L2 almost orthogonality of the dyadic pieces

The Schwartz case. Let $f\in \mathcal{S}$. For every $j$, [F1] gives ${\mathrm{\Delta }}_{j}f\in \mathcal{S}$ with $\widehat{{\mathrm{\Delta }}_{j}f}={\phi }_j\widehat{f}$; by Plancherel [F2] applied to $g={\mathrm{\Delta }}_{j}f$ and to $f$, $\mathrm{\parallel }{\mathrm{\Delta }}_{j}f{\mathrm{\parallel }}_2^2={\int }_{{\mathbb{R}}^n}\mathrm{\mid }\widehat{{\mathrm{\Delta }}_{j}f}{\mathrm{\mid }}^2={\int }_{{\mathbb{R}}^n}{\phi }_j^2\mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2$ and $\mathrm{\parallel }f{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2$.

Summing the Schwartz identity. With $f\in \mathcal{S}$ as in step 1.1, the series ${\sum }_j{\phi }_j(\xi {)}^2$ is locally finite with ${\sum }_j{\phi }_j^2\in [1\mathrm{/}3,1]$ pointwise by [F1], so Tonelli's theorem [F6] applied to the nonnegative functions $\mathrm{\mid }{\phi }_j\widehat{f}{\mathrm{\mid }}^2$ gives ${\sum }_{j\ge 0}\mathrm{\parallel }{\mathrm{\Delta }}_{j}f{\mathrm{\parallel }}_2^2={\int }_{{\mathbb{R}}^n}({\sum }_{j\ge 0}{\phi }_j(\xi {)}^2)\mathrm{\mid }\widehat{f}(\xi ){\mathrm{\mid }}^{2}d\xi$, and the pointwise bounds sandwich this between $\frac{1}{3}\int \mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2=\frac{1}{3}\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$ and $\int \mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. This proves the two-sided estimate, and the finiteness of ${\sum }_j\mathrm{\parallel }{\mathrm{\Delta }}_{j}f{\mathrm{\parallel }}_2^2$, for Schwartz $f$.

The identity $\widehat{{\mathrm{\Delta }}_{j}g}={\phi }_j\widehat{g}$ passes to $L^2$. For $j\ge 0$ both maps $g\mapsto {\mathrm{\Delta }}_{j}g=g\ast K_j$ and $g\mapsto {\mathcal{F}}_2^{-1}({\phi }_j{\mathcal{F}}_{2}g)$ are bounded linear operators on $L^2$ by [F3] and [F4], and they agree on the dense subspace $\mathcal{S}$ by step 1.1 and [F3]; given $g\in L^2$ and a sequence $g_k\in \mathcal{S}$ with $g_k\to g$ from [F5], both operators applied to $g_k$ converge in $L^2$ to their values at $g$, so ${\mathrm{\Delta }}_{j}g={\mathcal{F}}_2^{-1}({\phi }_j{\mathcal{F}}_{2}g)$ and hence $\mathrm{\parallel }{\mathrm{\Delta }}_{j}g{\mathrm{\parallel }}_2^2=\int {\phi }_j^2\mathrm{\mid }{\mathcal{F}}_{2}g{\mathrm{\mid }}^2$ for every $j$.

Conclusion. For $g\in L^2$, step 2.2 gives $\mathrm{\parallel }{\mathrm{\Delta }}_{j}g{\mathrm{\parallel }}_2^2=\int {\phi }_j^2\mathrm{\mid }{\mathcal{F}}_{2}g{\mathrm{\mid }}^2$ for every $j$, and Tonelli [F6] then gives ${\sum }_{j\ge 0}\mathrm{\parallel }{\mathrm{\Delta }}_{j}g{\mathrm{\parallel }}_2^2=\int \left({\sum }_j{\phi }_j^2\right)\mathrm{\mid }{\mathcal{F}}_{2}g{\mathrm{\mid }}^2$; the pointwise bounds $\frac{1}{3}\le {\sum }_j{\phi }_j^2\le 1$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=\int \mathrm{\mid }{\mathcal{F}}_{2}g{\mathrm{\mid }}^2$ from Plancherel [F2] yield $\frac{1}{3}\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\le {\sum }_j\mathrm{\parallel }{\mathrm{\Delta }}_{j}g{\mathrm{\parallel }}_2^2\le \mathrm{\parallel }g{\mathrm{\parallel }}_2^2$, which is the stated two-sided estimate, and the finiteness of the sum.