Convolution between $L^1$ function and a singular integral kernel

Question

I meet a problem when reading Modern Fourier Analysis(3rd. Edition) written by L.Grafakos. On pg.82 he writes:

Fix $L\in\mathbb{Z}^+$. Suppose that $\{K_j(x)\}_{j=1}^L$ is a family of functions defined on $\mathbb{R}^n\backslash\{0\}$ with the following properties: There exist constant $A,B<\infty$ and an integer $N$ such that for all multi-indices $\alpha$ with $|\alpha|\le N$ and $x\ne0$ we have

$$ \sum\limits_{j=1}^L|\partial^{\alpha}K_j(x)|\le A|x|^{-n-|\alpha|}<\infty $$

and also

$$ \sup\limits_{\xi\in\mathbb{R}^n}\sum\limits_{j=1}^L|\widehat{K_j}(\xi)|\le B<\infty. $$ Note that for $h\in L^1(\mathbb{R}^n)$, $K_j*h$ is a well-defined function in $L^{1,\infty}(\mathbb{R}^n)$.

Here is my question: How to prove $K_j*h\in L^{1,\infty}(\mathbb{R}^n)$? At first I wanted to use Young's inequality given in Classical Fourier Analysis (3rd Edition)(pg.23), saying when $1\le p<\infty,1<q,r<\infty$ satisfy $$ \frac{1}{q}+1=\frac{1}{p}+\frac{1}{r}, $$ then for all $f\in L^p(G),g\in L^{r,\infty}(G)$ we have $$ \|f*g\|_{L^{q,\infty}(G)}\le C_{p,q,r}\|g\|_{L^{r,\infty}(G)}\|f\|_{L^p(G)}. $$ But I suddenly realized that the inequality above cannot handle the situation when $q=1$, hence it cannot be used to prove $K_j*h\in L^{1,\infty}(\mathbb{R}^n)$.

I also tried to use Calderon-Zygmund theorem, but $K_j$ may not satisfy the cancellation condition: $$ \sup\limits_{0<R_1<R_2<\infty}|\int_{R_1<|x|<R_2}K(x)dx|=A_3<\infty. $$

Shall I use definition of $L^{1,\infty}$ to begin the proof, showing $K_j*h$ satisfies that definition, but not using inequalities or theorems? What's the common strategy when we need to show convolution between a $L^1$ function and a singular integral kernel is in $L^{1,\infty}$?

Answer 1

As long as $N\geq 1$, the two listed conditions ensure that $T:f\mapsto K_j*f$ is a Calderon-Zygmund operator; see Stein’s Singular Integrals page 29. From an inspection of Stein’s proof (which directly uses the Calderon-Zygmund decomposition) one gets just from the first condition that $T$ is of weak type (1,1), e.g. $T$ is bounded from $L^1$ to weak $L^1$. You can also see a proof of this in Grafakos Classical Fourier Analysis page 359, theorem 5.3.3 (here Grafakos uses the weaker Hormander condition, which follows immediately from your first stated condition with $N=1$ using the mean value theorem).