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?)