[proofplan]
By the [Coifman--Latter Atomic Decomposition](/theorems/3177), every $f \in H^1$ admits a representation $f = \sum_j \lambda_j a_j$ with $H^1$-atoms $a_j$ and $\sum |\lambda_j| \asymp \|f\|_{H^1}$. We reduce the bound to a uniform $L^1$ estimate $\|Ta\|_{L^1} \le C$ for atoms $a$, then split $\mathbb{R}^n = 2B \sqcup (2B)^c$ where $B$ is the supporting ball of $a$. On $2B$ we use $L^2$-boundedness of $T$ together with Cauchy--Schwarz to obtain a dimensional bound. On $(2B)^c$ we use the cancellation $\int a\,d\mathcal{L}^n = 0$ to subtract a constant from the kernel and invoke the Hörmander condition. Both regions contribute constants depending only on $A$, $\|T\|_{\mathcal{L}(L^2)}$, and $n$, and the atomic series then assembles the global $H^1 \to L^1$ bound.
[/proofplan]
[step:Reduce to a uniform $L^1$ bound on atoms via the atomic decomposition]
Let $f \in H^1(\mathbb{R}^n)$. By the [Coifman--Latter Atomic Decomposition](/theorems/3177), there exist $H^1$-atoms $\{a_j\}_{j=1}^\infty$ and scalars $\{\lambda_j\}_{j=1}^\infty \subseteq \mathbb{C}$ with
\begin{align*}
f = \sum_{j=1}^\infty \lambda_j a_j \quad \text{in } \mathcal{S}'(\mathbb{R}^n), \qquad \sum_{j=1}^\infty |\lambda_j| \le C_n^{\mathrm{at}}\,\|f\|_{H^1(\mathbb{R}^n)}.
\end{align*}
Since $H^1 \hookrightarrow L^1$, the same series converges in $L^1(\mathbb{R}^n)$.
Suppose we have shown
\begin{align*}
\|Ta\|_{L^1(\mathbb{R}^n)} \le C_0 \tag{$\star$}
\end{align*}
for some constant $C_0 = C_0(A, \|T\|_{\mathcal{L}(L^2)}, n)$ and every $H^1$-atom $a$. Then the partial sums $S_N := \sum_{j=1}^N \lambda_j a_j$ satisfy
\begin{align*}
\|TS_N\|_{L^1} \le \sum_{j=1}^N |\lambda_j|\,\|Ta_j\|_{L^1} \le C_0\sum_{j=1}^\infty|\lambda_j| < \infty,
\end{align*}
and the partial sums $TS_N$ form a Cauchy sequence in $L^1$ by the same estimate restricted to tails. So $TS_N \to g$ in $L^1$ for some $g$. Define $Tf := g$; this is consistent with the original definition on $L^2$ by density. Then
\begin{align*}
\|Tf\|_{L^1(\mathbb{R}^n)} \le C_0\sum_{j=1}^\infty|\lambda_j| \le C_0 C_n^{\mathrm{at}}\,\|f\|_{H^1(\mathbb{R}^n)},
\end{align*}
which is the conclusion. It remains to establish ($\star$).
[/step]
[step:Bound $\|Ta\|_{L^1(2B)}$ via $L^2$-boundedness and Cauchy--Schwarz]
Fix an $H^1$-atom $a: \mathbb{R}^n \to \mathbb{C}$ supported in a ball $B = B(x_B, r_B)$ with $\|a\|_{L^\infty} \le |B|^{-1}$ and $\int a\,d\mathcal{L}^n = 0$. Write $2B = B(x_B, 2r_B)$ and $|2B| = 2^n|B|$.
Since $a \in L^2(\mathbb{R}^n)$ — indeed, $\|a\|_{L^2}^2 = \int_B|a|^2\,d\mathcal{L}^n \le |B|\,|B|^{-2} = |B|^{-1}$ — and $T: L^2 \to L^2$ is bounded with operator norm $\|T\|_{\mathcal{L}(L^2)}$,
\begin{align*}
\|Ta\|_{L^2(\mathbb{R}^n)} \le \|T\|_{\mathcal{L}(L^2)}\,\|a\|_{L^2(\mathbb{R}^n)} \le \|T\|_{\mathcal{L}(L^2)}\,|B|^{-1/2}.
\end{align*}
Restricting to $2B$ and applying the [Cauchy--Schwarz inequality](/theorems/???) with weight $\mathbb{1}_{2B}$,
\begin{align*}
\|Ta\|_{L^1(2B)} = \int_{\mathbb{R}^n}\mathbb{1}_{2B}(x)\,|Ta(x)|\,d\mathcal{L}^n(x) \le \|\mathbb{1}_{2B}\|_{L^2(\mathbb{R}^n)}\,\|Ta\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Since $\|\mathbb{1}_{2B}\|_{L^2}^2 = |2B| = 2^n|B|$, $\|\mathbb{1}_{2B}\|_{L^2} = 2^{n/2}|B|^{1/2}$. Therefore
\begin{align*}
\|Ta\|_{L^1(2B)} \le 2^{n/2}|B|^{1/2}\cdot\|T\|_{\mathcal{L}(L^2)}\,|B|^{-1/2} = 2^{n/2}\|T\|_{\mathcal{L}(L^2)}.
\end{align*}
This bound depends only on $\|T\|_{\mathcal{L}(L^2)}$ and $n$.
[/step]
[step:Bound $\|Ta\|_{L^1((2B)^c)}$ via the cancellation of the atom and the Hörmander condition]
For $x \in (2B)^c$, the kernel representation $Ta(x) = \int K(x,y)a(y)\,d\mathcal{L}^n(y)$ holds because $a \in L^2$ has compact support in $B$ and $x \notin B$, so $K(x,y)$ is locally bounded for $y \in B$ (using the size condition $|K(x,y)| \le A|x-y|^{-n}$ with $|x - y| \ge r_B > 0$). The integral $\int_B K(x,y)a(y)\,d\mathcal{L}^n(y)$ converges absolutely.
Using $\int_B a(y)\,d\mathcal{L}^n(y) = 0$, we may subtract the constant $K(x, x_B)\int a\,d\mathcal{L}^n = 0$:
\begin{align*}
Ta(x) = \int_B [K(x,y) - K(x,x_B)]\,a(y)\,d\mathcal{L}^n(y), \qquad x \in (2B)^c.
\end{align*}
Taking the absolute value and integrating over $(2B)^c$,
\begin{align*}
\|Ta\|_{L^1((2B)^c)} &= \int_{(2B)^c}\biggl|\int_B[K(x,y) - K(x,x_B)]\,a(y)\,d\mathcal{L}^n(y)\biggr|\,d\mathcal{L}^n(x) \\
&\le \int_{(2B)^c}\int_B|K(x,y) - K(x,x_B)|\,|a(y)|\,d\mathcal{L}^n(y)\,d\mathcal{L}^n(x).
\end{align*}
By [Tonelli's theorem](/theorems/???) applied to the non-negative integrand on the product measure space $((2B)^c, \mathcal{L}^n) \times (B, \mathcal{L}^n)$,
\begin{align*}
\|Ta\|_{L^1((2B)^c)} \le \int_B|a(y)|\biggl(\int_{(2B)^c}|K(x,y) - K(x,x_B)|\,d\mathcal{L}^n(x)\biggr)d\mathcal{L}^n(y).
\end{align*}
We bound the inner integral by the Hörmander condition. For $y \in B$ and $x \in (2B)^c$,
\begin{align*}
|x - x_B| \ge 2r_B \ge 2|y - x_B|,
\end{align*}
i.e., $x$ lies in the set $\{x : |x - x_B| > 2|y - x_B|\}$. (Here we set $z := x_B$ in the Hörmander condition's variables — the condition is symmetric in the second-argument variables, written with $y, z$, and we are using it with $z = x_B$ and "$y$" $= y$, so the condition reads $\int_{|x - y| > 2|x_B - y|}|K(x, x_B) - K(x, y)|d\mathcal{L}^n(x) \le A$. The inclusion $\{x : |x - x_B| > 2r_B\} \subseteq \{x : |x - y| > 2|x_B - y|\}$ holds because $|x - y| \ge |x - x_B| - |x_B - y| \ge |x - x_B| - r_B \ge \tfrac{1}{2}|x - x_B| > 2r_B \ge 2|x_B - y|$.)
Therefore
\begin{align*}
\int_{(2B)^c}|K(x,y) - K(x,x_B)|\,d\mathcal{L}^n(x) \le \int_{|x-y| > 2|x_B - y|}|K(x,y) - K(x,x_B)|\,d\mathcal{L}^n(x) \le A.
\end{align*}
Substituting back,
\begin{align*}
\|Ta\|_{L^1((2B)^c)} \le A\int_B|a(y)|\,d\mathcal{L}^n(y) = A\,\|a\|_{L^1(\mathbb{R}^n)} \le A\cdot|B|\cdot|B|^{-1} = A,
\end{align*}
using $\|a\|_{L^1} \le |B|\|a\|_{L^\infty} \le 1$.
[/step]
[step:Combine the two regions to obtain the uniform atom bound and conclude]
From the two preceding steps,
\begin{align*}
\|Ta\|_{L^1(\mathbb{R}^n)} = \|Ta\|_{L^1(2B)} + \|Ta\|_{L^1((2B)^c)} \le 2^{n/2}\|T\|_{\mathcal{L}(L^2)} + A =: C_0.
\end{align*}
The constant $C_0$ depends only on $A$, $\|T\|_{\mathcal{L}(L^2)}$, and $n$. This establishes ($\star$) from the first step, and combined with the atomic decomposition gives
\begin{align*}
\|Tf\|_{L^1(\mathbb{R}^n)} \le C_0 C_n^{\mathrm{at}}\,\|f\|_{H^1(\mathbb{R}^n)},
\end{align*}
which is the bound claimed by the theorem with $C_{A,n} := C_0 C_n^{\mathrm{at}}$. The $L^2$-defined operator $T$ is hereby extended to a bounded $T: H^1(\mathbb{R}^n) \to L^1(\mathbb{R}^n)$, completing the proof.
[/step]
