The Weighted H^p Estimates for Commutators of Fractional Integrals

Abstract

For 0 < α < n, let I_α be the Riesz potential operator, $b\in L_{\text {loc}}(\mathbb {R}^{n})$. Harboure et al. (Illinois J. Math. 41(4), 676–700 1997) showed that when $b\in BMO(\mathbb {R}^{n})$, the commutator [b,I_α] may be not bounded from $H^{1}(\mathbb {R}^{n})$ to $L^{n/(n-\alpha )}(\mathbb {R}^{n})$. In this paper, the authors show that there are nontrivial subspaces of $BMO(\mathbb {R}^{n})$, when b belongs to these subspaces, such that [b,I_α] is bounded from $H_{\omega ^{p}}^{p}(\mathbb {R}^{n})$ to $L_{\omega ^{q}}^{q}(\mathbb {R}^{n})$, or from $H_{\omega ^{p}}^{p}(\mathbb {R}^{n})$ to $H_{\omega ^{q}}^{q}(\mathbb {R}^{n})$ for certain 0 < p ≤ 1 with 1/q = 1/p − α/n and $\omega \in A_{\infty }$. The corresponding results for the commutators of fractional integrals with general homogeneous kernels and Hörmander type kernels are also given.

Appendix

In this section, we give the proofs of Theorems 5.10-5.13.

Proof of Theorem 5.10

By Lemma 2.5, it suffices to show that, for any continuous $(H_{\omega ^{p}}^{p}(\mathbb {R}^{n}),\infty )$-atom a, there exists a constant C > 0 independent of a such that $\|T_{\alpha } a\|_{L_{\omega ^{q}}^{q}(\mathbb {R}^{n})}\leq C$. Let a be an $(H_{\omega ^{p}}^{p}(\mathbb {R}^{n}),\infty )$-atom with supp a ∈ B(x₀,r). We write

$$ \begin{array}{@{}rcl@{}} \|T_{\alpha}a\|_{L_{\omega^{q}}^{q}(\mathbb R^{n})} &=&\Big({\int}_{\mathbb R^{n}}|T_{\alpha} a(x)|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq&\Big({\int}_{2B}|T_{\alpha} a(x)|^{q}\omega(x)^{q}dx\Big)^{1/q} +\Big({\int}_{(2B)^{c}}|T_{\alpha} a(x)|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &=:&K_{1}+K_{2}. \end{array} $$

For K₁, it is known that ω^{ns/(ns−n−sα)} ∈ A₁ implies ω^{ns(1+η)/(ns−n−sα)} ∈ A₁ for some η > 0. Let

$$\frac{1}{q}=\frac{1}{q_{1}}+\frac{1}{p}-\frac{1}{p_{1}},\quad\frac{1}{q_{1}} =\frac{1}{p_{1}}-\frac{\alpha}{n},~\text{and}~ \frac{1}{p_{1}}=\frac{1}{s}+\frac{\alpha}{n}+\frac{ns-n-s\alpha}{sn(1+\eta)}.$$

Since q₁ = ns(1 + η)/(ns + nη − sα) < ns(1 + η)/(ns − n − sα), we have $\omega ^{q_{1}}\in A_{1}\subset A_{1+q_{1}/p_{1}^{\prime }}$. It follows from Lemma 2.3 that $\omega \in A_{p_{1},q_{1}}$. Applying Hölder’s inequality and Theorem I, we get

$$ \begin{array}{@{}rcl@{}} K_{1} &\leq& C\|T_{\alpha} a\|_{L_{\omega^{q_{1}}}^{q_{1}}(\mathbb{R}^{n})}r^{n(1/p-1/p_{1})}\\ &\leq& C\|a\|_{L_{\omega^{p_{1}}}^{p_{1}}(\mathbb{R}^{n})}r^{n(1/p-1/p_{1})}\\ &\leq& C\|a\|_{L^{\infty}(\mathbb{R}^{n})}(\omega^{p_{1}}(B))^{1/p_{1}}r^{n(1/p-1/p_{1})}\\ &\leq& C(\omega^{p}(B))^{-1/p}(\omega^{p_{1}}(B))^{1/p_{1}}r^{n(1/p-1/p_{1})}. \end{array} $$

It follows from Lemma 2.1 that

$$\omega^{p}\in RH_{ns(1+\eta)/p(ns-n-s\alpha)}\subset RH_{p_{1}/p}.$$

By the definition of the reverse Hölder condition,

$$\big(\omega^{p}(B)\big)^{-1/p}\big(\omega^{p_{1}}(B)\big)^{1/p_{1}}\leq C r^{n(1/p_{1}-1/p)},$$

which implies K₁ ≤ C. For K₂, by the cancellation condition of a and the Minkowski inequality,

$$ \begin{array}{@{}rcl@{}} K_{2} &=&\Big({\int}_{(2B)^{c}}\Big|{\int}_{B}(K_{\alpha}(x-y)-K_{\alpha}(x))a(y)dy\Big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq &{\int}_{B}|a(y)|{\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{q}\omega(x)^{q}dx\Big)^{1/q}dy. \end{array} $$

Noting that s > q, Hölder’s inequality gives

$$ \begin{array}{@{}rcl@{}} &&{\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq& {\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}\Big({\int}_{2^{j+1}B\backslash2^{j}B} \omega(x)^{sq/(s-q)}dx\Big)^{1/q-1/s}. \end{array} $$

Since ns(1 + η)/(ns − n − sα) = sq₁/(s − q₁) > sq/(s − q), we have ω^sq/(s−q) ∈ A₁. Then by Lemma 2.2, we get

$$ \begin{array}{@{}rcl@{}} K_{2} &\leq& {\int}_{B}{\sum}_{j=1}^{\infty}(2^{j} r)^{n-\alpha}2^{j\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B}|K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times (\omega^{sq/(s-q)}(2^{j} B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s} 2^{-j\alpha}\\ &\leq& C {\int}_{B}{\sum}_{j=1}^{\infty}(2^{j}r)^{n-\alpha}2^{j\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times 2^{j(n/q-n/s)}(\omega^{sq/(s-q)} (B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s}2^{-j\alpha}. \end{array} $$

Due to ω^sq/(s−q) ∈ A₁ gives ω^p ∈ RH_sq/p(s−q), thus

$$(\omega^{sq/(s-q)} (B))^{1/q-1/s}(\omega^{p}(B))^{-1/p}\leq C r^{-n/s-\alpha}.$$

Then,

$$ \begin{array}{@{}rcl@{}} K_{2} & \leq& C {\int}_{B}{\sum}_{j=1}^{\infty}(2^{j}r)^{n-\alpha}2^{j\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\times 2^{j(n/q-n/s)}(2^{j}r)^{\alpha-n+n/s}2^{-j\alpha} \big(\omega^{p}(B)\big)^{1/p}r^{-n/s-\alpha}\\ &\leq& C{\int}_{B}|a(y)|dy\big(\omega^{p}(B)\big)^{1/p}r^{-n}\leq C. \end{array} $$

This completes the proof of Theorem 5.10. □

Proof of Theorem 5.11

We use the same argument as in the proof of Theorem 5.10. Let a be a continuous $(H_{\omega }^{1}(\mathbb {R}^{n}),\infty )$-atom supported in the ball B(x₀,r). It suffices to check that for q = n/(n − α),

$$\|T_{\alpha} a\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq C.$$

We write

$$\|T_{\alpha} a\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq K_{1}+K_{2},$$

where K₁ and K₂ are the same as in the proof of Theorem 1.1. It is easy to know that K₁ ≤ C. For K₂, by the cancellation condition of a and the Minkowski inequality,

$$ \begin{array}{@{}rcl@{}} K_{2} &\leq& {\int}_{B} {\sum}_{j=1}^{\infty}(2^{j}r)^{n-\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times (\omega^{sq/(s-q)}(2^{j} B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s}\\ &\leq &C{\int}_{B} {\sum}_{j=1}^{\infty}(2^{j}r)^{n-\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times 2^{j(n/q-n/s)}(\omega^{sq/(s-q)} (B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s}\\ &\leq& C{\int}_{B} {\sum}_{j=1}^{\infty}(2^{j}r)^{n-\alpha}\Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B}|K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times 2^{j(n/q-n/s)}(2^{j}r)^{\alpha-n+n/s} r^{-n/s-\alpha}\omega(B)\\ &\leq& C{\int}_{B}|a(y)|dy r^{-n}\omega(B). \end{array} $$

This completes the proof of Theorem 5.11. □

Before proving Theorem 5.12, we first establish the following lemma.

Lemma A.1

Let 0 < α ≤ 1, n/(n + α) ≤ p < 1, 1/q = 1/p − α/n, and n/(n − α) < s. Suppose that $K_{\alpha }\in \tilde H_{\alpha ,s}^{\alpha }\cap S_{\alpha }$, and ω^{ns/(ns−n−sα)} ∈ A₁. Then for $b\in \text {BMO}(\mathbb {R}^{n})$ and any continuous $(H_{\omega ^{p}}^{p}(\mathbb {R}^{n}),\infty )$-atom a related to a ball $B:=B(x_{0},r)\subset \mathbb {R}^{n}$, we have

$$\big\|(b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq C\|b\|_{BMO(\mathbb{R}^{n})}.$$

Proof

We write

$$ \begin{array}{@{}rcl@{}} \big\|(b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})} &\leq& \Big({\int}_{2B}\big|(b(x)-b_{B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &&\qquad+\Big({\int}_{(2B)^{c}}\big|(b(x)-b_{B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &=:&K_{1}+K_{2}. \end{array} $$

For K₁, it is known that ω^{ns/(ns−n−sα)} ∈ A₁ implies ω^{ns(1+η)/(ns−n−sα)} ∈ A₁ for some η > 0. By the definition of p₁, q₁ in Theorem 5.10 and Lemma 2.3, we have $\omega \in A_{p_{1},q_{1}}$. Applying Hölder’s inequality, Theorem 5.10 and Theorem I, we get

$$ \begin{array}{@{}rcl@{}} K_{1} &=&\Big({\int}_{2B}\big|(b(x)-b_{B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq& |b_{B}-b_{2B}|\big\|T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})} +\Big({\int}_{2B}\big|(b(x)-b_{2B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq& C |b_{B}-b_{2B}|\| a\|_{H_{\omega^{p}}^{p}(\mathbb{R}^{n})} +\Big({\int}_{2B}\big|(b(x)-b_{2B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q} \end{array} $$

$$ \begin{array}{@{}rcl@{}} &\leq& C\|b\|_{BMO(\mathbb{R}^{n})}+\Big({\int}_{2B} \big|b(x)-b_{2B}\big|^{p_{1}p/(p_{1}-p)}dx\Big)^{1/p-1/p_{1}} \|T_{\alpha} a\|_{L_{\omega^{q_{1}}}^{q_{1}}(\mathbb{R}^{n})}\\ &\leq& C\|b\|_{BMO(\mathbb{R}^{n})}+C|2B|^{1/p-1/p_{1}} \|b\|_{BMO(\mathbb{R}^{n})}\|a\|_{L_{\omega^{p_{1}}}^{p_{1}}(\mathbb{R}^{n})}\\ &\leq& C\|b\|_{BMO(\mathbb{R}^{n})}+C|2B|^{1/p-1/p_{1}} \|b\|_{BMO(\mathbb{R}^{n})}\big(\omega^{p}(B)\big)^{-1/p}\big(\omega^{p_{1}}(B)\big)^{1/p_{1}}. \end{array} $$

Due to ω^{ns(1+η)/(ns−n−sα)} ∈ A₁, then

$$\omega^{p}\in RH_{ns(1+\eta)/p(ns-n-s\alpha)}\subset RH_{p_{1}/p}.$$

By the definition of the reverse Hölder condition,

$$\big(\omega^{p}(B)\big)^{-1/p}\big(\omega^{p_{1}}(B)\big)^{1/p_{1}}\leq C r^{n(1/p_{1}-1/p)},$$

which implies K₁ ≤ C.

We now turn to estimate the other term K₂. By the cancellation condition of a and the Minkowski inequality, we have

$$ \begin{array}{@{}rcl@{}} K_{2}&=&\Big({\int}_{(2B)^{c}}\Big|(b(x)-b_{B}){\int}_{B}\big(K_{\alpha}(x-y) -K_{\alpha}(x)\big)a(y)dy\Big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq& {\int}_{B}|a(y)|\Big({\int}_{(2B)^{c}}\Big|K_{\alpha}(x-y)-K_{\alpha}(x)\Big|^{q} \big|b(x)-b_{B}\big|^{q}\omega(x)^{q}dx\Big)^{1/q}dy\\ &\leq& {\int}_{B}|a(y)|{\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\setminus2^{j}B} \Big|K_{\alpha}(x-y)-K_{\alpha}(x)\Big|^{q}\big|b(x)-b_{B}\big|^{q}\omega(x)^{q}dx\Big)^{1/q}dy. \end{array} $$

Since ns(1 + η)/(ns − n − sα) = sq₁/(s − q₁) > sq/(s − q), we have ω^sq/(s−q) ∈ A₁. Then by Hölder’s inequality, Lemmas 2.2 and 2.4, we get,

$$ \begin{array}{@{}rcl@{}} &&{\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{q}\big|b(x)-b_{B}\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &\leq& {\sum}_{j=1}^{\infty}\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}\\ &&\qquad\times\Big({\int}_{2^{j+1}B\backslash2^{j}B} \big|b(x)-b_{B}\big|^{sq/(s-q)}\omega(x)^{sq/(s-q)}dx\Big)^{1/q-1/s}\\ &\leq& C{\sum}_{j=1}^{\infty} j\Big({\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}\\ &&\qquad\times\|b\|_{BMO(\mathbb{R}^{n})} (\omega^{sq/(s-q)}(2^{j} B))^{1/q-1/s}\\ &\leq& C{\sum}_{j=1}^{\infty} j(2^{j} r)^{n-\alpha}2^{j\alpha} \Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B}|K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}\\ &&\qquad\times 2^{j(n/q-n/s)}(\omega^{sq/(s-q)}(B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s} 2^{-j\alpha}\|b\|_{BMO(\mathbb{R}^{n})}. \end{array} $$

Thus,

$$ \begin{array}{@{}rcl@{}} K_{2} &\leq& C{\int}_{B}{\sum}_{j=1}^{\infty} j(2^{j} r)^{n-\alpha}2^{j\alpha} \Big(\frac{1}{(2^{j}r)^{n}} {\int}_{2^{j+1}B\backslash2^{j}B}|K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times 2^{j(-n+n/q)}(\omega^{sq/(s-q)}(B))^{1/q-1/s}r^{\alpha-n+n/s} \|b\|_{BMO(\mathbb{R}^{n})}\\ &\leq& C{\int}_{B}|a(y)|dy r^{-n} \big(\omega^{p}(B)\big)^{1/p}\|b\|_{BMO(\mathbb{R}^{n})} \leq C\|b\|_{BMO(\mathbb{R}^{n})}. \end{array} $$

This, together with the estimate of K₁, leads to

$$\big\| (b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq C\| b\|_{\text{BMO}(\mathbb{R}^{n})}.$$

This completes the proof of Lemma A.1. □

We are now in a position to prove Theorem 5.12.

Proof of Theorem 5.12

By Lemmas 2.5 and A.1, employing the same arguments as in the proof of Theorem 1.1, we can deduce the conclusion of Theorem 5.12. The details are omitted here. □

Similarly to the proof of Theorem 5.12, in order to prove Theorem 5.13, we first establish the following auxiliary lemma.

Lemma A.2

Let 0 < α < n, n/(n − α) < s. Suppose that $K_{\alpha }\in \tilde H_{\alpha ,s}\cap S_{\alpha }$, and ω^{ns/(ns−n−sα)} ∈ A₁. Then for $b\in \text {BMO}(\mathbb {R}^{n})$ and any continuous $(H_{\omega }^{1}(\mathbb {R}^{n}),\infty )$-atom a related to a ball $B:=B(x_{0},r)\subset \mathbb {R}^{n}$, we have

$$\big\|(b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq C\|b\|_{BMO(\mathbb{R}^{n})},$$

where q = n/(n − α).

Proof

Using the same arguments as in Lemma A.1, we write

$$ \begin{array}{@{}rcl@{}} \big\|(b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})} &\leq &\Big({\int}_{2B}\big|(b(x)-b_{B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &&\quad+\Big({\int}_{(2B)^{c}}\big|(b(x)-b_{B})T_{\alpha} a(x)\big|^{q}\omega(x)^{q}dx\Big)^{1/q}\\ &=&:K_{1}+K_{2}. \end{array} $$

It is easy to know that $K_{1}\leq C\|b\|_{BMO(\mathbb {R}^{n})}.$ For K₂, by the cancellation condition of a and the Minkowski inequality,

$$ \begin{array}{@{}rcl@{}} K_{2} &\leq& C{\int}_{B}{\sum}_{j=1}^{\infty} j(2^{j} r)^{n-\alpha}\Big(\frac{1}{(2^{j}r)^{n}}{\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times (\omega^{sq/(s-q)}(2^{j} B))^{1/q-1/s}(2^{j}r)^{\alpha-n+n/s} \|b\|_{BMO(\mathbb{R}^{n})}\\ &&\leq C{\int}_{B}{\sum}_{j=1}^{\infty} j(2^{j} r)^{n-\alpha}\Big(\frac{1}{(2^{j}r)^{n}}{\int}_{2^{j+1}B\backslash2^{j}B} |K_{\alpha}(x-y)-K_{\alpha}(x)|^{s}dx\Big)^{1/s}|a(y)|dy\\ &&\qquad\times (\omega^{sq/(s-q)}(B))^{1/q-1/s}r^{\alpha-n+n/s} \|b\|_{BMO(\mathbb{R}^{n})}\\ &&\leq C\|b\|_{BMO(\mathbb{R}^{n})}{\int}_{B}|a(y)|dyr^{-n}\omega(B). \end{array} $$

This, together with the estimate of K₁, leads to

$$\big\| (b-b_{B})T_{\alpha} a\big\|_{L_{\omega^{q}}^{q}(\mathbb{R}^{n})}\leq C\| b\|_{\text{BMO}(\mathbb{R}^{n})}.$$

This completes the proof of Lemma A.2. □

We are now in a position to prove Theorem 5.13.

Proof of Theorem 5.13

By Lemmas 2.5 and A.2, employing the same arguments as in the proof of Theorem 1.1, we can deduce the conclusion of Theorem 5.13. The details are omitted here. □