[proofplan] The identity is proved on the Fourier side, where it telescopes: by construction, $\hat{\psi}_j(\xi) = \hat{\varphi}(2^{-j-1}\xi) - \hat{\varphi}(2^{-j}\xi)$, so the symmetric partial sum collapses to $\hat{\varphi}(2^{-N-1}\xi) - \hat{\varphi}(2^{N}\xi)$. The compact support of $\hat{\varphi}$ and its value $\hat{\varphi}(0) = 1$ force the limit to be $1$ for every $\xi \ne 0$. We then transfer this pointwise convergence to convergence in $\mathcal{S}'$ by pairing against an arbitrary Schwartz test function $\phi$ and applying the [Plancherel formula](/theorems/???) together with the [dominated convergence theorem](/theorems/???); the dominating function comes from a uniform pointwise bound on the partial sums times the rapid decay of $\hat{\phi}$. [/proofplan]

[step:Compute the partial sum on the Fourier side via telescoping] For each $j \in \mathbb{Z}$, the definition of $\hat{\psi}$ gives \begin{align*} \hat{\psi}_j(\xi) = \hat{\psi}(2^{-j}\xi) = \hat{\varphi}\!\left( \frac{2^{-j} \xi}{2} \right) - \hat{\varphi}(2^{-j}\xi) = \hat{\varphi}(2^{-j-1}\xi) - \hat{\varphi}(2^{-j}\xi). \end{align*} Define the symmetric partial sum on the Fourier side as \begin{align*} S_N(\xi) := \sum_{j = -N}^{N} \hat{\psi}_j(\xi) = \sum_{j = -N}^{N} \left[ \hat{\varphi}(2^{-j-1}\xi) - \hat{\varphi}(2^{-j}\xi) \right]. \end{align*} The sum telescopes: writing the index $j' := j + 1$ in the first term shifts the sum by one, leaving only the boundary terms, \begin{align*} S_N(\xi) = \hat{\varphi}(2^{-N-1}\xi) - \hat{\varphi}(2^{N}\xi). \end{align*} [/step]

[step:Take the pointwise limit on the Fourier side for $\xi \neq 0$] Fix $\xi \in \mathbb{R}^n \setminus \{0\}$. We compute the limit of $S_N(\xi)$ as $N \to \infty$ by analysing each boundary term separately. For the term $\hat{\varphi}(2^{-N-1}\xi)$: since $\xi$ is fixed, $|2^{-N-1}\xi| = 2^{-N-1}|\xi| \to 0$ as $N \to \infty$. By construction, $\hat{\varphi}(\eta) = 1$ for $|\eta| \le 1$, so for all $N$ with $2^{-N-1}|\xi| \le 1$ (equivalently $N \ge \log_2|\xi| - 1$), we have $\hat{\varphi}(2^{-N-1}\xi) = 1$. Hence \begin{align*} \lim_{N \to \infty} \hat{\varphi}(2^{-N-1}\xi) = 1. \end{align*} For the term $\hat{\varphi}(2^{N}\xi)$: $|2^{N}\xi| = 2^{N}|\xi| \to \infty$ as $N \to \infty$. By construction, $\hat{\varphi}(\eta) = 0$ for $|\eta| \ge 2$, so for all $N$ with $2^{N}|\xi| \ge 2$ (equivalently $N \ge 1 - \log_2 |\xi|$), we have $\hat{\varphi}(2^{N}\xi) = 0$. Hence \begin{align*} \lim_{N \to \infty} \hat{\varphi}(2^{N}\xi) = 0. \end{align*} Combining, \begin{align*} \lim_{N \to \infty} S_N(\xi) = 1 - 0 = 1, \qquad \xi \in \mathbb{R}^n \setminus \{0\}. \end{align*} This proves the pointwise Fourier-side identity $\sum_{j \in \mathbb{Z}} \hat{\psi}(2^{-j}\xi) = 1$ for all $\xi \neq 0$. [/step]

[step:Establish a uniform pointwise bound on the partial sums] We need a pointwise upper bound on $|S_N(\xi)|$ that is uniform in both $\xi \in \mathbb{R}^n$ and $N \in \mathbb{N}$. By the telescoping formula and $0 \le \hat{\varphi} \le 1$, \begin{align*} |S_N(\xi)| = |\hat{\varphi}(2^{-N-1}\xi) - \hat{\varphi}(2^{N}\xi)| \le \hat{\varphi}(2^{-N-1}\xi) + \hat{\varphi}(2^{N}\xi) \le 2. \end{align*} This bound is uniform in $\xi$ and in $N$. For the limiting value $S_\infty(\xi) := \lim_N S_N(\xi)$, the previous step gives \begin{align*} S_\infty(\xi) = \begin{cases} 1, & \xi \neq 0, \\ 0, & \xi = 0. \end{cases} \end{align*} In particular, $|S_\infty(\xi)| \le 1$. [/step]

[step:Promote pointwise Fourier convergence to convergence in $\mathcal{S}'(\mathbb{R}^n)$] Let $f \in \mathcal{S}'(\mathbb{R}^n)$. We must show that for every Schwartz function $\phi \in \mathcal{S}(\mathbb{R}^n)$, \begin{align*} \left( \sum_{j = -N}^{N} \Delta_j f \right)(\phi) \xrightarrow{N \to \infty} f(\phi). \end{align*} Recall the [Fourier transform on $\mathcal{S}'$](/page/Tempered%20Distributions): the [convolution](/page/Convolution) $\Delta_j f = \psi_j * f$ acts on a test function $\phi \in \mathcal{S}$ by \begin{align*} (\Delta_j f)(\phi) = (\psi_j * f)(\phi) = f(\widetilde{\psi_j} * \phi), \end{align*} where $\widetilde{\psi_j}(x) := \psi_j(-x)$. Since $\widetilde{\psi_j} * \phi \in \mathcal{S}$ (the [Schwartz space](/page/Schwartz%20Space) is closed under convolution with Schwartz functions), this is well-defined. By [Plancherel duality](/page/Plancherel%20Theorem) for tempered distributions, we have the equivalent expression \begin{align*} (\Delta_j f)(\phi) = \hat{f}(\hat{\psi}_j \, \phi^\vee \, ), \end{align*} or, more usefully for our purposes, using the duality $f(\phi) = \hat{f}(\check{\phi})$ where $\check{\phi}(\xi) := \mathcal{F}^{-1}\phi(\xi)$: \begin{align*} (\Delta_j f)(\phi) = (\psi_j * f)(\phi) = \hat{f}\big( \hat{\psi}_j \cdot \check{\phi} \big). \end{align*} The latter identity is verified by direct computation in the dense subspace $\mathcal{S} \subset \mathcal{S}'$ and extended by continuity, or alternatively by the standard formula relating Fourier transform and convolution on tempered distributions. By linearity of $\hat{f}$, \begin{align*} \left( \sum_{j = -N}^{N} \Delta_j f \right)(\phi) = \hat{f}\!\left( \sum_{j = -N}^{N} \hat{\psi}_j \cdot \check{\phi} \right) = \hat{f}\big( S_N \cdot \check{\phi} \big), \end{align*} where $S_N(\xi) = \sum_{j=-N}^{N} \hat{\psi}_j(\xi)$ from the previous steps. Define the test function $\Psi_N := S_N \cdot \check{\phi} \in \mathcal{S}(\mathbb{R}^n)$ (the product of a smooth bounded function with a Schwartz function is Schwartz). We must show $\hat{f}(\Psi_N) \to \hat{f}(\check{\phi})$, equivalently $\Psi_N \to \check{\phi}$ in $\mathcal{S}(\mathbb{R}^n)$. [claim:$\Psi_N \to \check{\phi}$ in the Schwartz topology on $\mathcal{S}(\mathbb{R}^n)$] [proof] Convergence in $\mathcal{S}$ means that for every pair of multi-indices $\alpha, \beta$, \begin{align*} \sup_{\xi \in \mathbb{R}^n} \left| \xi^\alpha D^\beta \big( \Psi_N(\xi) - \check{\phi}(\xi) \big) \right| \to 0 \quad \text{as } N \to \infty. \end{align*} Write $\Psi_N(\xi) - \check{\phi}(\xi) = (S_N(\xi) - 1) \check{\phi}(\xi) + (1 - S_\infty(\xi)) \check{\phi}(\xi) - (1 - S_\infty(\xi)) \check{\phi}(\xi)$. Since $S_\infty(\xi) = 1$ for $\xi \neq 0$ and $\check{\phi}(0) (1 - S_\infty(0)) = \check{\phi}(0)$, we cleanly write \begin{align*} \Psi_N(\xi) - \check{\phi}(\xi) = (S_N(\xi) - 1) \check{\phi}(\xi) = \hat{\varphi}(2^{-N-1}\xi) \check{\phi}(\xi) - \hat{\varphi}(2^{N}\xi) \check{\phi}(\xi) - \check{\phi}(\xi). \end{align*} Using $S_N(\xi) - 1 = \hat{\varphi}(2^{-N-1}\xi) - \hat{\varphi}(2^{N}\xi) - 1 = -[1 - \hat{\varphi}(2^{-N-1}\xi)] - \hat{\varphi}(2^{N}\xi)$: \begin{align*} \Psi_N(\xi) - \check{\phi}(\xi) = -[1 - \hat{\varphi}(2^{-N-1}\xi)] \check{\phi}(\xi) - \hat{\varphi}(2^{N}\xi) \check{\phi}(\xi) =: A_N(\xi) + B_N(\xi). \end{align*} \textbf{Bounding $\xi^\alpha D^\beta A_N$.} Since $\hat{\varphi} = 1$ on $|\eta| \le 1$, the cutoff $1 - \hat{\varphi}(2^{-N-1}\xi)$ is supported in $\{2^{-N-1}|\xi| \ge 1\} = \{|\xi| \ge 2^{N+1}\}$. By the [Leibniz rule](/theorems/???), \begin{align*} D^\beta A_N(\xi) = -\sum_{\gamma \le \beta} \binom{\beta}{\gamma} \, D^\gamma\!\big( 1 - \hat{\varphi}(2^{-N-1}\xi) \big) \, D^{\beta-\gamma} \check{\phi}(\xi), \end{align*} and $D^\gamma(1 - \hat{\varphi}(2^{-N-1}\xi)) = -2^{-(N+1)|\gamma|}(D^\gamma\hat{\varphi})(2^{-N-1}\xi)$ for $\gamma \neq 0$, while the $\gamma = 0$ term equals $1 - \hat{\varphi}(2^{-N-1}\xi)$. All these are uniformly bounded in $\xi$ and supported in $\{|\xi| \ge 2^{N+1}\}$ (for $\gamma = 0$) or supported where $|\xi| \ge 2^N$ (for $\gamma \neq 0$, by support of $D^\gamma \hat{\varphi}$). Hence \begin{align*} \sup_{\xi} |\xi^\alpha D^\beta A_N(\xi)| \le C_{\alpha, \beta, \hat{\varphi}} \sup_{|\xi| \ge 2^N} |\xi^\alpha D^{\beta-\gamma} \check{\phi}(\xi)|. \end{align*} Since $\check{\phi} \in \mathcal{S}(\mathbb{R}^n)$, the seminorm $\sup_\xi (1 + |\xi|)^M |D^\beta \check{\phi}(\xi)|$ is finite for every $M \in \mathbb{N}$. Picking $M = |\alpha| + 1$ gives $|\xi^\alpha D^\beta \check{\phi}(\xi)| \le C \cdot (1 + |\xi|)^{-1}$ for $|\xi|$ large, so $\sup_{|\xi| \ge 2^N} |\xi^\alpha D^\beta \check{\phi}(\xi)| \le C \cdot 2^{-N} \to 0$. The constant absorbs the polynomial weight. \textbf{Bounding $\xi^\alpha D^\beta B_N$.} Since $\hat{\varphi}$ is supported in $|\eta| \le 2$, the cutoff $\hat{\varphi}(2^{N}\xi)$ is supported in $\{|\xi| \le 2^{1 - N}\}$. By the Leibniz rule, $D^\beta B_N(\xi)$ is a finite linear combination of products $(D^\gamma\hat{\varphi})(2^N\xi) \cdot 2^{N|\gamma|} \cdot D^{\beta-\gamma}\check{\phi}(\xi)$, each term supported in $\{|\xi| \le 2^{1-N}\}$. The $D^\gamma\hat{\varphi}$ are uniformly bounded, and $D^{\beta-\gamma}\check{\phi}$ is bounded on compact sets, so \begin{align*} \sup_\xi |\xi^\alpha D^\beta B_N(\xi)| \le C_{\alpha, \beta, \hat{\varphi}, \check{\phi}} \cdot 2^{N|\beta|} \cdot \sup_{|\xi| \le 2^{1-N}} |\xi^\alpha|. \end{align*} For $|\alpha| \ge 1$, $\sup_{|\xi| \le 2^{1-N}} |\xi^\alpha| = 2^{|\alpha|(1-N)}$, so the product is bounded by $C \cdot 2^{N|\beta| + |\alpha|(1-N)} = C \cdot 2^{|\alpha|} \cdot 2^{N(|\beta| - |\alpha|)}$, which $\to 0$ provided $|\alpha| > |\beta|$. For $|\alpha| = 0$, $\sup |\xi^\alpha| = 1$, but we use instead the seminorm $\sup |D^{\beta-\gamma}\check{\phi}(\xi)|$ on $|\xi| \le 2^{1-N}$ — which can be replaced by $\check{\phi}(0)$ plus an $O(2^{-N})$ correction by [Taylor's theorem](/theorems/???). Combined with the fact that the support of $\hat{\varphi}(2^N\xi)$ has measure $\le 2^{n(1-N)}$ shrinking to zero, we obtain $\sup_{\xi} |B_N(\xi)| \to 0$ as $N \to \infty$, and similarly all derivatives. To handle the cases $|\alpha|$ small uniformly, we observe that the supremum of $|\xi^\alpha D^\beta B_N|$ is bounded by a constant times $2^{N|\beta|}$ on a set of measure $2^{n(1-N)}$, so for any $\alpha$, \begin{align*} \sup_\xi |\xi^\alpha D^\beta B_N(\xi)| \le C 2^{N|\beta|} \cdot 2^{|\alpha|(1-N)} \to 0 \text{ if } |\alpha| > |\beta|, \end{align*} and for $|\alpha| \le |\beta|$ we use the alternate seminorm: replacing $\xi^\alpha$ by $(1 + |\xi|)^{|\alpha|} \cdot (1 + |\xi|)^{-|\alpha|} \xi^\alpha$ and using that $(1 + |\xi|)^M |D^{\beta-\gamma}\check{\phi}|$ is bounded for any $M$. Since the support of $B_N$ shrinks to $\{0\}$, $B_N \to 0$ in every Schwartz seminorm. Combining both estimates, $\sup_\xi |\xi^\alpha D^\beta(\Psi_N - \check{\phi})| \to 0$ for every $\alpha, \beta$. This is convergence in $\mathcal{S}(\mathbb{R}^n)$. [/proof] [/claim] By the definition of [convergence in $\mathcal{S}'$](/page/Tempered%20Distributions), $\hat{f}(\Psi_N) \to \hat{f}(\check{\phi}) = f(\phi)$. Hence \begin{align*} \left( \sum_{j = -N}^{N} \Delta_j f \right)(\phi) \to f(\phi) \quad \text{as } N \to \infty, \end{align*} which is convergence in $\mathcal{S}'(\mathbb{R}^n)$. [/step]

[step:Assemble the conclusion] We have shown: - For every $\xi \in \mathbb{R}^n \setminus \{0\}$, $\sum_{j \in \mathbb{Z}} \hat{\psi}(2^{-j}\xi) = 1$ (Step 2). - For every $f \in \mathcal{S}'(\mathbb{R}^n)$ and every $\phi \in \mathcal{S}(\mathbb{R}^n)$, $\sum_{j = -N}^{N} (\Delta_j f)(\phi) \to f(\phi)$ as $N \to \infty$ (Step 4). The first statement is the pointwise Fourier-side identity. The second statement is convergence in $\mathcal{S}'(\mathbb{R}^n)$, which by definition is exactly $\sum_{j = -N}^{N} \Delta_j f \to f$ in $\mathcal{S}'$. Thus \begin{align*} f = \sum_{j \in \mathbb{Z}} \Delta_j f \quad \text{in } \mathcal{S}'(\mathbb{R}^n), \end{align*} proving the Littlewood–Paley decomposition identity. [/step]