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I was going through the proof on Stein's Harmonic Analysis: Real-Variable Methods, Orthogonality and Oscillatory Integrals of the fact that BMO is the dual of $H^1$ (this is on Chapter IV, section 1.2., my question is more concretely about section 1.2.1). To prove this, he uses the atomic decomposition of $H^1$, and defines $H^1_\alpha$ as the space of finite linear combinations of atoms. As this space is dense, it would suffice to show $$\left|\int_{\mathbb R^N}f(x)g(x)dx\right|\leq c\lVert f\rVert \lVert g\rVert$$ for $f$ in BMO and $g\in H^1_\alpha$ in order to show every element in BMO yields a bounded linear functional on $H^1$ (as $H^1_\alpha\subset H^1$ is dense). The norms are, of course, the BMO norm for $f$ and some $H^1$ norm for $g$. In what follows, there are two things I don't understand. First, he wants to prove the inequality for $f\in L^\infty$ and then extend it to the general case, but I don't see why this is necessary. To be precisse, given $f\in L^\infty$, $g\in H^1_\alpha$ with an atomic decomposition $$g(x)=\sum_{k=1}^n \lambda_k a_k(x)$$ we may write $$\int fg=\sum_k \lambda_k\int fa_k=\sum_k \lambda_k\int_{B_k}(f-f_{B_k})a_k.$$ Here $B_k$ is the support of $a_k$ and $f_{B_k}$ is the integral average of $f$ in $B_k$. The last equality follows from the cancellation properties of $H^1$ atoms. So far I don't see where we might be using the fact that $f$ is bounded, as the first equality follows the linearity of the integral, and for the second the only assumption we need on $f$ is that it is locally integrable (so that the averages indeed exist). Then, he writes $$\left|\int fg\right|\leq\sum_k\frac{|\lambda_k|}{|B_k|}\int |f-f_{B_k}|\leq\sum_k |\lambda_k| \|f\|,$$ where the first inequality follows from the properties of $H^1$ atoms. This proves the desired inequality for the case in which $f\in L^\infty$. So: what do we need $f$ to be bounded for? And also, why is the sum of coefficients $\sum_k |\lambda_k|\leq c\|g\|$?

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You are missing an important detail: $g\in H^1_a$, so $g=\sum_{k=1}^n\tilde{\lambda}_k \tilde{a}_k$, but the proof is using another atomic decomposition of $g$, not that one.

In fact, remember that $H^1$ norm is defined as follow: $$ \left\lVert g\right\rVert_{H^{1}}:=\inf\left\{\sum_{j\in\mathbb{N}} \left\lvert{\lambda_j}\right\rvert: \exists\{a_j\}_{j\in\mathbb{N}}\text{ atoms s.t. }g=\sum_{j\in\mathbb{N}}\lambda_j a_j\text{ in } L^1\right\} $$ So you can always choose a "good" atomic representation of $g$ s.t. $g=\sum_{j\in\mathbb{N}}\lambda_j a_j$ and $\sum_k \left\lvert \lambda_k \right\rvert<c\left\lVert g \right\rVert_{H^1}$ is satisfied. The atoms and the coefficients of this representation are no longer the same of the previous and the problem is that the sum in this representation can be an infinite series, even if $g\in H^1_a$.

In the rest of the proof you use this "good" representation of $g$ and no more the other. For this reason you need $f\in L^\infty$ to be sure that $fg\in L^1$. In particular when you write: $$\int fg=\int f\sum_k \lambda_k a_k=\sum_k \lambda_k\int fa_k$$ The second equality follows from dominated convergence theorem and not from the linearity of integral, because you are using the "good" atomic representation. This should answer both your questions

Edit: About $H^1$ norm. It is easy to check that $\left\lVert\cdot\right\rVert_{H^1}$(as defined above) is a norm on $H^1$. Let's write $\left\lvert g\right\rvert_{H^1}:=\left\lVert M_{\Phi}g\right\rVert_{L^1}$. In Stein you can find the proof that exists a (possibly infinite atomic decomposition) of $g$ s.t.: $$c_1\left\lvert g\right\rvert_{H^1}\leq \sum_{k\in\mathbb{N}}\left\lvert\lambda_k\right\rvert\leq c_2\left\lvert g\right\rvert_{H^1} $$ The first inequality is always true, for every atomic decomposition of $g$. The second inequality instead (if you carefully read the statement of Theorem 2 in Stein's Harmonic Hanalysis, Ch. III) is true only for a particular decomposition(the one constructed in the proof of that theorem), which could be infinite. In Stein's book yu can also find a proof of the first inequality a few lines above that theorem. From this chain of inequalities it follows that $\left\lVert\cdot\right\rVert_{H^1}$ is equivalent to $\left\lvert\cdot\right\rvert_{H^1}$.

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  • $\begingroup$ Okay, that makes a lot of sense. However, Stein defines the $H^p$ norm of $f$ as $||M_\Phi f||_{L^p}$, where $\Phi$ is a suitable Schwartz function and $M_\Phi f(x)=\sup_{t>0}|\Phi_t*f(x)|$. I suppose the norm you define is equivalent to any of these(?) I also realized it is on the proof of the atomic decomposition that $\sum |\lambda_k|\leq c||f||_{H^1}$, so why do we have to pick a particular, possibly infinite decomposition? $\endgroup$ Jan 2 at 16:51
  • $\begingroup$ Yes, it is an equivalent norm, I've added a few lines about this fact $\endgroup$
    – Luca.b
    Jan 3 at 8:41

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