Proof for weak (1,1)-estimate for Hilbert Transform I am following the proof given in the book Fourier Analysis by Javier Duoandikoetxea and I am unsure about some elements of the proof given for the weak $(1,1)$ estimate for Hilbert Transform.
Firstly, the theorem is stated for a Schwartz $f$. So is it correct to say that proving "$H$ is weak $(1,1)$" is equivalent to saying that we are only proving the $(1,1)$ estimate when $f \in L^1(\mathbb{R}) \cap \mathcal{S}(\mathbb{R})$?
Secondly, I am unsure why is the bad part $b$ of the function $f$ is in $L^2(\mathbb{R})$. It is not hard to see that the good part $g$ is in $L^1(\mathbb{R}) \cap L^\infty(\mathbb{R})$ since its norm equals that of $f$ and it is above bounded by the bound set by the Calderon-Zygmund decomposition. But I am unsure why the bad part
$$b (x):= \sum_{j} b_j(x)  :=  \sum_{j} \left(f(x) - \frac{1}{ \lvert I_j \rvert } \int_{I_j} f\right) \chi _{I_j}(x),$$
is in $L^2(\mathbb{R})$, where the collection $\{I_j\}$ is of disjoint intervals given by the Calderon-Zygmund decomposition.
Lastly, why is the following statement true?

Even though $b_j \not\in \mathcal{S}$, when $x \not\in 2I_j$ the formula $$Hb_j(x) = \int_{I_j} \frac{b_j(y)}{x-y} \mathrm{\, d}y$$ is still valid.

 A: I believe the following answers:

*

*$b$ is in $L^2(\mathbb{R})$ since $b = f- g$ where $f \in \mathcal{S} \subset L^2(\mathbb{R})$ and $g \in L^2(\mathbb{R})$


*To see why the statement holds true, let $b_j^\varepsilon$ be the mollification such that $b_j^\varepsilon|_{I_j} = b_j, \operatorname{supp} b_j^\varepsilon \subset 2I_j.$ Then $b_j^\varepsilon \xrightarrow{L^2(\mathbb{R}) }  b_j$ and so $Hb_j^\varepsilon \to Hb_j$, by boundedness of $H$ on $L^2(\mathbb{R}). $ Now we have $\forall  x \not\in 2I_j: $
$$ Hb_j(x) = \lim_{\varepsilon \to 0^+} Hb_j^\varepsilon(x) = \lim_{\varepsilon \to 0^+} \lim_{s \to 0^+}  \int_{ \lvert y \rvert > s}\frac{1}{\pi} \frac{b_j^\varepsilon(y)}{x - y} = \lim_{\varepsilon \to 0^+}   \int_{\operatorname{supp} b_j^\varepsilon}\frac{1}{\pi} \frac{b_j^\varepsilon(y)}{x - y} =\int_{I_j} \frac{1}{\pi} \frac{b_j(y)}{x - y},$$
where the latter equality follows from $b^\varepsilon _j \xrightarrow{a.e.}_s b_j $ and $ \left\lvert \tfrac{b_j^\varepsilon}{x-y} \right\rvert \leq \frac{2\lVert b_j^\varepsilon  \rVert_{L^\infty(2I_j) }}{ \lvert I_j \rvert } \in L^2(2I_j)  $. Then by dominated convergence, the limit goes inside, and we have the required equality.
