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 $\cup_{j \leq n} \{x_j =0 \}$. )
The fundamental ingredient identity f = \sum_{j \in Z^n} Δ_j^# Δ_j^# f easily holds whenever f is a function in S_0 = Schwartz functions whose Fourier transform is compactly supported in R^n-{coordinate planes}. Such functions are dense in L^p for 1<p<∞ and so the proof proceeds as in the book, except that Schwartz functions are replaced by S_0 (special Schwartz functions).