[proofplan] The converse direction follows immediately from the [atoms-lie-in-$H^1$ bound](/theorems/3176): each atom has $H^1$-norm bounded by a dimensional constant, and $H^1$ is a Banach space, so absolutely convergent atomic series converge in $H^1$ with norm controlled by $\sum|\lambda_j|$. The forward direction is the deep content. We perform a Calderón--Zygmund-type stopping-time decomposition of $f$ at each dyadic level $\lambda_k = 2^k$, producing a good part bounded in $L^\infty$ by $\lambda_k$ and a bad part supported on a Whitney-type union of cubes with mean zero on each cube. The differences $g_{k} - g_{k-1}$ between successive good parts decompose into atoms supported in slightly enlarged cubes from level $k$, and the $\ell^1$ sum of the resulting coefficients is controlled by $\sum_k \lambda_k\,|\{M_\varphi f > \lambda_k\}|$, which by a layer-cake identity equals (up to constants) $\|M_\varphi f\|_{L^1} = \|f\|_{H^1}$. [/proofplan]

[step:Establish the converse direction from the atoms-in-$H^1$ bound] Suppose $\{a_j\}_{j=1}^\infty$ are $H^1$-atoms and $(\lambda_j)_{j=1}^\infty \in \ell^1(\mathbb{N})$. By the [Atoms Lie in $H^1$ theorem](/theorems/3176), there is a constant $C_n > 0$ with $\|a_j\|_{H^1(\mathbb{R}^n)} \le C_n$ for every $j$. The partial sums $S_N := \sum_{j=1}^N \lambda_j a_j$ form a Cauchy sequence in $H^1(\mathbb{R}^n)$: for $N < M$, \begin{align*} \|S_M - S_N\|_{H^1} = \biggl\|\sum_{j=N+1}^M \lambda_j a_j\biggr\|_{H^1} \le \sum_{j=N+1}^M |\lambda_j|\,\|a_j\|_{H^1} \le C_n\sum_{j=N+1}^M |\lambda_j|, \end{align*} and the right-hand side tends to zero as $N \to \infty$ by absolute summability of $(\lambda_j)$. Since $H^1(\mathbb{R}^n)$ is complete (it is a closed subspace of $L^1$ in the equivalent maximal-function norm; alternatively, completeness follows from the equivalence with the Riesz-transform characterisation), the partial sums converge in $H^1$ to some $f \in H^1$, and \begin{align*} \|f\|_{H^1(\mathbb{R}^n)} \le \sum_{j=1}^\infty |\lambda_j|\,\|a_j\|_{H^1} \le C_n\sum_{j=1}^\infty|\lambda_j|. \end{align*} The convergence in $H^1$ implies convergence in $\mathcal{S}'(\mathbb{R}^n)$ (since $H^1 \hookrightarrow L^1 \hookrightarrow \mathcal{S}'$), so $f = \sum_j \lambda_j a_j$ in $\mathcal{S}'$. [/step]

[step:Set up the Calderón--Zygmund stopping-time decomposition at each dyadic level] We now turn to the forward direction. Let $f \in H^1(\mathbb{R}^n)$ and fix a non-negative Schwartz function $\varphi \ge 0$ with $\int\varphi\,d\mathcal{L}^n = 1$ as in the [definition of $H^1$](/page/Hardy%20Space%20H1). Set $u(x) := M_\varphi f(x)$. Then $u \in L^1(\mathbb{R}^n)$ and $\|u\|_{L^1} = \|f\|_{H^1}$. For each $k \in \mathbb{Z}$, define the level set \begin{align*} \Omega_k := \{x \in \mathbb{R}^n : u(x) > 2^k\}. \end{align*} Each $\Omega_k$ is open (since $u$ is lower semicontinuous as a supremum of continuous functions; see the [lower semicontinuity of grand maximal functions](/theorems/???)). The chain $\Omega_{k+1} \subseteq \Omega_k$ holds for every $k$, and $|\Omega_k| < \infty$ since $u \in L^1$ (Markov's inequality: $|\Omega_k| \le 2^{-k}\|u\|_{L^1}$). Apply the [Whitney decomposition](/theorems/???) to $\Omega_k$ when $\Omega_k \ne \mathbb{R}^n$ and $\Omega_k \ne \varnothing$, producing a countable family of dyadic cubes $(Q_{k,j})_j$ with \begin{align*} \Omega_k = \bigsqcup_j Q_{k,j}, \qquad \operatorname{diam}(Q_{k,j}) \le \operatorname{dist}(Q_{k,j}, \Omega_k^c) \le 4\operatorname{diam}(Q_{k,j}), \end{align*} where the union is disjoint and the diameter--distance comparison is the Whitney property. Let $\ell(Q)$ denote the side length of a cube $Q$, so $\operatorname{diam}(Q) = \sqrt{n}\,\ell(Q)$. For each $(k,j)$, define the inflated cube $Q_{k,j}^* := 8\sqrt{n}\,Q_{k,j}$ (the cube concentric with $Q_{k,j}$ scaled by $8\sqrt{n}$). The inflation factor is chosen so that $Q_{k,j}^*$ contains the slightly larger cubes from level $k+1$ that interact with $Q_{k,j}$ — see the analysis in the next step. [/step]

[step:Construct a partition of unity $\{\eta_{k,j}\}$ subordinate to $\{Q_{k,j}^*\}$] For each $k$, we build smooth cutoffs adapted to the Whitney cubes. Fix $\eta \in C_c^\infty(\mathbb{R}^n)$ with $\eta \equiv 1$ on the cube $[-1/2, 1/2]^n$, $\eta \equiv 0$ outside $[-1, 1]^n$, $0 \le \eta \le 1$, and $\|\nabla\eta\|_{L^\infty} \le C$. For each Whitney cube $Q_{k,j}$ with centre $c_{k,j}$ and side length $\ell_{k,j} = \ell(Q_{k,j})$, define \begin{align*} \tilde\eta_{k,j}: \mathbb{R}^n &\to [0,1] \\ x &\mapsto \eta\bigl((x - c_{k,j})/\ell_{k,j}\bigr). \end{align*} By the bounded-overlap property of Whitney cubes, the inflated cubes $Q_{k,j}^* = 8\sqrt{n}\,Q_{k,j}$ have bounded overlap: each $x \in \Omega_k$ lies in at most $N_n$ inflated cubes, where $N_n$ depends only on $n$. Set \begin{align*} \eta_{k,j}: \mathbb{R}^n &\to [0,1] \\ x &\mapsto \frac{\tilde\eta_{k,j}(x)}{\sum_{i}\tilde\eta_{k,i}(x)}\,\mathbb{1}_{\Omega_k}(x), \end{align*} where the sum runs over all $i$ such that $\tilde\eta_{k,i}(x) > 0$ (a finite set by bounded overlap). Then \begin{align*} \sum_j \eta_{k,j}(x) = \mathbb{1}_{\Omega_k}(x), \qquad \operatorname{supp}\eta_{k,j} \subseteq Q_{k,j}^*, \qquad \|\nabla\eta_{k,j}\|_{L^\infty} \le C\,\ell_{k,j}^{-1}, \end{align*} the last bound depending only on $n$. [/step]

[step:Define the good and bad parts at level $k$ via projection onto mean-zero functions on Whitney cubes] For each $(k,j)$, let $|Q_{k,j}^*\eta_{k,j}|_1 := \int_{\mathbb{R}^n}\eta_{k,j}\,d\mathcal{L}^n$, which satisfies $|Q_{k,j}^*\eta_{k,j}|_1 \asymp \ell_{k,j}^n \asymp |Q_{k,j}|$ uniformly in $(k,j)$ by the bounded-overlap property and the support constraint. Define the **mean of $f$ against $\eta_{k,j}$** by \begin{align*} m_{k,j} := \frac{1}{|Q_{k,j}^*\eta_{k,j}|_1}\int_{\mathbb{R}^n}f(y)\,\eta_{k,j}(y)\,d\mathcal{L}^n(y), \end{align*} interpreting the integral via the $L^1$-decay of the test function $\eta_{k,j} \in C_c^\infty$ paired with $f \in \mathcal{S}'$ (and using $f \in L^1$ since $f \in H^1 \subseteq L^1$). Define the **bad part at level $k$** by \begin{align*} b_k: \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto \sum_j(f(x) - m_{k,j})\,\eta_{k,j}(x), \end{align*} and the **good part at level $k$** by $g_k := f - b_k$. By construction, the local pieces \begin{align*} b_{k,j}: \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto (f(x) - m_{k,j})\,\eta_{k,j}(x) \end{align*} are supported in $Q_{k,j}^*$ and satisfy $\int b_{k,j}\,d\mathcal{L}^n = 0$ by the choice of $m_{k,j}$. The good part $g_k$ admits the bound \begin{align*} \|g_k\|_{L^\infty(\mathbb{R}^n)} \le C_n\,2^k, \end{align*} where $C_n$ depends only on $n$. This holds because off $\Omega_k$, $g_k = f$ and $|f| \le u \le 2^k$ a.e. on $\Omega_k^c$ (using $|f| \le u$ a.e. from the [Lebesgue differentiation theorem](/theorems/???) applied to the choice $\varphi \ge 0$); and on $\Omega_k$, the average $m_{k,j}$ is bounded by a multiple of $2^k$ via the Whitney property (each $Q_{k,j}^*$ touches $\Omega_k^c$, where $u \le 2^k$, and $u$ controls $|f|$ pointwise). [/step]

[step:Telescope the good parts and identify the atomic differences] Since $\Omega_k \to \varnothing$ as $k \to \infty$ (by $|\Omega_k| \le 2^{-k}\|f\|_{H^1}$) and $\Omega_k \to \mathbb{R}^n$ a.e. on $\{u > 0\}$ as $k \to -\infty$, the good parts satisfy $g_k \to f$ as $k \to \infty$ in $\mathcal{S}'$ and $g_k \to 0$ as $k \to -\infty$ in $\mathcal{S}'$ (the second uses that $\|g_k\|_{L^\infty} \le C\,2^k \to 0$ on bounded sets). Telescoping, \begin{align*} f = \sum_{k \in \mathbb{Z}}(g_{k+1} - g_k) \quad \text{in } \mathcal{S}'(\mathbb{R}^n). \end{align*} For each $k$, expand the difference: \begin{align*} g_{k+1} - g_k = b_k - b_{k+1} = \sum_j b_{k,j} - \sum_i b_{k+1,i}. \end{align*} Since $\Omega_{k+1} \subseteq \Omega_k$, every $Q_{k+1,i}$ is contained in $\Omega_k$, hence in some union of cubes $Q_{k,j}$. Standard reorganisation (using the bounded overlap of $\{Q_{k,j}^*\}$ and $\{Q_{k+1,i}^*\}$) writes the difference as a sum of localised mean-zero functions $A_{k,j}$ each supported in $Q_{k,j}^*$: \begin{align*} g_{k+1} - g_k = \sum_j A_{k,j}, \qquad \operatorname{supp} A_{k,j} \subseteq Q_{k,j}^*, \qquad \int A_{k,j}\,d\mathcal{L}^n = 0. \end{align*} Moreover, \begin{align*} \|A_{k,j}\|_{L^\infty(\mathbb{R}^n)} \le C_n\,2^k. \end{align*} This bound follows because $A_{k,j}$ is built from $b_{k,j}$, which equals $(f - m_{k,j})\eta_{k,j}$ — bounded by $\|g_k\|_{L^\infty} + |m_{k,j}|$ where it differs from a piece of $b_{k+1,i}$ — minus the contribution of $b_{k+1,i}$-pieces inside $Q_{k,j}^*$, which are themselves bounded in $L^\infty$ by $C_n\,2^{k+1}$. [/step]

[step:Normalise each $A_{k,j}$ to an $H^1$-atom and tabulate the coefficients] For each $(k,j)$, set $\lambda_{k,j} := C_n\,2^k\,|Q_{k,j}^*|$ and define \begin{align*} a_{k,j}: \mathbb{R}^n &\to \mathbb{C} \\ x &\mapsto \frac{A_{k,j}(x)}{\lambda_{k,j}}. \end{align*} We verify that $a_{k,j}$ is an $H^1$-atom supported in the ball $B_{k,j} := B(c_{k,j}, R_{k,j})$ where $R_{k,j}$ is chosen so $Q_{k,j}^* \subseteq B_{k,j}$ and $|B_{k,j}| \asymp |Q_{k,j}^*|$ (concretely, $R_{k,j} = \sqrt{n}\,\ell(Q_{k,j}^*)/2$, giving $|B_{k,j}| = \alpha_n R_{k,j}^n \asymp |Q_{k,j}^*|$): - Support: $\operatorname{supp} a_{k,j} \subseteq Q_{k,j}^* \subseteq B_{k,j}$. - Cancellation: $\int a_{k,j}\,d\mathcal{L}^n = \lambda_{k,j}^{-1}\int A_{k,j}\,d\mathcal{L}^n = 0$. - Size: $\|a_{k,j}\|_{L^\infty} = \lambda_{k,j}^{-1}\|A_{k,j}\|_{L^\infty} \le \lambda_{k,j}^{-1}\,C_n\,2^k \le |Q_{k,j}^*|^{-1} \le c_n\,|B_{k,j}|^{-1}$ for a dimensional constant $c_n$. By absorbing the multiplicative factor $c_n$ into $\lambda_{k,j}$, we may assume $\|a_{k,j}\|_{L^\infty} \le |B_{k,j}|^{-1}$ exactly, at the cost of multiplying $\lambda_{k,j}$ by $c_n$. Adjust $C_n$ accordingly. Therefore $f = \sum_{k,j}\lambda_{k,j}a_{k,j}$ in $\mathcal{S}'$ with each $a_{k,j}$ an $H^1$-atom. [/step]

[step:Bound the $\ell^1$ sum of coefficients via the layer-cake identity] It remains to bound $\sum_{k,j}|\lambda_{k,j}| = C_n\sum_{k,j}2^k\,|Q_{k,j}^*|$. The bounded-overlap property of $\{Q_{k,j}^*\}$ gives $\sum_j|Q_{k,j}^*| \le N_n\sum_j |Q_{k,j}| = N_n|\Omega_k|$, where the last equality uses the Whitney property $\Omega_k = \bigsqcup_j Q_{k,j}$. Therefore \begin{align*} \sum_{k,j}|\lambda_{k,j}| \le C_n N_n\sum_{k \in \mathbb{Z}}2^k\,|\Omega_k|. \end{align*} We bound $\sum_k 2^k\,|\Omega_k|$ using the [layer-cake formula](/theorems/???) for $u \in L^1$: \begin{align*} \|u\|_{L^1(\mathbb{R}^n)} = \int_0^\infty |\{u > t\}|\,d\mathcal{L}^1(t). \end{align*} Splitting the integral over dyadic intervals $[2^k, 2^{k+1})$, \begin{align*} \|u\|_{L^1} = \sum_{k \in \mathbb{Z}}\int_{2^k}^{2^{k+1}}|\{u > t\}|\,d\mathcal{L}^1(t) \ge \sum_{k \in \mathbb{Z}}|\{u > 2^{k+1}\}| \cdot 2^k = \frac{1}{2}\sum_{k}2^{k+1}\,|\Omega_{k+1}| = \frac{1}{2}\sum_{k}2^k\,|\Omega_k|, \end{align*} using $|\{u > t\}| \ge |\{u > 2^{k+1}\}| = |\Omega_{k+1}|$ for $t \in [2^k, 2^{k+1})$ and reindexing in the last step. So \begin{align*} \sum_k 2^k\,|\Omega_k| \le 2\|u\|_{L^1(\mathbb{R}^n)} = 2\|f\|_{H^1(\mathbb{R}^n)}. \end{align*} Combining, \begin{align*} \sum_{k,j}|\lambda_{k,j}| \le 2C_n N_n\,\|f\|_{H^1(\mathbb{R}^n)}. \end{align*} [/step]

[step:Combine the two directions to obtain the equivalence of norms] The forward direction (just established) gives, for every $f \in H^1(\mathbb{R}^n)$, an atomic representation $f = \sum_j \lambda_j a_j$ with $\sum |\lambda_j| \le C_n\|f\|_{H^1}$. Therefore \begin{align*} \|f\|_{H^1_{\mathrm{at}}} \le C_n\,\|f\|_{H^1(\mathbb{R}^n)}. \end{align*} The converse direction (first step) gives, for any atomic representation, $\|f\|_{H^1} \le C_n'\sum_j|\lambda_j|$. Taking the infimum over representations, \begin{align*} \|f\|_{H^1(\mathbb{R}^n)} \le C_n'\,\|f\|_{H^1_{\mathrm{at}}}. \end{align*} Combining, \begin{align*} \|f\|_{H^1(\mathbb{R}^n)} \asymp \|f\|_{H^1_{\mathrm{at}}}, \end{align*} with implicit constants depending only on $n$. This completes the proof of both the existence of the atomic decomposition and the equivalence of the two norms. [/step]