Proof We first consider the $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{\frac {1}{2}}(\mathbb {R})$ estimate for $H_{\alpha ,\gamma ,j}$ . If $j<j_0$ , then

$$ \begin{align*}\left|{\frac{\partial(u,v)}{\partial(x,t)}}\right|=|1-\gamma'(t)|\geq 1-|\gamma'(t)|>1-|\gamma'(2^{j_0})|\gtrsim 1,\end{align*} $$

and furthermore, the change of variables $u=x-t$ and $v=x-\gamma (t)$ yields that

(3.3) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\right\|_{L^{1}(\mathbb{R})}\lesssim \int_{\mathbb{R}}\int_{\mathbb{R}} f(u)g(v){\frac{1}{|1-\gamma'(t)|}}\,\textrm{d}u\textrm{d}v\lesssim \|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align} $$

On the other hand, for any measurable set E with finite measure, apply Hölder’s inequality and (3.3) to deduce that

(3.4) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\leq |E|\left\|H_{\alpha,\gamma,j}(f,g)\right\|_{L^{1}(\mathbb{R})}\lesssim |E|\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align} $$

Let the interval $I_k:=((k-1)2^j,k2^j]$ with $k\in \mathbb {Z}$ . It is easy to check that $H_{\alpha ,\gamma ,j}(f \chi _{I_{k}},g\chi _{I_{k+d}})(x)$ is supported in a finitely many terms of these intervals and $H_{\alpha ,\gamma ,j}(f \chi _{I_{k}},g\chi _{I_{k+d}})(x)=0$ when $|d|>3$ for every fixed $k\in \mathbb {Z}$ . Therefore, by (3.4), we conclude that

$$ \begin{align*} \left\|H_{\alpha,\gamma,j}(f \chi_{I_{k}},g\chi_{I_{k+d}})\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim 2^j\left\|f \chi_{I_{k}}\right\|_{L^{1}(\mathbb{R})}\left\|g\chi_{I_{k+d}}\right\|_{L^{1}(\mathbb{R})}, \end{align*} $$

which, together with the triangle inequality and the Cauchy–Schwarz inequality, shows that

(3.5) $$ \begin{align} &\left\|H_{\alpha,\gamma,j}(f,g)\right\|_{L^{\frac{1}{2}}(\mathbb{R})} =\left\|\sum\limits_{d\in\mathbb{Z},~|d|\leq 3}\sum\limits_{k\in\mathbb{Z}}H_{\alpha,\gamma,j}(f \chi_{I_{k}},g\chi_{I_{k+d}})\right\|_{L^{\frac{1}{2}}(\mathbb{R})} \\ &\quad\leq \sum\limits_{d\in\mathbb{Z},~|d|\leq 3} \left[\sum\limits_{k\in\mathbb{Z}} \left\|H_{\alpha,\gamma,j}(f \chi_{I_{k}},g\chi_{I_{k+d}})\right\|^{\frac{1}{2}}_{L^{\frac{1}{2}}(\mathbb{R})} \right]^2\nonumber\\ &\quad\lesssim 2^j \sum\limits_{d\in\mathbb{Z},~|d|\leq 3} \left[\sum\limits_{k\in\mathbb{Z}} \left(\left\|f \chi_{I_{k}}\right\|_{L^{1}(\mathbb{R})}\left\|g\chi_{I_{k+d}}\right\|_{L^{1}(\mathbb{R})} \right)^{\frac{1}{2}} \right]^2\lesssim 2^j\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}.\nonumber \end{align} $$

This is the desired $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{\frac {1}{2}}(\mathbb {R})$ estimate for $H_{\alpha ,\gamma ,j}$ in the case of $j<j_0$ .

We now turn to the case of $j>j_{0}+2$ . An easy calculation gives

$$ \begin{align*}\left|{\frac{\partial(u,v)}{\partial(x,t)}}\right|\geq|\gamma'(t)|-1\geq|\gamma'(2^{j-1})|-1\geq|\gamma'(2^{j_0+2})|-|\gamma'(2^{j_0+1})|\gtrsim 1.\end{align*} $$

Taking this into account one can see that

$$ \begin{align*} \left\|H_{\alpha,\gamma,j}(f,g)\right\|_{L^{1}(\mathbb{R})}\lesssim \frac{1}{|\gamma'(2^{j-1})|-1}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align*} $$

As in (3.4), noticing that $|\gamma '(2^{j-1})|-1 \gtrsim |\gamma '(2^{j-1})|$ for $j>j_{0}+2$ , thus we have an estimate

(3.6) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim \frac{|E|}{|\gamma'(2^{j-1})|}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align} $$

From $\lim \limits _{t\rightarrow 0^+}\gamma (0)=0$ and (1.3), it shows that $|\gamma (t)|$ is strictly increasing on ${t\in \mathbb {R}^+}$ . We also from (2.3) have $\lim \limits _{t\rightarrow \infty }|\gamma (t)|=\infty $ . Then, we can split $\mathbb {R}$ as a union of intervals $J_k:=((k-1)|\gamma (2^j)|,k|\gamma (2^j)|]$ with $k\in \mathbb {Z}$ . Furthermore, one sees that $H_{\alpha ,\gamma ,j}(f \chi _{J_{k}},g\chi _{J_{k+d}})(x)$ is supported in a finitely many terms of these intervals, and there exists a universal constant C such that $H_{\alpha ,\gamma ,j}(f \chi _{J_{k}},g\chi _{J_{k+d}})(x)=0$ when $|d|>C$ for every fixed $k\in \mathbb {Z}$ . Indeed, we may take $C:=\max \{-2+\frac {C^{(1)}_2}{2},1+2^{{j_0}+2}\}$ . Therefore, by (3.6), (2.3) and (1.3), we obtain

$$ \begin{align*} &\left\|H_{\alpha,\gamma,j}(f \chi_{J_{k}},g\chi_{J_{k+d}})\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\\&\quad\lesssim \frac{|\gamma(2^j)|}{|\gamma'(2^{j-1})|} \left\|f \chi_{J_{k}}\right\|_{L^{1}(\mathbb{R})}\left\|g\chi_{J_{k+d}}\right\|_{L^{1}(\mathbb{R})}\lesssim 2^j\left\|f \chi_{J_{k}}\right\|_{L^{1}(\mathbb{R})}\left\|g\chi_{J_{k+d}}\right\|_{L^{1}(\mathbb{R})}. \end{align*} $$

Furthermore, as in (3.5), we also assert that

(3.7) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim2^j\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

holds for all $j>j_{0}+2$ .

Putting things together, for the non-critical case $j\neq j_0, j_0+1, j_0+2$ , we have established the $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{\frac {1}{2}}(\mathbb {R})$ estimate for $H_{\alpha ,\gamma ,j}$ . The remaining $L^p(\mathbb {R})\times L^q(\mathbb {R}) \rightarrow L^r(\mathbb {R})$ estimates follow by interpolation with the following three trivial estimates: $L^\infty (\mathbb {R})\times L^\infty (\mathbb {R}) \rightarrow L^\infty (\mathbb {R})$ , $L^\infty (\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^1(\mathbb {R})$ and $L^1(\mathbb {R})\times L^\infty (\mathbb {R}) \rightarrow L^1(\mathbb {R})$ . Therefore, we finish the proof of Proposition 3.1.

Proof We only consider the case $j= j_0$ , the proofs of the cases $j=j_0+1, j_0+2$ are similar to the case $j= j_0$ and we omit the details. Since $j_0$ is a negative integer, we can bound $H_{\alpha ,\gamma ,j_0}(f,g)(x)$ by

$$ \begin{align*}\int_{0}^{1}f(x-t)g(x-\gamma(t))\,\textrm{d}t.\end{align*} $$

Moreover, we can restrict x into the interval $[0,1]$ . Indeed, let $f_N(x):=f(x+N)$ and $g_N(x):=g(x+N)$ if we have the restriction of x, then $\|H_{\alpha ,\gamma ,j_0}(f,g)\|^r_{L^{r}(\mathbb {R})}$ can be dominated by a constant times

$$ \begin{align*} &\sum\limits_{N\in\mathbb{Z}}\int_{0}^{1}\left(\int_{0}^{1}f_N(x-t)g_N(x-\gamma(t))\,\textrm{d}t\right)^r\,\textrm{d}x \\&\quad\lesssim \sum_{N\in\mathbb{Z}}\left\|f_{N}\right\|_{L^{p}(\mathbb{R})}^{r}\left\|g_{N}\right\|_{L^{q}(\mathbb{R})}^{r} \lesssim \|f\|_{L^{p}(\mathbb{R})}^{r}\|g\|_{L^{q}(\mathbb{R})}^{r}, \end{align*} $$

where the last inequality follows from Hölder’s inequality.

Therefore, it suffices to show that

(3.8) $$ \begin{align} \int_{0}^{1}\left(\int_{0}^{1}f(x-t)g(x-\gamma(t))\,\textrm{d}t\right)^r\,\textrm{d}x\lesssim \|f\|_{L^{p}(\mathbb{R})}^{r}\|g\|_{L^{q}(\mathbb{R})}^{r} \end{align} $$

for all $p,q,r$ satisfying $\frac {1}{p}+\frac {1}{q}=\frac {1}{r}$ , $p, q \in [1, \infty ]$ and $(p,q)\neq (1,1)$ .

When $r\geq 1$ , it is easy to see that (3.8) follows from Minkowski’s integral inequality and Hölder’s inequality. When $\frac {1}{2}<r<1$ , we need some more complicated calculations. Indeed, let $\psi $ be a smooth function supported on $\{t\in \mathbb {R}:\ \frac {1}{2}\leq |t|\leq 2\}$ with the property that $0\leq \psi (t)\leq 1$ and $\Sigma _{k\in \mathbb {Z}} \psi _k(t)=1$ , where $\psi _k(t):=\psi (2^{-k}t)$ . In order to obtain (3.8) with $\frac {1}{2}<r<1$ , it suffices to establish the following estimate: there exists a positive constant $\varepsilon $ such that

(3.9) $$ \begin{align} \int_{0}^{1}\left[\int_{0}^{1}f(x-t)g(x-\gamma(t))\psi\left(2^{k}(t-t_{0})\right)\,\textrm{d}t\right]^r\,\textrm{d}x\lesssim 2^{-\varepsilon k}\|f\|_{L^{p}(\mathbb{R})}^{r}\|g\|_{L^{q}(\mathbb{R})}^{r} \end{align} $$

for all $k>0$ , where $p,q,r$ satisfy $\frac {1}{p}+\frac {1}{q}=\frac {1}{r}$ , $p, q \in [1, \infty ]$ and $\frac {1}{2}<r<1$ .

We establish (3.9) by considering the following two cases:

1. • The case of $2^{-k+2}\leq t_0$ . Notice that ${\frac {1}{2}}\cdot 2^{-k}\leq |t-t_0|\leq 2\cdot 2^{-k}$ , it implies $|t-t_0|\leq {\frac {1}{2}}t_0$ and we can then rewrite the left hand side of (3.9) as

   (3.10) $$ \begin{align} \int_{0}^{1}\left[\int_{-\frac{t_0}{2}}^{\frac{t_0}{2}}f_{t_0}(x-t)g_{t_0}\left(x-\Upsilon_{t_0}(t)\right)\psi\left(2^{k}t\right)\,\textrm{d}t\right]^r\,\textrm{d}x, \end{align} $$
   where
   $$ \begin{align*}f_{t_0}(x):=f(x-t_0), ~~g_{t_0}(x):=g(x-\gamma(t_0))~~ \textrm{and} ~~\Upsilon_{t_0}(t):=\gamma(t+t_{0})-\gamma(t_{0}).\end{align*} $$

   On the other hand, we obviously have $|t|\leq 2^{-k+1}$ . Moreover, by Lagrange’s mean value theorem, $|\Upsilon _{t_0}(t)|\leq |\gamma '(\zeta )||t|\leq 2|\gamma '(2)|2^{-k}$ , where $\zeta \in ({\frac {t_0}{2}},{\frac {3}{2}}t_0)\subset (0,2)$ . Then, if we set

   $$ \begin{align*} \Omega_{N}:=\left(2N|\gamma'(2)|2^{-k}, 2(N+1)|\gamma'(2)|2^{-k}\right] \end{align*} $$
   with $N=0,1,\ldots ,\lceil \frac {2^k}{2|\gamma '(2)|}\rceil $ , which trivially leads to $x-t, x-\Upsilon _{t_0}(t)\in \Omega _{N-1}\cup \Omega _{N} \cup \Omega _{N+1}$ for all $x\in \Omega _{N}$ .

   Therefore, to obtain (3.9), it is enough to show that, there exists a positive constant $\varepsilon $ such that

   (3.11) $$ \begin{align} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{r}(\mathbb{R})}^{r}\lesssim 2^{-\varepsilon k}\|f_{t_0, N}\|_{L^{p}(\mathbb{R})}^{r}\|g_{t_0, N}\|_{L^{q}(\mathbb{R})}^{r} \end{align} $$
   for all $k>0$ , where $f_{t_0, N}:=f_{t_0}\chi _{\Omega _{N}}$ , $g_{t_0, N}:=g_{t_0}\chi _{\Omega _{N}}$ and
   $$ \begin{align*} \Lambda_N(f_{t_0, N}, g_{t_0, N})(x):=\chi_{\Omega_{N}}(x)\int_{-\frac{t_0}{2}}^{\frac{t_0}{2}}f_{t_0, N}(x-t)g_{t_0, N}\left(x-\Upsilon_{t_0}(t)\right)\psi\left(2^{k}t\right)\,\textrm{d}t. \end{align*} $$

   Indeed, notice that $\frac {1}{2}<r<1$ , from (3.10), by the triangle inequality, we can bound the left hand side of (3.9) by

   $$ \begin{align*}\sum_{N\in [0, \lceil\frac{2^k}{2|\gamma'(2)|}\rceil]} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{r}(\mathbb{R})}^{r}\end{align*} $$
   essentially. This, in combination with (3.11) and Hölder’s inequality, leads to the desired estimate (3.9).

   We now turn to the proof of (3.11) by establishing the following estimates:

   1. ∘ The $L^1(\mathbb {R})\times L^\infty (\mathbb {R}) \rightarrow L^1(\mathbb {R})$ boundedness of $\Lambda _N(f_{t_0, N}, g_{t_0, N})$ . A calculation gives

      $$ \begin{align*} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{1}(\mathbb{R})}&\leq \|g_{t_0, N}\|_{L^{\infty}(\mathbb{R})}\int_{0}^{1}\chi_{\Omega_{N}}(x)\int_{-\frac{t_0}{2}}^{\frac{t_0}{2}}f_{t_0, N}(x-t)\psi\left(2^{k}t\right)\,\textrm{d}t\,\textrm{d}x\\ & \lesssim 2^{- k}\|f_{t_0, N}\|_{L^{1}(\mathbb{R})}\|g_{t_0, N}\|_{L^{\infty}(\mathbb{R})}. \end{align*} $$

   2. ∘ The $L^{\infty }(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^1(\mathbb {R})$ boundedness of $\Lambda _N(f_{t_0, N}, g_{t_0, N})$ . One easily sees that

      $$ \begin{align*} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{1}(\mathbb{R})}&\leq \|f_{t_0, N}\|_{L^{\infty}(\mathbb{R})}\kern-1.5pt\int_{-\frac{t_0}{2}}^{\frac{t_0}{2}}\kern-1.5pt\int_{0}^{1}\chi_{\Omega_{N}}(x)g_{t_0, N}\left(x-\Upsilon_{t_0}(t)\right)\,\textrm{d}x ~\psi\left(2^{k}t\right) \,\textrm{d}t\\ & \lesssim 2^{- k}\|f_{t_0, N}\|_{L^{\infty}(\mathbb{R})}\|g_{t_0, N}\|_{L^{1}(\mathbb{R})}. \end{align*} $$

   3. ∘ The $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{\frac {1}{2}}(\mathbb {R})$ boundedness of $\Lambda _N(f_{t_0, N}, g_{t_0, N})$ . As in Proposition 3.1, we need a $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{1}(\mathbb {R})$ estimate for $\Lambda _N(f_{t_0, N}, g_{t_0, N})$ . By Lagrange’s mean value theorem, it follows from the change of variables $u=x-t$ and $v=x-\Upsilon _{t_0}(t)$ that

      $$ \begin{align*}\left|{\frac{\partial(u,v)}{\partial(x,t)}}\right|=|1-\Upsilon_{t_0}'(t)|=|\gamma'(t+t_{0})-\gamma'(t_{0})|=|\gamma"(\vartheta)||t|\gtrsim 2^{-k},\end{align*} $$
      where $\vartheta \in [{\frac {1}{2}}t_0, {\frac {3}{2}}t_0]$ . Then,
      $$ \begin{align*} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{1}(\mathbb{R})}&\lesssim \int_{\mathbb{R}}\int_{\mathbb{R}} f_{t_0, N}(u)g_{t_0, N}(v)\left|{\frac{\partial(x,t)}{\partial(u,v)}}\right|\,\textrm{d}u\textrm{d}v\\ &\lesssim 2^{k} \|f_{t_0, N}\|_{L^{1}(\mathbb{R})}\|g_{t_0, N}\|_{L^{1}(\mathbb{R})}. \end{align*} $$

      Moreover, one may use the Cauchy–Schwartz inequality to deduce that

      $$ \begin{align*} \left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim |\Omega_{N}|\left\|\Lambda_N(f_{t_0, N}, g_{t_0, N})\right\|_{L^{1}(\mathbb{R})} \lesssim \|f_{t_0, N}\|_{L^{1}(\mathbb{R})}\|g_{t_0, N}\|_{L^{1}(\mathbb{R})}. \end{align*} $$

   Putting things together, it follows from an interpolation that (3.11) holds for all $p,q,r$ satisfying $\frac {1}{p}+\frac {1}{q}=\frac {1}{r}$ , $p, q \in [1, \infty ]$ and $\frac {1}{2}<r<1$ .

2. • The case of $2^{-k+2}> t_0$ . Let

   $$ \begin{align*}\Pi_k(f,g)(x):=\int_{0}^{1}f(x-t)g(x-\gamma(t))\psi\left(2^{k}(t-t_{0})\right)\,\textrm{d}t,\end{align*} $$
   it follows from $2^{-k+2}> t_0$ that $k< \log _2 (\frac {4}{t_0})$ , therefore to prove (3.9) it will suffice to prove
   $$ \begin{align*} \left\|\Pi_k(f,g)\right\|_{L^{r}([0, 1])}\lesssim \|f\|_{L^{p}(\mathbb{R})}\|g\|_{L^{q}(\mathbb{R})} \end{align*} $$
   for all $p,q,r$ satisfying $\frac {1}{p}+\frac {1}{q}=\frac {1}{r}$ , $p, q \in [1, \infty ]$ and $\frac {1}{2}<r<1$ . This can be proved by an interpolation with the following estimates:

   1. ∘ The $L^1(\mathbb {R})\times L^\infty (\mathbb {R}) \rightarrow L^1([0, 1])$ boundedness of $\Pi _k(f,g)$ . It is easy to get

      $$ \begin{align*} &\left\|\Pi_k(f,g)\right\|_{L^{1}([0, 1])}\\&\quad\leq \|g\|_{L^{\infty}(\mathbb{R})}\int_{0}^{1}\int_{0}^{1} f(x-t)\psi\left(2^{k}(t-t_{0})\right)\,\textrm{d}t\,\textrm{d}x \lesssim \|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\infty}(\mathbb{R})}. \end{align*} $$

   2. ∘ The $L^{\infty }(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^1([0, 1])$ boundedness of $\Pi _k(f,g)$ . One obtains

      $$ \begin{align*} &\left\|\Pi_k(f,g)\right\|_{L^{1}([0, 1])}\\&\quad\leq \|f\|_{L^{\infty}(\mathbb{R})}\int_{0}^{1}\int_{0}^{1}g(x-\gamma(t))\psi\left(2^{k}(t-t_{0})\right)\,\textrm{d}t\,\textrm{d}x \lesssim \|f\|_{L^{\infty}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align*} $$

   3. ∘ The $L^1(\mathbb {R})\times L^1(\mathbb {R}) \rightarrow L^{\frac {1}{2}}([0, 1])$ boundedness of $\Pi _k(f,g)$ . From ${\frac {1}{2}}\cdot 2^{-k}\leq |t-t_0|\leq 2\cdot 2^{-k}$ , $2^{-k+2}> t_0$ and $0<t<1$ , which means that $0<t\leq \frac {7}{8}t_0$ or $ \frac {9}{8}t_0\leq t<1$ . Furthermore, by the facts that $\gamma '(t_0)=1$ and $|\gamma '(t)|$ is strictly increasing on ${t\in \mathbb {R}^+}$ , the change of variables $u=x-t$ and $v=x-\gamma (t)$ implies that

      $$ \begin{align*} \left|{\frac{\partial(u,v)}{\partial(x,t)}}\right|&=|1-\gamma'(t)|=|\gamma'(t)-\gamma'(t_{0})|\\ &\geq \begin{cases}|\gamma'(t)|-|\gamma'(t_{0})|\geq \left|\gamma'\left(\frac{9}{8}t_0\right)\right|-\left|\gamma'(t_{0})\right|\gtrsim 1, ~~~\textrm{if} ~\frac{9}{8}t_0\leq t<1;\\ |\gamma'(t_{0})|-|\gamma'(t)|\geq \left|\gamma'(t_{0})\right|-\left|\gamma'\left(\frac{7}{8}t_0\right)\right|\gtrsim 1, ~~~\textrm{if} ~0<t\leq \frac{7}{8}t_0. \end{cases} \end{align*} $$
      Hence, we have
      $$ \begin{align*} \left\|\Pi_k(f,g)\right\|_{L^{1}([0, 1])}&\lesssim \int_{\mathbb{R}}\int_{\mathbb{R}} f(u)g(v)\left|{\frac{\partial(x,t)}{\partial(u,v)}}\right|\,\textrm{d}u\textrm{d}v\lesssim \|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align*} $$
      This, together with the Cauchy–Schwartz inequality, leads to
      $$ \begin{align*} \left\|\Pi_k(f,g)\right\|_{L^{\frac{1}{2}}([0, 1])}\lesssim\left\|\Pi_k(f,g)\right\|_{L^{1}([0, 1])}\lesssim \|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align*} $$

Therefore, we finish the proof of the Proposition 3.2.

Proof For $H_{\alpha ,\gamma ,j}$ defined in (3.1), let E be a measurable set with finite measure, a calculation gives

$$ \begin{align*} \left \|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{1}(\mathbb{R})} \leq\|g\|_{L^{\infty}(\mathbb{R})}\int_{E}\int_{2^{j-1}}^{2^j}f(x-t)\,\textrm{d}t\,\textrm{d}x \leq|E|\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\infty}(\mathbb{R})} \end{align*} $$

for all $j\in \mathbb {Z}$ . This, combined with Propositions 3.1 and 3.2, shows that

(4.2) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{1}(\mathbb{R})}\lesssim \min\left\{2^j,|E|\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\infty}(\mathbb{R})} \end{align} $$

for all $j\in \mathbb {Z}$ . From (3.4) and (3.5), we can write

(4.3) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim \min\left\{2^j,|E|\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

for all $j<j_0$ , and by (3.6) and (3.7), it follows that

(4.4) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{2}}(\mathbb{R})}\lesssim \min\left\{2^j,\frac{|E|}{|\gamma'(2^{j-1})|}\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

for all $j>j_0+2$ .

Now, for any fixed $f\in L^1(\mathbb {R})$ , the operator $H_{\alpha ,\gamma ,j}(f,g)\chi _{E}$ may be viewed as a linear operator about g. Interpolation (4.2) with (4.3) implies that

(4.5) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_1}}(\mathbb{R})}\lesssim \min\left\{2^j,|E|\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})} \end{align} $$

for all $j<j_0$ with $\theta _1\in (0,1)$ . From Proposition 3.2, reiterating this interpolation process, we also have

(4.6) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_1}}(\mathbb{R})}\lesssim \min\left\{2^{j_0},2^{{j_0}\theta_{1}}|E|^{1-\theta_{1}}\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})} \end{align} $$

for all $j=j_0, j_0+1, j_0+2$ . Using the interpolation argument on (4.2) with (4.4), we claim that

(4.7) $$ \begin{align} &\left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_1}}(\mathbb{R})}\\&\quad\lesssim \min\left\{2^j,\frac{2^{j(1-\theta_{1})}|E|^{\theta_1}}{|\gamma'(2^{j-1})|^{\theta_{1}}},2^{j\theta_1} |E|^{1-\theta_1},\frac{|E|}{|\gamma'(2^{j-1})|^{\theta_1}}\right\}\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})} \nonumber\end{align} $$

for all $j>j_0+2$ .

Now pick $E_{\lambda }:=\{x\in {\mathbb {R}}:\ |H_{\alpha ,\gamma }(f,g)(x)|>\lambda \}$ with nonnegative functions $f, g \in L^1(\mathbb {R}) \cap L^{\infty }(\mathbb {R})$ . One may use Chebyshev’s inequality and (4.1), (4.5)–(4.7) to deduce that

(4.8) $$ \begin{align} \lambda^{\frac{1}{1+\theta_{1}}}|E_{\lambda}|&\lesssim \int_{E_{\lambda}}\left(\sum\limits_{j\in\mathbb{Z}}2^{(\alpha-1)j}H_{\alpha,\gamma,j}(f,g)(x)\right)^{\frac{1}{1+\theta_{1}}}\,\textrm{d}x \lesssim\sum\limits_{j\in\mathbb{Z}}2^{{\frac{\alpha-1}{1+\theta_{1}}}j}\int_{E_{\lambda}}H_{\alpha,\gamma,j}(f,g)(x))^{\frac{1}{1+\theta_{1}}}\,\textrm{d}x\\ \nonumber &\lesssim \Delta_{1}+\Delta_{2}+\Delta_{3}, \end{align} $$

where

$$ \begin{align*}\begin{cases} \Delta_{1}:=\sum\limits_{j<j_{0}}2^{{\frac{\alpha-1}{1+\theta_{1}}j}}\min\left\{2^j,|E_\lambda|\right\}^{\frac{1}{1+\theta_{1}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}; \\ \Delta_{2}:=2^{{\frac{\alpha-1}{1+\theta_{1}}}j_{0}}\min\left\{2^{j_0},2^{{j_0}\theta_{1}}|E_\lambda|^{1-\theta_{1}}\right\}^{\frac{1}{1+\theta_{1}}}\|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}};\\ \Delta_{3}:=\sum\limits_{j>j_{0}+2}2^{{\frac{\alpha-1}{1+\theta_{1}}}j}\min\left\{2^j,\frac{2^{j(1-\theta_{1})}|E_\lambda|^{\theta_1}}{|\gamma'(2^{j-1})|^{\theta_{1}}},2^{j\theta_1} |E_\lambda|^{1-\theta_1},\frac{|E_\lambda|}{|\gamma'(2^{j-1})|^{\theta_1}}\right\}^{\frac{1}{1+\theta_{1}}}\|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}. \end{cases}\end{align*} $$

These three terms are treated separately in the following section. We are ready to calculate $\Delta _{1}$ , which can be written as

$$ \begin{align*} \left[\sum\limits_{j<j_{0},~{|E_{\lambda}|\geq2^j}}2^{{\frac{\alpha}{1+\theta_{1}}}j}+ \sum\limits_{j<j_{0},~{|E_{\lambda}|<2^j}}2^{{\frac{\alpha-1}{1+\theta_{1}}}j}|E_\lambda|^{\frac{1}{1+\theta_{1}}}\right] \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}. \end{align*} $$

Notice that

$$ \begin{align*}\sum\limits_{j<j_{0},~{E_{\lambda}\geq2^j}}2^{{\frac{\alpha}{1+\theta_{1}}}j}\lesssim \begin{cases} 2^{{\frac{\alpha}{1+\theta_{1}}}{j_0}} , & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha}{1+\theta_{1}}}, & |E_\lambda|<2^{j_0}, \end{cases}\end{align*} $$

and

$$ \begin{align*}\sum\limits_{j<j_{0},~{|E_{\lambda}|<2^j}}2^{{\frac{\alpha-1}{1+\theta_{1}}}j}|E_\lambda|^{\frac{1}{1+\theta_{1}}}\lesssim \begin{cases} 0, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha}{1+\theta_{1}}}, & |E_\lambda|<2^{j_0}. \end{cases}\end{align*} $$

Hence we have the estimate

(4.9) $$ \begin{align} \Delta_{1}\lesssim \begin{cases} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha}{1+\theta_{1}}}\|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|<2^{j_0}. \end{cases} \end{align} $$

Now we turn to the second term $\Delta _{2}$ , it is easy to see that

(4.10) $$ \begin{align} \Delta_{2}\lesssim \begin{cases} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{1-\theta_{1}}{1+\theta_{1}}}\|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|<2^{j_0}. \end{cases} \end{align} $$

Our next goal is to consider $\Delta _{3}$ , a calculation gives

$$ \begin{align*} \min\left\{2^j,\frac{2^{j(1-\theta_{1})}|E_\lambda|^{\theta_1}}{|\gamma'(2^{j-1})|^{\theta_{1}}},2^{j\theta_1} |E_\lambda|^{1-\theta_1},\frac{|E_\lambda|}{|\gamma'(2^{j-1})|^{\theta_1}}\right\}= \begin{cases} 2^j, & |E_\lambda|\geq2^j|\gamma'(2^{j-1})|; \\ \frac{2^{j(1-\theta_{1})}|E_\lambda|^{\theta_1}}{|\gamma'(2^{j-1})|^{\theta_{1}}},&2^j\leq|E_\lambda|<2^j|\gamma'(2^{j-1})|;\\ \frac{|E_\lambda|}{|\gamma'(2^{j-1})|^{\theta_1}} , & |E_\lambda|<2^j. \end{cases} \end{align*} $$

We can then write $\Delta _{3}=\Delta _{3,1}+\Delta _{3,2}+\Delta _{3,3}$ , where

$$ \begin{align*}\begin{cases} \Delta_{3,1}:= \sum\limits_{ \substack{j>j_{0}+2 \\|E_\lambda|\geq2^j|\gamma'(2^{j-1})| }} 2^{{\frac{\alpha}{1+\theta_{1}}}j} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}; \\ \Delta_{3,2}:=\sum\limits_{ \substack{j>j_{0}+2 \\2^j\leq|E_\lambda|<2^j|\gamma'(2^{j-1})| }} {\frac{2^{{\frac{\alpha-\theta_{1}}{1+\theta_{1}}}j}|E_\lambda|^{\frac{\theta_{1}}{1+\theta_{1}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{1}}{1+\theta_{1}}}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}};\\ \Delta_{3,3}:=\sum\limits_{ \substack{j>j_{0}+2 \\|E_\lambda|<2^j}} {\frac{2^{{\frac{\alpha-1}{1+\theta_{1}}}j}|E_\lambda|^{\frac{1}{1+\theta_{1}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{1}}{1+\theta_{1}}}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}. \end{cases}\end{align*} $$

Corresponding to $\Delta _{3,1}$ , for any $\varepsilon>0$ , it follows from Lemma 2.1 that

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\|E_\lambda|\geq2^j|\gamma'(2^{j-1})| }} 2^{{\frac{\alpha}{1+\theta_{1}}}j}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha}{(\omega_{\infty,1}-\varepsilon)(1+\theta_{1})}}, & |E_\lambda|\geq2^{j_0+2}|\gamma'(2^{j_0+1})|; \\ 0, & 2^{j_0+2}\leq|E_{\lambda}|<2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ 0, &|E_{\lambda}|<2^{j_0+2}. \end{cases}\end{align*} $$

On the other hand, let us set $\varepsilon \in (0,\omega _{\infty ,1})$ , and we take that $\alpha <\theta _{1}(\omega _{\infty ,1}-\varepsilon )$ , then by Lemma 2.1 one has

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\2^j\leq|E_\lambda|<2^j|\gamma'(2^{j-1})| }} {\frac{2^{{\frac{\alpha-\theta_{1}}{1+\theta_{1}}}j}|E_\lambda|^{\frac{\theta_{1}}{1+\theta_{1}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{1}}{1+\theta_{1}}}}} \lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}}}, & |E_\lambda|\geq2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ |E_\lambda|^{{\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}}}, & 2^{j_0+2}\leq|E_{\lambda}|<2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ 0, &|E_{\lambda}|<2^{j_0+2}. \end{cases}\end{align*} $$

This is the estimate about $\Delta _{3,2}$ . Furthermore, for estimating $\Delta _{3,3}$ , we need to take $\alpha -1+\theta _{1}>0$ . It follows from $\omega _{\infty ,1}>1$ that $\alpha -1+\theta _{1}-\theta _{1}(\omega _{\infty ,1}-\varepsilon )<0$ for all ${\varepsilon \in (0,\omega _{\infty ,1}-1]}$ . From Lemma 2.1, we then have

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\|E_\lambda|<2^j}} {\frac{2^{{\frac{\alpha-1}{1+\theta_{1}}}j}|E_\lambda|^{\frac{1}{1+\theta_{1}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{1}}{1+\theta_{1}}}}}\lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_{1}}}}, & |E_\lambda|\geq2^{j_0+2}|\gamma'(2^{j_{0}+1})|; \\ |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_{1}}}}, & 2^{j_0+2}\leq|E_{\lambda}|<2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ |E_\lambda|^{{\frac{1}{1+\theta_1}}}, &|E_{\lambda}|<2^{j_0+2}. \end{cases}\end{align*} $$

Notice that

$$ \begin{align*}|E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_{1}}}}\leq|E_\lambda|^{\frac{\alpha}{(\omega_{\infty,1}-\varepsilon)(1+\theta_{1})}} \leq |E_\lambda|^{{\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}}}\end{align*} $$

holds for $|E_\lambda |\geq 2^{j_0+2}$ if $\varepsilon \in (0,\min \{\omega _{\infty ,1}, \omega _{\infty ,1}-1\})$ . Consequently,

(4.11) $$ \begin{align} \Delta_{3}\lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0+2};\\ |E_\lambda|^{{\frac{1}{1+\theta_{1}}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, &|E_{\lambda}|<2^{j_0+2}. \end{cases} \end{align} $$

It follows from $\alpha -1+\theta _{1}>0$ and $\theta _1\in (0,1)$ that $ |E_\lambda |^{{\frac {1}{1+\theta _{1}}}}\leq |E_\lambda |^{\frac {\alpha }{1+\theta _{1}}}\leq |E_\lambda |^{\frac {1-\theta _{1}}{1+\theta _{1}}}$ holds for $|E_{\lambda }|<2^{j_0+2}$ . This, combined with (4.8)–(4.11), we conclude that

(4.12) $$ \begin{align} \lambda^{\frac{1}{1+\theta_{1}}}|E_{\lambda}|\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0+2}; \\ |E_\lambda|^{\frac{1-\theta_{1}}{1+\theta_{1}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, &|E_{\lambda}|<2^{j_0+2}. \end{cases} \end{align} $$

By a calculation, for $\varepsilon $ small enough, we have that $\frac {\omega _{\infty ,2}-\omega _{\infty ,1}}{\omega _{\infty ,2}-1}<\frac {\alpha }{1-\alpha }<\frac {\omega _{\infty ,2}}{\omega _{\infty ,2}-\omega _{\infty ,1}+1}$ implies

(4.13) $$ \begin{align} \max\left\{1-\alpha, \alpha, \frac{\alpha}{\omega_{\infty,1}-\varepsilon}\right\}<\frac{\omega_{\infty,2}-\alpha+\varepsilon}{2\omega_{\infty,2}-\omega_{\infty,1}+3\varepsilon}. \end{align} $$

Then, we can take

(4.14) $$ \begin{align} \theta_1\leq \frac{\omega_{\infty,2}-\alpha+\varepsilon}{2\omega_{\infty,2}-\omega_{\infty,1}+3\varepsilon} \end{align} $$

with $\varepsilon $ small enough, which further implies that

$$ \begin{align*}\frac{\alpha+\theta_1(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(\omega_{\infty,2}+\varepsilon)(1+\theta_{1})}\leq \frac{1-\theta_{1}}{1+\theta_{1}},\end{align*} $$

This, together with (4.12), gives

(4.15) $$ \begin{align} \lambda|E_{\lambda}|^{\frac{\omega_{\infty,2}+ \theta_1(\omega_{\infty,1}-\varepsilon)-\alpha+\varepsilon}{\omega_{\infty,2}+\varepsilon}}\lesssim \|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})} \end{align} $$

for all $0<|E_\lambda |<\infty $ , where $\varepsilon $ small enough and $\theta _1$ satisfies (4.14). Therefore, we have established the $L^{1}(\mathbb {R})\times L^{\frac {1}{\theta _{1}}}(\mathbb {R})\rightarrow L^{\frac {\omega _{\infty ,2}+\varepsilon }{\omega _{\infty ,2}+ \theta _1(\omega _{\infty ,1}-\varepsilon )-\alpha +\varepsilon }, \infty }(\mathbb {R})$ boundedness for $H_{\alpha ,\gamma }$ .

It is easy to see that $\frac {\omega _{\infty ,2}+ \theta _1(\omega _{\infty ,1}-\varepsilon )-\alpha +\varepsilon }{\omega _{\infty ,2}+\varepsilon }$ takes the maximal value $\frac {2(\omega _{\infty ,2}-\alpha +\varepsilon )}{2\omega _{\infty ,2}-\omega _{\infty ,1}+3\varepsilon }$ when $\theta _1= \frac {\omega _{\infty ,2}-\alpha +\varepsilon }{2\omega _{\infty ,2}-\omega _{\infty ,1}+3\varepsilon }$ . On the other hand, from (4.13), it follows $\frac {\alpha }{\omega _{\infty ,1}-\varepsilon }<\frac {\omega _{\infty ,2}-\alpha +\varepsilon }{2\omega _{\infty ,2}-\omega _{\infty ,1}+3\varepsilon }$ , which equals to $2\alpha +\varepsilon <\omega _{\infty ,1}$ , we then have $2\alpha <\omega _{\infty ,1}$ . This, together with $\alpha <1<\omega _{\infty ,1}\leq \omega _{\infty ,2}$ , shows $3\alpha <\omega _{\infty ,1}+\omega _{\infty ,2}$ . Therefore,

$$ \begin{align*}\frac{\omega_{\infty,2}-\alpha+\varepsilon}{2\omega_{\infty,2}-\omega_{\infty,1}+3\varepsilon}< \frac{\omega_{\infty,2}-\alpha}{2\omega_{\infty,2}-\omega_{\infty,1}}\end{align*} $$

holds for all $\varepsilon $ small enough. Putting things together, we conclude that $H_{\alpha ,\gamma }$ maps $L^{1}(\mathbb {R})\times L^{2r( \varepsilon ) }(\mathbb {R})$ to $L^{r(\varepsilon ), \infty }(\mathbb {R})$ , where $\frac {1}{r( \varepsilon ) }:=\frac {2(\omega _{\infty ,2}-\alpha )}{2\omega _{\infty ,2}-\omega _{\infty ,1}}-\varepsilon $ with $\varepsilon $ small enough. Therefore, we finish the proof of Proposition 4.1.

Proof The main difference here lies in the establishing of $\Delta _{3,2}$ . In fact, let ${\varepsilon \in (0,\omega _{\infty ,1})}$ , and we take $\alpha>\theta _{1}(\omega _{\infty ,1}-\varepsilon )$ . By Lemma 2.1, one immediately has

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\2^j\leq|E_\lambda|<2^j|\gamma'(2^{j-1})| }} {\frac{2^{{\frac{\alpha-\theta_{1}}{1+\theta_{1}}}j}|E_\lambda|^{\frac{\theta_{1}}{1+\theta_{1}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{1}}{1+\theta_{1}}}}} \lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_1}}}, & |E_\lambda|\geq2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_1}}}, & 2^{j_0+2}\leq|E_{\lambda}|<2^{j_0+2}|\gamma'(2^{j_{0}+1})|;\\ 0, &|E_{\lambda}|<2^{j_0+2}. \end{cases}\end{align*} $$

The estimation of $\Delta _{3,3}$ is the same as in Proposition 4.1, and we also need to take ${\alpha -1+\theta _{1}>0}$ , which can be realized by the condition that $\frac {\alpha }{1-\alpha }\geq \omega _{\infty ,1}>1$ . Notice that $\alpha>\theta _{1}(\omega _{\infty ,1}-\varepsilon )$ , which implies $|E_\lambda |^{\frac {\alpha }{(\omega _{\infty ,1}-\varepsilon )(1+\theta _{1})}}\leq |E_\lambda |^{{\frac {\alpha -\theta _1(\omega _{\infty ,1}-1-\varepsilon )}{1+\theta _1}}}$ for $|E_\lambda |\geq 2^{j_0+2}$ . Thus one can write

$$ \begin{align*} \Delta_{3}\lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_1}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0+2}; \\ |E_\lambda|^{{\frac{1}{1+\theta_{1}}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, &|E_{\lambda}|<2^{j_0+2}. \end{cases} \end{align*} $$

As in Proposition 4.1, we have an estimate

(4.16) $$ \begin{align} \lambda^{\frac{1}{1+\theta_{1}}}|E_{\lambda}|\lesssim \begin{cases} |E_\lambda|^{{\frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_1}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, & |E_\lambda|\geq2^{j_0+2}; \\ |E_\lambda|^{\frac{1-\theta_{1}}{1+\theta_{1}}} \|f\|_{L^{1}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})}^{\frac{1}{1+\theta_{1}}}, &|E_{\lambda}|<2^{j_0+2}. \end{cases} \end{align} $$

We now want to obtain that

(4.17) $$ \begin{align} \frac{\alpha-\theta_1(\omega_{\infty,1}-1-\varepsilon)}{1+\theta_1}\leq \frac{1-\theta_{1}}{1+\theta_{1}}, \end{align} $$

which equals to $\theta _1(2-\omega _{\infty ,1}+\varepsilon )\leq 1-\alpha $ . If $2-\omega _{\infty ,1}+\varepsilon \leq 0$ , (4.17) is obvious. If $2-\omega _{\infty ,1}+\varepsilon>0$ , (4.17) equals to

(4.18) $$ \begin{align} \theta_1\leq\frac{1-\alpha}{2-\omega_{\infty,1}+\varepsilon}. \end{align} $$

Clearly $1-\alpha \leq \frac {1-\alpha }{2-\omega _{\infty ,1}+\varepsilon }$ holds for all $\varepsilon \in (0,\omega _{\infty ,1}-1]$ , which shows (4.18) can be realized. This, combined with (4.16) and (4.17), means that

$$ \begin{align*} \lambda|E_{\lambda}|^{1-\alpha+\theta_1(\omega_{\infty,1}-\varepsilon)}\lesssim\|f\|_{L^{1}(\mathbb{R})}\|g\|_{L^{\frac{1}{\theta_1}}(\mathbb{R})} \end{align*} $$

for all $0<|E_\lambda |<\infty $ , where

(4.19) $$ \begin{align} \begin{cases} \theta_1\leq\frac{1-\alpha}{2-\omega_{\infty,1}+\varepsilon}, &\omega_{\infty,1}<2\alpha+\varepsilon;\\ \theta_1< \frac{\alpha}{\omega_{\infty,1}-\varepsilon}, & 2\alpha+\varepsilon\leq\omega_{\infty,1}, \end{cases} \end{align} $$

and $\varepsilon>0$ small enough. Therefore, we have established the $L^{1}(\mathbb {R})\times L^{\frac {1}{\theta _{1}}}(\mathbb {R})\rightarrow L^{\frac {1}{1-\alpha +\theta _1(\omega _{\infty ,1}-\varepsilon )}, \infty }(\mathbb {R})$ boundedness for $H_{\alpha ,\gamma }$ .

On the other hand, by (4.19), for $\varepsilon>0$ small enough, one easily sees that

$$ \begin{align*} \begin{cases} 1-\alpha+\theta_1(\omega_{\infty,1}-\varepsilon)\leq\frac{2(1-\alpha)}{2-\omega_{\infty,1}+\varepsilon}<\frac{2(1-\alpha)}{2-\omega_{\infty,1}}, &\omega_{\infty,1}<2\alpha+\varepsilon;\\ 1-\alpha+\theta_1(\omega_{\infty,1}-\varepsilon)< 1, & 2\alpha+\varepsilon\leq\omega_{\infty,1}. \end{cases} \end{align*} $$

Therefore, we conclude that $H_{\alpha ,\gamma }$ maps $L^{1}(\mathbb {R})\times L^{\bar {q}( \varepsilon ) }(\mathbb {R})$ to $L^{\bar {r}(\varepsilon ), \infty }(\mathbb {R})$ with $\varepsilon>0$ small enough. Here, $\frac {1}{\bar {r}( \varepsilon )}$ tends to maximal value as $\varepsilon \rightarrow 0$ .

Notice that $\frac {\alpha }{\omega _{\infty ,1}-\varepsilon }$ increases as $\varepsilon $ increases when $2\alpha +\varepsilon \leq \omega _{\infty ,1}$ , then $\frac {1}{\bar {q}( \varepsilon )}$ doesn’t tend to maximal value as $\varepsilon \rightarrow 0$ . We now want $\frac {1}{\bar {q}( \varepsilon )}$ to take maximal value. In fact, from the process of our proofs, we can find that $\varepsilon $ need only satisfies $\varepsilon \leq \omega _{\infty ,1}-2\alpha $ and $\varepsilon \leq \omega _{\infty ,1}-1$ when $\omega _{\infty ,1}>2\alpha $ . Moreover, using $2\alpha <\omega _{\infty ,1}\leq \frac {\alpha }{1-\alpha }$ , we have $\frac {1}{2}<\alpha $ . Therefore, we should only to take $\varepsilon \leq \omega _{\infty ,1}-2\alpha $ . It is easy to check that $\frac {1}{\bar {q}( \varepsilon )}$ tends to the maximal value $\frac {1}{2}$ when $\varepsilon =\omega _{\infty ,1}-2\alpha $ , and $\frac {1}{\bar {r}(\varepsilon )}$ tends to $1$ . Then, $H_{\alpha ,\gamma }$ maps $L^{1}(\mathbb {R})\times L^{\frac {2}{1-\varepsilon }}(\mathbb {R})$ to $L^{\frac {1}{1-\varepsilon }, \infty }(\mathbb {R})$ with $\varepsilon>0$ small enough when $\omega _{\infty ,1}>2\alpha $ .

Putting things together, we finish the proof of Proposition 4.3.

Proof Let E be a measurable set with finite measure, and recall that $H_{\alpha ,\gamma ,j}$ defined in (3.1) with $j\in \mathbb {Z}$ , by the fact that $|\gamma '(t)|$ is strictly increasing on $t\in \mathbb {R}^+$ , a routine calculation gives

$$ \begin{align*} &\left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{1}(\mathbb{R})} \\&\quad\leq\|f\|_{L^{\infty}(\mathbb{R})}\int_E\int_{2^{j-1}}^{2^j}g(x-\gamma(t))\,\textrm{d}t\,\textrm{d}x\leq \frac{|E|}{|\gamma'(2^{j-1})|} \|f\|_{L^{\infty}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align*} $$

Combining Propositions 3.1 and 3.2, for any $j\in \mathbb {Z}$ , we immediately have

(5.1) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{1}(\mathbb{R})} \leq \min\left\{2^j, \frac{|E|}{|\gamma'(2^{j-1})|} \right\} \|f\|_{L^{\infty}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align} $$

Notice that $H_{\alpha ,\gamma ,j}(f,g)\chi _{E}$ can be viewed as a linear operator about f for any fixed $g\in L^1(\mathbb {R})$ , by interpolation between (5.1) with (4.3), we conclude that

$$ \begin{align*} &\left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_2}}(\mathbb{R})} \\&\quad\lesssim \min\left\{2^j,2^{j(1-\theta_{2})}|E|^{\theta_{2}},{\frac{2^{j\theta_2}|E|^{1-\theta_{2}}}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}} ,{\frac{|E|}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}}\right\}\|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align*} $$

for all $j<j_0$ with $\theta _2\in (0,1)$ . From Proposition 3.2, reiterating this interpolation process, then the following estimate is valid:

$$ \begin{align*} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_1}}(\mathbb{R})}\lesssim \min\left\{2^{j_0},\frac{2^{{j_0}\theta_{2}}|E|^{1-\theta_{2}}}{ |\gamma'(2^{j_0-1})|^{1-\theta_{2}} }\right\}\|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align*} $$

for all $j=j_0, j_0+1, j_0+2$ . A further interpolation between (5.1) with (4.4) gives

(5.2) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_2}}(\mathbb{R})} \leq \min\left\{2^j, \frac{|E|}{|\gamma'(2^{j-1})|} \right\} \|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

for all $j>j_0+2$ . Moreover, in order to establish the desired estimate, we bound

(5.3) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_2}}(\mathbb{R})} \leq \min\left\{2^j, \frac{2^{j(1-\theta_2)}|E|^{\theta_2}}{|\gamma'(2^{j-1})|^{\theta_2}} \right\} \|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

for all $j>j_0+2$ .

Recall that $E_{\lambda }=\{x\in {\mathbb {R}}:\ |H_{\alpha ,\gamma }(f,g)(x)|>\lambda \}$ . Then, by a similar argument as in (4.8), we conclude that

(5.4) $$ \begin{align} \lambda^{\frac{1}{1+\theta_2}}|E_{\lambda}|\lesssim \Upsilon_{1}+\Upsilon_{2}+\Upsilon_{3}, \end{align} $$

where

$$ \begin{align*}\begin{cases} \Upsilon_{1}:=\sum\limits_{j<j_{0}}2^{{\frac{\alpha-1}{1+\theta_{2}}j}}\min\left\{2^j,2^{j(1-\theta_{2})}|E_{\lambda}|^{\theta_{2}},{\frac{2^{j\theta_2}|E_{\lambda}|^{1-\theta_{2}}}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}} ,{\frac{|E_{\lambda}|}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}}\right\}^{\frac{1}{1+\theta_2}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}; \\ \Upsilon_{2}:=2^{{\frac{\alpha-1}{1+\theta_{2}}}j_{0}}\min\left\{2^{j_0},\frac{2^{{j_0}\theta_{2}}|E_{\lambda}|^{1-\theta_{2}}}{ |\gamma'(2^{j_0-1})|^{1-\theta_{2}} }\right\}^{\frac{1}{1+\theta_2}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})};\\ \Upsilon_{3}:=\sum\limits_{j>j_{0}+2}2^{{\frac{\alpha-1}{1+\theta_{2}}}j}\min\left\{2^j, \frac{2^{j(1-\theta_2)}|E_{\lambda}|^{\theta_2}}{|\gamma'(2^{j-1})|^{\theta_2}} \right\}^{\frac{1}{1+\theta_2}} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}. \end{cases}\end{align*} $$

Consider $\Upsilon _1$ , in analogy to the calculation in $\Delta _{3}$ , and observe that

$$ \begin{align*} &\min\left\{2^j,2^{j(1-\theta_{2})}|E_{\lambda}|^{\theta_{2}},{\frac{2^{j\theta_2}|E_{\lambda}|^{1-\theta_{2}}}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}} ,{\frac{|E_{\lambda}|}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}}\right\}\\&\quad\leq \begin{cases} 2^j, & |E_{\lambda}|\geq 2^j; \\ 2^{j(1-\theta_{2})}|E_{\lambda}|^{\theta_{2}}, & 2^j{|\gamma'(2^{j-1})|}\leq|E_{\lambda}|< 2^j; \\ {\frac{|E_{\lambda}|}{|\gamma'(2^{j-1})|^{1-\theta_{2}}}}, & |E_{\lambda}|<2^j|\gamma'(2^{j-1})|. \end{cases} \end{align*} $$

Hence we can write

$$ \begin{align*} \Upsilon_{1}=\Upsilon_{1,1}+\Upsilon_{1,2}+\Upsilon_{1,3} \end{align*} $$

with

$$ \begin{align*}\begin{cases} \Upsilon_{1,1}:=\sum\limits_{ \substack{j<j_{0} \\ |E_{\lambda}|\geq2^j }} 2^{{\frac{\alpha}{1+\theta_{2}}j}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}; \\ \Upsilon_{1,2}:=\sum\limits_{ \substack{j<j_{0} \\ 2^j{|\gamma'(2^{j-1})|}\leq|E_{\lambda}|<2^j }} 2^{{\frac{\alpha-\theta_{2}}{1+\theta_{2}}}j}|E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}; \\ \Upsilon_{1,3}:=\sum\limits_{ \substack{j<j_{0} \\ |E_{\lambda}|<2^j|\gamma'(2^{j-1})| }} \frac{2^{{\frac{\alpha-1}{1+\theta_{2}}j}}|E_\lambda|^{\frac{1}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}. \end{cases}\end{align*} $$

Let’s now consider $\Upsilon _{1,1}$ . For any $\varepsilon>0$ , it follows from Lemma 2.1 that

$$ \begin{align*}\sum\limits_{ \substack{j<j_{0} \\ |E_{\lambda}|\geq2^j }} 2^{{\frac{\alpha}{1+\theta_{2}}j}}\lesssim \begin{cases} 2^{\frac{\alpha}{1+\theta_{2}}j_0}, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}, & 2^{j_0}|\gamma'(2^{j_{0}-1})|\leq|E_{\lambda}|<2^{j_0};\\ |E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

To see what happens corresponding to $\Upsilon _{1,2}$ and $\Upsilon _{1,3}$ , due to a minor technical issue we will assume that $\alpha <\theta _2$ . For any $\varepsilon>0$ , by Lemma 2.1, it is easy to see that

$$ \begin{align*}\sum\limits_{ \substack{j<j_{0} \\ 2^j{|\gamma'(2^{j-1})|}\leq|E_{\lambda}|<2^j }} 2^{{\frac{\alpha-\theta_{2}}{1+\theta_{2}}}j}|E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}\lesssim \begin{cases} 0, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}, & 2^{j_0}|\gamma'(2^{j_{0}-1})|\leq|E_{\lambda}|<2^{j_0};\\ |E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|, \end{cases}\end{align*} $$

and

$$ \begin{align*}\sum\limits_{ \substack{j<j_{0} \\ |E_{\lambda}|<2^j|\gamma'(2^{j-1})| }} \frac{2^{{\frac{\alpha-1}{1+\theta_{2}}j}}|E_\lambda|^{\frac{1}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}\lesssim \begin{cases} 0, & |E_\lambda|\geq2^{j_0}; \\ 0, & 2^{j_0}|\gamma'(2^{j_{0}-1})|\leq|E_{\lambda}|<2^{j_0};\\ |E_\lambda|^{{\frac{\alpha-\theta_{2}}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}+{\frac{\theta_{2}}{1+\theta_{2}}}}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

The above implies

(5.5) $$ \begin{align} \Upsilon_1\lesssim \begin{cases} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ \left(|E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}+|E_\lambda|^{{\frac{\alpha-\theta_{2}}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}+{\frac{\theta_{2}}{1+\theta_{2}}}}\right) \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

For $\Upsilon _2$ , we obviously have the estimate

(5.6) $$ \begin{align} \Upsilon_{2}\lesssim \begin{cases} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_{\lambda}|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

For $\Upsilon _3$ , which can be written as

$$ \begin{align*}\left[\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|\geq2^j{|\gamma'(2^{j-1})|} }} 2^{{\frac{\alpha}{1+\theta_{2}}}j}+ \sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|<2^j{|\gamma'(2^{j-1})|} }} \frac{2^{{\frac{\alpha-\theta_{2}}{1+\theta_{2}}}j} |E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{2}}{1+\theta_{2}}}}\right] \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}.\end{align*} $$

For any $\varepsilon>0$ , by Lemma 2.1, we can find

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|\geq2^j{|\gamma'(2^{j-1})|} }} 2^{{\frac{\alpha}{1+\theta_{2}}}j}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha}{(1+\theta_{2})(\omega_{\infty,1}-\varepsilon)}}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|;\\ 0, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

Moreover, for $\varepsilon \in (0, \omega _{\infty ,1}-1]$ , by Lemma 2.1, then the following estimate is valid:

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|<2^j{|\gamma'(2^{j-1})|} }} \frac{2^{{\frac{\alpha-\theta_{2}}{1+\theta_{2}}}j} |E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{\theta_{2}}{1+\theta_{2}}}}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+\theta_2(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

Because of $|E_\lambda |^{\frac {\alpha }{(1+\theta _{2})(\omega _{\infty ,1}-\varepsilon )}}< |E_\lambda |^{\frac {\alpha +\theta _{2}(\omega _{\infty ,2}-\omega _{\infty ,1}+2\varepsilon )}{(1+\theta _{2})(\omega _{\infty ,2}+\varepsilon )}}$ holding for $|E_\lambda |\geq 2^{j_0}|\gamma '(2^{j_{0}-1})|$ , which comes from the fact that $\alpha <\theta _2(\omega _{\infty ,1}-\varepsilon )$ with $\varepsilon \in (0, \omega _{\infty ,1}-1]$ , we then arrive at

(5.7) $$ \begin{align} \Upsilon_{3}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+\theta_{2}(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

Notice that $\omega _{0,1}>1$ , we can set $\varepsilon \in (0, \omega _{0,1}-1]$ , and it follows from $\alpha <\theta _2$ that

$$ \begin{align*}|E_\lambda|^{{\frac{\alpha-\theta_{2}}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}+{\frac{\theta_{2}}{1+\theta_{2}}}}\leq|E_\lambda|^{\frac{\alpha}{1+\theta_{2}}} ~~\textrm{and} ~~|E_\lambda|^{\frac{\theta_2}{1+\theta_{2}}}<|E_\lambda|^{\frac{\alpha}{1+\theta_{2}}}\end{align*} $$

for $|E_{\lambda }|<2^{j_0}|\gamma '(2^{j_{0}-1})|$ . Furthermore, since $\alpha>1-\theta _2$ , one can check that $|E_\lambda |^{\frac {\alpha }{1+\theta _{2}}}\leq |E_\lambda |^{\frac {1-\theta _{2}}{1+\theta _{2}}}$ holds for $|E_{\lambda }|<2^{j_0}|\gamma '(2^{j_{0}-1})|$ directly. These estimates, combined with the estimates (5.4)–(5.7) and the fact that $ |E_\lambda |^{\frac {\alpha +\theta _{2}(\omega _{\infty ,2}-\omega _{\infty ,1}+2\varepsilon )}{(1+\theta _{2})(\omega _{\infty ,2}+\varepsilon )}}\geq 1$ for $|E_\lambda |\geq 2^{j_0}|\gamma '(2^{j_{0}-1})|$ , lead to

(5.8) $$ \begin{align} \lambda^{\frac{1}{1+\theta_2}}|E_{\lambda}|\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+\theta_{2}(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

Therefore, we obtain a similar estimate with (4.12). Here, we used (4.13). As in Proposition 4.1, we conclude that $H_{\alpha ,\gamma }$ maps $L^{2r( \varepsilon )}(\mathbb {R})\times L^{1 }(\mathbb {R})$ to $L^{r(\varepsilon ), \infty }(\mathbb {R})$ , where $\frac {1}{r( \varepsilon ) }=\frac {2(\omega _{\infty ,2}-\alpha )}{2\omega _{\infty ,2}-\omega _{\infty ,1}}-\varepsilon $ with $\varepsilon $ small enough. Therefore, we finish the proof of Proposition 5.1.

Proof Repeating the proof of Proposition 5.1, we then arrive at (5.4). Consider $\Upsilon _{1}$ , we now assume that $\alpha>\theta _2$ . For any $\varepsilon>0$ , by Lemma 2.1, one sees

$$ \begin{align*}\sum\limits_{ \substack{j<j_{0} \\ 2^j{|\gamma'(2^{j-1})|}\leq|E_{\lambda}|<2^j }} 2^{{\frac{\alpha-\theta_{2}}{1+\theta_{2}}}j}|E_\lambda|^{\frac{\theta_{2}}{1+\theta_{2}}}\lesssim \begin{cases} 0, & |E_\lambda|\geq2^{j_0}; \\ |E_\lambda|^{\frac{\alpha+\theta_2(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,2}+\varepsilon)(1+\theta_{2})}}, & 2^{j_0}|\gamma'(2^{j_{0}-1})|\leq|E_{\lambda}|<2^{j_0};\\ |E_\lambda|^{\frac{\alpha+\theta_2(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,2}+\varepsilon)(1+\theta_{2})}}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

Notice that $\alpha -\omega _{0,2}-\varepsilon +\theta _2(\omega _{0,2}-1+\varepsilon )<0$ , this along with Lemma 2.1 yields

$$ \begin{align*}\sum\limits_{ \substack{j<j_{0} \\ |E_{\lambda}|<2^j|\gamma'(2^{j-1})| }} \frac{2^{{\frac{\alpha-1}{1+\theta_{2}}j}}|E_\lambda|^{\frac{1}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}\lesssim \begin{cases} 0, & |E_\lambda|\geq2^{j_0}; \\ 0, & 2^{j_0}|\gamma'(2^{j_{0}-1})|\leq|E_{\lambda}|<2^{j_0};\\ |E_\lambda|^{\frac{\alpha+\omega_{0,1}-\omega_{0,2}-2\varepsilon+\theta_{2}(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

It is easy to check that

$$ \begin{align*}|E_\lambda|^{{\frac{\alpha}{1+\theta_{2}}}}\leq|E_\lambda|^{\frac{\alpha+\theta_2(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,2}+\varepsilon)(1+\theta_{2})}}\leq |E_\lambda|^{\frac{\alpha+\omega_{0,1}-\omega_{0,2}-2\varepsilon+\theta_{2}(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}\end{align*} $$

holds for $|E_{\lambda }|<2^{j_0}|\gamma '(2^{j_{0}-1})|$ . We then have

(5.10) $$ \begin{align} \Upsilon_1\lesssim \begin{cases} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|;\\ |E_\lambda|^{\frac{\alpha+\omega_{0,1}-\omega_{0,2}-2\varepsilon+\theta_{2}(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

$\Upsilon _2$ has the same estimate as in Proposition 5.1, and we turn to $\Upsilon _3$ . It is different from (5.3), we here need to bound (5.2) as

(5.11) $$ \begin{align} \left\|H_{\alpha,\gamma,j}(f,g)\chi_{E}\right\|_{L^{\frac{1}{1+\theta_2}}(\mathbb{R})} \leq \min\left\{2^j, \frac{2^{j\theta_2}|E|^{1-\theta_2}}{|\gamma'(2^{j-1})|^{1-\theta_2}} \right\} \|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})}. \end{align} $$

Then, we can write

$$ \begin{align*} \Upsilon_{3}=\sum\limits_{j>j_{0}+2}2^{{\frac{\alpha-1}{1+\theta_{2}}}j}\min\left\{2^j, \frac{2^{j\theta_2}|E_{\lambda}|^{1-\theta_2}}{|\gamma'(2^{j-1})|^{1-\theta_2}} \right\}^{\frac{1}{1+\theta_2}} \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, \end{align*} $$

which can be rewritten as

$$ \begin{align*}\left[\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|\geq2^j{|\gamma'(2^{j-1})|} }} 2^{{\frac{\alpha}{1+\theta_{2}}}j}+ \sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|<2^j{|\gamma'(2^{j-1})|} }} \frac{2^{{\frac{\alpha-1+\theta_{2}}{1+\theta_{2}}}j} |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}\right] \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}.\end{align*} $$

For any $\varepsilon>0$ , by Lemma 2.1, we also have

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|\geq2^j{|\gamma'(2^{j-1})|} }} 2^{{\frac{\alpha}{1+\theta_{2}}}j}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha}{(1+\theta_{2})(\omega_{\infty,1}-\varepsilon)}}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|;\\ 0, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

We assume that $\alpha <(1-\theta _2)(\omega _{\infty ,1}-\varepsilon )$ with $\varepsilon \in (0, \omega _{\infty ,1})$ , it follows from Lemma 2.1 that

$$ \begin{align*}\sum\limits_{ \substack{j>j_{0}+2 \\ |E_{\lambda}|<2^j{|\gamma'(2^{j-1})|} }} \frac{2^{{\frac{\alpha-1+\theta_{2}}{1+\theta_{2}}}j} |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}{|\gamma'(2^{j-1})|^{\frac{1-\theta_{2}}{1+\theta_{2}}}}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+(1-\theta_2)(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}}, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases}\end{align*} $$

Based on the fact $|E_\lambda |^{\frac {\alpha }{(1+\theta _{2})(\omega _{\infty ,1}-\varepsilon )}}< |E_\lambda |^{\frac {\alpha +(1-\theta _{2})(\omega _{\infty ,2}-\omega _{\infty ,1}+2\varepsilon )}{(1+\theta _{2})(\omega _{\infty ,2}+\varepsilon )}}$ holds for $|E_\lambda |\geq 2^{j_0}|\gamma '(2^{j_{0}-1})|$ , we deduce that

(5.12) $$ \begin{align} \Upsilon_{3}\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+(1-\theta_{2})(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|; \\ |E_\lambda|^{\frac{1-\theta_{2}}{1+\theta_{2}}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align} $$

Combining (5.10), (5.6), and (5.12), then the following estimate is valid:

$$ \begin{align*} \lambda^{\frac{1}{1+\theta_2}}|E_{\lambda}|\lesssim \begin{cases} |E_\lambda|^{\frac{\alpha+(1-\theta_{2})(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}}\|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, & |E_\lambda|\geq2^{j_0}|\gamma'(2^{j_{0}-1})|;\\ \left(|E_\lambda|^{\frac{\alpha+\omega_{0,1}-\omega_{0,2}-2\varepsilon+\theta_{2}(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})}}+|E_\lambda|^{\frac{1-\theta_2}{1+\theta_2}}\right) \|f\|^{\frac{1}{1+\theta_2}}_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|^{\frac{1}{1+\theta_2}}_{L^{1}(\mathbb{R})}, &|E_{\lambda}|<2^{j_0}|\gamma'(2^{j_{0}-1})|. \end{cases} \end{align*} $$

If we have

$$ \begin{align*}\frac{\alpha+(1-\theta_{2})(\omega_{\infty,2}-\omega_{\infty,1}+2\varepsilon)}{(1+\theta_{2})(\omega_{\infty,2}+\varepsilon)}< \frac{1-\theta_2}{1+\theta_2}\leq \frac{\alpha+\omega_{0,1}-\omega_{0,2}-2\varepsilon+\theta_{2}(\omega_{0,2}-1+\varepsilon)}{(\omega_{0,1}-\varepsilon)(1+\theta_{2})},\end{align*} $$

which equals to $\frac {\omega _{0,2}-\alpha +\varepsilon }{\omega _{0,2}+\omega _{0,1}-1}\leq \theta _2<\frac {\omega _{\infty ,1}-\alpha -\varepsilon }{\omega _{\infty ,1}-\varepsilon }$ for $\varepsilon>0$ small enough, we then obtain that

(5.13) $$ \begin{align} \lambda|E_{\lambda}|^{\frac{\omega_{\infty,1}-\alpha-\varepsilon+ \theta_2(2\omega_{\infty,2}- \omega_{\infty,1} +3\varepsilon)}{\omega_{\infty,2}+\varepsilon}}\lesssim \|f\|_{L^{\frac{1}{\theta_2}}(\mathbb{R})}\|g\|_{L^{1}(\mathbb{R})} \end{align} $$

holds for all $0<|E_\lambda |<\infty $ . Indeed, $\theta _2<\frac {\omega _{\infty ,1}-\alpha -\varepsilon }{\omega _{\infty ,1}-\varepsilon }$ comes from the assumption in estimating $\Upsilon _{3}$ . In order to show that we can take $\theta _2\geq \frac {\omega _{0,2}-\alpha +\varepsilon }{\omega _{0,2}+\omega _{0,1}-1}$ , it suffices to show that

$$ \begin{align*}\frac{\omega_{0,2}-\alpha+\varepsilon}{\omega_{0,2}+\omega_{0,1}-1}<\frac{\omega_{\infty,1}-\alpha-\varepsilon}{\omega_{\infty,1}-\varepsilon}\leq \alpha\end{align*} $$

holds for $\varepsilon>0$ small enough, which is trivially verified by the condition that $\frac {\alpha }{1-\alpha }\geq \omega _{\infty ,1}\geq \omega _{0,2}$ and $\omega _{0,1}>1$ . Thus, we obtain (5.13) for $\theta _2<\frac {\omega _{\infty ,1}-\alpha -\varepsilon }{\omega _{\infty ,1}-\varepsilon }$ with $\varepsilon>0$ small enough.

Notice that

$$ \begin{align*}\frac{\omega_{\infty,1}-\alpha-\varepsilon+ \theta_2(2\omega_{\infty,2}- \omega_{\infty,1} +3\varepsilon)}{\omega_{\infty,2}+\varepsilon}<\frac{2(\omega_{\infty,1}-\alpha-\varepsilon)}{\omega_{\infty,1}-\varepsilon}\end{align*} $$

when $\theta _2<\frac {\omega _{\infty ,1}-\alpha -\varepsilon }{\omega _{\infty ,1}-\varepsilon }$ with $\varepsilon>0$ small enough, and $\frac {2(\omega _{\infty ,1}-\alpha -\varepsilon )}{\omega _{\infty ,1}-\varepsilon }<\frac {2(\omega _{\infty ,1}-\alpha )}{\omega _{\infty ,1}}$ , we conclude that $H_{\alpha ,\gamma }$ maps $L^{2\tilde {r}( \varepsilon )}(\mathbb {R})\times L^{1 }(\mathbb {R})$ to $L^{\tilde {r}(\varepsilon ), \infty }(\mathbb {R})$ , where $\frac {1}{\tilde {r}( \varepsilon ) }=\frac {2(\omega _{\infty ,1}-\alpha )}{\omega _{\infty ,1}}-\varepsilon $ with $\varepsilon $ small enough. Therefore, we finish the proof of Proposition 5.3.