**4. Weighted Lebesgue and** *BMO<sup>γ</sup>* **norm inequalities for** *S<sup>α</sup>* **and** *H<sup>α</sup>*

Before beginning our study of the generalized Calderón operator, we notice that *Sαf* can be identically infinite for some functions *<sup>f</sup>* belonging to *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ or *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* . For example, for *<sup>ω</sup>* � 1 and *<sup>α</sup>* <sup>&</sup>gt;0, if *f x*ð Þ¼ j j *<sup>x</sup>* �*<sup>α</sup> χBc* ð Þ 0,1 ð Þ *<sup>x</sup>* and *<sup>n</sup>=<sup>α</sup>* <sup>&</sup>lt;*p*, then *<sup>f</sup>* <sup>∈</sup>*Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ but *<sup>S</sup>α<sup>f</sup>* � <sup>∞</sup>. For the case *<sup>n</sup>=<sup>α</sup>* <sup>¼</sup> *<sup>p</sup>*, if *g x*ð Þ¼ j j *<sup>x</sup>* �*<sup>α</sup>* ð Þ log <sup>j</sup>*x*<sup>j</sup> �ð Þ <sup>1</sup>þ1*=<sup>p</sup> <sup>=</sup>*<sup>2</sup> *χBc* ð Þ 0,2 ð Þ *x* , then *<sup>g</sup>* <sup>∈</sup>*L<sup>p</sup> <sup>ω</sup>*�*<sup>p</sup>* ð Þ but *<sup>S</sup>α<sup>g</sup>* � <sup>∞</sup>. Also, if *h x*ð Þ¼ *<sup>χ</sup>Bc* ð Þ 0,1 ð Þ *<sup>x</sup>* , then *<sup>h</sup>*∈*BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* but *Sαh* � ∞ for all <sup>0</sup>≤*α*<sup>&</sup>lt; *<sup>n</sup>*. However, in Lemma 4.7 we will show that if *<sup>f</sup>* belongs to *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ∪*BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* and *<sup>S</sup>αf x*ð Þ is finite for some *<sup>x</sup>* ¼6 0, then *<sup>S</sup>α<sup>f</sup>* is finite on *<sup>n</sup>*nf g<sup>0</sup> . This also happens for the generalized Hilbert operator since the comparison (1).

Therefore, throughout the following sections we shall consider *S<sup>α</sup>* and *H<sup>α</sup>* defined on functions *<sup>f</sup>* belonging to *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ or *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that *Sαf* and *Hαf* are finite for some *x* 6¼ 0.

Also, note that *<sup>S</sup>α<sup>f</sup>* is finite on *<sup>n</sup>*nf g<sup>0</sup> for all compactly supported functions *<sup>f</sup>* <sup>∈</sup>*L*<sup>∞</sup> *<sup>ω</sup>*�<sup>1</sup> ð Þ, and the same holds for *<sup>H</sup>αf*. These functions belongs to *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ and those such that zero is not in their support belongs to *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* .

The operator *P* is naturally bounded from *BM*<sup>0</sup> into *L*<sup>∞</sup> and analogously, *Q* is naturally bounded from *BM*<sup>0</sup> into *BMO* (see Proposition 3.5 in [13]). So, immediately the Calderón operator is bounded from *BM*<sup>0</sup> into *BMO*. This natural boundedness is our motivation in order to consider the *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* spaces and obtain Theorems 1.5 and 1.7. Likewise, since *<sup>L</sup>*<sup>∞</sup> *<sup>ω</sup>*�<sup>1</sup> ð Þ<sup>⊂</sup> *BM*0ð Þ *<sup>ω</sup>* , we get Corollaries 4.1 and 4.2.

We now state the main results of this chapter.

Theorem 1.4 Suppose *α* >0, *n=α* ≤*p*< *n=*ð Þ *α* � 1 <sup>þ</sup>, *η* ¼ 1 þ 1*=n* þ 1*=p* � *α=n* and *<sup>δ</sup>* <sup>¼</sup> *<sup>α</sup>=<sup>n</sup>* � <sup>1</sup>*=p*. The operator *<sup>S</sup><sup>α</sup>* is bounded from *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ into *BMO<sup>δ</sup>* ð Þ *<sup>ω</sup>* and *<sup>ω</sup><sup>p</sup>*<sup>0</sup> ∈ *D*<sup>0</sup> if and only if *ω*∈*RH*<sup>0</sup> *p*<sup>0</sup> ð Þ ∩ *Dη*.

Theorem 1.5 Suppose 0 ≤*α*< 1, 0 ≤*γ* <1*=n* � *α=n*, *η* ¼ 1 þ 1*=n* � *α=n* � *γ* and *δ* ¼ *<sup>α</sup>=<sup>n</sup>* <sup>þ</sup> *<sup>γ</sup>*. The operator *<sup>S</sup><sup>α</sup>* is bounded from *BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* into *BMO<sup>δ</sup>* ð Þ *ω* and *ω*∈ *D*<sup>0</sup> if and only if *ω*∈ *Dη*.

**Corollary 4.1.** *Let <sup>η</sup>* <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>1</sup>*=n. Then S is bounded from L*<sup>∞</sup> *<sup>ω</sup>*�<sup>1</sup> ð Þ *into BMO*ð Þ *<sup>ω</sup> and ω*∈ *D*<sup>0</sup> *if and only if ω*∈ *Dη.*

Theorem 1.6 Suppose *α*>0, *n=α*≤ *p*<*n=*ð Þ *α* � 1 <sup>þ</sup>, *η* ¼ 1 þ 1*=n* þ 1*=p* � *α=n* and *<sup>δ</sup>* <sup>¼</sup> *<sup>α</sup>=<sup>n</sup>* � <sup>1</sup>*=p*. The operator *<sup>H</sup><sup>α</sup>* is bounded from *<sup>L</sup><sup>p</sup> <sup>ω</sup>*�*<sup>p</sup>* ð Þ into *BMO<sup>δ</sup>* ð Þ *ω* if and only if *ω*∈ H0ð Þ *α*, *p* ∩ *RH*<sup>0</sup> *p*<sup>0</sup> ð Þ ∩ *Dη*.

Theorem 1.7 Suppose 0 ≤*α*< 1, 0 ≤*γ* <1*=n* � *α=n*, *η* ¼ 1 þ 1*=n* � *α=n* � *γ* and *δ* ¼ *<sup>α</sup>=<sup>n</sup>* <sup>þ</sup> *<sup>γ</sup>*. The operator *<sup>H</sup><sup>α</sup>* is bounded from *BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* into *BMO<sup>δ</sup>* ð Þ *ω* if and only if *ω*∈ H0ð Þ *α* þ *nγ*, ∞ ∩ *Dη*.

**Corollary 4.2.** *Let <sup>η</sup>* <sup>¼</sup> <sup>1</sup> <sup>þ</sup> <sup>1</sup>*=n. Then H is bounded from L*<sup>∞</sup> *<sup>ω</sup>*�<sup>1</sup> ð Þ *into BMO*ð Þ *<sup>ω</sup> if and only if ω*∈ H0ð Þ 0, ∞ ∩ *Dη.*

*Remark* 4.3. It is classic the study of the boundedness of operators between *L*<sup>∞</sup> and *BMO* spaces. In [10], the results obtained in Corollaries 4.1 and 4.2 are originals, even in the unweighted case for *H*. The unweighted case for *S* is contained in Proposition 3.5 of [13].

*Remark* 4.4. The limit case *p* ¼ ∞ (*p*<sup>0</sup> ¼ 1) of Theorem 1.4 is contained in Theorem 1.5 with *γ* ¼ 0, since the hypotheses on the weights coincide. This also is true to Theorems 1.6 and 1.7.

Let *α*, *p* and *η* be as in Theorems 1.4 and 1.6. It is not difficult to show that if *ωp*<sup>0</sup> <sup>∈</sup> *<sup>A</sup>*1,0 then *<sup>ω</sup>*<sup>∈</sup> H0ð Þ *<sup>α</sup>*, *<sup>p</sup>* <sup>∩</sup> *RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ <sup>∩</sup> *<sup>D</sup>η*. Also, if *<sup>ω</sup>*ð Þ¼ *<sup>x</sup>* j j *<sup>x</sup> <sup>β</sup>* with *<sup>β</sup>* <sup>∈</sup>ð Þ 0, 1 <sup>þ</sup> *<sup>n</sup>=<sup>p</sup>* � *<sup>α</sup>* , then *<sup>ω</sup><sup>p</sup>*<sup>0</sup> ∉ *A*1,0 but *ω*∈ H0ð Þ *α*, *p* ∩ *RH*<sup>0</sup> *p*<sup>0</sup> ð Þ ∩ *Dη*. Furthermore, if *<sup>ω</sup>*ð Þ¼ *<sup>x</sup>* j j *<sup>x</sup> <sup>β</sup>* with *<sup>β</sup>* <sup>¼</sup> <sup>1</sup> <sup>þ</sup> *<sup>n</sup>=<sup>p</sup>* � *<sup>α</sup>*, then *<sup>ω</sup>*<sup>∈</sup> *RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ <sup>∩</sup> *<sup>D</sup><sup>η</sup>* but *<sup>ω</sup>* <sup>∉</sup> H0ð Þ *<sup>α</sup>*, *<sup>p</sup>* . Now, if in addition 0 <sup>&</sup>lt;*<sup>α</sup>* <sup>&</sup>lt;1 and *<sup>p</sup>*<sup>0</sup> <sup>&</sup>gt; *<sup>n</sup>=*ð Þ <sup>1</sup> � *<sup>α</sup>* , we have that if *<sup>ω</sup><sup>p</sup>*<sup>0</sup> ∈ *Ap*<sup>0</sup> <sup>þ</sup>1,0 then *ω*∈ H0ð Þ *α*, *p* ∩ *RH*<sup>0</sup> *p*<sup>0</sup> ð Þ ∩ *Dη*. In fact, the H0ð Þ *α*, *p* -condition is obtained directly from the *Ap*<sup>0</sup> <sup>þ</sup>1,0-condition, and by Hölder inequality we have that

$$\left(\frac{\alpha^{p'}(B(0,2(|\mathbf{x\_0}|+r)))}{|B(0,2(|\mathbf{x\_0}|+r))|}\right)^{1/p'} \le C \frac{|B(0,2(|\mathbf{x\_0}|+r))|}{\alpha^{-1}(B(\mathbf{x\_0},r))} \le C \left(\frac{|\mathbf{x\_0}|+r}{r}\right)^n \frac{\alpha(B(\mathbf{x\_0},r))}{|B(\mathbf{x\_0},r)|}$$

$$\le C \left(\frac{|\mathbf{x\_0}|+r}{r}\right)^{1-a+n/p} \frac{\alpha(B(\mathbf{x\_0},r))}{|B(\mathbf{x\_0},r)|}$$

for all balls *B x*ð Þ 0,*<sup>r</sup>* <sup>⊂</sup> *<sup>n</sup>*. Thus, the *RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ and *<sup>D</sup><sup>η</sup>* conditions follow from the last expression.

On the other hand, suppose that *α*, *γ* and *η* be as in Theorems 1.5 and 1.7. If *ω* ∈ *A*1,0 then *<sup>ω</sup>*<sup>∈</sup> H0ð Þ *<sup>α</sup>* <sup>þ</sup> *<sup>n</sup>γ*, <sup>∞</sup> <sup>∩</sup> *<sup>D</sup>η*. Also, if *<sup>ω</sup>*ð Þ¼ *<sup>x</sup>* j j *<sup>x</sup> <sup>β</sup>* with *<sup>β</sup>* <sup>∈</sup> ð Þ 0, 1 � *<sup>α</sup>* � *<sup>n</sup><sup>γ</sup>* , then *<sup>ω</sup>* <sup>∉</sup> *<sup>A</sup>*1,0 but *<sup>ω</sup>*<sup>∈</sup> H0ð Þ *<sup>α</sup>* <sup>þ</sup> *<sup>n</sup>γ*, <sup>∞</sup> <sup>∩</sup> *<sup>D</sup>η*. Finally, if *<sup>ω</sup>*ð Þ¼ *<sup>x</sup>* j j *<sup>x</sup> <sup>β</sup>* with *<sup>β</sup>* <sup>¼</sup> <sup>1</sup> � *<sup>α</sup>* � *<sup>n</sup>γ*, then *ω*∈ *D<sup>η</sup>* but *ω* ∉ H0ð Þ *α* þ *nγ*, ∞ .

We shall denote by *A x*ð Þ ,*r*, *R* with 0<*r*<*R* the annulus centered at *x* with radii *r* and *R*, and by *C* and *c* positive constants not necessarily the same at each occurrence.

Before proceeding to the proofs of the main theorems we give some previous lemmas.

Suppose that 1<*p* < ∞ and *ω* ∈*RH*<sup>0</sup> *p*<sup>0</sup> ð Þ, then it is easy to see that there exists *C* such that

$$\int\_{B} |f| \leq \mathcal{C} \frac{\alpha(B)}{|B|^{1}/p} \||f||\_{L^{p}}(\boldsymbol{\alpha}^{-p}) \tag{8}$$

for all *<sup>f</sup>* <sup>∈</sup>*L<sup>p</sup> <sup>ω</sup>*�*<sup>p</sup>* ð Þand for every ball *<sup>B</sup>*<sup>⊂</sup> *<sup>n</sup>* centered at the origin. **Lemma 4.6.** ð Þ*i Let* 0< *α*<*n and* 1< *p*< ∞*. If ω*∈ H0ð Þ *α*, *p then there exists C such that*

$$\int\_{B^\cdot} \frac{|f(\boldsymbol{y})|}{|\boldsymbol{y}|^{n-a+1}} d\boldsymbol{y} \le C \frac{o(B)}{|B|^{1+1/p-a/n+1/n}} \left\|\boldsymbol{f}\right\|\_{L^p(\boldsymbol{o}^{-p})}$$

for all *<sup>f</sup>* <sup>∈</sup>*L<sup>p</sup> <sup>ω</sup>*�*<sup>p</sup>* ð Þ and for every ball *<sup>B</sup>*<sup>⊂</sup> *<sup>n</sup>* centered at the origin.

ð Þ *ii Let* 0≤*α* <1*,* 0≤*γ* < 1*=n* � *α=n and η* ¼ 1 þ 1*=n* � *α=n* � *γ. If ω*∈ H0ð Þ *α* þ *nγ*, ∞ ∩ *D<sup>η</sup> then there exists C such that*

*A Brief Look at the Calderón and Hilbert Operators DOI: http://dx.doi.org/10.5772/intechopen.106027*

$$\int\_{B^c} \frac{|f(\boldsymbol{y})|}{|\boldsymbol{y}|^{n-a+1}} d\boldsymbol{y} \le C \frac{\alpha(B)}{|B|^{\eta}} \||\boldsymbol{f}||\_{BM\_0^{\boldsymbol{\eta}}(\boldsymbol{o})}.$$

for all *f* ∈*BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* and for every ball *<sup>B</sup>*<sup>⊂</sup> *<sup>n</sup>* centered at the origin.

**Proof:** The part ð Þ*i* is immediate from Hölder's inequality and Definition 3.11. For ð Þ *ii* , since the hypothesis on *ω* and (3.10), for *B* ¼ *B*ð Þ 0,*r* we have

$$\begin{split} \int\_{B^{r}} \frac{|f(\boldsymbol{y})|}{|\boldsymbol{y}|^{n-a+1}} d\boldsymbol{y} &\leq C \sum\_{k=0}^{\infty} \frac{1}{\left(2^{k}r\right)^{n-a+1}} \int\_{2^{k}r \leq |\boldsymbol{y}| < 2^{k+1}r} |f(\boldsymbol{y})| d\boldsymbol{y} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{r}(\boldsymbol{\alpha})} \sum\_{k=0}^{\infty} \frac{\alpha\left(B\left(0, 2^{k+1}r\right)\right)}{\left(2^{k}r\right)^{n-a+1-n\eta}} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{r}(\boldsymbol{\alpha})} \sum\_{k=0}^{\infty} \frac{\alpha\left(B\left(0, 2^{k+1}r\right)\right)B\left(0, 2^{k}r\right)}{\left(2^{k}r\right)^{n-(a+n\eta)+1}} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{r}(\boldsymbol{\alpha})} \frac{\alpha(B)}{|B|^{\eta}}. \end{split}$$

**Lemma 4.7.** ð Þ*<sup>i</sup> Let <sup>α</sup>* <sup>&</sup>gt;0*,* <sup>1</sup><*<sup>p</sup>* <sup>&</sup>lt; <sup>∞</sup> *and <sup>ω</sup>*∈*RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ*. If f* <sup>∈</sup> *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ *and there exists <sup>x</sup>* 6¼ <sup>0</sup> *such that Sαf x*ð Þ *is finite, then Sαf is finite on <sup>n</sup>*nf g<sup>0</sup> *and Sα<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*<sup>1</sup> *loc <sup>n</sup>* ð Þ*. The claim also holds for Hα.*

ð Þ *ii Let <sup>ω</sup>*<sup>∈</sup> *<sup>D</sup>η. If f* <sup>∈</sup>*BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω and there exists x* 6¼ 0 *such that Sαf x*ð Þ *is finite, then Sαf is finite on <sup>n</sup>*nf g<sup>0</sup> *and Sα<sup>f</sup>* <sup>∈</sup>*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ*. The claim also holds for Hα.*

**Proof:** Since (3.1) we will only consider the operator *Sα*. Suppose *f* is a nonnegative function in *L*<sup>1</sup> *loc <sup>n</sup>* ð Þ such that *<sup>S</sup>αf x*ð Þ<sup>0</sup> <sup>&</sup>lt; <sup>∞</sup> for some *<sup>x</sup>*<sup>0</sup> 6¼ 0. Then *<sup>Q</sup>αf x*ð Þ<sup>&</sup>lt; <sup>∞</sup> for ∣*x*∣≥∣*x*0∣, and if 0< ∣*x*∣<∣*x*0∣ then

$$Q\_{\mathfrak{q}}f(\mathbf{x}) \le \frac{1}{|\mathfrak{x}|^{n-\alpha}} \int\_{|\mathbf{x}| < |\mathfrak{y}| < |\mathfrak{x}\_0|} f(\mathbf{y}) d\mathbf{y} + Q\_{\mathfrak{q}} f(\mathbf{x}\_0) < \infty.$$

Furthermore, since

$$\int\_{B(0,r)} (Q\_q f(\mathbf{x}) - Q\_q f(\boldsymbol{\nu})) d\mathbf{x} \le \int\_{B(0,r)} f(\boldsymbol{\nu}) r^a d\boldsymbol{\nu} < \infty,$$

where <sup>∣</sup>*ν*<sup>∣</sup> <sup>¼</sup> *<sup>r</sup>*, then *<sup>Q</sup>α<sup>f</sup>* <sup>∈</sup>*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ.

If *α* >0 it is immediate that *Pαf* ∈*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ. Therefore, ð Þ*<sup>i</sup>* follows from (4.5). For ð Þ *ii* it remains to show that *<sup>P</sup>α<sup>f</sup>* <sup>∈</sup>*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ in the case *<sup>α</sup>* <sup>¼</sup> 0. Let *<sup>B</sup> <sup>j</sup>* <sup>¼</sup> *<sup>B</sup>* 0, 2�*<sup>j</sup> <sup>r</sup>* � �, *<sup>j</sup>* <sup>¼</sup> 0, 1, … , by (3.10) we have

$$\begin{split} \int\_{B\_{0}} \frac{1}{|\mathcal{X}|^{n}} \int\_{B(0, |\varkappa|)} f(\boldsymbol{y}) d\boldsymbol{y} d\boldsymbol{x} &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{\ell}(\boldsymbol{\alpha})} \int\_{B\_{0}} \frac{\alpha(B(0, |\boldsymbol{x}|))}{|\boldsymbol{x}|^{n - \eta \gamma}} d\boldsymbol{x} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{\ell}(\boldsymbol{\alpha})} \sum\_{j=0}^{\infty} \frac{r^{n\gamma - n}}{2^{j(\boldsymbol{w} \boldsymbol{y} - \boldsymbol{n}})} \int\_{B\_{j} \backslash B\_{j+1}} \alpha(B\_{j}) d\boldsymbol{x} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{\ell}(\boldsymbol{\alpha})} r^{n\gamma} \sum\_{j=0}^{\infty} \frac{\alpha(B\_{j} \backslash B\_{j+1})}{2^{j\eta \gamma}} \\ &\leq C \|\boldsymbol{f}\|\_{BM\_{0}^{\ell}(\boldsymbol{\alpha})} r^{n\gamma} \alpha(B\_{0}). \end{split}$$

**Proof of Theorem 1.4:** We begin showing the sufficient condition. Let *B* ¼ *B x*ð Þ 0,*r* . If *x*<sup>0</sup> ¼ 0, let *u* ¼ *re*1*=*2 and *v* ¼ 3*re*1*=*4, where *e*<sup>1</sup> ¼ ð Þ 1, … , 0 . If *x*<sup>0</sup> 6¼ 0, let *u* ¼ jð Þ *x*0jþ*r=*2 *x*0*=*∣*x*0∣ and *v* ¼ jð Þ *x*0jþ3*r=*4 *x*0*=*∣*x*0∣. Thus, we consider the following two regions

$$\begin{aligned} U &= B(u, r/8) \cap \{ u + h : \text{sign}(u\_i) = \text{sign}(h\_i) \; i = 1, \dots, n \}, \\ V &= B(v, r/4) \cap \{ v + h : \text{sign}(v\_i) = \text{sign}(h\_i) \; i = 1, \dots, n \}, \end{aligned} \tag{9}$$

where *u* ¼ ð Þ *u*1, … , *un* , *v* ¼ ð Þ *v*1, … , *vn* and *h* ¼ ð Þ *h*1, … , *hn* . In the case *ui* ¼ 0 for some *i*, we choose *hi* >0. Clearly, we have the estimates distð Þ¼ *U*,*V Cr*,

$$|U| = \frac{1}{2^n} |B(u, r/8)| = \text{C}|B| \quad \text{and} \quad |V| = \frac{1}{2^n} |B(v, r/4)| = \text{C}|B|.$$

Let *<sup>f</sup>* a nonnegative function in *<sup>L</sup><sup>p</sup> <sup>ω</sup>*�*<sup>p</sup>* ð Þ such that suppð Þ*<sup>f</sup>* <sup>⊂</sup>*B*ð Þ 0, <sup>j</sup>*x*0jþ*r=*<sup>2</sup> , where suppð Þ*f* is the closure of the set f g *x* : *f x*ð Þ 6¼ 0 . Then

$$\begin{split} \|\mathbb{S}\_{q}f\|\_{BMO^{\delta}(w)} &\geq \frac{C}{\alpha(B)|B|} \int\_{B} \int\_{B} |\mathbb{S}\_{q}f(\boldsymbol{x}) - \mathbb{S}\_{q}f(\boldsymbol{z})| d\boldsymbol{z} d\boldsymbol{x} \\ &\geq \frac{C}{\alpha(B)|B|} \int\_{U} \int\_{V} \left| \left(\frac{1}{|\boldsymbol{x}|^{n-a}} - \frac{1}{|\boldsymbol{z}|^{n-a}}\right) \int\_{B(0, |\boldsymbol{x}\_{0}| + r/2)} f(\boldsymbol{y}) d\boldsymbol{y}| d\boldsymbol{z} d\boldsymbol{x} .\end{split}$$

Note that, for *x*∈ *U* and *z*∈*V* we have <sup>1</sup> j j *<sup>x</sup> <sup>n</sup>*�*<sup>α</sup>* � <sup>1</sup> j j *<sup>z</sup> <sup>n</sup>*�*<sup>α</sup>* <sup>≥</sup>*<sup>C</sup> <sup>r</sup>* ð Þ <sup>j</sup>*x*0jþ*<sup>r</sup> <sup>n</sup>*�*α*þ1. Then

$$\|\mathbb{S}\_{\mathfrak{q}}f\|\_{BMO^{\delta}(\alpha)} \geq \frac{Cr^{n+1}}{o(B)|B|^{\delta}(|\mathbb{x}\_{0}|+r)^{n-\alpha+1}} \int\_{B(0,|\mathbb{x}\_{0}|+r/2)} f(y) dy. \tag{10}$$

Thus, taking *f y*ð Þ¼ *<sup>ω</sup><sup>p</sup>*<sup>0</sup> ð Þ*y χ<sup>B</sup>*ð Þ 0,j*x*0jþ*r=*<sup>2</sup> ð Þ*y* in (10) and since the boundedness of *S<sup>α</sup>* and *ω<sup>p</sup>*<sup>0</sup> ∈ *D*0, we have

$$\left(\frac{a^{p'}(B(\mathbf{0},|\boldsymbol{\varkappa\_0}|+r))}{|B(\mathbf{0},|\boldsymbol{\varkappa\_0}|+r)|}\right)^{1/p'} \le C\left(\frac{|\boldsymbol{\varkappa\_0}|+r}{r}\right)^{1-a+n/p}\frac{o(B)}{|B|}\dots$$

Taking *x*<sup>0</sup> ¼ 0 in the last expression, we have that *ω*∈*RH*<sup>0</sup> *p*<sup>0</sup> ð Þ. Then, applying the Hölder's inequality, we obtain that *ω* satisfies the desired condition *Dη*.

Now, let us show the necessary condition. Let *<sup>f</sup>* <sup>∈</sup>*Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ such that *<sup>S</sup>αf x*ð Þ is finite for some *<sup>x</sup>* 6¼ 0 and let *<sup>ω</sup>*<sup>∈</sup> *RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ <sup>∩</sup> *<sup>D</sup>η*. It is immediate that *<sup>ω</sup><sup>p</sup>*<sup>0</sup> ∈ *D*0. Thus *Sαf* ∈*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ by ð Þ*<sup>i</sup>* of Lemma 4.7. First, we consider *<sup>B</sup>* <sup>¼</sup> *<sup>B</sup>*ð Þ 0,*<sup>r</sup>* , *<sup>x</sup>*<sup>∈</sup> *<sup>B</sup>* and *<sup>x</sup>* 6¼ 0. Let *ν* be such that ∣*ν*∣ ¼ *r*, and let

$$K\_{\nu}(x,y) = \min\left\{ \mathbf{1}, \frac{|y|^{n-a}}{|x|^{n-a}} \right\} - \min\left\{ \mathbf{1}, \frac{|y|^{n-a}}{|\nu|^{n-a}} \right\}.$$

Then, since *Kν*ð Þ¼ *x*, *y* 0 for ∣*y*∣> ∣*ν*∣, we have

$$S\_q f(\varkappa) - S\_q f(\nu) = \int\_{|\boldsymbol{\nu}| \le |\boldsymbol{\nu}|} K\_{\boldsymbol{\nu}}(\boldsymbol{\varkappa}, \boldsymbol{\jmath}) \frac{f(\boldsymbol{\jmath})}{|\boldsymbol{\jmath}|^{n-a}} d\boldsymbol{\jmath}.\tag{11}$$

*A Brief Look at the Calderón and Hilbert Operators DOI: http://dx.doi.org/10.5772/intechopen.106027*

If ∣*y*∣≤∣*ν*∣ then *Kν*ð Þ *x*, *y* ≥0, so

$$\begin{split} \frac{1}{\alpha(B)} \int\_{B} |S\_{\boldsymbol{\theta}}f(\mathbf{x}) - S\_{\boldsymbol{\theta}}f(\boldsymbol{\nu})| d\mathbf{x} &\leq \frac{1}{\alpha(B)} \int\_{B} \int\_{B} K\_{\boldsymbol{\nu}}(\mathbf{x}, \mathbf{y}) \frac{|f(\boldsymbol{\nu})|}{|\boldsymbol{y}|^{n-a}} d\mathbf{y} d\mathbf{x} \\ &= \frac{1}{\alpha(B)} \int\_{B} \int\_{|\boldsymbol{\nu}| \leq |\boldsymbol{\nu}|} K\_{\boldsymbol{\nu}}(\mathbf{x}, \mathbf{y}) \frac{|f'(\boldsymbol{\nu})|}{|\boldsymbol{y}|^{n-a}} d\mathbf{y} d\mathbf{x} + \frac{1}{\alpha(B)} \int\_{B} \int\_{|\boldsymbol{\nu}| < |\boldsymbol{\nu}| \leq \boldsymbol{\nu}} K\_{\boldsymbol{\nu}}(\mathbf{x}, \mathbf{y}) \frac{|f'(\boldsymbol{\nu})|}{|\boldsymbol{y}|^{n-a}} d\mathbf{y} d\mathbf{x}. \end{split} \tag{12}$$

Now we estimate each term in (12). If <sup>∣</sup>*y*∣≤∣*x*<sup>∣</sup> then *<sup>K</sup>ν*ð Þ *<sup>x</sup>*, *<sup>y</sup>* <sup>≤</sup>j j *<sup>y</sup> <sup>n</sup>*�*<sup>α</sup>* j j *<sup>x</sup>* �ð Þ *<sup>n</sup>*�*<sup>α</sup>* . So, by (8) we have

$$\begin{aligned} \frac{1}{\alpha(B)} \int\_{B} \int\_{|y| \le |x|} K\_{\nu}(x, y) \frac{|f(y)|}{|y|^{n-a}} dy dx & \le \frac{1}{\alpha(B)} \int\_{B} \frac{1}{|x|^{n-a}} \int\_{B} |f(y)| dy dx \\ & \le C \|f\|\_{L^{r}(\alpha^{-p})} |B|^{\delta} .\end{aligned}$$

For the second term, since 0≤ *Kν*ð Þ *x*, *y* ≤1 and (8), we have

$$\begin{split} \frac{1}{\omega(B)} \int\_{B} \int\_{|x| < |y| \le r} K\_{\nu}(x, y) \frac{|f(y)|}{|y|^{n-a}} dy dx &\leq \frac{1}{\omega(B)} \int\_{B} \frac{1}{|x|^{n-a}} \int\_{|x| < |y| \le r} |f(y)| dy dx \\ &\leq \frac{C}{\omega(B)} \int\_{B} |f(y)| |y|^{a} dy \\ &\leq C \|f\|\_{L^{p}(\alpha r)} |B|^{\delta} .\end{split} \tag{13}$$

Then, by (12) and (13), we have proved

$$\frac{1}{\alpha(B)|B|^{\delta}} \int\_{B} |\mathcal{S}\_{q}f(\boldsymbol{x}) - \mathcal{S}\_{q}f(\boldsymbol{\nu})| d\boldsymbol{x} \leq C \|f\|\_{L^{r}(\alpha^{-p})},\tag{14}$$

for every ball *B* centered at the origin.

We now consider *B* ¼ *B x*ð Þ 0,*r* with *r*<∣*x*0∣*=*8. By (14) it is enough to consider only these balls *B*. Let *x*∈ *B* and *ν* ¼ jð Þ *x*0jþ*r x*0*=*∣*x*0∣. In the same way as (11), we have

$$\mathcal{S}\_q f(\varkappa) - \mathcal{S}\_q f(\nu) = \int\_{|\boldsymbol{\nu}| \le |\boldsymbol{\nu}|} K\_{\boldsymbol{\nu}}(\boldsymbol{\varkappa}, \boldsymbol{\jmath}) \frac{f(\boldsymbol{\jmath})}{|\boldsymbol{\jmath}|^{\boldsymbol{n}-\boldsymbol{\alpha}}} d\boldsymbol{\jmath}.$$

Now, we note that if ∣*y*∣≤∣*ν*∣ then *Kν*ð Þ *x*, *y* ≥0. Applying the mean value theorem and using ∣*ν*∣ � ∣*x*∣, then

$$K\_{\nu}(\varkappa, \boldsymbol{y}) \le \frac{|\boldsymbol{y}|^{n-a}}{|\boldsymbol{\varkappa}|^{n-a}} - \frac{|\boldsymbol{y}|^{n-a}}{|\boldsymbol{\nu}|^{n-a}} \le C \frac{r|\boldsymbol{y}|^{n-a}}{|\boldsymbol{\nu}|^{n-a+1}}.\tag{15}$$

Thus, by (8) and *ω* ∈ *Dη*, we have

$$\begin{split} \frac{1}{\alpha(B)} \int\_{B} |S\_{\mathfrak{q}}f(\boldsymbol{x}) - S\_{\mathfrak{q}}f(\boldsymbol{\nu})| d\boldsymbol{x} &\leq C \frac{r}{\alpha(B)|\boldsymbol{\nu}|^{n-\alpha+1}} \int\_{B} \int\_{|\boldsymbol{\nu}| \leq |\boldsymbol{\nu}|} |f(\boldsymbol{\nu})| d\boldsymbol{y} d\boldsymbol{x} \\ &\leq C \|f\|\_{L^{r}(\boldsymbol{\alpha}^{-p})} \frac{r^{n+1}}{|\boldsymbol{\nu}|^{n-\alpha+1+n/p}} \frac{\alpha(B(\mathsf{O}, |\boldsymbol{\nu}|))}{\alpha(B)} \\ &\leq C \|f\|\_{L^{r}(\boldsymbol{\alpha}^{-p})} |B|^{\delta} .\end{split} \tag{16}$$

**45**

Therefore, (14) and (16) complete the proof of the theorem.

**Proof of Theorem 1.5:** We begin showing the sufficient condition. Let *B* ¼ *B x*ð Þ 0,*r* and let *u*, *v*, *U* and *V* as in (9) of the proof of Theorem 1.4. Then, we again have

$$\|\mathbb{S}\_{q}f\|\_{BMO^{\delta}(a)} \geq \frac{Cr^{n+1}}{o(B)|B|^{\delta}(|x\_{0}|+r)^{n-a+1}}\int\_{B(0,|x\_{0}|+r/2)} f(y)dy,\tag{17}$$

for every nonnegative function *f* in *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that suppð Þ*f* ⊂ *B*ð Þ 0, j*x*0jþ*r=*2 . Now, if *γ* ¼ 0 we take *f y*ð Þ¼ *ω*ð Þ*y χB*ð Þ 0,j*x*0jþ*r=*<sup>2</sup> ð Þ*y* in (17) and since ∥*f* ∥*BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* ≤1, the boundedness of *S<sup>α</sup>* and *ω*∈ *D*0, we have *ω* ∈ *Dη*.

$$\text{If } \gamma > 0 \text{, let} \\ f(\boldsymbol{y}) = P\_{\boldsymbol{n}\boldsymbol{\gamma}} \left( a \chi\_{B(0, |\boldsymbol{x}\_0| + r/2)} \right) (\boldsymbol{y}) \text{, then } \|\boldsymbol{f}\|\_{\operatorname{BM}\_0^r(o)} \le \mathcal{C} \text{ and}$$

$$\begin{split} \int\_{B(0,|\mathbf{x}\_0|+r/2)} f(\mathbf{y}) d\mathbf{y} &= \mathbf{C} \int\_{B(0,|\mathbf{x}\_0|+r/2)} a(\mathbf{t}) ((|\mathbf{x}\_0|+r/2)^{\eta \gamma} - |\mathbf{t}|^{\eta \gamma}) d\mathbf{t} \\ &\geq \mathbf{C} (|\mathbf{x}\_0|+r)^{\eta \gamma} a(\mathcal{B}(\mathbf{0}, (|\mathbf{x}\_0|+r/2)/2)) .\end{split} \tag{18}$$

Therefore, using this function *f* in (17), the boundedness of *Sα*, (18) and *ω*∈ *D*0, we have *ω*∈ *Dη*.

Now, let us show the necessary condition. Let *f* ∈ *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that *Sαf x*ð Þ is finite for some *<sup>x</sup>* 6¼ 0 and let *<sup>ω</sup>*<sup>∈</sup> *<sup>D</sup>η*. Thus *<sup>S</sup>α<sup>f</sup>* <sup>∈</sup>*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ by ð Þ *ii* of Lemma 4.7. We begin considering *B* ¼ *B*ð Þ 0,*r* , *x*∈ *B* and *x* 6¼ 0. Let *ν* be such that ∣*ν*∣ ¼ *r*. In the same way as we did in (12), we have

$$\begin{split} \frac{1}{\alpha(B)} \int\_{B} |\mathcal{S}\_{\eta}f(\boldsymbol{x}) - \mathcal{S}\_{\boldsymbol{a}}f(\boldsymbol{\nu})| d\boldsymbol{x} &\leq \frac{1}{\alpha(B)} \int\_{B} \int\_{|\boldsymbol{y}| \leq |\boldsymbol{x}|} K\_{\boldsymbol{\nu}}(\boldsymbol{x}, \boldsymbol{y}) \frac{|f(\boldsymbol{y})|}{|\boldsymbol{y}|^{\boldsymbol{n}-\alpha}} d\boldsymbol{y} d\boldsymbol{x} \\ &+ \frac{1}{\alpha(B)} \int\_{B} \int\_{|\boldsymbol{x}| < |\boldsymbol{y}| \leq r} K\_{\boldsymbol{\nu}}(\boldsymbol{x}, \boldsymbol{y}) \frac{|f(\boldsymbol{y})|}{|\boldsymbol{y}|^{\boldsymbol{n}-\alpha}} d\boldsymbol{y} d\boldsymbol{x}, \end{split} \tag{19}$$

where *<sup>K</sup>ν*ð Þ¼ *<sup>x</sup>*, *<sup>y</sup>* min 1, j j *<sup>y</sup> <sup>n</sup>*�*<sup>α</sup>* j j *<sup>x</sup> <sup>n</sup>*�*<sup>α</sup>* n o � min 1, j j *<sup>y</sup> <sup>n</sup>*�*<sup>α</sup>* j j *<sup>ν</sup> <sup>n</sup>*�*<sup>α</sup>* n o*:* We estimate the first term of (19). Let *<sup>B</sup> <sup>j</sup>* <sup>¼</sup> *<sup>B</sup>* 0, 2�*<sup>j</sup> <sup>r</sup>* � �, *<sup>j</sup>* <sup>¼</sup> 0, 1, … . Thus, since

*<sup>K</sup>ν*ð Þ *<sup>x</sup>*, *<sup>y</sup>* <sup>≤</sup>j j *<sup>y</sup> <sup>n</sup>*�*<sup>α</sup>* j j *<sup>x</sup>* �ð Þ *<sup>n</sup>*�*<sup>α</sup>* for <sup>∣</sup>*y*∣≤∣*x*<sup>∣</sup> and (5), we have

$$\frac{1}{\alpha(B)} \int\_{B} \int\_{|y| \le |x|} K\_{\nu}(x, y) \frac{|f(y)|}{|y|^{n - \alpha}} dy dx \le \frac{1}{\alpha(B)} \int\_{B} \frac{1}{|x|^{n - \alpha}} \int\_{|y| \le |x|} |f(y)| dy dx$$

$$\le C \|f\|\_{BM\_0^\circ(\alpha)} \frac{1}{\alpha(B)} \int\_{B} \frac{\alpha(B(0, |x|))}{|x|^{n - \alpha - \eta j}} dx$$

$$\le C \|f\|\_{BM\_0^\circ(\alpha)} \frac{r^{\eta \gamma + \alpha}}{\alpha(B)} \sum\_{j=0}^\infty \frac{\alpha(B\_j/B\_{j+1})}{2^{j(\eta \gamma + \alpha)}} \tag{20}$$

$$\le C \|f\|\_{BM\_0^\circ(\alpha)} \frac{|B|^\delta}{\alpha(B)} \sum\_{j=0}^\infty \alpha(B\_j/B\_{j+1})$$

$$= C \|f\|\_{BM\_0^\circ(\alpha)} |B|^\delta.$$

For the second term of (19), since 0 ≤*Kν*ð Þ *x*, *y* ≤ 1, we have

$$\begin{split} \frac{1}{\alpha(B)} \int\_{B} \int\_{|x| \le |y| \le r} K\_{\nu}(x, y) \frac{|f(y)|}{|y|^{n-\alpha}} dy dx &\le \frac{1}{\alpha(B)} \int\_{B} \frac{|f(y)|}{|y|^{n-\alpha}} \int\_{|x| \le |y|} \mathbf{1} dx dy\\ &\le C \|f\|\_{BM\_0^r(\alpha)} |B|^\delta. \end{split} \tag{21}$$

Therefore, by (19)-(21) we have proved

$$\frac{1}{\log(B)|B|} \int\_{B} |S\_{q}f(\boldsymbol{x}) - S\_{q}f(\boldsymbol{\nu})| d\boldsymbol{x} \leq C \|f\|\_{BM\_{0}^{r}(o)},\tag{22}$$

for every ball *B* centered at the origin.

We now consider *B* ¼ *B x*ð Þ 0,*r* with *r*< ∣*x*0∣*=*8. By (22) it is enough to consider only these balls *B*. Let *x*∈ *B* and *ν* ¼ jð Þ *x*0jþ*r x*0*=*∣*x*0∣. In the same way as we obtained (11) and (15) in the previous proof, we have

$$\mathcal{S}\_q f(\varkappa) - \mathcal{S}\_q f(\nu) = \int\_{|\nu| \le |\nu|} K\_\nu(\varkappa, \jmath) \frac{f(\jmath)}{|\jmath|^{n-a}} d\jmath$$

and *<sup>K</sup>ν*ð Þ *<sup>x</sup>*, *<sup>y</sup>* <sup>≤</sup>*Cr y*j j*<sup>n</sup>*�*<sup>α</sup>* j j *<sup>ν</sup>* �ð Þ *<sup>n</sup>*�*α*þ<sup>1</sup> . By *<sup>ω</sup>*<sup>∈</sup> *<sup>D</sup>η*, we have

$$\begin{split} \frac{1}{\alpha(B)} \int\_{B} |S\_{\mathfrak{g}}f(\boldsymbol{x}) - S\_{\mathfrak{g}}f(\boldsymbol{\nu})| d\boldsymbol{x} &\leq C \frac{r^{n+1}}{\alpha(B) |\boldsymbol{\nu}|^{n-\alpha+1}} \int\_{|\boldsymbol{\nu}| \leq |\boldsymbol{\nu}|} |f(\boldsymbol{\nu})| d\boldsymbol{y} \\ &\leq C \|f\|\_{BM\_{0}^{r}(\boldsymbol{\nu})} \frac{r^{n+1}}{|\boldsymbol{\nu}|^{n-\alpha+1-\eta\gamma}} \frac{\alpha(B(\mathsf{O}, |\boldsymbol{\nu}|))}{\alpha(B)} \\ &\leq C \|f\|\_{BM\_{0}^{r}(\boldsymbol{\nu})} |B|^{\delta}. \end{split} \tag{23}$$

Therefore, (22) and (23), complete the proof of the theorem. Let *<sup>x</sup>*, *<sup>ν</sup>*<sup>∈</sup> *<sup>n</sup>*, *<sup>ν</sup>* 6¼ 0, then

$$\begin{split} |H\_{\mathfrak{q}}f(\mathbf{x}) - H\_{\mathfrak{q}}f(\boldsymbol{\nu})| &\leq \int\_{|\boldsymbol{y}| \leq |\boldsymbol{\nu}|} |f(\boldsymbol{y})| \left| \frac{\mathbf{1}}{(|\boldsymbol{x}| + |\boldsymbol{y}|)^{n-a}} - \frac{\mathbf{1}}{(|\boldsymbol{\nu}| + |\boldsymbol{y}|)^{n-a}} \right| d\boldsymbol{y} \\ &+ \int\_{|\boldsymbol{y}| > |\boldsymbol{\nu}|} |f(\boldsymbol{y})| \left| \frac{\mathbf{1}}{(|\boldsymbol{x}| + |\boldsymbol{y}|)^{n-a}} - \frac{\mathbf{1}}{(|\boldsymbol{\nu}| + |\boldsymbol{y}|)^{n-a}} \right| d\boldsymbol{y}. \end{split} \tag{24}$$

**Proof of Theorem 1.6:** We begin showing the sufficient condition. Let *B* ¼ *B x*ð Þ 0,*r* and let *u*, *v*, *U* and *V* as in (9) of the proof of Theorem 1.4. Note that if *x*∈ *U*, *z*∈*V* then for all *y*∈ *<sup>n</sup>*,

$$\frac{1}{(|\boldsymbol{x}|+|\boldsymbol{y}|)^{n-a}} - \frac{1}{(|\boldsymbol{z}|+|\boldsymbol{y}|)^{n-a}} \ge \mathcal{C} \frac{r}{(|\boldsymbol{x}\_0|+r+|\boldsymbol{y}|)^{n-a+1}}.\tag{25}$$

Hence, if *<sup>f</sup>* is a nonnegative function in *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ such that suppð Þ*<sup>f</sup>* <sup>⊂</sup> *<sup>A</sup>*ð Þ 0,*r*, *<sup>m</sup>* and taking *x*<sup>0</sup> ¼ 0 in (25), we have

$$\|\|H\_q f\|\|\_{BMO^\delta(w)} \ge \frac{Cr^{n+1}}{o o(B) |B|} \int\_{A(0, r, m)} \frac{f(y)}{|y|^{n-a+1}} dy,\tag{26}$$

for every ball *B* centered at the origin.

Thus, taking *f <sup>m</sup>*,*<sup>j</sup>* ð Þ¼ *<sup>y</sup>* j j *<sup>y</sup>* �ð Þ *<sup>n</sup>*�*α*þ<sup>1</sup> *<sup>=</sup>*ð Þ *<sup>p</sup>*�<sup>1</sup> *<sup>ω</sup>p*<sup>0</sup> ð Þ*y χAm*,*<sup>j</sup>* ð Þ*y* in (26) where *Am*,*<sup>j</sup>* ¼ *A*ð Þ 0,*r*, *m* ∩ f g *y* : 1*=j*≤*ω*ð Þ*y* <*j* , *m*, *j* ¼ 1, 2, … , using the boundedness of *H<sup>α</sup>* and letting *m* ! ∞, *j* ! ∞ we obtain that *ω*∈ H0ð Þ *α*, *p* .

On the other hand, if *<sup>f</sup>* is a nonnegative function in *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ such that suppð Þ*f* ⊂*B*ð Þ 0, 2ð Þ j*x*0jþ*r* , then by (25)

$$\|H\_q f\|\_{BMO^\delta(w)} \ge \frac{Gr^{n+1}}{o(B)|B|^\delta(|x\_0|+r)^{n-a+1}} \int\_{B(0,2(|x\_0|+r))} f(y) dy. \tag{27}$$

Thus, taking *f <sup>j</sup>* ð Þ¼ *<sup>y</sup> <sup>ω</sup>p*<sup>0</sup> ð Þ*y χ<sup>A</sup> <sup>j</sup>* ð Þ*y* in (27) where *A <sup>j</sup>* ¼

*B*ð Þ 0, 2ð Þ j*x*0jþ*r* ∩ f g *y* : 1*=j*≤*ω*ð Þ*y* < *j* , *j* ¼ 1, 2, … , and using the boundedness of *Hα*, we have

$$\left(\int\_{A\_j} \alpha^{p'}(\mathbf{y}) d\mathbf{y}\right)^{1/p'} \le C \left(\frac{|\mathbf{x}\_0| + r}{r}\right)^{n-a+1} \frac{\alpha(B)}{|B|^{1/p}}.$$

Letting *j* ! ∞ and taking *x*<sup>0</sup> ¼ 0 in the last expression, we can obtain that *ω*∈ *RH*<sup>0</sup> *p*<sup>0</sup> ð Þ. Then, applying Hölder's inequality, we obtain *ω*∈ *Dη*.

Now, let us show the necessary condition. Let *<sup>f</sup>* <sup>∈</sup> *Lp <sup>ω</sup>*�*<sup>p</sup>* ð Þ such that *<sup>H</sup>αf x*ð Þ is finite for some *<sup>x</sup>* 6¼ 0 and let *<sup>ω</sup>* such that *<sup>ω</sup>*<sup>∈</sup> H0ð Þ *<sup>α</sup>*, *<sup>p</sup>* <sup>∩</sup> *RH*<sup>0</sup> *<sup>p</sup>*<sup>0</sup> ð Þ <sup>∩</sup> *<sup>D</sup>η*. Hence *<sup>H</sup>α<sup>f</sup>* <sup>∈</sup>*L*<sup>1</sup> *loc <sup>n</sup>* ð Þ since ð Þ*i* of Lemma 4.7. We begin considering *B* ¼ *B*ð Þ 0,*r* , *x*∈*B* and *x* 6¼ 0. Let *ν* be such that ∣*ν*∣ ¼ *r*. We estimate the two terms of (24). By (8), we have

$$\begin{split} \frac{1}{\nu(B)} \int\_{B} \int\_{|\boldsymbol{y}| \leq |\boldsymbol{\nu}|} \left| \frac{f(\boldsymbol{y})}{(|\boldsymbol{x}| + |\boldsymbol{y}|)^{n-a}} - \frac{f(\boldsymbol{y})}{(|\boldsymbol{\nu}| + |\boldsymbol{y}|)^{n-a}} \right| d\boldsymbol{y} d\boldsymbol{x} \\ &\leq \frac{C}{\nu(B)} \int\_{B} \int\_{B} \frac{|f(\boldsymbol{y})|}{|\boldsymbol{x}|^{n-a}} d\boldsymbol{y} d\boldsymbol{x} \\ &\leq C \|f\|\_{L^{r}(\boldsymbol{\alpha}^{-r})} \int\_{B} \frac{1}{|\boldsymbol{x}|^{n-a+n/p}} d\boldsymbol{x} = C \|f\|\_{L^{r}(\boldsymbol{\alpha}^{-r})} |B|^{\delta}. \end{split} \tag{28}$$

To analyze the second term of (24), we use the mean value theorem, then

$$\left| \frac{1}{(|\varkappa| + |\nu|)^{n-a}} - \frac{1}{(|\nu| + |\nu|)^{n-a}} \right| \le C \frac{r}{|\nu|^{n-a+1}}.$$

Thus, by ð Þ*i* of Lemma 4.6

$$\begin{split} \frac{1}{\alpha o(B)} \int\_{B} \int\_{|y| > |\nu|} \left| \frac{f(y)}{(|x| + |y|)^{n-a}} - \frac{f(y)}{(|\nu| + |y|)^{n-a}} \right| dy d\mathbf{x} &\leq \frac{Cr}{o(B)} \int\_{B} \int\_{B'} \frac{|f(y)|}{|y|^{n-a+1}} dy d\mathbf{x} \\ &\leq C \| f \|\_{L^{p}(a^{p})} |B|^{\delta} .\end{split} \tag{29}$$

Therefore, by (24)–(29), we have proved

$$\frac{1}{\alpha(B)|B|} \int\_{B} |H\_{q}f(\varkappa) - H\_{q}f(\nu)| d\varkappa \le C \|f\|\_{L^{p}(\alpha^{-p})},\tag{30}$$

*A Brief Look at the Calderón and Hilbert Operators DOI: http://dx.doi.org/10.5772/intechopen.106027*

for every ball *B* centered at the origin.

We now consider *B* ¼ *B x*ð Þ 0,*r* with *r*< ∣*x*0∣*=*8. By (28) it is enough to consider only these balls *B*. Let *x*∈ *B* and *ν* ¼ jð Þ *x*0jþ*r x*0*=*∣*x*0∣, then ∣*ν*∣ � ∣*x*∣ and ∣*x*∣ � ∣*x*0∣. Using ∣*y*∣≤∣*ν*∣ and the mean value theorem

$$\left|\frac{1}{(|\varkappa|+|\nu|)^{n-a}} - \frac{1}{(|\nu|+|\nu|)^{n-a}}\right| \le C \frac{r}{|\varkappa\_0|^{n-a+1}}.$$

Then, by (8) and *ω*∈ *D<sup>η</sup>*

$$\begin{split} &\frac{1}{\alpha(B)} \int\_{B} \int\_{|y| \le |\nu|} |f(y)| \left| \frac{1}{(|\varkappa| + |y|)^{n-a}} - \frac{1}{(|\nu| + |y|)^{n-a}} \right| d\jmath dx \\ &\le \frac{Cr^{n+1}}{\alpha(B) |\varkappa\_0|^{n-a+1}} \int\_{|y| \le |\nu|} |f(y)| d\jmath \\ &\le C \|f\|\_{L^p(\alpha - \mathbb{P})} \frac{r^{n+1}}{|\varkappa\_0|^{n-a+1+n/p}} o(B(\mathbf{0}, |\nu|)) \\ &= C \|f\|\_{L^p(\alpha - \mathbb{P})} |\mathcal{B}|^{\delta} .\end{split} \tag{31}$$

Now, using the mean value theorem

$$\left| \frac{1}{(|\varkappa| + |\nu|)^{n-a}} - \frac{1}{(|\nu| + |\nu|)^{n-a}} \right| \le C \frac{r}{|\nu|^{n-a+1}}.$$

Then, by ð Þ*i* of Lemma 4.6

$$\begin{split} \frac{1}{\alpha o(B)} \int\_{B} \int\_{|y|>|\nu|} \left| \frac{f(y)}{(|x|+|y|)^{n-a}} - \frac{f(y)}{(|\nu|+|y|)^{n-a}} \right| dy d\mathbf{x} &\leq \frac{Cr}{o(B)} \int\_{B} \int\_{B^{\prime}} \frac{|f(y)|}{|y|^{n-a+1}} dy d\mathbf{x} \\ &= C \|f\|\_{L^{p}(a^{-p})} |B|^{\delta} .\end{split} \tag{32}$$

Therefore, by (24) with *ν* ¼ jð Þ *x*0jþ*r x*0*=*∣*x*0∣, (31) and (32), we have

$$\frac{1}{\alpha(B)|B|^{\delta}} \int\_{B} |H\_{q}f(\boldsymbol{\omega}) - H\_{q}f(\boldsymbol{\nu})| d\boldsymbol{\omega} \leq C \|f\|\_{L^{r}(\boldsymbol{\alpha}^{-p})},$$

for every ball *B* ¼ *B x*ð Þ 0,*r* considered. This completes the proof of the theorem.

**Proof of Theorem 1.7:** We begin showing the sufficient condition. Let *B* ¼ *B x*ð Þ 0,*r* and let *u*, *v*, *U* and *V* as in (9) of the proof of Theorem 1.4. Then, as in (26) of the proof of Theorem 4 (with *x*<sup>0</sup> ¼ 0), we again have

$$\|H\_q f\|\_{BMO^\delta(\alpha)} \ge \frac{Cr^{n+1}}{o(B)|B|} \int\_{A(0,r,m)} \frac{f(y)}{|y|^{n-a+1}} dy \tag{33}$$

for every nonnegative function *f* in *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that suppð Þ*f* ⊂ *A*ð Þ 0,*r*, *m* and for every ball *B* centered at the origin.

Thus, taking *f y*ð Þ¼ j j *<sup>y</sup> <sup>n</sup><sup>γ</sup> ω*ð Þ*y χ<sup>A</sup>*ð Þ 0,*r*,*<sup>m</sup>* ð Þ*y* in (33), using that ∥*f* ∥*BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* ≤ 1, the boundedness of *H<sup>α</sup>* and letting *m* ! ∞, we have that *ω* ∈ H0ð Þ *α* þ *nγ*, ∞ .

On the other hand, as in (27) of the proof of Theorem 1.6 we again have

$$\|\|H\_qf\|\|\_{BMO^\delta(w)} \ge \frac{Cr^{n+1}}{o(B)|B|^\delta(|\mathbf{x}\_0|+r)^{n-a+1}} \int\_{B(0,2(|\mathbf{x}\_0|+r))} f(y) dy,\tag{34}$$

for every nonnegative function *f* in *BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that suppð Þ*f* ⊂ *B*ð Þ 0, 2ð Þ j*x*0jþ*r* and for every ball *B* ¼ *B x*ð Þ 0,*r* .

If *γ* ¼ 0, we take *f y*ð Þ¼ *ω*ð Þ*y χB*ð Þ 0,2ð Þ <sup>j</sup>*x*0jþ*<sup>r</sup>* ð Þ*y* in (4.34) and since ∥*f* ∥*BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* ≤ 1 and the boundedness of *Hα*, we have *ω*∈ *Dη*.

If *<sup>γ</sup>* <sup>&</sup>gt;0, let *f y*ð Þ¼ *Pn<sup>γ</sup> ωχB*ð Þ 0,2ð Þ <sup>j</sup>*x*0jþ*<sup>r</sup>* � �ð Þ*<sup>y</sup>* then <sup>∥</sup>*<sup>f</sup>* <sup>∥</sup>*BM<sup>γ</sup>* <sup>0</sup>ð Þ *<sup>ω</sup>* ≤*C* and as in (4.18) of the proof of Theorem 1.5, we have

$$\int\_{B(\mathbf{0},2(|\mathbf{x}\_0|+r))} f(\boldsymbol{\eta})d\boldsymbol{\eta} \ge \mathbf{C}(|\boldsymbol{\kappa}\boldsymbol{\alpha}|+r)^{\boldsymbol{\eta}\cdot\boldsymbol{\eta}}o(B(\mathbf{0},|\boldsymbol{\kappa}\boldsymbol{\alpha}|+r))\dots$$

Therefore, using this function *f* in (34) and the boundedness of *Hα*, we have *ω* ∈ *Dη*. Now, let us show the necessary condition. Let *f* ∈*BM<sup>γ</sup>* <sup>0</sup>ð Þ *ω* such that *Hαf x*ð Þ is finite

for some *<sup>x</sup>* 6¼ 0 and let *<sup>ω</sup>*<sup>∈</sup> H0ð Þ *<sup>α</sup>* <sup>þ</sup> *<sup>n</sup>γ*, <sup>∞</sup> <sup>∩</sup> *<sup>D</sup>η*. Hence *<sup>H</sup>α<sup>f</sup>* <sup>∈</sup> *<sup>L</sup>*<sup>1</sup> *loc <sup>n</sup>* ð Þ by ð Þ *ii* of Lemma 4.7. We begin considering *B* ¼ *B*ð Þ 0,*r* , *x*∈*B* and *x* 6¼ 0. Let *ν* be such that ∣*ν*∣ ¼ *r*. We estimate the two terms of (24). Then,

$$\begin{split} &\frac{1}{\alpha(B)} \int\_{B} \int\_{|y| \le |\nu|} |f(y)| \left| \frac{1}{\left( |\boldsymbol{x}| + |\boldsymbol{y}| \right)^{n-\alpha}} - \frac{1}{\left( |\boldsymbol{\nu}| + |\boldsymbol{y}| \right)^{n-\alpha}} \right| d\boldsymbol{y} d\boldsymbol{x} \\ &\leq \frac{C}{\alpha(B)} \int\_{B} \frac{1}{|\boldsymbol{x}|^{n-\alpha}} \int\_{|\boldsymbol{y}| \le |\boldsymbol{\nu}|} f(\boldsymbol{y}) d\boldsymbol{y} d\boldsymbol{x} \\ &\leq C \|f'\|\_{BM\_{0}^{\ell}(\alpha)} \frac{1}{\alpha(B)} \int\_{B} \frac{\alpha(B(\mathsf{O}, |\boldsymbol{\nu}|)) |\boldsymbol{\nu}|^{n\gamma}}{|\boldsymbol{x}|^{n-\alpha}} d\boldsymbol{x} \\ &\leq C \|f'\|\_{BM\_{0}^{\ell}(\alpha)} |\boldsymbol{B}|^{\delta}. \end{split} \tag{35}$$

For the second term of (24), using the mean value theorem

$$\left| \frac{1}{\left( |\varkappa| + |\nu| \right)^{n-a}} - \frac{1}{\left( |\nu| + |\nu| \right)^{n-a}} \right| \le C \frac{r}{|\nu|^{n-a+1}}.\tag{36}$$

Then, by ð Þ *ii* of Lemma 4.6

$$\begin{split} &\frac{1}{\alpha(B)} \int\_{B} \int\_{|y|>|\nu|} |f(y)| \left| \frac{1}{\left( |x|+|y| \right)^{n-\alpha}} - \frac{1}{\left( |\nu|+|y| \right)^{n-\alpha}} \right| dydx \\ &\leq \frac{Cr}{\alpha(B)} \int\_{B} \int\_{B^{c}} \frac{|f(y)|}{|y|^{n-\alpha+1}} dydx \\ &\leq C \|f'\|\_{BM\_0'(w)} |B|^{\delta}. \end{split} \tag{37}$$

Therefore, by (24) and (35)–(37), we have proved

$$\frac{1}{\left|\alpha(B)|B|^{\delta}}\right\}\_{B} |H\_{q}f(\infty) - H\_{q}f(\nu)|d\infty \le C \|f\|\_{BM\_{q}^{r}(o)},\tag{38}$$

for every ball *B* centered at the origin.

*A Brief Look at the Calderón and Hilbert Operators DOI: http://dx.doi.org/10.5772/intechopen.106027*

We now consider *B* ¼ *B x*ð Þ 0,*r* with *r*<∣*x*0∣*=*8. By (33) it is enough to consider only these balls *B*. Let *x*∈ *B* and *ν* ¼ jð Þ *x*0jþ*r x*0*=*∣*x*0∣, then ∣*ν*∣ � ∣*x*∣ and ∣*x*∣ � ∣*x*0∣. If ∣*y*∣≤∣*ν*∣, by the mean value theorem

$$\left| \frac{1}{(|\varkappa| + |\nu|)^{n-a}} - \frac{1}{(|\nu| + |\nu|)^{n-a}} \right| \le C \frac{r}{|\varkappa\_0|^{n-a+1}}.$$

Then, since *ω*∈ *Dη*, we have

$$\begin{split} &\frac{1}{\alpha(B)} \int\_{B} \int\_{|\boldsymbol{\nu}| \leq |\boldsymbol{\nu}|} |\boldsymbol{f}(\boldsymbol{\nu})| \Big| \frac{1}{(|\boldsymbol{\nu}| + |\boldsymbol{\nu}|)^{n-a}} - \frac{1}{(|\boldsymbol{\nu}| + |\boldsymbol{\nu}|)^{n-a}} \Big| d\boldsymbol{y} d\boldsymbol{\nu} \\ &\leq \frac{Cr^{n+1}}{\alpha(B) |\boldsymbol{x}\_0|^{n-a+1}} \int\_{|\boldsymbol{\nu}| \leq |\boldsymbol{\nu}|} |\boldsymbol{f}(\boldsymbol{\nu})| d\boldsymbol{y} \\ &\leq \|\boldsymbol{f}\|\_{BM\_0^t(\boldsymbol{\nu})} |\boldsymbol{B}|^\delta. \end{split} \tag{39}$$

On the other hand, using again the mean value theorem as in (36) and ð Þ *ii* of Lemma 4.6, we get

$$\begin{split} &\frac{1}{\alpha(B)} \int\_{B} \int\_{|y|>|\nu|} |f(y)| \left| \frac{1}{\left(|x|+|y|\right)^{n-a}} - \frac{1}{\left(|\nu|+|y|\right)^{n-a}} \right| dydx \\ &\leq \frac{Cr^{n+1}}{\alpha(B)} \int\_{|y|>|\nu|} \frac{|f(y)|}{|y|^{n-a+1}} dyd \\ &\leq C \|f\|\_{BM\_0^\ell(\alpha)} \frac{r^{n+1}}{\alpha(B)} \frac{\alpha(B(\mathbf{0}, |\nu|))}{|\nu|^{n\eta}} \\ &\leq C \|f\|\_{BM\_0^\ell(\alpha)} |B|^\delta. \end{split} \tag{40}$$

Thus, by (24) and (39)–(40), we have proved

$$\frac{1}{\alpha(B)|B|^{\delta}} \int\_{B} |H\_{q}f(\boldsymbol{\omega}) - H\_{q}f(\boldsymbol{\omega})| d\boldsymbol{\omega} \leq C \|f\|\_{BM\_{0}^{r}(\alpha)},$$

for every ball *B* ¼ *B x*ð Þ 0,*r* considered. This completes the proof of the theorem.
