The following Theorem is proved for all $n\in \mathbb{N}$ in Grafakos (Thm 6.2.7), but I believe that there is an issue in his proof, specifically when he asserts that $$\sum_{j\leq 0} \int_{\delta<|x|<2\delta} |K_j(x)|dx\leq \sum_{j\leq 0}(1+2^j 2\delta)^{1/4}<\infty$$ which is clearly not the case (the series diverges!). But it is not a serious mistake, I was able to adapt his proof to make it work for $n\geq 2$. I am wondering if anyone here could provide the proof for the case $n=1$. Here is the Theorem (and my proof for $n\geq 2$, which you may skip as it is not relevant for my question):
Thm: Suppose we are in $\mathbb{R}^n$ for $n\geq 2$ (because of a technicality, our proof doesn't work for $n=1$). If $m\in C^k(\mathbb{R}^n\setminus \{0\})\cap L^\infty(\mathbb{R}^n)$ for $k=\lfloor n/2\rfloor+1$, then it holds that for $1<p<\infty$, $ (m \widehat{f})^{\vee}$ extends to a unique bounded operator $T_p$ defined on all of $L^p(\mathbb{R}^n)$ provided $\sup_{\xi\not=0}\sup_{|\alpha|\leq \lfloor \frac{n}{2}\rfloor+1}|\xi|^{\alpha}|\partial_x ^\alpha m(\xi) | \leq A$. Furthermore, one has explicitly that: $$\|T_p f\|_{L^p(\mathbb{R}^n)}\leq C_{p,n}(1+A+\|m\|_{L^\infty(\mathbb{R}^n)}^2)^{A_{p,n}}\|f\|_{L^p(\mathbb{R}^n)} $$
Proof: (for $n\geq 2$ following Grafakos and introducing inequality K-j to fix the argument) We claim that if $f\in C_0^\infty(\mathbb{R}^n\setminus \{0\})$, there exists $K:\mathbb{R}^n\rightarrow \mathbb{C}$ which is integrable on any compact set that does not contain the origin such that: \begin{equation}\tag{1} \langle K,f\rangle=\langle m^{\vee},f\rangle \end{equation} where $m^{\vee}$ is to be interpreted in the tempered distribution sense. Proving this assertion will be difficult and we first consider a smooth $h:\mathbb{R}^n\rightarrow \mathbb{R}$ supported on $\text{int}(B_2)$ such that $0\leq h(x)\leq 1$ and $h|_{B_1}=1$. We then have that for $j\in \mathbb{N}$::
$$h(2^{-j}x)-h(2^{j+1}x)=\begin{cases} 0 &|x|\in (2^{j+2},\infty)\\ 1 & |x|\in (2^{-j},2^j)\\ 0 & |x|\in [0,2^{-j-1})\\ g(x) & \text{otherwise} \end{cases}$$ where $|g(x)|\leq 1 $. Therefore, we obtain that: $$\lim_{j\rightarrow \infty} \sum_{k=-j}^j (h(2^{-k}x)-h(2^{-(k+1)}x) )=\lim_{j\rightarrow \infty} (h(2^{-j}x)-h(2^{j+1} x))=\chi_{\{0\}^C}(x)$$ Because $h(2^{-k}x)-h(2^{-(k+1)}x)$ is smooth and has compact support, then it is attained as the image of a member of the Schwartz space via the Fourier transform, namely, we may write that $h(\xi)-h(2\xi)=\widehat{\zeta}(\xi)$, which will be supported on $[1/2,2]$. Furthermore, we have that the sum above converges to: $$\lim_{j\rightarrow \infty }\sum_{k=-j}^j \widehat{\zeta}(2^{-k}\xi)=\chi_{\{0\}^C}(\xi)$$ Now we define $K_k(\xi)=(m(\xi)\widehat{\zeta}(2^{-k}x))^{\vee}$. I claim that it holds that: \begin{equation}\tag{K} \sup_{j\in \mathbb{Z}}\int_{\mathbb{R}^n}|K_j(x)|(1+2^j|x|)^{t}dx\leq C_{t,n} A \end{equation} \begin{equation}\tag{gradK} \sup_{j\in \mathbb{Z}}2^{-j}\int_{\mathbb{R}^n}|\nabla K_j(x)|(1+2^j|x|)^{t}dx\leq C_{t,n} A \end{equation}
for $t<1/2$. And also, we have that for $1/2>t\geq (1-n)/2$:
\begin{equation}\tag{K-j} \sup_{k\in \mathbb{Z}}\int_{\mathbb{R}^n}|K_j(x)|(2^j|x|)^{t}dx\leq C_{t,n} A \end{equation}
Let us prove all of these inequalities. We have that by H"older: \begin{multline*} \int_{\mathbb{R}^n}|K_j(x)|(1+2^j|x|)^{t}dx\leq \left(\int_{\mathbb{R}^n}|(m(2^{-j}\xi)\widehat{\zeta}(\xi))^{\vee}(x)|^2(1+2^j|x|)^{2\lfloor n/2\rfloor+2}dx\right)^{1/2}\\ \left(\int_{\mathbb{R}^n}(1+2^j |x|)^{-2(\lfloor n/2\rfloor -2+2t}\right)^{1/2} \end{multline*} We will deal with these integrals separately. We have that because of radiality of the second integrand, it follows that: $$ \int_{\mathbb{R}^n}(1+2^j |x|)^{-2\lfloor n/2\rfloor-2+1/2}dx\lesssim \int_0^\infty \frac{r^{n-1}}{(1+2^j r)^{2\lfloor n/2 \rfloor+2-2t}}dr= 2^{-jn}\int_0^\infty \frac{r^{n-1}}{(1+r)^{2\lfloor n/2 \rfloor+2-2t}}dr$$ Now we notice that: $$n-1-2\lfloor n/2\rfloor-2+2t=\begin{cases} 2t-3 & \text{$n$ even}\\ 2t-2 & \text{$n$ odd} \end{cases}<-1$$ whenever $t<1/2$ which implies that the integral above is finite and we are left with: $$\int_{\mathbb{R}^n}|K_j(x)|(1+2^j|x|)^{t}dx\leq 2^{-jn/2}\left(\int_{\mathbb{R}^n}|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)|^2(1+2^j|x|)^{2\lfloor n/2\rfloor+2}dx\right)^{1/2}$$ In the second case, we have that: \begin{multline*} \int_{\mathbb{R}^n}|K_j(x)|(2^j|x|)^{t}dx\leq \left(\int_{\mathbb{R}^n}|(m(2^{-j}\xi)\widehat{\zeta}(\xi))^{\vee}(x)|^2(1+2^j|x|)^{2\lfloor n/2\rfloor+2}dx\right)^{1/2}\\ \left(\int_{\mathbb{R}^n}(1+2^j |x|)^{-2(\lfloor n/2\rfloor -2}(2^j|x|)^{2t}\right)^{1/2} \end{multline*} as before, the integral to the right is upper bounded (up to multiplicative constants) in polar coordinates via: $$ 2^{-jn}\int_0^\infty\frac{r^{n-1+2t}}{(1+r)^{2\lfloor n/2\rfloor+2 }}dr$$ which is as before a multiplicative constant, provided $ n-1+2t\geq 0$ and $t<1/2$. Both of these conditions are the ones presented, so that one concludes: $$\int_{\mathbb{R}^n}|K_j(x)|(2^j|x|)^{t}dx\leq 2^{-jn/2}\left(\int_{\mathbb{R}^n}|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)|^2(1+2^j|x|)^{2\lfloor n/2\rfloor+2}dx\right)^{1/2}$$ From now on the computations are identical to prove both inequalities K and K-j. We expand the integrand, which yields that: $$(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)(1+2^j|x|)^{2\lfloor n/2\rfloor+2}=\sum_{k=0}^{2\lfloor n/2\rfloor+2}{2\lceil n/2\rceil+2 \choose k}2^{jk}(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}|x|^k$$ Now, turning our attention to $|x|^k$, we have that: $$|x|^k=\sqrt{\sum_{\alpha\in (\mathbb{N}\cup\{0\})^n:|\alpha|= k}{k\choose \alpha_1,\ldots \alpha_n} x^{2\alpha_1}_1\ldots x_n^{2\alpha_n} }$$ We estimate the sum by its largest term multiplied by the number of multi-indices with size $k$, that is ${k+n-1\choose k}$, then, estimating the maximum by all terms: $$|x|^k\leq \sum_{\alpha\in (\mathbb{N}\cup\{0\})^n:|\alpha|= k}\sqrt{{k+n-1\choose k}{k\choose \alpha_1,\ldots \alpha_n}} |x^{\alpha_1}_1\ldots x_n^{\alpha_n} | $$ Therefore, $$|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)(1+2^j|x|)^{2\lfloor n/2\rfloor+2}|\lesssim \sum_{k=0}^{2\lfloor n/2\rfloor+2}2^{jk}\sum_{|\alpha|= k}|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)x^{\alpha}|$$
Now, because $\widehat{\zeta}(2^{-j}\xi)$ is supported away from the origin and has compact support with $m(\xi)$ being continuously differentiable there up to order $2\lfloor n/2\rfloor+2$, we obtain that: $$(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)x^{\alpha}=\frac{(\partial^{\alpha }_\xi (m(\xi)\widehat{\zeta}(2^{-j}\xi ) ))^{\vee}}{(-2\pi i)^{|\alpha|}}$$ Returning with this above, using Minkowski's inequality and also the fact of the Fourier transform being an isometry in $L^2(\mathbb{R}^n)$: $$\begin{cases}\int_{\mathbb{R}^n}|K_j(x)|(1+2^j |x|)^{t}dx & t<1/2\\\int_{\mathbb{R}^n}|K_j(x)|(2^j |x|)^{t}dx& (1-n)/2 \leq t<1/2\end{cases}\lesssim $$ $$2^{-jn/2} \sum_{k=0}^{2\lfloor n/2\rfloor +2}2^{{jk} }\sum_{|\alpha|= k}\left(\int_{\mathbb{R}^n}\left|\sum_{\beta<\alpha}{\alpha \choose\beta}\partial^{\beta}_{\xi}(m)(\xi)\partial^{\alpha-\beta}(\widehat{\zeta})(2^{-j }\xi)2^{-j|\alpha-\beta|}\right|^2\right)^{1/2}\leq $$ $$2^{-jn/2}\sum_{k=0}^{2\lfloor n/2 \rfloor+2}2^{jk}\sum_{|\alpha|= k}\sum_{\beta\leq \alpha} 2^{-j|\alpha-\beta|}\left(\int_{\text{supp}(\widehat{\zeta}(2^{-j}x))}|\partial^{\beta}(m)(\xi)|^2\right)^{1/2} $$
Now, because the support of this function lies in $[1/2,2]$, our integral will effectively depend on $x$ satisfying $1/2\leq 2^{-j}x\leq 2$, which translates as:
$$2^{-jn/2}\sum_{k=0}^{2\lfloor n/2\rfloor+2 }2^{jk}\sum_{|\alpha|= k}\sum_{\beta\leq \alpha}2^{-j|\alpha-\beta|}\left(\int_{2^j/2}^{2^{j+1}}\frac{A^2}{|\xi|^{2|\beta|}}d\xi \right)^{1/2}$$ Now, if $n\not=2|\beta|$, the integral of $1/|\xi|^{2\beta}$ may be estimated above by: $$\frac{2^{(j+1)(n-2|\beta|)}}{n-2|\beta|}\lesssim 2^{j(n-2|\beta|) }$$ and when $n=2\beta$, we obtain a logarithm which becomes an absolute constant, so that the bound $ 2^{j(n-2|\beta|) }$ continues to hold. All in all, we obtain that: $$2^{-jn/2}\sum_{k=0}^{2\lfloor n/2\rfloor+2 }2^{jk}\sum_{|\alpha|= k}\sum_{\beta\leq \alpha}2^{-j|\alpha-\beta|/2}\left(A^2 2^{jn}2^{-2j |\beta |} \right)^{1/2}\leq A2^{-jn/2}\sum_{k=0}^{2\lfloor n/2\rfloor+2 }2^{jk}\sum_{|\alpha|= k}\sum_{\beta\leq \alpha}2^{-j|\alpha-\beta|} 2^{jn/2}2^{-j |\beta |} $$ As $|\alpha-\beta|=|\alpha|-|\beta|$, we obtain that: $$\int_{\mathbb{R}^n}|K_j(x)|(1+2^j |x|)^{t}dx\lesssim A\sum_{k=0}^{2\lfloor n/2\rfloor+2}2^{jk}\sum_{|\alpha|= k}2^{-j |\alpha|}\lesssim A$$ An identical computation as the one performed above shows us that: $$2^{-j}\int_{\mathbb{R}^n}|\partial_ rK_j(x)|(1+2^j|x|)^{t}dx\leq 2^{-j }2^{-jn/2}\left(\int_{\mathbb{R}^n}|\partial_r(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)|^2(1+2^j|x|)^{2\lfloor n/2\rfloor+2}dx\right)^{1/2}$$ where the integrand is estimated by: $$|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)(1+2^j|x|)^{2\lfloor n/2\rfloor+2}|\lesssim \sum_{k=0}^{2\lfloor n/2\rfloor+2}2^{jk}\sum_{|\alpha|= k}|(m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)x^{\alpha}|$$
Now, because $\widehat{\zeta}(2^{-j}\xi)$ is supported away from the origin and has compact support with $m(\xi)$ being continuously differentiable there up to order $2\lfloor n/2\rfloor+2$, we obtain that: $$\partial_r ( m(\xi)\widehat{\zeta}(2^{-j}\xi))^{\vee}(x)x^{\alpha}=(-1)^{|\alpha|}\frac{(\partial^{\alpha }_\xi (m(\xi)\xi_r\widehat{\zeta}(2^{-j}\xi ) ))^{\vee}}{(2\pi i)^{|\alpha|-1}}$$ The same computations follow through with $\widehat{\tilde{\zeta}}(\xi)=\xi_r \zeta(\xi)$, so that above we obtain $2^{j}\widehat{\tilde{\zeta}}(2^{-j}\xi)$. Repeating the argument done previously verbatim, we are left with: $$\int_{\mathbb{R}^n}|\partial_r K_j(x)|(1+2^j |x|)^{t}dx\lesssim 2^{-j}2^{-jn/2}\sum_{k=0}^{2\lfloor n/2 \rfloor+2}2^{jk}\sum_{|\alpha|= k}\sum_{\beta\leq \alpha} 2^{-j|\alpha-\beta|}\left(2^j\int_{\text{supp}(\widehat{\tilde{\zeta}}(2^{-j}x))}|\partial^{\beta}(m)(\xi)|^2\right)^{1/2} $$ Because $\widehat{\tilde{\zeta}}(\xi)$ and $\tilde{\zeta}$ have the same support and the $2^j$ factor cancels, we obtain an upper bound depending on $A$ as before, which finishes the proof of inequalities K, gradK and K-j above. Now that we have them, I claim this implies that $\sum_{j\in \mathbb{Z}} |K_j|$ is integrable on any compact set away from the origin. Indeed, we have that as $1/4<1/2$ and $(1-n)/2\leq -1/2<-1/4<1/2$ (we did ask that $n\geq2$ in the statement), we may take them as values of $t$ in the respective inequalities to obtain: $$\sum_{j\in \mathbb{Z}}\int_{\delta<|x|<1/\delta } |K_j(x)| dx\leq \sum_{j\geq 0}\frac{1}{(1+2^j\delta)^{1/4}}\int_{\delta <|x|}|K_j(x)|(1+2^j|x|)^{1/4}dx+$$ $$ \sum_{j\leq 0}(2^j \delta^{-1})^{1/4}\int_{|x|<1/\delta }|K_j(x)|(2^j|x|)^{-1/4}dx \lesssim \frac{A}{\delta^{1/4}}$$
So it follows that by the monotone convergence theorem $\sum_{j\in \mathbb{Z}} |K_j|$ is integrable on any compact set that does not contain the origin, so absolutely convergent there and we are well within our rights to define $K:=\sum_{j\in \mathbb{Z}}K_j$ almost everywhere and if $\varphi\in C_0^\infty(\mathbb{R}^n\setminus \{0\})$ we obtain that by the dominated convergence theorem (with $\sum_{j\in \mathbb{Z}}|K_j|\varphi$ as the dominating function in $L^1(\mathbb{R}^n)$): $$ \int_{\mathbb{R}^n} K(x)\varphi(x) dx=\lim_{j\rightarrow \infty}\sum_{k=-j}^j\int_{\mathbb{R}^n} (m(\xi)\widehat{\zeta}(2^{-k }\xi ))^{\vee}(x)\varphi(x)dx=\lim_{j\rightarrow \infty }\sum_{k=-j}^j\int_{\mathbb{R}^n} m(\xi)\widehat{\zeta}(2^{-k}\xi ) \varphi^{\vee}(\xi)dx $$ Now, because $m(\xi)\varphi^{\vee}(\xi)$ is integrable and by our construction $\sum_{k=-j}^j|\widehat{\zeta}(2^{-k}\xi)|\leq 1$, then it follows that we may apply Lebesgue's Dominated convergence once more to yield that: $$\langle K,\varphi \rangle=\int_{\mathbb{R}^n}K(x)\varphi(x) dx=\int_{\mathbb{R}^n} m(\xi )\varphi^{\vee}(\xi)dx=\langle m^{\vee},\varphi\rangle$$ This proves relation 1 and allows us to prove that our operator $(m \widehat{f})^{\vee}$ is defined on a dense subset $\mathcal{F}$ of $L^1(\mathbb{R}^n)$ such that: $$\sup_{\alpha>0}\alpha |\{x:|(m \widehat{f})^{\vee}(x)|>\alpha\}|\leq \|f\|_{L^1(\mathbb{R}^n)}$$
Let us see why this is the case. First we define the set $\mathcal{F}$ as the set of finite linear combinations of characteristic functions of cubes of side length $1/2^L$ whose endpoints are dyadics, i.e, numbers of the form $\mathbb{Z}/2^L$ where we allow $L$ to vary in the integers.
If we use the Calderon-Zygmund Lemma on a member $f\in \mathcal{F}$ at level $\alpha$, then we obtain that $f=g+\sum_{j=1}^n b_j$, where there are only finitely many $b_j$. This follows because there are only finitely many chosen sets at any generation (due to our function having compact support) and if we fix $L$ the greatest integer for which we used a dyadic cube of length $1/2^L$, then our function will be constant at any cube in that generation and subdivision will not alter the value of $\frac{1}{|Q|}\int_Q|f|dy$. Therefore, no new cubes are selected after large enough of a generation, so we conclude $n<\infty$.
Another observation is that $b_j$ is a member of $\mathcal{F}$ as $f\chi_{Q_j}$ is a member of $\mathcal{F}$ and adding members of $\mathcal{F}$ results in a member of $\mathcal{F}$ by subdividing the cubes if necessary and disregarding the boundary of dyadic cubes which has measure zero. We denote by $Q_j$ the dyadic cubes where $b_j$ has support, $q_j$ is the center of $Q_j$ and $Q_j^*$ is a cube with the same center as $Q_j$, but with sidelength $\ell(Q_j^*)=2\sqrt{n}\ell(Q_j)$. We then obtain that: $$\alpha|\{x:|(m\widehat{f})^{\vee}(x)|>\alpha\}|\leq \alpha |\{x:|(m\widehat{g})^{\vee}(x)|>\alpha/2\}|+\alpha\left|\left\{x: \left|\sum_{i=1}^n(m \widehat{b}_i)^{\vee}(x)\right|>\alpha/2\right\}\right|\leq$$ $$ \alpha |\{x:|(m\widehat{g})^{\vee}(x)|>\alpha/2\}|+\alpha\left|\left\{x\in (\cup_{j=1}^n Q_j^*)^C: \left|\sum_{i=1}^n(m \widehat{b}_i)^{\vee}(x)\right|>\alpha/2\right\}\right|+\alpha \sum_{j=1}^n|Q_j^*|$$ The middle term will prove to be the most troublesome to estimate. In the meantime we may estimate the other terms via: $$\sum_{j=1}^n |Q_j^*|=(2\sqrt{n})^n\sum_{j=1}^n |Q_j|<(2\sqrt{n})^n\sum_{j=1}^n\frac{\int_{Q_j}|f|dx}{\alpha}\leq \frac{(2\sqrt{n})^n \|f\|_{L^1(\mathbb{R}^n)}}{\alpha}$$ And also: $$|\{x:|(m\widehat{g})^{\vee}(x)|>\alpha/2\}|\leq \frac{4}{\alpha^2}\|(m\widehat{g})\|_{L^2(\mathbb{R}^n)}^2\leq \frac{4\|m\|_{L^\infty(\mathbb{R}^n)}^2\|g\|_{L^2(\mathbb{R}^n)}^2}{\alpha^2}\leq \frac{4\|m\|_{L^\infty(\mathbb{R}^n)}^2 \|g\|_{L^1(\mathbb{R}^n)}\|g\|_{L^\infty(\mathbb{R}^n)}}{\alpha^2} $$ One of the properties of the Calderon-Zygmund Lemma is that $\|g\|_{L^\infty(\mathbb{R}^n)}\leq 2^n \alpha$ and also $\|g\|_{L^1(\mathbb{R}^n)}\leq \|f\|_{L^1(\mathbb{R}^n)}$, so that the bound above simplifies to: $$|\{x:(m\widehat{g})^{\vee}>\alpha/2\}|\leq2^n \frac{4\|m\|_{L^\infty(\mathbb{R}^n)}^2\|f\|_{L^1(\mathbb{R}^n)}}{\alpha }$$ Finally, we will work with the middle bound, but prior to doing that, we notice that if we mollify $b_i$ we obtain a sequence of smooth $b_{i,k}$ of compact support such that $b_{i,k}\rightarrow b_i$ in $L^2(\mathbb{R}^n)$. Because the Fourier transform is an isometry and $m$ lies in $L^\infty(\mathbb{R}^n)$, the convergence $(m\widehat{b_{i,k}})^{\vee}\rightarrow (m\widehat{b_i})^{\vee}$ also takes place in $L^2(\mathbb{R}^n)$ and we may take a subsequence (which we denote the same), such that there is pointwise convergence almost everywhere. In the sense of tempered distributions it holds that $(m \widehat{b_{i,k}})^{\vee}=m^{\vee}*b_{i,k}$. To prove this it suffices to prove that they coincide at any $\widehat{\varphi}$ where $\varphi\in \mathcal{S}(\mathbb{R}^n)$, but this is the case, if we use Proposition 1.3 in Linares' textbook: $$ \langle m^\vee * b_{i,k},\widehat{\varphi} \rangle=\langle \mathcal{F}(m^{\vee}* b_{i,k}),\varphi \rangle=\langle m^{\vee \wedge }, \widehat{b_{i,k}}\varphi\rangle=\langle m, \widehat{b_{i,k}} \varphi \rangle=\langle (m\widehat{b_{ik}})^{\vee}, \widehat{\varphi}\rangle$$
But as $m^\vee *b_{i,k}$ and $(m\widehat{b_{i,k}})^{\vee}$ are both tempered distribution functions, they must coincide almost everywhere. Now, for almost every $x\in (\cup_{j=1}^n Q_j^*)^C$:
$$(m \widehat{b}_i)^{\vee}(x)=\lim_k (m\widehat{b_{i,k}})^{\vee}(x)=\lim_km^\vee *b_{i,k}(x) $$ Now the magic occurs. Because $x\in (\cup_{j=1}^n Q_j^*)^C\subseteq (Q_i^*)^C$, I claim that $b_{i,k }(x-\cdot )$ is supported on a compact set which does not contain the origin. This will follow provided there exists a ball centered at the origin that does not intersect: $$ \{z: x-z\in Q_i\}\subseteq (Q_{i}^*)^C-Q_i$$ But this is clearly the case as if $a\in Q_i$ and $b\in (Q_i^*)^C$: \begin{equation} |a-b|\geq |b-q_i|-|a-q_i|\geq 2\sqrt{n}\ell(Q_i)-\sqrt{n} \ell(Q_i)>0 \end{equation} Therefore, as $m^{\vee}* b_{i,k }(x):=m^{\vee}(b_{i,k}(x-\cdot ))$ and we have just proved that $b_{i,k}(x-\cdot )$ is a smooth function whose support does not contain zero for large enough $k$, we conclude that we are in the conditions to apply equation 1 which we have already established, so almost everywhere we have that for $x\in (\cup Q_i^*)^C$: $$(m\widehat{b_i})^{\vee}(x)=\lim_k \int_{\mathbb{R}^n}K(y) b_{i,k}(x-y)dy=\int_{\mathbb{R}^n} K(y) b_i (x-y) dy$$ where we are using Lebesgue's Dominated Convergence Theorem in this last passage, as one has $|K(y)b_{i,k}(x-y)|\leq \|b_i\|_{L^\infty(\mathbb{R}^n)}|K(y) \chi_{(\delta,1/\delta)}(y)|$ for sufficiently small $\delta$. Therefore, one has that as the mean of the $b_i$ is zero: $$\left|\left\{x\in (\cup_{j=1}^n Q_j^*)^C: \left|\sum_{j=1}^n(m \widehat{b}_j)^{\vee}(x)\right|>\alpha/2\right\}\right|\leq\frac{2}{\alpha} \sum_{j=1}^n \int_{(\cup_{j=1}^n Q_j^* )^C} \left|\int_{Q_j} (K(x-y)-K(x-q_j))b_j(y)dy\right|dx $$ By Tonelli and the fact that $(\cup_{j=1}^nQ_j^*)^C\subseteq Q_i^*$ for every $i$, one may estimate this further by: \begin{equation}\tag{PR} \frac{2}{\alpha}\sum_{j=1}^n \int_{(Q_j)}|b_j(y)|\int_{(Q_j^*)^C}|K(x-y)-K(x-q_j)|dx dy \end{equation} We estimate this integral: $$\int_{(Q_i^*)^C}|K(x-y)-K(x-q_j)|dx=\int_{(Q_i^*)^C-q_j} |K(q_j+x-y)-K(x)|dx\leq \sum_{k\in \mathbb{Z}}\int_{(Q_i^*)^C-q_j} |K_k(q_j+x-y)-K_k(x)|dx $$ Let us estimate the summands separately. We have that: $$\int_{(Q_j^*)^C-q_j}\int_0^1 |\nabla K_k(x+\theta (q_j-y))\cdot (q_j-y)|d\theta dx $$ Now we notice that $(Q_j^*)^C-q_j\subseteq \{x:|x|\geq 2\sqrt{n}\ell(Q_j)\}$ and our $y$ lies in $Q_j$, so $|y-q_j|\leq \sqrt{n}\ell(Q_j)$, so we obtain that $$(Q_j^*)^C-q_j\subseteq \{x:|x|\geq 2|y-q_j|\}$$ and the integral above may be estimated via: $$ \int_0^1 \int_{|x|\geq 2|y-q_j|} |\nabla K_k(x+\theta(q_j-y))||q_j-y|dx d\theta $$ Now we notice that there exists $k$ such that $2^{K-1}\leq |q_j-y|\leq 2^K$ so that one obtains these last integrals may be estimated further by use of gradK: \begin{equation}\tag{FR} 2^K\int_0^1 \int_{|x|\geq 2|y-q_j|}|\nabla K_k(x+\theta(q_j-y))|(1+2^k |x+\theta(q_j-y)|)^{1/4}dxd\theta\lesssim 2^K 2^{k}A \end{equation} On the other hand, we could have alternatively bounded for $\delta$ sufficiently small: $$\int_{(Q_i^*)^C-q_j}|K_k(q_j+x-y)-K_k(x)|dx\leq$$ $$\int_{(Q_i^*)^C-y}|K_k(x) |dx+\int_{(Q_i^*)^C-q_k }|K_k(x)|dx\leq 2\int_{(Q_i^*)^C-Q_i}|K_k(x)| dx\leq$$ \begin{equation}\tag{SR} \frac{2}{(1+2^k|y-q_j|)^{1/4}}\int_{|x|> \sqrt{n}\ell(Q_i)\geq |y-q_j|}|K_k(x) |(1+2^k|x|)^{1/4}dx\lesssim\frac{A}{(1+2^k 2^{K-1})^{1/4}} \end{equation}
Returning with inequality FR and SR into PR, we obtain that the integrand in PR may be estimated to yield:
$$\frac{2}{\alpha}\sum_{j=1}^n \int_{(Q_j)}|b_j(y)|\int_{(Q_j^*)^C}|K(x-y)-K(x-q_j)|dx dy \leq $$ $$\frac{1}{\alpha}\sum_{j=1}^n \int_{Q_j}|b_j(y)|\left(\sum_{k\leq K } 2^K 2^{k}A+\sum_{k> K} \frac{A}{(1+2^k 2^{K-1})^{2/4}} \right)dx\lesssim \frac{A\|f\|_{L^1(\mathbb{R}^n)}}{\alpha }$$
Combining all bounds shows us that: $$\sup_{\alpha>0}\alpha|\{x:|(m\widehat{f})^{\vee}(x)|>\alpha\}|\lesssim\left(2^n4\|m\|_{L^\infty(\mathbb{R}^n)}^2+A+(2\sqrt{n})^n\right)\|f\|_{L^1}\lesssim(1+A+\|m\|_{L^\infty(\mathbb{R}^n)}^2)\|f\|_{L^1(\mathbb{R}^n)}$$ Now, this implies that Cauchy sequences $\{f_n\}_{n=1}^\infty$ in $ \mathcal{F}$ are mapped via $f\mapsto (m\widehat{f})^{\vee}$ to Cauchy sequences in measure, which then must converge to a measurable $g$. This allows us to extend our operator to $L^1(\mathbb{R}^n)$ to say $T_1$ and we notice that $T_1$ remains of weak $(1,1)$ type as if $f_n\rightarrow f$ in $L^1(\mathbb{R}^n)$ we have that: $$ \alpha|\{|g|>\alpha\}|\leq\limsup_n \alpha|\{|g-T_1(f_n)|>\alpha/2\}|+$$ $$\limsup_n \alpha |\{|T_1(f_n)|>\alpha/2\}|\leq 2 C_n(A+1+\|m\|_{L^\infty(\mathbb{R}^n)}^2)\|f\|_{L^1(\mathbb{R}^n)} $$ where $g=T_1(f)$ by definition. Furthermore, by Plancherel we know that our operator is also defined on $L^2(\mathbb{R}^n)$. We need an operator defined on $L^1(\mathbb{R}^n)+L^2(\mathbb{R}^n)$ to apply Marcinkiewicz's Interpolation Theorem. Given $ =f_1+f_2$ with $f_1\in L^1(\mathbb{R}^n)$, $f_2\in L^2(\mathbb{R}^n)$, then we define: $$T(g)=T_1(f_1)+(m\widehat{f_2})^{\vee}$$ Of course we need to verify that this operator is well defined. Suppose that $f_1+f_2=\tilde{f}_1+\tilde{f}_2$, where $f_1,\tilde{f}_1$, $f_2,\tilde{f}_2\in L^2$. Then we obtain that $f_1-\tilde{f}_1=\tilde{f_2}-f_2$ will lie in $L^1(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$. If we take a smooth sequence of functions approaching $\tilde{f_2}-f_2$ in both $L^1(\mathbb{R}^n)$ and $L^2(\mathbb{R}^n)$, we may then further approach members of this sequence uniformly by linear combinations of finite characteristic functions supported on dyadics. Let us call this sequence $d_k$. But then it follows that: $$ T_1(f_1)-T_1(\tilde{f}_1)=T_1(f_1-\tilde{f}_1)\xleftarrow[\mathcal{M}]{k\rightarrow \infty}T_1(d_k)=(m \widehat{d_k})^{\vee}\xrightarrow[L^2(\mathbb{R}^n)]{k\rightarrow \infty}(m \widehat{f_1})^{\vee}+(m\widehat{f_2})^{\vee} $$ Now, because $T_1$ is of weak $(1,1)$ type it follows that $T_1(f_1)$ and $T_1(\tilde{f}_1)$ are both finite almost everywhere. Therefore, we may rearrange $T_1(f_1)-T_1(\tilde{f}_1)=(m \widehat{f_1})^{\vee}+(m\widehat{f_2})^{\vee}$ to conclude that our operator is well defined on $L^1(\mathbb{R}^n)+L^2(\mathbb{R}^n)$ and by Marcinkiewicz's Theorem, we conclude that: $$\|T(f)\|_{L^p(\mathbb{R}^n)}\lesssim 2p^{1/p}\left(\frac{1+A+\|m\|_{L^\infty(\mathbb{R}^n)}^2}{p-1}+\frac{2}{p-2}\right)^{1/p}\|f\|_{L^p(\mathbb{R}^n)}\quad 1<p<2$$ Now, when $\infty>p>2$ we notice that if $f\in L^p(\mathbb{R}^n)\cap L^2(\mathbb{R}^n)$, then it holds that: $$\|(m\widehat{f})^{\vee}\|_{L^p(\mathbb{R}^n)}=\sup_{g\in \mathcal{S}(\mathbb{R}^n)\:,\:\|g\|_{L^{p/(p-1)}(\mathbb{R}^n)}\leq1}\int_{\mathbb{R}^n} (m\widehat{f})^{\vee} \widehat{\widehat{g}} dx= $$ $$\sup_{g\in \mathcal{S}(\mathbb{R}^n)\:,\:\|g\|_{L^{p/(p-1)}(\mathbb{R}^n)}\leq1}\int_{\mathbb{R}^n}f (m\widehat{g})^{\vee}dx\leq \|f\|_{L^p(\mathbb{R}^n)}\sup_{g\in \mathcal{S}(\mathbb{R}^n)\:,\:\|g\|_{L^{p/(p-1)}(\mathbb{R}^n)}\leq 1}\|(m \widehat{g})^{\vee}\|_{L^\frac{p}{p-1}(\mathbb{R}^n)}$$ Now we notice that because $\infty>p>2$ it holds that $1<p/(p-1)<2$ and we obtain that: $$ \|(m\widehat{f})^{\vee}\|_{L^p(\mathbb{R}^n)}\leq 2(p/(p-1))^{(p-1)/p}\left(\frac{1+A+\|m\|_{L^\infty(\mathbb{R}^n)}^2}{p/(p-1)-1}+\frac{2}{p/(p-1)-2}\right)^{(p-1)/p}\|f\|_{L^p(\mathbb{R}^n)} $$
