1 Introduction

The Calderón–Zygmund theorem is a well-known tool to investigate the \(L^p\) boundedness of singular integral operators. It was originally developed by Calderón and Zygmund [2] and later improved by Hörmander [6]. Today it is usually stated as follows:

Theorem A

Let T be a singular integral operator with a kernel K. Suppose that T is bounded from \(L^{p_0}(\mathbb {R}^d)\) to \(L^{p_0, \infty }(\mathbb {R}^d)\) for some \(1< p_0 < \infty \) and its kernel K satisfies the Hörmander condition;

$$\begin{aligned} {[}K]_{H} := \sup _{B \subset \mathbb {R}^d} \sup _{y \in B} \int _{x \in \mathbb {R}^d \setminus 2B} |K(x, y) - K(x, c(B))| \, dx < \infty , \end{aligned}$$
(1.1)

where the supremum \(\sup _{B \subset \mathbb {R}^d}\) is taken over all balls B in \(\mathbb {R}^d\), c(B) is the center of B, 2B denotes the ball with the same center as B and whose radius is twice as long. Then T is bounded from \(L^1(\mathbb {R}^d)\) to \(L^{1, \infty }(\mathbb {R}^d)\). It follows that T is bounded on \(L^p(\mathbb {R}^d)\) for all \(1< p < p_0\).

In 2019, Grafakos and Stockdale [5] introduced an \(L^q\) mean Hörmander condition (\(H_q\) condition for short);

$$\begin{aligned}{}[K]_{H_q} := \sup _{B \subset \mathbb {R}^d} \left( \frac{1}{|B|} \int _{ y \in B } \left( \int _{x \in \mathbb {R}^d \setminus 2B} |K(x, y) - K(x, c(B))| \, dx\right) ^q dy \right) ^{1/q} < \infty \nonumber \\ \end{aligned}$$
(1.2)

and proved the following:

Theorem B

[5] Let T be a singular integral with a kernel K. Suppose that T is bounded from \(L^{p_0}(\mathbb {R}^d)\) to \(L^{p_0, \infty }(\mathbb {R}^d)\) for some \(1< p_0 < \infty \) and its kernel K satisfies the \(H_{q'}\) condition for some \(1 \le q < p_0\) where \(q'\) denotes the Hölder conjugate of q. Then T is bounded from \(L^{q}(\mathbb {R}^d)\) to \(L^{q, \infty }(\mathbb {R}^d)\). It follows that T is bounded on \(L^p(\mathbb {R}^d)\) for all \(q< p < p_0\).

Note that \([K]_{q_1} \le [K]_{q_2}\) if \(1 \le q_1 \le q_2 \le \infty \) and the \(H_\infty \) condition is the same as the classical Hörmander condition (1.1). They named Theorem B ‘limited-range Calderón–Zygmund theorem’ because it implies the \(L^p\) boundedness not for all \(1< p < p_0\) but for the ‘limited-range’; \(q< p < p_0\). However, as stated in [5], they did not find any operators that satisfy the assumption of Theorem B and not bounded on \(L^q\). In this sense, there is no evidence that it is truly a limited-range theorem. In this paper, we show that it is not actually limited-ranged. In fact, the \(H_q\) condition is enough for the \(L^1 \rightarrow L^{1, \infty }\) boundedness even in the worst case \(q=1\).

Theorem 1

Let T be a singular integral operator with a kernel K. Suppose that T is bounded from \(L^{p_0}(\mathbb {R}^d)\) to \(L^{p_0, \infty }(\mathbb {R}^d)\) for some \(1< p_0 < \infty \) and its kernel K satisfies the \(H_{1}\) condition;

$$\begin{aligned} {[}K]_{H_1} = \sup _{B \subset \mathbb {R}^d} \frac{1}{|B|} \int _{ y \in B } \int _{x \in \mathbb {R}^d \setminus 2B} |K(x, y) - K(x, c(B))| \, dx \, dy < \infty . \end{aligned}$$
(1.3)

Then T is bounded from \(L^1(\mathbb {R}^d)\) to \(L^{1, \infty }(\mathbb {R}^d)\) with a constant proportional to \(\Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }} + [K]_{H_1}\).

Our proof is motivated by Fefferman’s proof of the \(L^1 \rightarrow L^{1, \infty }\) boundedness of strongly singular integral operators (see [4, Theorem 2’]). In the proof, we form the Calderón–Zygmund decomposition of f; \(f = g+b\), and approximate the bad part b by a certain function \(\widetilde{b}\).

Also we will give a criterion of the \(L^2\) boundedness for convolution type singular integral operators under the \(H_1\) condition.

Theorem 2

Let \(K \in \mathcal {S}'(\mathbb {R}^d) \cap { L^1_{\text {loc}} } (\mathbb {R}^d \setminus \{ 0 \})\) be such that

$$\begin{aligned}&A := \sup _{0< a< b< \infty } \left| \int _{a< |x|< b} K(x) \, dx \, \right| < \infty , \end{aligned}$$
(1.4)
$$\begin{aligned}&B := \sup _{ a > 0 } \frac{1}{a} \int _{|x|< a} |x| |K(x)| \, dx < \infty , \end{aligned}$$
(1.5)
$$\begin{aligned}&{[}K]_{H_1} := \sup _{r>0} \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r} |K(x-y) - K(x)| \, dx \, dy < \infty , \end{aligned}$$
(1.6)

where \(V_d\) denotes the volume of the d dimensional unit ball, and define \(K_{\varepsilon , R} := K \chi _{ \{\varepsilon< |x| < R \} }\) for any \(0< \varepsilon< R < \infty \). Then \(K_{\varepsilon , R}\) satisfies

$$\begin{aligned} \sup _{ 0< \varepsilon< R< \infty } \sup _{\xi \in \mathbb {R}^d} | \widehat{K_{\varepsilon , R}} (\xi ) | < \infty \end{aligned}$$

with a constant proportional to \(A + B + [K]_{H_1}\).

This is a natural generalization of the classical result stated by using the \(H_\infty \) condition (see [1, Theorem 3], [3, Proposition 5.5]).

Note that it remains an open question: is the \(H_1\) condition actually weaker than the classical one? As of this writing, we have no examples of K such that \([K]_{H_\infty } = \infty \) but \([K]_{H_1} < \infty \).

This paper is organized as follows. We prove Theorem 1 in Sect. 2 and Theorem 2 in Sect. 3. In Sect. 4, we will remark on the \(H^1 \rightarrow L^1\) boundedness under the assumption of Theorem 1.

2 Proof of Theorem 1

We use the following lemma:

Lemma A

[4, Decomposition Lemma] Let \(f \in L^1(\mathbb {R}^d)\) and \(\lambda > 0\). Then there exists a family of disjoint dyadic cubes \(\{Q_j\}_j\) such that

$$\begin{aligned} |f(x)| \le \lambda \text { a.e. } x \in \mathbb {R}^d \setminus \Omega , \nonumber \\ \frac{1}{|B_j|} \int _{B_j} |f(x)| \, dx \le 9^d \lambda , \nonumber \\ |\Omega ^*| \le d^{d/2} V_d |\Omega | \le C_d d^{d/2} V_d \frac{ \Vert f\Vert _1 }{ \lambda } , \end{aligned}$$
(2.1)
$$\begin{aligned} \sum _j \chi _{2B_j} \le (33 \sqrt{d} / 2)^d V_d , \end{aligned}$$
(2.2)

where

  • \(C_d\) denotes a constant which depends only on the dimension d,

  • \(B_j\) denotes the smallest ball circumscribing \(Q_j\),

  • \(\Omega := \bigcup _j Q_j\),

  • \(\Omega ^* := \bigcup _j 2B_j\).

Furthermore, if we define

$$\begin{aligned} g := f \chi _{\mathbb {R}^d \setminus \Omega }, \end{aligned}$$
(2.3)
$$\begin{aligned} b_j := f \chi _{Q_j} , \quad b := f \chi _{\Omega } = \sum _j b_j , \end{aligned}$$
(2.4)

then immediately it follows that

$$\begin{aligned} \Vert g\Vert _{p_0}^{p_0} \le \lambda ^{p_0 - 1} \Vert f\Vert _1 , \end{aligned}$$
(2.5)
$$\begin{aligned} \frac{1}{|B_j|} \Vert b_j\Vert _1 \le 9^d \lambda , \end{aligned}$$
(2.6)
$$\begin{aligned} \Vert b\Vert _1 = \sum _{j} \Vert b_j\Vert _1 \le \Vert f\Vert _1 . \end{aligned}$$
(2.7)

Lemma A is essentially the Whitney decomposition of \(\{ \, x \in \mathbb {R}^d \, : \, Mf(x) > \lambda \, \}\), where M is the Hardy–Littlewood maximal function with uncentered balls.Footnote 1

Note that our good part g (2.3) and bad part b (2.4) are different from usual ones. Ordinarily, they are defined by

$$\begin{aligned} g := f \chi _{\mathbb {R}^d \setminus \Omega } + \sum _j \left( \frac{1}{|Q_j|} \int _{Q_j} f \right) \chi _{Q_j} , \\ b_j := \left( f - \frac{1}{|Q_j|} \int _{Q_j} f \right) \chi _{Q_j} , \quad b := \sum _j b_j \end{aligned}$$

to guarantee the zero mean condition \(\int b_j = 0\). However, our proof does not require it, hence we use our simpler definition.

Now we are going to give the proof of Theorem 1.

Proof of Theorem 1

Fix \(f \in L^1(\mathbb {R}^d) \cap L^\infty (\mathbb {R}^d)\), \(t, \lambda > 0\) and form the Calderón–Zygmund decomposition of f at height \(t \lambda \) (where t is given later to set appropriate estimates). In addition, fix \(\varphi \in C_c^\infty (\mathbb {R}^d)\) such that

$$\begin{aligned} {{\,\mathrm{supp}\,}}\varphi \subset B(0, 1) , \quad \int \varphi = 1 , \quad \varphi \ge 0 \end{aligned}$$
(2.8)

and write \(\varphi _j(x) := s_j^{-d} \varphi (s_j^{-1} x)\) where \(r_j\) is the radius of \(B_j\) and \(s_j := r_j / 2\). We approximate \(b_j\) by \(\widetilde{b}_j := b_j * \varphi _j\) and b by \(\widetilde{b} := \sum _j \widetilde{b}_j\).

Now we have \(f = g - (\widetilde{b} - b) + \widetilde{b}\) and it suffices to show the following inequalities.

$$\begin{aligned} \lambda | \{ \, x \in \mathbb {R}^d \, : \, |Tg(x)| > \lambda \, \} | \lesssim \Vert f\Vert _1 , \end{aligned}$$
(2.9)
$$\begin{aligned} \lambda |\{ \, x \in \mathbb {R}^d \, : \, |T(\widetilde{b} - b)(x)| > \lambda \, \}| \lesssim \Vert f\Vert _1 , \end{aligned}$$
(2.10)
$$\begin{aligned} \lambda | \{ \, x \in \mathbb {R}^d \, : \, |T\widetilde{b}(x)| > \lambda \, \} | \lesssim \Vert f\Vert _1 . \end{aligned}$$
(2.11)

Proof of (2.9)

Since T is bounded from \(L^{p_0}(\mathbb {R}^d)\) to \(L^{p_0, \infty }(\mathbb {R}^d)\), it follows that

$$\begin{aligned} \lambda | \{ \, x \in \mathbb {R}^d \, : \, |Tg(x)|> \lambda \, \} |&= \frac{1}{\lambda ^{p_0 - 1}} \lambda ^{p_0} | \{ \, x \in \mathbb {R}^d \, : \, |Tg(x)| > \lambda \, \} | \\&\le \frac{1}{\lambda ^{p_0 - 1}} \Vert Tg\Vert _{p_0, \infty }^{p_0} \\&\le \frac{1}{\lambda ^{p_0 - 1}} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \Vert g\Vert _{p_0}^{p_0} \\&\underset{(2.5)}{\le } \frac{1}{\lambda ^{p_0 - 1}} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} (t \lambda )^{p_0-1} \Vert f\Vert _1 \\&= t^{p_0 - 1} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \Vert f\Vert _1 . \end{aligned}$$

\(\square \)

Proof of (2.10)

Since

$$\begin{aligned} \{ \, x \in \mathbb {R}^d \, : \, |T(\widetilde{b} - b)(x)|> \lambda \, \} \subset \Omega ^* \cup \{ \, x \in \mathbb {R}^d \setminus \Omega ^* \, : \, |T(\widetilde{b} - b)(x)| > \lambda \, \}, \end{aligned}$$

it follows that

$$\begin{aligned}&\lambda |\{ \, x \in \mathbb {R}^d \, : \, |T(\widetilde{b} - b)(x)|> \lambda \, \}| \\&\quad \le \lambda | \Omega ^* | + \lambda |\{ \, x \in \mathbb {R}^d \setminus \Omega ^* \, : \, |T(\widetilde{b} - b)(x)| > \lambda \, \}| \\&\quad \underset{(2.1)}{\le } t^{-1} C_d d^{d/2} V_d \Vert f\Vert _1 + \Vert T(\widetilde{b} - b)\Vert _{L^1(\mathbb {R}^d \setminus \Omega ^*)} . \end{aligned}$$

We will estimate the second term by the \(H_1\) condition (1.3). For each j and \(x \in \mathbb {R}^d \setminus 2B_j\), we write

$$\begin{aligned} T b_j (x)&= \int _{y \in \mathbb {R}^d} K(x, y) b_j(y) \, dy \\&= \int _{z \in B(0, s_j)} \left( \int _{y \in B_j} K(x, y) b_j(y) \, dy \right) \varphi _j(z) \, dz \end{aligned}$$

and

$$\begin{aligned} T \tilde{b}_j (x)&= \int _{y \in \mathbb {R}^d} K(x, y) \tilde{b}_j(y) \, dy \\&= \int _{y \in \mathbb {R}^d} K(x, y) \int _{z \in B(0, s_j)} b_j(y-z) \varphi _j(z) \, dz \, dy \\&= \int _{z \in B(0, s_j)} \int _{y \in \mathbb {R}^d} K(x, y) b_j(y-z) \, dy \, \varphi _j(z) \, dz \\&= \int _{z \in B(0, s_j)} \left( \int _{y \in B_j} K(x, y+z) b_j(y) \, dy \right) \varphi _j(z) \, dz \end{aligned}$$

since T is a singular integral operator with a kernel K. Therefore, for each j, we have

$$\begin{aligned}&\Vert T(\tilde{b}_j - b_j)\Vert _{L^1(\mathbb {R}^d \setminus \Omega ^*)} \\&\quad \le \int _{x \in \mathbb {R}^d \setminus 2B_j} | T \tilde{b}_j (x) - Tb_j(x) | \, dx \\&\quad = \int _{x \in \mathbb {R}^d \setminus 2B_j} \left| \int _{z \in B(0, s_j)} \left( \int _{y \in B_j} ( K(x, y+z) - K(x, y) ) b_j(y) \, dy \right) \varphi _j(z) \, dz \, \right| dx \\&\quad \le \int _{y \in B_j} \left( \int _{z \in B(0, s_j)} \int _{x \in \mathbb {R}^d \setminus 2B_j} |K(x, y + z) - K(x, y)| \, dx \, \varphi _j(z) \, dz \right) |b_j(y)| \, dy \\&\quad \le \Vert \varphi _j\Vert _\infty \int _{y \in B_j} \left( \int _{z \in B(0, s_j)} \int _{x \in \mathbb {R}^d \setminus B(y , r_j)} |K(x, y + z) - K(x, y)| \, dx \, dz \right) |b_j(y)| \, dy \\&\quad = V_d \Vert \varphi \Vert _\infty \int _{y \in B_j} \left( \frac{1}{V_d s_j^d} \int _{z \in B(y, s_j)} \int _{x \in \mathbb {R}^d \setminus B(y , 2 s_j)} |K(x, z) - K(x, y)| \, dx \, dz \right) |b_j(y)| \, dy \\&\quad \le V_d \Vert \varphi \Vert _\infty \int _{y \in B_j} [K]_{H_1} |b_j(y)| \, dy \\&\quad = V_d \Vert \varphi \Vert _\infty [K]_{H_1} \Vert b_j\Vert _1 . \end{aligned}$$

It follows that

$$\begin{aligned} \Vert T(\widetilde{b} - b)\Vert _{L^1(\mathbb {R}^d \setminus \Omega ^*)} \le \sum _j \Vert T(\tilde{b}_j - b_j)\Vert _{L^1(\mathbb {R}^d \setminus \Omega ^*)} \le V_d \Vert \varphi \Vert _\infty [K]_{H_1} \Vert f\Vert _1. \end{aligned}$$

\(\square \)

Proof of (2.11)

By the same argument as in the proof of (2.9), we have

$$\begin{aligned} \lambda | \{ \, x \in \mathbb {R}^d \, : \, |T\widetilde{b}(x)| > \lambda \, \} |&\le \frac{1}{\lambda ^{p_0 - 1}} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \Vert \widetilde{b}\Vert _{p_0}^{p_0} \\&\le \frac{1}{\lambda ^{p_0 - 1}} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \Vert \widetilde{b}\Vert _{\infty }^{p_0 - 1} \Vert \widetilde{b}\Vert _1 . \end{aligned}$$

Since it is obvious that

$$\begin{aligned} \Vert \widetilde{b}\Vert _1 \le \sum _j \Vert b_j * \varphi _j\Vert _1 \le \sum _j \Vert b_j\Vert _1 \underset{(2.7)}{\le } \Vert f\Vert _1, \end{aligned}$$

it is enough to show that \(\Vert \widetilde{b}\Vert _\infty \lesssim \lambda \). For each j, we have

$$\begin{aligned} \Vert b_j * \varphi _j\Vert _\infty \le \Vert b_j\Vert _1 \Vert \varphi _j\Vert _\infty = \Vert \varphi \Vert _\infty \frac{\Vert b_j\Vert _1}{s_j^d} \underset{(2.6)}{\le } 18^d V_d \Vert \varphi \Vert _\infty t \lambda . \end{aligned}$$

Therefore, it follows from the bounded overlap property (2.2),

$$\begin{aligned} \Vert \widetilde{b}\Vert _\infty \le 297^d d^{d/2} V_d^2 \Vert \varphi \Vert _\infty t \lambda . \end{aligned}$$

Hence we conclude that

$$\begin{aligned} \lambda | \{ \, x \in \mathbb {R}^d \, : \, |T\widetilde{b}(x)| > \lambda \, \} | \le t^{p_0 - 1} ( 297^d d^{d/2} V_d^2 \Vert \varphi \Vert _\infty )^{p_0 - 1} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \Vert f\Vert _1. \end{aligned}$$

\(\square \)

Combining estimates above, we obtain

$$\begin{aligned} \Vert Tf\Vert _{1, \infty }&\le 3 \{ t^{ p_0 - 1 } \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} + ( t^{-1} C_d d^{d/2} V_d + V_d \Vert \varphi \Vert _\infty [K]_{H_1} ) \\&\quad + t^{p_0 - 1} ( 297^d d^{d/2} V_d^2 \Vert \varphi \Vert _\infty )^{p_0 - 1} \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} \} \Vert f\Vert _1 . \end{aligned}$$

Finally, remember that t and \(\varphi \) are arbitrary. Since \(\inf _{\varphi satisfies (2.8) } \Vert \varphi \Vert _\infty = V_d^{-1}\), we conclude that

$$\begin{aligned}&\Vert T\Vert _{L^1 \rightarrow L^{1, \infty }} \\&\quad \le 3 \inf _{t>0} \, ( t^{-1} C_d d^{d/2} V_d + [K]_{H_1} + t^{p_0 - 1} ( 1 + ( 297^d d^{d/2} V_d )^{ p_0 - 1 } ) \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }}^{p_0} ) \\&\quad = 3 \left( p_0 ( 1 + ( 297^d d^{d/2} V_d )^{ p_0 - 1 } )^{1/{p_0}} \left( \frac{C_d d^{d/2} V_d}{p_0 - 1} \right) ^{ 1 - 1/p_0 } \Vert T\Vert _{L^{p_0} \rightarrow L^{p_0, \infty }} + [K]_{H_1} \right) . \end{aligned}$$

\(\square \)

3 The Proof of Theorem 2

We use the following lemma:

Lemma 1

If \(K \in { L^1_{\text {loc}} } (\mathbb {R}^d \setminus \{ 0 \})\) satisfies (1.5) and (1.6), then

$$\begin{aligned} \sup _{ 0< \varepsilon< R < \infty } [K_{\varepsilon , R}]_{H_1} \le [K]_{H_1} + 7B. \end{aligned}$$
(3.1)

Proof of Lemma 1

It is obvious that

$$\begin{aligned}&\frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r} |K_{\varepsilon , R}(x-y) - K_{\varepsilon , R}(x)| \, dx \, dy \\&\quad = \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r, \varepsilon< |x-y|< R,\varepsilon< |x|< R} |K(x-y) - K(x)| \, dx \, dy \\&\qquad + \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r, \varepsilon< |x-y|< R, |x| \le \varepsilon } |K(x-y)| \, dx \, dy \\&\qquad + \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r, \varepsilon< |x-y|< R, |x| \ge R} |K(x-y)| \, dx \, dy \\&\qquad + \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r, |x-y| \le \varepsilon , \varepsilon< |x|< R} |K(x)| \, dx \, dy \\&\qquad + \frac{1}{V_d r^d} \int _{|y| \le r} \int _{|x| \ge 2r, |x-y| \ge R, \varepsilon< |x| < R} |K(x)| \, dx \, dy \end{aligned}$$

and the first term is bounded by \([K]_{H_1}\). To estimate other terms, note that (1.5) implies

$$\begin{aligned} \sup _{a>0} \int _{a< |x|< ca} |K(x)| \, dx \le \sup _{a>0} \int _{a< |x| < ca} \frac{|x|}{a} |K(x)| \, dx \le cB \end{aligned}$$

for any \(c>1\). Since we have

$$\begin{aligned}&\varepsilon< |x-y|< R, |x| \le \varepsilon\Rightarrow & {} |x-y| \le |x| + |y| \le 3|x| / 2 \le 3 \varepsilon / 2,&\\&\varepsilon< |x-y|< R, |x| \ge R\Rightarrow & {} |x-y| \ge |x| - |y| \ge |x| / 2 \ge R/2,&\\&|x-y| \le \varepsilon , \varepsilon< |x|< R\Rightarrow & {} |x| \le 2(|x-y| + |y|) - |x| \le 2 |x-y|< 2\varepsilon ,&\\&|x-y| \ge R, \varepsilon< |x| < R\Rightarrow & {} |x| \ge 2(|x-y| - |y| )/3 + |x|/3 \ge 2|x-y| / 3 \ge 2R/3&\end{aligned}$$

under the condition \(2|y| \le 2r \le |x|\), the second and fifth terms are bounded by 3B/2, the third and fourth terms are bounded by 2B. \(\square \)

Proof of Theorem 2

Fix \(0< \varepsilon< R < \infty \) and \(\xi \in \mathbb {R}^d\). Since it is obvious that

$$\begin{aligned} | \widehat{K_{\varepsilon , R}} (0) | = \left| \int _{\varepsilon< |x| < R} K(x) \, dx \, \right| \underset{(1.4)}{\le } A, \end{aligned}$$

we assume \(\xi \ne 0\) and write \(s := |\xi |^{-1}\). If we decompose \(\widehat{K_{\varepsilon , R}} (\xi ) \) as

$$\begin{aligned}&\widehat{K_{\varepsilon , R}} (\xi ) \\&\quad = \int _{x \in \mathbb {R}^d} K_{\varepsilon , R}(x) e^{- 2 \pi i x \cdot \xi } \, dx \\&\quad = \int _{ |x|< 2s } K_{\varepsilon , R}(x) ( e^{- 2 \pi i x \cdot \xi } - 1 ) \, dx + \int _{ |x| < 2s } K_{\varepsilon , R}(x) \, dx + \int _{2s \le |x| } K_{\varepsilon , R}(x) e^{- 2 \pi i x \cdot \xi } \, dx \\&\quad =: I_1 + I_2 + I_3, \end{aligned}$$

then we easily get

$$\begin{aligned} |I_1|\le & {} \int _{|x|< 2s} |K_{\varepsilon , R}(x)| | e^{- 2 \pi i x \cdot \xi } - 1 | \, dx \le 4 \pi \frac{1}{2s} \int _{ |x|< 2s} |x| |K_{\varepsilon , R}(x)| \, dx \underset{(1.5)}{\le } 4 \pi B , \\ |I_2|= & {} \left| \int _{\varepsilon< |x| < 2s} K_{\varepsilon , R}(x) \, dx \, \right| \underset{(1.4)}{\le } A. \end{aligned}$$

To estimate \(I_3\), fix a radial function \(\varphi \in C_c^\infty (\mathbb {R}^d)\) such that

$$\begin{aligned} {{\,\mathrm{supp}\,}}\varphi \subset B(0, 1) , \quad \int \varphi = 1 , \quad \varphi \ge 0, \quad |\widehat{\varphi } (1)| < 1 \end{aligned}$$

and define \(\varphi _s(x) := s^{-d} \varphi (s^{-1} x)\). Moreover, rewrite

$$\begin{aligned} I_3&\ = \int _{|x| \ge 2s} \int _{|y| \le s} K_{\varepsilon , R}(x) \varphi _s(y) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx \end{aligned}$$
(3.2)

and introduce

$$\begin{aligned} I_4:= & {} \int _{|x| \ge 2s} \int _{|y| \le s} K_{\varepsilon , R}(x-y) \varphi _s(y) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx , \end{aligned}$$
(3.3)
$$\begin{aligned} I_5:= & {} \int _{|x| < 2s} \int _{|y| \le s} K_{\varepsilon , R}(x-y) \varphi _s(y) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx \end{aligned}$$
(3.4)
$$\begin{aligned}= & {} \int _{|x|< 2s} \int _{|x-y| \le s} K_{\varepsilon , R}(y) \varphi _s(x-y) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx \nonumber \\&\ =&\int _{|x| < 2s} \int _{|y| \le 3s} K_{\varepsilon , R}(y) \varphi _s(x-y) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx , \end{aligned}$$
(3.5)
$$\begin{aligned} I_6:= & {} \int _{|x| < 2s} \int _{|y| \le 3s} K_{\varepsilon , R}(y) \varphi _s(x) \, dy \, e^{- 2 \pi i x \cdot \xi } \, dx \end{aligned}$$
(3.6)
$$\begin{aligned}&\ =&\widehat{\varphi _s}(\xi ) \int _{|y| \le 3s} K_{\varepsilon , R}(y) \, dy. \end{aligned}$$
(3.7)

We decompose \(I_3\) into \((I_3 - I_4) + (I_4 + I_5) - (I_5 - I_6) - I_6\). By (3.2), (3.3) and Lemma 1, we get

$$\begin{aligned} |I_4 - I_3|&\\ \underset{(3.2), (3.3)}{=}&\left| \int _{|x| \ge 2s} \int _{ |y| \le s } ( K_{\varepsilon , R}(x-y) - K_{\varepsilon , R}(x) ) \varphi _s(y) \, e^{- 2 \pi i x \cdot \xi } \, dy \, dx \, \right| \\&\le \int _{ |y| \le s } \int _{|x| \ge 2s} | K_{\varepsilon , R}(x-y) - K_{\varepsilon , R}(x) | \, \varphi _s(y) \, dx \, dy \\&\le V_d \Vert \varphi \Vert _\infty \frac{1}{V_d s^d} \int _{ |y| \le s } \int _{|x| < 2s} | K_{\varepsilon , R}(x-y) - K_{\varepsilon , R}(x) | \, dx \, dy \\&\le V_d \Vert \varphi \Vert _\infty [K_{\varepsilon , R}]_{H_1} \\ \underset{(3.1)}{\le }&V_d \Vert \varphi \Vert _\infty ( [K]_{H_1} + 7B ). \end{aligned}$$

For \(I_5 - I_6\), use (3.5), (3.6) and the mean value theorem to obtain

$$\begin{aligned}&|I_5 - I_6| \\&\quad \underset{(3.5), (3.6)}{=} \left| \int _{|x|< 2s} \int _{|y| \le 3s} K_{\varepsilon , R}(y) ( \varphi _s(x-y) - \varphi _s(x) ) e^{-2\pi i x \cdot \xi } \, dy \, dx \, \right| \\&\qquad \le \int _{|x|< 2s} \int _{|y| \le 3s} |K_{\varepsilon , R}(y) | | \varphi _s(x-y) - \varphi _s(x) | \, dy \, dx \\&\qquad \le \int _{|x|< 2s} \int _{|y| \le 3s} |K_{\varepsilon , R}(y) | s^{-d-1} |y| \Vert \nabla \varphi \Vert _\infty \, dy \, dx \\&\qquad = 3 s^{-d} \Vert \nabla \varphi \Vert _\infty \int _{|x| < 2s} \left( \frac{1}{3s} \int _{|y| \le 3s} |y| |K_{\varepsilon , R}(y) | \, dy \right) dx \\&\qquad \underset{(1.5)}{\le } 3 s^{-d} \Vert \nabla \varphi \Vert _\infty \int _{|x| \le 2s} B \, dx \\&\qquad = 3 \cdot 2^d V_d \Vert \nabla \varphi \Vert _\infty B. \end{aligned}$$

For \(I_4 + I_5\) and \(I_6\), remark that \(\widehat{\varphi _s}(\xi ) = \widehat{\varphi }(s\xi ) = \widehat{\varphi }(1)\) because \(\varphi \) is radial and \(s = |\xi |^{-1}\). Then it follows immediately that

$$\begin{aligned} I_4 + I_5 \underset{(3.3), (3.4)}{=} \widehat{ K_{\varepsilon , R} * \varphi _s}(\xi ) = \widehat{\varphi }(1) \widehat{K_{\varepsilon , R}}(\xi ), \\ |I_6| \underset{(3.7)}{=} \left| \widehat{\varphi _s}(\xi ) \int _{|x| \le 3s} K_{\varepsilon , R}(y) \, dy \, \right| \underset{(1.4)}{\le } | \widehat{\varphi }(1) | A. \end{aligned}$$

Now we have

$$\begin{aligned}&| \widehat{K_{\varepsilon , R}} (\xi ) | \\&\quad \le |I_1| + |I_2| + |I_3 - I_4| + |I_4 + I_5| + |I_5 - I_6| + |I_6| \\&\quad \le 4 \pi B + A + V_d \Vert \varphi \Vert _\infty ( [K]_{H_1} + 7B ) + |\widehat{\varphi }(1)| |\widehat{K_{\varepsilon , R}}(\xi )| \\&\qquad + 3 \cdot 2^d V_d \Vert \nabla \varphi \Vert _\infty B + | \widehat{\varphi }(1) | A \end{aligned}$$

for any \(\xi \in \mathbb {R}^d\) (it is still valid in the case \(\xi = 0\)). Finally, remember \(|\widehat{\varphi }(1)| < 1\) to conclude that

$$\begin{aligned} | \widehat{K_{\varepsilon , R}} (\xi ) | \le \frac{ ( 1 + |\widehat{\varphi }(1)| ) A + (4 \pi + V_d (7 \Vert \varphi \Vert _\infty + 3 \cdot 2^d \Vert \nabla \varphi \Vert _\infty ) )B + V_d \Vert \varphi \Vert _{\infty } [K]_{H_1} }{ 1 - | \widehat{\varphi }(1) | }. \end{aligned}$$

\(\square \)

4 Remark

We can also obtain the \(H^1 \rightarrow L^1\) boundedness under the assumption of Theorem 1.

Theorem 3

Let T be a singular integral operator with a kernel K. Suppose that T is bounded from \(L^{p_0}(\mathbb {R}^d)\) to \(L^{p_0, \infty }(\mathbb {R}^d)\) for some \(1< p_0 < \infty \) and its kernel K satisfies the \(H_{1}\) condition. Then T is bounded from \(H^1(\mathbb {R}^d)\) to \(L^1(\mathbb {R}^d)\).

To see this, note that Theorem 1 implies that T is bounded on \(L^p(\mathbb {R}^d)\) for any \(1< p < p_0\). Hence we assume that T is bounded on \(L^{p_0}(\mathbb {R}^d)\) for some \(1< p_0 < \infty \) without loss of generality. Now we can show the \(H^1(\mathbb {R}^d) \rightarrow L^1(\mathbb {R}^d)\) boundedness. We do not prove it here because its proof is the almost same as that of the classical theorem (see [3, Proposition 6.2, Corollary 6.3]).