Action of Calderon-Zygmund operator on $H^{1,\infty}$

126 Views Asked by At

I am trying to understand the theorem in Jean-Lin Journe's book Calderon-Zygmund Operators, Pseudo-Differential Operators and the Cauchy Integral of Calderon. On page 49, the theorem says:

Let $T$ be associated to a standard kernel. Then the following are equivalent:

  1. $T$ satisfies: for $Q \subset \mathbb{R}^n$, $a \in L^\infty(\mathbb{R}^n)$ that is supported in $Q$, we have $$\int_{\overline{Q}} |Ta|\, dx \leq C||a||_\infty |Q|.$$ Here $Q$ is a cube and $\overline{Q}$ is the concentric cube with twice the radius.
  2. $T$ is bounded map from $H^{1,\infty}$ into $L^1$.
  3. $T$ is a bounded map from $L_c^\infty$ to BMO.

My question lies in the implication from 3 to 2. I understand that for any $\infty$-atom $a$,i.e., $supp(a) \subset Q$ and $||a||_\infty \leq \frac{1}{|Q|}$, $$||Ta||_1 \leq C.$$ However, I do not see how this implies $||Ta||_1 \leq C||a||_{H^{1,\infty}}$.

More precisely, since $||a||_{H^{1,\infty}} \leq 1$ for this $\infty$-atom, how do we know that $C \leq C'\cdot||a||_{H^{1,\infty}}$?