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I posted this question on Math.SE where I unfortunately received no answers even after a bounty. As such, I am putting it here, in hopes to receive a response.

For the proof of Theorem 6.1.6 in Classical Fourier Analysis 3rd Edition, we want to establish the inequality $$ \frac{1}{C_n( p + \frac{1}{p-1})^{2n}} \lVert f \rVert_{L^p} \leq \left\lVert \lVert \Delta _j ^\# f \rVert_{\ell^2(\mathbb{Z}^n)} \right\rVert_{L^p(\mathbb{R}^n)}.$$ where $\widehat{\Delta _j^\# f} = \mathbf{1}_{R_j} \hat{f} $ for some tiling $$R_j = \prod_{i \leq n } I_{j_i} = \prod_{i \leq n} [2^{j_i} , 2^{j_i + 1}) \cup ( - 2^{j_i + 1} ,- 2^{j_i}], \quad j = (j_1, \ldots, j_n) \in \mathbb{Z}^n. $$

Now, the proof (last paragraph page 429) asserts that the fundamental ingredient is to show $$f = \sum_{j \in \mathbb{Z}^n} \Delta _j^\# \Delta _j ^\# f, \quad f \in \mathcal{S}(\mathbb{R}^n). $$ But I am at a loss as to how this can be seen. In the previous proofs (which the book refers to consult for this proof), it is usually argued by showing the Fourier transform of the difference has support equal to $\{0\}$ and thus the difference is a polynomial which we could establish to be 0, as these two quantities are in $L^p$. However, note $$\hat{f} - \sum_{j \in \mathbb{Z}^n} \widehat{\Delta _j^\# \Delta _j^\# f} = \hat{f} - \sum_{j \in \mathbb{Z}^n} \mathbf{1}_{R_j}^2 \hat{f} = \hat{f} (1 - \mathbf{1}_{\cup_{j \in \mathbb{Z}^n} R_j} ) = \hat{f} \mathbf{1}_{\cup \{x_i = 0\}}. $$ Here the support doesn't equal $\{0\}$ and so the argument won't work.

(I believe there is an error in the book as the current versions of the book suggests that $\cup_{j \in \mathbb{Z}^n } R_j = \mathbb{R}^n \backslash \{0\}$ (and so the argument can proceed) but this is revised in the errata to be $\cup_{j \leq n} \{x_j =0 \}$. Would this be a correct assessment?)

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Apologies if I've misunderstood the question; Grafakos comments in this paragraph that

The fundamental ingredient in the proof is that $f=\sum_{\mathbf{j}\in\mathbb{Z}^n}\Delta_\mathbf{j}^\#\Delta_\mathbf{j}^\#f$ for all Schwartz functions $f$, where the sum is interpreted as the $L^2$-limit of a sequence of partial sums. Thus the series converges in $\mathscr{S}'$...

Since $f$ is Schwartz, $\hat{f}$ is Schwartz (hence $L^2$), so the sum of the Fourier transforms converges in $L^2$ to $\hat{f}$ (since the missing set is null), so (by Plancherel) the original sum converges in $L^2$ to $f$. Thus the identity holds in $L^2$, and he uses this to conclude convergence in $\mathscr{S}'$ (which is a weaker topology than $L^2$).

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  • $\begingroup$ Thanks, that was simpler than I thought. I got confused as previous theorems required $f \in \mathcal{S}'$ for the lower inequality, and as such we needed the Fourier transform of the difference to be supported at $\{0\},$ whereas here $f \in \mathcal{S}$ so no such trickery is required. $\endgroup$
    – newbie
    Commented Nov 24, 2023 at 7:56

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