[proofplan] The proof has two halves linked by the [atomic decomposition](/theorems/???) of $H^1(\mathbb{R}^n)$. **Forward direction (BMO acts on $H^1$):** for any $H^1$-atom $a$ supported in a ball $B$ with $\int a\, d\mathcal{L}^n = 0$ and $\|a\|_\infty \le |B|^{-1}$, the cancellation $\int a = 0$ lets us replace $g$ by $g - g_B$ and bound the pairing $|\int a g| \le \|g\|_{\mathrm{BMO}}$ uniformly over atoms. Linearity and the atomic decomposition propagate this to a bound on general $f \in H^1$, yielding $\|\Lambda_g\|_{(H^1)^*} \le C_n\|g\|_{\mathrm{BMO}}$. **Converse direction (every functional comes from BMO):** restrict $\Lambda \in (H^1)^*$ to the dense subspace $H^1\cap L^2$, represent it via the [$L^2$ Riesz representation theorem](/theorems/???) by an element $g \in L^2_{\mathrm{loc}}$, and verify $g \in \mathrm{BMO}$ by testing $\Lambda$ against atoms supported in a fixed cube $Q$ with zero mean — these are the dual mean-zero $L^2$ functions on $Q$. The boundedness of $\Lambda$ on $H^1$ controls the $L^2$ oscillation of $g$ over $Q$ uniformly, hence via the [equivalence of BMO $L^p$ norms](/theorems/3181) gives $g \in \mathrm{BMO}$. [/proofplan] [step:Recall atomic structure of $H^1$ and the pairing of an atom with a BMO function] Recall the [atomic characterisation of $H^1(\mathbb{R}^n)$](/theorems/???): there exists a constant $C_n > 0$ such that every $f \in H^1(\mathbb{R}^n)$ admits a decomposition \begin{align*} f = \sum_{j = 1}^\infty \lambda_j a_j \qquad \text{in } H^1(\mathbb{R}^n), \end{align*} where each $a_j: \mathbb{R}^n \to \mathbb{R}$ is a **$(1, \infty)$-atom**, namely there is a ball $B_j \subset \mathbb{R}^n$ with \begin{align*} \operatorname{supp} a_j \subseteq B_j, \qquad \|a_j\|_{L^\infty(\mathbb{R}^n)} \le |B_j|^{-1}, \qquad \int_{\mathbb{R}^n}a_j\, d\mathcal{L}^n = 0, \end{align*} and $\lambda_j \in \mathbb{R}$ with $\sum_j |\lambda_j| \le C_n\|f\|_{H^1}$. Conversely, any series $f := \sum_j\lambda_j a_j$ of atoms with $\sum|\lambda_j| < \infty$ converges in $H^1$ with $\|f\|_{H^1} \le \sum|\lambda_j|$. For an atom $a$ supported in a ball $B$ and a function $g \in \mathrm{BMO}(\mathbb{R}^n)$, define $g_B := \frac{1}{|B|}\int_B g\, d\mathcal{L}^n$. We claim \begin{align*} \left|\int_{\mathbb{R}^n}a(x)g(x)\, d\mathcal{L}^n(x)\right| \le \|g\|_{\mathrm{BMO}}. \tag{Atom-BMO} \end{align*} The key observation is that since $\int a = 0$, \begin{align*} \int_{\mathbb{R}^n}a(x)g(x)\, d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n}a(x)\bigl(g(x) - g_B\bigr)\, d\mathcal{L}^n(x) \\ &= \int_B a(x)\bigl(g(x) - g_B\bigr)\, d\mathcal{L}^n(x), \end{align*} where the second equality uses $\operatorname{supp}a \subseteq B$. Subtracting $g_B$ is legitimate because $\int_{\mathbb{R}^n}a\cdot g_B\, d\mathcal{L}^n = g_B\int_{\mathbb{R}^n}a\, d\mathcal{L}^n = 0$ — the cancellation property of the atom is exactly what makes this subtraction permissible. Applying $|ab| \le |a||b|$ pointwise and the bound $\|a\|_{L^\infty} \le |B|^{-1}$, \begin{align*} \left|\int_B a(x)\bigl(g(x) - g_B\bigr)\, d\mathcal{L}^n(x)\right| \le \|a\|_{L^\infty}\int_B |g(x) - g_B|\, d\mathcal{L}^n(x) \le \frac{1}{|B|}\int_B |g - g_B|\, d\mathcal{L}^n. \end{align*} By the definition of the BMO seminorm applied to the **ball** $B$ — the BMO definition by axis-parallel cubes and the BMO definition by balls are equivalent up to a dimensional constant; we absorb this constant into $\|g\|_{\mathrm{BMO}}$ — the right-hand side is at most $\|g\|_{\mathrm{BMO}}$, proving (Atom-BMO). [/step] [step:Construct $\Lambda_g$ on $H^1$ via the atomic decomposition and bound its operator norm] Let $g \in \mathrm{BMO}(\mathbb{R}^n)$ be fixed. We construct a bounded linear functional $\Lambda_g: H^1(\mathbb{R}^n)\to\mathbb{R}$ as follows. **Definition on atoms and finite atomic sums.** For every atom $a$ supported in $B$, the integral $\int_{\mathbb{R}^n}a g\, d\mathcal{L}^n$ is well-defined: $a \in L^\infty$ has compact support and $g \in L^1_{\mathrm{loc}}$, so $a g \in L^1(\mathbb{R}^n)$ with $\int |ag|\, d\mathcal{L}^n \le \|a\|_\infty\int_B|g|\, d\mathcal{L}^n < \infty$. Therefore, for any finite atomic sum $f = \sum_{j = 1}^N\lambda_j a_j$, set \begin{align*} \Lambda_g(f) := \sum_{j = 1}^N \lambda_j \int_{\mathbb{R}^n}a_j g\, d\mathcal{L}^n. \end{align*} This is linear in $f$ over finite atomic sums. **Bound on finite atomic sums.** By (Atom-BMO), \begin{align*} |\Lambda_g(f)| \le \sum_{j = 1}^N|\lambda_j|\cdot \left|\int a_j g\, d\mathcal{L}^n\right| \le \sum_{j = 1}^N|\lambda_j|\cdot \|g\|_{\mathrm{BMO}}. \end{align*} Taking the infimum over atomic decompositions of $f$ (the atomic norm $\|f\|_{H^1}^{\mathrm{at}} := \inf\{\sum|\lambda_j| : f = \sum\lambda_j a_j \text{ atomically}\}$ is equivalent to $\|f\|_{H^1}$ up to the dimensional constant $C_n$ from step 1), \begin{align*} |\Lambda_g(f)| \le C_n\|g\|_{\mathrm{BMO}}\|f\|_{H^1}. \tag{$\star$} \end{align*} **Extension to all of $H^1$.** Finite atomic sums are dense in $H^1$. For a general $f \in H^1$ with atomic decomposition $f = \sum_{j = 1}^\infty \lambda_j a_j$, the partial sums $f_N := \sum_{j = 1}^N\lambda_j a_j$ converge to $f$ in $H^1$, and the sequence $(\Lambda_g(f_N))_{N\ge 1}$ is Cauchy in $\mathbb{R}$ because \begin{align*} |\Lambda_g(f_N) - \Lambda_g(f_M)| = |\Lambda_g(f_N - f_M)| \le C_n\|g\|_{\mathrm{BMO}}\|f_N - f_M\|_{H^1} \to 0 \end{align*} as $M, N \to \infty$. Define $\Lambda_g(f) := \lim_{N\to\infty}\Lambda_g(f_N)$. The limit is independent of the atomic decomposition: if $f = \sum_j\lambda_j a_j = \sum_j\mu_j b_j$ are two decompositions, the difference $0 = \sum_j\lambda_j a_j - \sum_j\mu_j b_j$ has atomic norm zero, and the bound (\star) applied to truncations gives $|\Lambda_g(\text{difference})_N| \to 0$. Hence the two limits agree. **Identification with the truncation formula.** When $f \in H^1\cap L^2$, define $f_j(x) := f(x)\mathbb{1}_{|f|\le j}\mathbb{1}_{B(0,j)}(x)$ for $j \ge 1$. Then $f_j \in L^2(\mathbb{R}^n)$ with compact support and $f_j \to f$ in $H^1$ (a standard density argument using continuity of the maximal function characterisation of $H^1$). The atomic decomposition can be chosen so that $\Lambda_g(f_j) = \int_{\mathbb{R}^n}f_j g\, d\mathcal{L}^n$, an absolutely convergent integral for each $j$ since $f_j \in L^2_{\mathrm{c}}$ and $g \in L^2_{\mathrm{loc}}$. Continuity of $\Lambda_g$ in $H^1$ then gives $\Lambda_g(f) = \lim_{j\to\infty}\int f_j g\, d\mathcal{L}^n$ for $f \in H^1\cap L^2$, and by density on all of $H^1$. **Linearity and operator norm.** Linearity of $\Lambda_g$ is inherited from linearity of integration on each atomic piece. The operator norm satisfies $\|\Lambda_g\|_{(H^1)^*} \le C_n\|g\|_{\mathrm{BMO}}$ by passing to the limit in (\star). [/step] [step:Reverse direction — represent $\Lambda \in (H^1)^*$ on $H^1\cap L^2$ via Riesz] Let $\Lambda \in (H^1)^*$. We construct $g \in \mathrm{BMO}(\mathbb{R}^n)$ with $\Lambda = \Lambda_g$, and bound $\|g\|_{\mathrm{BMO}}\le C_n\|\Lambda\|_{(H^1)^*}$. **Step (i) — Define $\Lambda$ on $H^1\cap L^2$.** Recall $H^1\cap L^2$ is dense in $H^1$ (atoms are in $L^2$, and finite atomic sums are dense). The restriction $\Lambda|_{H^1\cap L^2}$ is bounded by $\|\Lambda\|_{(H^1)^*}\|\cdot\|_{H^1}$. **Step (ii) — Localise to a cube.** Fix a cube $Q \subset \mathbb{R}^n$. Define \begin{align*} \mathcal{V}_Q := \left\{\phi \in L^2(\mathbb{R}^n) : \operatorname{supp}\phi \subseteq Q,\ \int_Q\phi\, d\mathcal{L}^n = 0\right\}. \end{align*} This is a closed subspace of $L^2(\mathbb{R}^n)$ — closed because the supremum-support condition is closed under $L^2$ limits and the linear functional $\phi \mapsto \int_Q\phi\, d\mathcal{L}^n$ is continuous on $L^2(Q)$. We claim $\mathcal{V}_Q\subseteq H^1$ with \begin{align*} \|\phi\|_{H^1} \le C_n |Q|^{1/2}\|\phi\|_{L^2(Q)} \qquad \text{for all } \phi \in \mathcal{V}_Q. \end{align*} To see this, normalise $\phi$ to have $\|\phi\|_{L^2(Q)} = 1$ and observe that $a := \phi/(|Q|^{1/2})$ is a $(2, \infty)$-atom on a $(1, 2)$-atom in the relaxed sense (depending on the convention): $\operatorname{supp}a\subseteq Q$, $\int_Q a = 0$, and $\|a\|_{L^2} \le |Q|^{-1/2}$. By the [equivalence of $H^1$ via $(1, q)$-atoms for any $q \in (1, \infty]$](/theorems/???), $a$ is an $H^1$-atom (up to a dimensional constant), so $\|a\|_{H^1} \le C_n$, hence $\|\phi\|_{H^1} \le C_n|Q|^{1/2}$. **Step (iii) — Apply Riesz representation on $\mathcal{V}_Q$.** The map $\phi \mapsto \Lambda(\phi)$ is linear on $\mathcal{V}_Q$ and bounded: \begin{align*} |\Lambda(\phi)| \le \|\Lambda\|_{(H^1)^*}\|\phi\|_{H^1} \le C_n\|\Lambda\|_{(H^1)^*}|Q|^{1/2}\|\phi\|_{L^2(Q)}. \end{align*} Hence $\Lambda|_{\mathcal{V}_Q}$ is a bounded linear functional on the closed subspace $\mathcal{V}_Q$ of $L^2(Q)$, with norm at most $C_n\|\Lambda\|_{(H^1)^*}|Q|^{1/2}$. By the [Hahn–Banach theorem](/theorems/???) we extend $\Lambda|_{\mathcal{V}_Q}$ to a bounded linear functional on $L^2(Q)$ with the same norm; by the [Riesz representation theorem on $L^2$](/theorems/???), there exists $g_Q \in L^2(Q)$ with \begin{align*} \Lambda(\phi) = \int_Q \phi(x)g_Q(x)\, d\mathcal{L}^n(x) \qquad \text{for all } \phi \in \mathcal{V}_Q, \end{align*} and $\|g_Q\|_{L^2(Q)} \le C_n\|\Lambda\|_{(H^1)^*}|Q|^{1/2}$. **Step (iv) — Glue to a global $g \in L^2_{\mathrm{loc}}$.** The function $g_Q$ is determined modulo constants on $Q$ — adding any constant to $g_Q$ leaves the integral $\int_Q\phi g_Q$ unchanged for $\phi$ with mean zero. We pick the canonical representative with $(g_Q)_Q = 0$, i.e. $\frac{1}{|Q|}\int_Q g_Q\, d\mathcal{L}^n = 0$. For two cubes $Q_1\subseteq Q_2$, the functions $g_{Q_1}$ and $g_{Q_2}|_{Q_1}$ both represent $\Lambda$ on $\mathcal{V}_{Q_1}$, hence differ by a constant on $Q_1$. Define $g$ on $\mathbb{R}^n = \bigcup_kB(0, k)$ by gluing the canonical representatives on a nested sequence $Q_k \nearrow \mathbb{R}^n$ of cubes (after adjusting by additive constants); the gluing is consistent because the differences are constant on overlaps. The result is a function $g \in L^2_{\mathrm{loc}}(\mathbb{R}^n)$, well-defined modulo a single global additive constant. [/step] [step:Verify $g \in \mathrm{BMO}$ by testing on atomic functions] Continuing from step 3, we show $g \in \mathrm{BMO}(\mathbb{R}^n)$ with $\|g\|_{\mathrm{BMO}}\le C_n\|\Lambda\|_{(H^1)^*}$. Fix any cube $Q$. We compute $\frac{1}{|Q|}\int_Q|g - g_Q|^2\, d\mathcal{L}^n$. By [duality on $L^2(Q)$ for the closed subspace of mean-zero functions](/theorems/???) — equivalently, the $L^2$ Riesz representation applied to $\mathcal{V}_Q$ — the $L^2(Q)$-norm of $g - g_Q$ (the orthogonal projection of $g_Q$ onto $\mathcal{V}_Q$, since $g - g_Q$ has mean zero on $Q$) equals \begin{align*} \|g - g_Q\|_{L^2(Q)} = \sup_{\substack{\phi \in \mathcal{V}_Q \\ \|\phi\|_{L^2(Q)} = 1}}\int_Q (g - g_Q)\phi\, d\mathcal{L}^n = \sup_{\substack{\phi \in \mathcal{V}_Q \\ \|\phi\|_{L^2(Q)} = 1}}\int_Q g\phi\, d\mathcal{L}^n, \end{align*} where the last equality uses $\int_Q\phi = 0$ to remove the constant $g_Q$ from the inner product. Because $\Lambda(\phi) = \int_Q g\phi\, d\mathcal{L}^n$ for $\phi \in \mathcal{V}_Q$ by step 3, \begin{align*} \|g - g_Q\|_{L^2(Q)} = \sup_{\substack{\phi \in \mathcal{V}_Q \\ \|\phi\|_{L^2(Q)} = 1}}|\Lambda(\phi)| \le \|\Lambda\|_{(H^1)^*}\sup_{\substack{\phi \in \mathcal{V}_Q \\ \|\phi\|_{L^2(Q)} = 1}}\|\phi\|_{H^1} \le C_n\|\Lambda\|_{(H^1)^*}|Q|^{1/2}, \end{align*} where the last inequality is the bound from step 3 step (ii). Squaring and dividing by $|Q|$, \begin{align*} \frac{1}{|Q|}\int_Q|g - g_Q|^2\, d\mathcal{L}^n = \frac{\|g - g_Q\|_{L^2(Q)}^2}{|Q|} \le C_n^2\|\Lambda\|_{(H^1)^*}^2. \end{align*} Taking the supremum over $Q$, \begin{align*} \|g\|_{\mathrm{BMO}_2}^2 = \sup_Q \frac{1}{|Q|}\int_Q|g - g_Q|^2\, d\mathcal{L}^n \le C_n^2\|\Lambda\|_{(H^1)^*}^2. \end{align*} By the [Equivalence of BMO $L^p$ Norms](/theorems/3181) with $p = 2$, \begin{align*} \|g\|_{\mathrm{BMO}} \le \|g\|_{\mathrm{BMO}_2} \le C_n\|\Lambda\|_{(H^1)^*}. \end{align*} Hence $g \in \mathrm{BMO}(\mathbb{R}^n)$ with the claimed bound on its BMO norm. [/step] [step:Verify $\Lambda = \Lambda_g$ globally and uniqueness] We have $\Lambda(\phi) = \int_{\mathbb{R}^n}\phi g\, d\mathcal{L}^n$ for every mean-zero compactly supported $L^2$ function $\phi$ — namely, $\Lambda$ and $\Lambda_g$ agree on $\bigcup_Q\mathcal{V}_Q$. Atomic finite sums in $H^1$ are exactly such mean-zero compactly supported $L^2$ functions (each atom is mean-zero $L^\infty\subseteq L^2$ on a ball, and a finite sum of these is mean-zero compactly supported $L^2$ if all the supports lie in a common cube — which can always be arranged by enlarging). By density of atomic finite sums in $H^1$ and continuity of both $\Lambda$ and $\Lambda_g$ (the latter from step 2), $\Lambda = \Lambda_g$ on all of $H^1(\mathbb{R}^n)$. For uniqueness modulo constants: suppose $\Lambda_{g_1} = \Lambda_{g_2}$ on $H^1$. Then $\Lambda_{g_1 - g_2} = 0$, so for every atom $a$ supported in any ball $B$, \begin{align*} 0 = \int_{\mathbb{R}^n}a(x)(g_1 - g_2)(x)\, d\mathcal{L}^n(x) = \int_B a(x)\bigl((g_1 - g_2) - (g_1 - g_2)_B\bigr)\, d\mathcal{L}^n(x), \end{align*} since the cancellation of atoms allows the constant subtraction. Taking the supremum over atoms supported in $B$ — this supremum is the $L^\infty$-dual of $L^1$ on $B$, which equals $\|(g_1 - g_2) - (g_1 - g_2)_B\|_{L^1(B)}/|B|^{-1}$ up to constants — gives that $(g_1 - g_2)|_B$ is a.e. constant for every ball $B$. By coverings, $g_1 - g_2$ is a.e. constant on $\mathbb{R}^n$. This is the desired uniqueness modulo constants. **Combining both directions.** We have shown: - (Step 2) For every $g \in \mathrm{BMO}(\mathbb{R}^n)$, the functional $\Lambda_g$ is in $(H^1)^*$ with $\|\Lambda_g\|_{(H^1)^*} \le C_n\|g\|_{\mathrm{BMO}}$. - (Steps 3–4) For every $\Lambda \in (H^1)^*$, there exists $g \in \mathrm{BMO}(\mathbb{R}^n)$, unique modulo constants, with $\Lambda = \Lambda_g$ and $\|g\|_{\mathrm{BMO}} \le C_n\|\Lambda\|_{(H^1)^*}$. Therefore the linear map \begin{align*} \mathrm{BMO}(\mathbb{R}^n)/\{\text{constants}\} &\to (H^1(\mathbb{R}^n))^* \\ [g] &\mapsto \Lambda_g \end{align*} is a bounded bijection with bounded inverse, with comparable norms $C_n^{-1}\|g\|_{\mathrm{BMO}/\mathrm{const}} \le \|\Lambda_g\|_{(H^1)^*}\le C_n\|g\|_{\mathrm{BMO}/\mathrm{const}}$. This is the asserted isomorphism $(H^1)^*\cong \mathrm{BMO}$. [/step]