-decomposition of ${f}$ w.r.t. ${\lambda}$, which yields ${\Omega= \bigcup_{Q\in\mathcal{Q}}{Q}}$ and write

$\displaystyle f=f_{\mathrm{good}}+f_{\mathrm{bad}},$

where

$\displaystyle f_{\mathrm{good}} := f(x) \qquad \text{ for } x \in \Omega^c$

and

$\displaystyle f_{\mathrm{good}} :=\frac{1}{|Q|}\int_{Q}{f} \qquad \text{ for } x \in Q \in \mathcal{Q}$

${f_{\mathrm{bad}}}$ is defined as a consequence. Notice ${|f_{\mathrm{good}}|\lesssim \lambda}$ everywhere. Then it is enough to estimate

$\displaystyle w\left(\{|Tf_{\mathrm{good}}|>\lambda\}\right) \quad \text{ and }\quad w\left(\{|Tf_{\mathrm{bad}}|>\lambda\}\right).$

Since ${T}$ is ${L^2 (w)}$ bounded (${w \in A_1}$ and thus ${w \in A_2}$ too), one has

$\displaystyle w\left(\{|Tf_{\mathrm{good}}|>\lambda\}\right) \lesssim \frac{1}{\lambda^2} \int{|f_{\mathrm{good}}|^2 w}$

$\displaystyle \leq \frac{1}{\lambda} \int_{\Omega^c}{|f_{\mathrm{good}}| w} + \sum_{Q\in\mathcal{Q}}{\frac{1}{\lambda^2}w(Q) \left(\frac{1}{|Q|}\int_{Q}{|f|}\right)^2}$

$\displaystyle \lesssim \frac{1}{\lambda} \int_{\Omega^c}{|f| w} + \sum_{Q\in\mathcal{Q}}{\frac{1}{\lambda}w(Q)\frac{1}{|Q|}\int_{Q}{|f|}},$

and since ${w \in A_1}$ we can estimate

$\displaystyle \frac{w(Q)}{|Q|}\int_{Q}{|f|} \leq \int_{Q}{|f|w},$

and thus ${f_{\mathrm{good}}}$ is alright.

As for ${f_{\mathrm{bad}}}$, we exploit cancellation on every ${Q}$. Write ${f_{\mathrm{bad}}^Q}$ for ${ \chi_Q f_{\mathrm{bad}}}$ (so ${\int{f_{\mathrm{bad}}^Q}=0}$ and ${f_{\mathrm{bad}} = \sum_{Q\in\mathcal{Q}}{f_{\mathrm{bad}}^Q}}$) and ${\Omega_\ast}$ for ${\bigcup_{Q\in\mathcal{Q}}{2Q}}$ (i.e. points close to any ${Q \subset \Omega}$); we only need to estimate the contribution from outside this set, since it’s small

$\displaystyle w\left(\Omega_\ast\right) \leq \sum_{Q\in \mathcal{Q}}{w(2Q)} \lesssim 2^{d}\sum_{Q\in \mathcal{Q}}{w(Q)}$

$\displaystyle \lesssim \sum_{Q\in \mathcal{Q}}{\frac{w(Q)}{|Q|}\frac{1}{\lambda}\int_{Q}{|f|}}\leq \frac{1}{\lambda}\sum_{Q\in \mathcal{Q}}{\int_{Q}{|f|w}}.$

Thus, since ${Tf_{\mathrm{bad}}^Q(x) = \int_{Q}{(K(x,y)-K(x,y'))f_{\mathrm{bad}}^Q (y)}\,dy}$ for any ${y' \in Q}$, estimate

$\displaystyle w\left(\{|Tf_{\mathrm{bad}}|>\lambda\}\cap (\Omega_\ast)^c \right)\leq \frac{1}{\lambda}\int_{(\Omega_\ast)^c}{|Tf_{\mathrm{bad}}| w}$

$\displaystyle \lesssim \frac{1}{\lambda} \sum_{Q\in\mathcal{Q}}{\int_{(2Q)^c}{\int_{Q}{|K(x,y) - K(x,c_Q)||f_{\mathrm{bad}}^Q (y)|w(x)}\,dy}\,dx},$

where ${c_Q}$ is the center of cube ${Q}$, and by the hypotheses on the kernel this quantity is bounded by

$\displaystyle \frac{1}{\lambda}\sum_{Q\in\mathcal{Q}}{\sum_{k \in \mathbb{N}}{\int_{|x-c_Q|\sim \ell(Q)2^k}{\int_{Q}{\frac{|y-c_Q|^\delta}{(|x-y|+|x-c_Q|)^{d+\delta}} |f_{\mathrm{bad}}^Q(y)| w(x)}\,dy}\,dx}}.$

Now ${|x-c_Q| \sim |x-y| \sim \ell(Q)2^k}$ and ${|y-c_Q|\lesssim \ell(Q)}$, thus this is bounded by

$\displaystyle \frac{1}{\lambda}\sum_{Q\in\mathcal{Q}}{\sum_{k\in\mathbb{N}}{2^{-k(d+\delta)}\ell(Q)^{-d} w(2^k Q) \int_{Q}{|f_{\mathrm{bad}}^Q|}}}.$

Since ${2^{-kd}\ell(Q)^{-d} = |2^k Q|^{-1}}$, the factor in front of the integral is

$\displaystyle 2^{-k\delta} \frac{w(2^kQ)}{|2^k Q|}$

and as done before, by the ${A_1}$ condition we can bound the expression by

$\displaystyle \frac{1}{\lambda}\sum_{Q\in\mathcal{Q}}{\sum_{k\in\mathbb{N}}{2^{-k\delta} \int_{Q}{|f_{\mathrm{bad}}^Q| w}}} \sim_\delta \frac{1}{\lambda}\int_{\Omega}{|f_{\mathrm{bad}}|w},$

and since ${|f_{\mathrm{bad}}|\leq |f| + |f_{\mathrm{good}}|}$ and ${|f_{\mathrm{good}}| \lesssim \lambda}$, the last term is controlled by ${\lambda^{-1}\int{|f|w}}$ and we’re done.

Remark 4 We’ve seen ${T}$ is bounded on ${L^p (A_p)}$ for any ${p>1}$, but how does the operator norm ${\|T\|_{L^p (w) \rightarrow L^p (w)}}$ depend on ${w}$? for the case ${p=2}$ it was conjectured that

$\displaystyle \|T\|_{L^2 (w) \rightarrow L^2 (w)} \leq C_T [w]_{A_2}\,,$

and this was known as ${A_2}$ conjecture. It was solved positively by Hyt\”{o}nen in 2010, and is now known as ${A_2}$ theorem. One of the presentations at last year’s Summer School in Bonn was exactly about the solution to the ${A_2}$ conjecture. I can’t remember the details but I’m planning to read the whole paper, hopefully I’ll get back on this.

5. Quick recap

• the ${A_p}$ condition is equivalent to strong boundedness ${L^p (w) \rightarrow L^p (w)}$ of ${M}$ for ${p>1}$, and to weak boundedness for ${p=1}$;
• For every ${w \in A_p}$ one can find a smaller ${\tilde{p} s.t. ${w \in A_{\tilde{p}}}$ too; in particular, every class is exactly the union of the previous classes: ${A_p = \bigcup_{q. This property turns out to be fundamental;
• functions of the form ${(Mf)^\delta}$ for ${0<\delta<1}$ are ${A_1}$ weights (and all ${A_1}$ weights are like this, modulus a factor ${K(x) \sim 1}$);
• Once an operator ${T}$ is bounded on all ${L^r (A_r)}$ for just one exponent ${r}$, it is bounded on all ${L^p(A_p)}$ for every exponent ${p>1}$;
• if ${T}$ is Calderón-Zygmund in particular, then it is bounded on every ${L^p(A_p)}$ for ${p>1}$, and weakly bounded w.r.t. ${A_1}$ weights.

6. Applications: an elegant proof of Marcinkiewicz multiplier theorem

As a treat to the reader who made it so far, I want to show the usefulness of weights theory by a remarkable example. We’ll prove Marcinkiewicz multipliers are ${L^p}$ bounded for ${1 with very little effort. Remember ${m}$ is said to be a Marcinkiewicz multiplier if ${m \in L^{\infty} (\mathbb{R})}$ and the total variation of ${m}$ on every dyadic interval is uniformly bounded by a constant. Then we can state formally

Theorem 9 (Marcinkiewicz multipliers) Let ${m: \mathbb{R}\rightarrow \mathbb{R}}$ be a Marcinkiewicz multiplier. Then the operator ${T_m}$ defined by

$\displaystyle \widehat{T_m f} = m \widehat{f}$

is ${L^p (\mathbb{R})}$ bounded for every ${1.

The proof goes through an easy lemma:

Lemma 10 Let ${\{T_j\}_{j \in\mathbb{Z}}}$ be a family of operators bounded on ${L^2 (w)}$ for any ${w \in A_2}$ with uniform constants (possibly depending on ${[w]_{A_2}}$). Then for ${1

$\displaystyle \left\|\left(\sum_{j \in \mathbb{Z}}{|T_j f_j|^2}\right)^{1/2}\right\|_{L^p} \lesssim_p \left\|\left(\sum_{j \in \mathbb{Z}}{|f_j|^2}\right)^{1/2}\right\|_{L^p}$

Proof: The result is trivial for ${p=2}$. Now, for ${p>2}$, the square of the norm on the left is

$\displaystyle \left\|\sum_{j \in \mathbb{Z}}{|T_j f_j|^2}\right\|_{L^{p/2}} = \int{\sum_{j \in \mathbb{Z}}{|T_j f_j|^2 u}}$

for some ${u \in L^{(p/2)'}}$ s.t. ${\|u\|_{L^{(p/2)'}}=1}$. Now ${u \leq M(u^{1/\nu})^{\nu}}$, and the latter is in ${A_1}$ for ${0< \nu <1}$ and thus is in ${A_2}$ too. Therefore the above is bounded by (remember the ${A_1}$ constant only depends on ${\nu}$, not on ${u}$)

$\displaystyle \sum_{j\in\mathbb{Z}}{\int{|T_j f_j|^2 M(u^{1/\nu})^{\nu}}} \lesssim_{\nu} \sum_{j\in\mathbb{Z}}{\int{|f_j|^2 M(u^{1/\nu})^{\nu}}},$

and by Hölder’s inequality

$\displaystyle \int{\sum_{j\in\mathbb{Z}}{|f_j|^2} M(u^{1/\nu})^{\nu}} \leq \left\|\sum_{j\in\mathbb{Z}}{|f_j|^2}\right\|_{L^{p/2}}\|M(u^{1/\nu})^{\nu}\|_{L^{(p/2)'}}$

$\displaystyle =\left\|\left(\sum_{j\in\mathbb{Z}}{|f_j|^2}\right)^{1/2}\right\|_{L^{p}}^2\|M(u^{1/\nu})\|_{L^{\nu(p/2)'}}^{1/\nu},$

and by boundedness of ${M}$, if you choose ${\nu}$ close enough to ${1}$ s.t. ${\nu(p/2)'>1}$,

$\displaystyle \|M(u^{1/\nu}\|_{L^{\nu(p/2)'}}^{1/\nu}\lesssim \|u^{1/\nu}\|_{L^{\nu(p/2)'}}^{1/\nu} = \|u\|_{L^{(p/2)'}}= 1.$

The remaining case ${1 follows by duality applying the previous part to the adjoints. $\Box$

We know for sure by a previous section that Calderón-Zygmund operators are ${L^p (A_p)}$ bounded for ${1 and in particular for ${p=2}$. We want to prove that the localized ${T_j =}$ operator with multiplier ${m_j:=m \chi_{I_j}}$, where ${I_j}$ is the dyadic interval ${\{x\,:\, 2^j < |x|< 2^{j+1}\}}$, is a family of bounded operators as required in the lemma. Then, by standard Littlewood-Paley theory it will follow (here ${S_j}$ is the operator associated to multiplier ${\chi_{I_j}}$)

$\displaystyle \|T_m f\|_{L^p} \sim \left\|\left(\sum_{j \in \mathbb{Z}}{|S_j T_m f|^2}\right)^{1/2}\right\|_{L^p}$

$\displaystyle = \left\|\left(\sum_{j \in \mathbb{Z}}{|T_j S_j f|^2}\right)^{1/2}\right\|_{L^p} \lesssim \left\|\left(\sum_{j \in \mathbb{Z}}{|S_j f|^2}\right)^{1/2}\right\|_{L^p} \sim \|f\|_{L^p},$

therefore proving the theorem.

Now, by the bounded variation condition we know there must exists ${dm}$ s.t. for ${\xi \in I_j}$

$\displaystyle m_j (\xi) = m(2^j)\chi_j(\xi) + \int_{2^j}^{\xi}{dm(t)},$

and with this we can rewrite formally

$\displaystyle T_j f (x) = \int{e^{2\pi i x \xi}m_j(\xi) \widehat{f}(\xi)}\,d\xi$

$\displaystyle = m(2^j) S_j f(x) + \int_{\mathbb{R}}{e^{2\pi i x \xi} \widehat{f}(\xi) \int_{2^j}^{\xi}{dm(t)}}\,d\xi$

$\displaystyle = m(2^j)S_j f(x) + \int_{\mathbb{R}}\int_{2^j}^{2^{j+1}}{e^{2\pi i x \xi} \widehat{f}(\xi) \chi_{]2^j,\xi[}(t) dm(t)}\,d\xi,$

and since ${\chi_{]2^j,\xi[}(t) = \chi_{[t,2^{j+1}[}(\xi)}$,

$\displaystyle = m(2^j)S_j f(x) + \int_{2^j}^{2^{j+1}}{S_{t,j} f(x)}\,dm(t),$

where ${S_{t,j}}$ is the operator with multiplier ${\chi_{[t,2^{j+1}[}}$. Now, we know that both ${S_j}$ and ${S_{t,j}}$ can be written in terms of the Hilbert transform\footnote{In general ${S_{]a,b[} = \frac{i}{2}\left(M_a H M_{-a} - M_b H M_{-b}\right)}$, where ${M_\eta}$ is modulation, i.e. ${M_\eta f(x) = e^{2\pi i \eta x} f(x)}$.}, and are thus ${L^2 (A_2)}$ bounded with comparable constants (independent of ${j}$). Therefore for ${w \in A_2}$

$\displaystyle \|T_j f\|_{L^2 (w)} \lesssim \|m\|_{L^\infty} \|S_j f\|_{L^2 (w)} + \left(\int{\left(\int_{2^j}^{2^{j+1}}{S_{t,j} f(x)}\,dm(t)\right)^2 w(x)}\,dx\right)^{1/2},$

which by Minkowski’s inequality is bounded by

$\displaystyle \lesssim \|m\|_{L^\infty} \|f\|_{L^2 (w)} + \int_{2^j}^{2^{j+1}}{\left(\int{|S_{t,j}f (x)|^2 w(x)}\,dx\right)^{1/2}}\,d|m|(t)$

$\displaystyle \lesssim \|m\|_{L^\infty} \|f\|_{L^2 (w)} + \|f\|_{L^2 (w)}\int_{2^j}^{2^{j+1}}{d|m|(t)} \lesssim_m \|f\|_{L^2 (w)}$

by the hypothesis of bounded variation of ${m}$ on dyadic intervals. This concludes the proof.

7. Appendix: proofs of Lemma 7 and Lemma 8

Proof: For a fixed cube ${Q}$ and ${x \in Q}$ it is enough to show that for some ${c\in \mathbb{R}}$

$\displaystyle \frac{1}{|Q|}\int_{Q}{|Tf - c|} \lesssim M(|f|^s)^{1/s}(x),$

since this is an equivalent characterization\footnote{Indeed, ${M_\# f(x) \leq 2 \sup_{Q\ni x}{\inf_{c \in \mathbb{R}}{\frac{1}{|Q|}\int_{Q}{|f-c|}}}}$ because ${|Q|^{-1}\int_{Q}{|f-f_Q|} \leq |Q|^{-1}\int_{Q}{|f-c|} + |c -f_Q| \leq |Q|^{-1}\int_{Q}{|f-c|} + \left||Q|^{-1}\int_{Q}{(f-c)}\right|}$, and the viceversa (with constant 1) is trivial.}. The strategy is to estimate separately the contributions from ${f}$ around the cube and ${f}$ far from it. Thus write ${f = f_1 + f_2}$ where ${f_1 = f \chi_{2Q}}$. We claim that ${c = Tf_2 (x)}$ is a good constant. Indeed

$\displaystyle \frac{1}{|Q|}\int_{Q}{|Tf(y) - Tf_2(x)|}\,dy \leq \frac{1}{|Q|}\int_{Q}{|Tf_1|} + \frac{1}{|Q|}\int_{Q}{|Tf_2(y) - Tf_2 (x)|}\,dy.$

By ${L^s}$ boundedness of ${T}$,

$\displaystyle \frac{1}{|Q|}\int_{Q}{|Tf_1(y)|}\,dy \leq \left(\frac{1}{|Q|}\int_{Q}{|Tf_1|^s}\right)^{1/s} \lesssim\left(\frac{1}{|Q|}\int_{Q}{|f_1|^s}\right)^{1/s}$

$\displaystyle =\left(\frac{1}{|Q|}\int_{Q}{|f|^s}\right)^{1/s} \leq M(|f|^s)^{1/s} (x).$

As for the second term, it is bounded by

$\displaystyle \frac{1}{|Q|}\int_{Q}{\int_{(2Q)^c}{|K(y,z) - K(x,z)||f(z)|}\,dz}\,dy,$

and since ${K}$ is a CZ-kernel this is bounded by

$\displaystyle \frac{1}{|Q|}\int_{Q}{\int_{(2Q)^c}{\frac{|y-z|^\delta}{(|y-z|+|x-z|)^{d+\delta}}|f(z)|}\,dz}\,dy$

$\displaystyle \lesssim \frac{1}{|Q|}\int_{Q}{\sum_{k \in \mathbb{N}}{\int_{|z-c_Q|\sim 2^k \ell (Q)}{\frac{\ell(Q)^\delta}{(2^k \ell(Q))^{d+\delta}}|f(z)|}\,dz}}\,dy$

$\displaystyle \leq \frac{1}{|Q|}\int_{Q}{\sum_{k \in \mathbb{N}}{2^{-k\delta} Mf(x)}}\,dy = C_\delta Mf(x) \leq C_\delta M(|f|^s)^{1/s} (x).$

$\Box$

Before proving the remaining lemma we show how to prove inequalities of the form

$\displaystyle \int{|Tf|^p} \lesssim \int{|Sf|^p}$

by using what’s known as a good lambda inequality, i.e. for some ${\delta>0}$ and for every ${\gamma, \lambda>0}$

$\displaystyle |\{|Tf|>2\lambda, |Sf|\leq \gamma\lambda \}| \lesssim \gamma^\delta |\{|Tf|>\lambda\}| \ \ \ \ \ (10)$

(of course we’ll be using a different measure, but it can be adapted easily). Heuristically, such an inequality is saying that ${|Sf|}$ can’t be much smaller than ${|Tf|}$ in a big proportion of the volume where ${|Tf|}$ is large – the portion must shrink as the ratio ${|Sf|/|Tf|}$ decreases. With (10), one has

$\displaystyle \int{|Tf|^p} = p \int_{0}^{\infty}{\lambda^{p-1}|\{|Tf|>\lambda\}|}\,d\lambda,$

and

$\displaystyle p \int_{0}^{2R}{\lambda^{p-1}|\{|Tf|>\lambda\}|}\,d\lambda = 2^p p \int_{0}^{R}{\lambda^{p-1}|\{|Tf|>2\lambda\}|}\,d\lambda$

$\displaystyle \leq 2^p p \int_{0}^{R}{\lambda^{p-1}\left(|\{|Tf|>2\lambda, |Sf|\leq \gamma\lambda\}|+ |\{|Sf|>\gamma\lambda\}|\right)}\,d\lambda$

$\displaystyle \lesssim 2^p p \gamma^\delta \int_{0}^{R}{\lambda^{p-1} |\{|Tf|>\lambda\}|}\,d\lambda + 2^p p \frac{1}{\gamma^{p-1}}\int_{0}^{\gamma R}{\lambda^{p-1}|\{|Sf|>\lambda\}|}\,\gamma^{-1}d\lambda,$

thus for ${\gamma}$ small enough one has

$\displaystyle p \int_{0}^{2R}{\lambda^{p-1}|\{|Tf|>\lambda\}|}\,d\lambda \lesssim p \int_{0}^{\gamma R}{\lambda^{p-1}|\{|Sf|>\lambda\}|}\,d\lambda,$

and the ${L^p}$ inequality follows by taking the limit for ${R\rightarrow \infty}$. One minor point: you have to verify the partial integrals are finite, and this is true for ${p\geq p_0}$ if ${Tf \in L^{p_0}}$ (easy).

Proof: By the remark just made, we can prove a good lambda inequality:

$\displaystyle w\left(\{|M_{\mathrm{dyad}}f|>2\lambda, |M_\# f|\leq \gamma \lambda\}\right) \lesssim \gamma^\delta w\left(\{|M_{\mathrm{dyad}}f|>\lambda\}\right).$

The sets ${\{|M_{\mathrm{dyad}}f|>\lambda\}}$ are the union of disjoint dyadic cubes (by definition of ${|M_{\mathrm{dyad}}}$). Take a maximal ${Q}$ in such union, i.e. ${|Q|^{-1} \int_{Q}{|f|}>\lambda}$ but if ${Q^\ast}$ is the parent dyadic cube then ${|Q^\ast|^{-1} \int_{Q^\ast}{|f|}\leq \lambda}$. It is enough to prove the good lambda inequality on each ${Q}$, so our goal is

$\displaystyle w\left(\{x\in Q\,: \, |M_{\mathrm{dyad}}f|>2\lambda, |M_\# f|\leq \gamma \lambda\}\right) \lesssim \gamma^\delta w\left(Q\right).$

We can make a further reduction thanks to the reverse Hólder inequality: for any ${w \in A_p}$ there exists a ${\delta>0}$ s.t. for ${S \subset Q}$

$\displaystyle w(S) \lesssim \left(\frac{|S|}{|Q|}\right)^\delta w(Q);$

indeed, by Hólder’s inequality

$\displaystyle w(S) = \int{w \chi_S}\leq \left(\int{w^{1+\varepsilon}}\right)^{1/(1+\varepsilon)}|S|^{\varepsilon/(1+\varepsilon)}$

$\displaystyle \lesssim \frac{w(Q)}{|Q|}|Q|^{1/(1+\varepsilon)}|S|^{\varepsilon/(1+\varepsilon)} = w(Q) \frac{|S|^{\varepsilon/(1+\varepsilon)}}{|Q|^{\varepsilon/(1+\varepsilon)}},$

and you can choose ${\delta:=\frac{\varepsilon}{1+\varepsilon}}$. Thus the good lambda inequality will be proved if we can prove that

$\displaystyle \left|\{x\in Q\,: \, |M_{\mathrm{dyad}}f|>2\lambda, |M_\# f|\leq \gamma \lambda\}\right| \lesssim \gamma |Q|.$

Take ${x}$ in the set on the left side of the inequality. Then, since ${|f_{Q^\ast}|\leq |Q^\ast|^{-1} \int_{Q^\ast}{|f|}\leq \lambda}$,

$\displaystyle M_{\mathrm{dyad}}((f-f_{Q^\ast})\chi_Q) (x)> M_{\mathrm{dyad}}(f\chi_Q) (x) -|f_{Q^\ast}| > 2\lambda -\lambda = \lambda.$

This implies

$\displaystyle \{x\in Q\,: \, |M_{\mathrm{dyad}}f|>2\lambda, |M_\# f|\leq \gamma \lambda\} \subset \{x\in Q\,: \, |M_{\mathrm{dyad}}((f-f_{Q^\ast})\chi_Q)|>\lambda\},$

and by weak ${(1,1)}$ boundedness of ${M_{\mathrm{dyad}}}$

$\displaystyle |\{x\in Q\,: \, |M_{\mathrm{dyad}}((f-f_{Q^\ast})\chi_Q)|>\lambda\}| \lesssim \lambda^{-1} \int_{Q}{|f-f_{Q^\ast}|}$

$\displaystyle \leq \lambda^{-1} \frac{|Q^\ast|}{|Q^\ast|}\int_{Q^\ast}{|f-f_{Q^\ast}|} \leq \lambda^{-1} 2^d |Q| \inf_{x \in Q} M_\# f (x) \lesssim_d \gamma |Q|,$

which concludes the proof if the set is non-empty (if it is, there’s nothing to prove). $\Box$

References:

[Duo] J. Duoandikoetxea, “Fourier Analysis“, Graduate Studies in Mathematics, vol. 29, AMS Providence, Rhode Island.