In the book Fourier Analysis by Javier Duoandikoetxea, the Calderon-Zygmund theorem is stated as follows:
Theorem $5.1$ (Calderón-Zygmund). Let $K$ be a tempered distribution in $\mathbb{R}^n$ which coincides with a locally integrable function on $\mathbb{R}^n \backslash\{0\}$ and is such that $$ \begin{gathered} |\hat{K}(\xi)| \leq A \\ \int_{|x|>2|y|}|K(x-y)-K(x)| d x \leq B, \quad y \in \mathbb{R}^n \end{gathered} $$ Then for $1<p<\infty$, $$ \|K * f\|_p \leq C_p\|f\|_p $$ and $$ \left|\left\{x \in \mathbb{R}^n:|K * f(x)|>\lambda\right\}\right| \leq \frac{C}{\lambda}\|f\|_1. $$
Earlier in the book, a tempered distribution is defined to be a continuous linear functional on Schwartz functions. As such, I have some questions regarding the terms here:
- By coinciding, does it mean $K(f) = \int_{\mathbb{R}^n \backslash \{0\}} gf$ where $f \in \mathcal{S}$ and $g \in L^1_{loc}(\mathbb{R}^n \backslash \{0\})$?
- Are the first and second conditions on the Fourier transform and integral really a condition on $g$? Because, it doesn't make sense to say $|\hat{K}(\xi)| \leq A$ since $\xi \in \mathbb{R}^n$ and $\hat{K}$ is defined by $\hat{K}(f) = K(\hat{f})$ for $f \in \mathcal{S}$? Similarly, for the second condition.
- The proof begins by suggesting that the first inequality implies $K * -$ is $(2,2)$-bounded by Plancharel theorem. However, we first need to know $K *f$ is $L^2$ and $L^p$ (for the remaining part of the proof) given $f \in \mathcal{S}$. This won't work out as usual since $g$ is only locally integrable so Young's won't work, and approximating $f$ by compactly supported smooth function won't work out easily because the estimate we can obtain is $$ \| K * f_\epsilon \|_p \leq \| g\| _{L^1({\rm supp}\, f_\epsilon)} \| f_\epsilon \|_p$$ where $f_\epsilon$ is a compactly supported smooth approximation. However, $ \|g\| _{L^1({\rm supp}\, f_\epsilon)} $ can blow up as $\epsilon \to \infty$.
Edit: For 3, the idea for $p = 2$ is as follows: let $Tf = K \star f$ and so \begin{align*} \lVert Tf \rVert_{2} &= \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle Tf, h \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle Tf, \hat{h} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle K \star f, \hat{h} \rangle \rvert \\ &= \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle K, \hat{h} \star \widetilde{f} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle K, (\hat{h} \star \tilde{f})^{\hat{\hat{\hat{\hat{\phantom{,}}}}}} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{K}, (\hat{h} \star \tilde{f})^{\hat{\hat{\hat{\phantom{,}}}}} \rangle \rvert \\& = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{K}, \left(\tilde{\hat{h} \star \tilde{f}}\right)^{\hat{\phantom{,}}} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{K}, \left(\tilde{\hat{h}} \star f \right)^{\hat{\phantom{,}}} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{K}, \hat{\tilde{\hat{h}}} \hat{f} \rangle \rvert \\&=\sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{K}, h \hat{f} \rangle \rvert = \sup_{\lVert h \rVert_{2 } = 1} \lvert \langle \hat{f} \hat{K}, h \rangle \rvert = \lVert \hat{f} \hat{K} \rVert_{2} \leq A \lVert f \rVert_{2}., \end{align*} The remaining part of proving that $K \star f$ is $L^p$ is obtained either by interpolation between (1,1)-weak inequality or duality.