I am reading GTM249 Classical Fourier Analysis, I have a question about the proof of Theorem 5.2.10, which is as follows:
Theorem 5.2.10. Let $n \geq 2$ and let $\Omega$ be an even integrable function on $\mathbf{S}^{n-1}$ with mean value zero that satisfies (5.2.24). Then the corresponding singular integral $T_{\Omega}$ is bounded on $L^{p}\left(\mathbf{R}^{n}\right), 1<p<\infty$, with norm at most a dimensional constant multiple of the quantity $\max \left((p-1)^{-2}, p^{2}\right)c_{\Omega}$, where (5.2.24) is as follows $$c_{\Omega}=\int_{\mathbf{S}^{n-1}}|\Omega(\theta)| \log(e+|\Omega(\theta)|) d \theta<\infty .$$ Define $$W_{\Omega}^{1}(x)=\frac{\Omega(x /|x|)}{|x|^{n}} \chi_{\frac{1}{2} \leq|x| \leq 2},$$ and $$K_{j}^{1}=\left(-i \frac{\xi_{j}}{|\xi|} \widehat{W_{\Omega}^{1}}(\xi)\right)^{\vee} .$$ In the proof, the author says since the Riesz transform $R_{j}$ is countably subadditive and maps $L^{p}$ to $L^{p}$ with norm at most $4(p-1)^{-1}$ for $1<p<2$, it follows from Exercise $1.3 .7$ that $K_{j}^{1}=R_{j}\left(W_{\Omega}^{1}\right)$ is integrable over the ball $|x| \leq 3 / 2$. The following is exercise $1.3.7$:
1.3.7.Let $(X, \mu)$ and $(Y, v)$ be two measure spaces with $\mu(X)<\infty$ and $v(Y)<\infty$. Let $T$ be a countably subadditive operator that maps $L^{p}(X)$ to $L^{p}(Y)$ for every $1<p \leq 2$ with norm $\|T\|_{L^{p} \rightarrow L^{p}} \leq A(p-1)^{-\alpha}$ for some fixed $A, \alpha>0$. (Countably subadditive means that $\left|T\left(\sum_{j} f_{j}\right)\right| \leq \sum_{j}\left|T\left(f_{j}\right)\right|$ for all $f_{j}$ in $L^{p}(X)$ with $\sum_{j} f_{j} \in L^{p}$.) Prove that for all $f$ measurable on $X$ we have $$ \int_{Y}|T(f)| d v \leq 6 A(1+v(Y))^{\frac{1}{2}}\left[\int_{X}|f|\left(\log _{2}^{+}|f|\right)^{\alpha} d \mu+C_{\alpha}+\mu(X)^{\frac{1}{2}}\right] .$$
I don't know how to get $K_{j}^{1}=R_{j}\left(W_{\Omega}^{1}\right)$, where the equality is in the sense of distribution. More precisely, let $A=\left\{x \in \mathbf{R}^{n}: 2 / 3<|x|<3 / 2\right\}$, for any $\phi\in C_{c}^{\infty}(A)$, I hope to prove $$\int -i \frac{\xi_{j}}{|\xi|} \widehat{W_{\Omega}^{1}}(\xi) \widehat{\phi(\xi)}d\xi=\int R_{j}\left(W_{\Omega}^{1}\right)(\xi)\phi(\xi)d\xi$$
Any help is really appreciated!
