[proofplan] The $L^2$ case is exact: by [Plancherel's theorem](/theorems/???) and the bounded overlap of the Fourier supports of the $\hat{\psi}_j$, the square function norm equals the $L^2$ norm up to a constant computable from the partition of unity. For $p \neq 2$, we view $f \mapsto (\Delta_j f)_{j \in \mathbb{Z}}$ as an $\ell^2(\mathbb{Z})$-valued operator and show it is a [vector-valued Calderón–Zygmund operator](/theorems/???): its kernel is the $\ell^2$-valued tempered distribution $K(x) = (\psi_j(x))_{j \in \mathbb{Z}}$, satisfying the size and Hörmander conditions in the operator-norm sense. The vector-valued Calderón–Zygmund theorem then promotes the $L^2$ bound to weak-$(1,1)$ and $L^p$ bounds for $1 < p < \infty$. The lower bound $\|f\|_{L^p} \lesssim \|S(f)\|_{L^p}$ follows by a dual argument: pair $f$ against $g \in L^{p'}$ via $\int f \bar{g} = \sum_j \int \Delta_j f \overline{\widetilde{\Delta}_j g}$ for a slight enlargement $\widetilde{\Delta}_j$, then apply Cauchy–Schwarz pointwise and Hölder. [/proofplan] [step:Establish the upper bound $\|S(f)\|_{L^2} \asymp \|f\|_{L^2}$ via Plancherel] By [Plancherel's theorem](/theorems/???), for any $h \in L^2(\mathbb{R}^n)$, $\|h\|_{L^2(\mathbb{R}^n)}^2 = (2\pi)^{-n} \|\hat{h}\|_{L^2(\mathbb{R}^n)}^2$. Apply this to $\Delta_j f = \psi_j * f$, whose Fourier transform is $\widehat{\Delta_j f}(\xi) = \hat{\psi}(2^{-j}\xi) \hat{f}(\xi)$. Compute \begin{align*} \|S(f)\|_{L^2(\mathbb{R}^n)}^2 &= \int_{\mathbb{R}^n} \sum_{j \in \mathbb{Z}} |\Delta_j f(x)|^2 \, d\mathcal{L}^n(x) \\ &= \sum_{j \in \mathbb{Z}} \|\Delta_j f\|_{L^2(\mathbb{R}^n)}^2 \\ &= (2\pi)^{-n} \sum_{j \in \mathbb{Z}} \int_{\mathbb{R}^n} |\hat{\psi}(2^{-j}\xi)|^2 |\hat{f}(\xi)|^2 \, d\mathcal{L}^n(\xi) \\ &= (2\pi)^{-n} \int_{\mathbb{R}^n} \left( \sum_{j \in \mathbb{Z}} |\hat{\psi}(2^{-j}\xi)|^2 \right) |\hat{f}(\xi)|^2 \, d\mathcal{L}^n(\xi). \end{align*} The first equality is the definition of $S(f)$. The second is [Tonelli's theorem](/theorems/???), valid because the integrand is non-negative. The third is Plancherel applied to each $\Delta_j f$. The fourth is Tonelli again, switching the sum and integral, valid by non-negativity. Define $\Phi(\xi) := \sum_{j \in \mathbb{Z}} |\hat{\psi}(2^{-j}\xi)|^2$ for $\xi \in \mathbb{R}^n$. By construction $\hat{\psi}$ is supported in $\{1/2 \le |\eta| \le 2\}$, so $\hat{\psi}(2^{-j}\xi) \neq 0$ requires $2^{j-1} \le |\xi| \le 2^{j+1}$. For each $\xi \neq 0$, this happens for at most three consecutive values of $j$ (the integers $j$ in $[\log_2|\xi| - 1, \log_2 |\xi| + 1]$). Thus the sum has at most three non-zero terms at each $\xi$, giving the upper bound \begin{align*} \Phi(\xi) \le 3 \|\hat{\psi}\|_{L^\infty(\mathbb{R}^n)}^2 \quad \text{for all } \xi \in \mathbb{R}^n \setminus \{0\}. \end{align*} For the lower bound, recall from the [Littlewood–Paley decomposition identity](/theorems/3186) that $\sum_{j \in \mathbb{Z}} \hat{\psi}(2^{-j}\xi) = 1$ for $\xi \neq 0$. Since at most three terms are non-zero at each $\xi$, the [Cauchy–Schwarz inequality](/theorems/???) on $\mathbb{R}^3$ gives \begin{align*} 1 = \left( \sum_{j} \hat{\psi}(2^{-j}\xi) \right)^2 \le 3 \sum_{j} |\hat{\psi}(2^{-j}\xi)|^2 = 3 \Phi(\xi), \end{align*} so $\Phi(\xi) \ge 1/3$ for $\xi \neq 0$. Substituting these bounds: \begin{align*} \frac{(2\pi)^{-n}}{3} \|\hat{f}\|_{L^2(\mathbb{R}^n)}^2 \le \|S(f)\|_{L^2(\mathbb{R}^n)}^2 \le 3 (2\pi)^{-n} \|\hat{\psi}\|_{L^\infty}^2 \|\hat{f}\|_{L^2(\mathbb{R}^n)}^2. \end{align*} Applying Plancherel once more, $\|\hat{f}\|_{L^2}^2 = (2\pi)^n \|f\|_{L^2}^2$, giving \begin{align*} \frac{1}{3} \|f\|_{L^2(\mathbb{R}^n)}^2 \le \|S(f)\|_{L^2(\mathbb{R}^n)}^2 \le 3 \|\hat{\psi}\|_{L^\infty}^2 \|f\|_{L^2(\mathbb{R}^n)}^2. \end{align*} Setting $A_2 := 1/\sqrt{3}$ and $B_2 := \sqrt{3} \, \|\hat{\psi}\|_{L^\infty}$ proves the $L^2$ inequality. [/step] [step:Reformulate as a vector-valued operator $T: f \mapsto (\Delta_j f)_j$ and identify its kernel] For $1 < p < \infty$ with $p \neq 2$, we obtain the bounds via the [vector-valued Calderón–Zygmund theorem](/theorems/???). Define the operator \begin{align*} T: L^2(\mathbb{R}^n) &\to L^2(\mathbb{R}^n; \ell^2(\mathbb{Z})), \\ f &\mapsto (\Delta_j f)_{j \in \mathbb{Z}}, \end{align*} where $L^2(\mathbb{R}^n; \ell^2(\mathbb{Z}))$ denotes the Bochner space of square-integrable functions valued in the Hilbert space $\ell^2(\mathbb{Z})$, with norm $\|F\|_{L^2(\mathbb{R}^n; \ell^2)}^2 := \int_{\mathbb{R}^n} \|F(x)\|_{\ell^2}^2 \, d\mathcal{L}^n(x)$. Note that $\|T f\|_{L^2(\mathbb{R}^n; \ell^2)} = \|S(f)\|_{L^2(\mathbb{R}^n)}$, so by Step 1, $T$ is bounded $L^2 \to L^2(\mathbb{R}^n; \ell^2)$ with norm $B_2 = \sqrt{3} \|\hat{\psi}\|_{L^\infty}$. The convolution kernel of $T$ is the $\ell^2(\mathbb{Z})$-valued tempered distribution \begin{align*} K: \mathbb{R}^n \setminus \{0\} &\to \ell^2(\mathbb{Z}), \\ x &\mapsto (\psi_j(x))_{j \in \mathbb{Z}}. \end{align*} Concretely, for $f \in L^2(\mathbb{R}^n)$ with compact support and $x$ outside the support, \begin{align*} T f(x) = (\Delta_j f(x))_j = \left( \int_{\mathbb{R}^n} \psi_j(x - y) f(y) \, d\mathcal{L}^n(y) \right)_j = \int_{\mathbb{R}^n} K(x - y) f(y) \, d\mathcal{L}^n(y), \end{align*} where the integral is interpreted as Bochner integration in $\ell^2(\mathbb{Z})$. [/step] [step:Verify the $\ell^2$-valued size estimate $\|K(x)\|_{\ell^2} \lesssim |x|^{-n}$] We show that there exists a constant $C_1 = C_1(n, \psi) > 0$ such that for all $x \in \mathbb{R}^n \setminus \{0\}$, \begin{align*} \|K(x)\|_{\ell^2(\mathbb{Z})}^2 = \sum_{j \in \mathbb{Z}} |\psi_j(x)|^2 \le \frac{C_1^2}{|x|^{2n}}. \end{align*} Recall $\psi_j(x) = 2^{jn} \psi(2^j x)$ where $\psi = \mathcal{F}^{-1}(\hat{\psi}) \in \mathcal{S}(\mathbb{R}^n)$ (the [Schwartz class](/page/Schwartz%20Space) is preserved by inverse Fourier transform). For any $M \in \mathbb{N}$, the Schwartz seminorm bound \begin{align*} |\psi(y)| \le \frac{C_M(\psi)}{(1 + |y|)^M} \quad \text{for all } y \in \mathbb{R}^n \end{align*} holds, where $C_M(\psi) > 0$ depends only on $M$ and $\psi$. Fix $x \in \mathbb{R}^n \setminus \{0\}$. Pick $M = 2n + 1$. Compute, for each $j \in \mathbb{Z}$, \begin{align*} |\psi_j(x)| = 2^{jn} |\psi(2^j x)| \le 2^{jn} \cdot \frac{C_M}{(1 + 2^j |x|)^M}. \end{align*} Split the sum at the threshold $j_0 := -\log_2 |x|$ (so $2^{j_0} |x| = 1$): \begin{align*} \sum_{j \in \mathbb{Z}} |\psi_j(x)|^2 = \sum_{j \le j_0} |\psi_j(x)|^2 + \sum_{j > j_0} |\psi_j(x)|^2. \end{align*} \textbf{Sum over $j \le j_0$.} Here $2^j |x| \le 1$, so $(1 + 2^j |x|)^{-M} \le 1$ and \begin{align*} |\psi_j(x)|^2 \le 2^{2jn} C_M^2. \end{align*} The geometric sum gives \begin{align*} \sum_{j \le j_0} 2^{2jn} = 2^{2 j_0 n} \cdot \sum_{k \ge 0} 2^{-2kn} = 2^{2 j_0 n} \cdot \frac{1}{1 - 2^{-2n}} = \frac{|x|^{-2n}}{1 - 2^{-2n}}. \end{align*} Hence the first part is bounded by $C_M^2 (1 - 2^{-2n})^{-1} |x|^{-2n}$. \textbf{Sum over $j > j_0$.} Here $2^j |x| > 1$, so $(1 + 2^j|x|)^M \ge (2^j |x|)^M$ and \begin{align*} |\psi_j(x)|^2 \le 2^{2jn} \cdot \frac{C_M^2}{(2^j |x|)^{2M}} = \frac{C_M^2}{|x|^{2M}} \cdot 2^{2j(n - M)}. \end{align*} Since $M = 2n + 1$, $n - M = -(n+1) < 0$, so the geometric sum converges: \begin{align*} \sum_{j > j_0} 2^{2j(n - M)} = 2^{2 j_0 (n - M)} \cdot \frac{2^{2(n - M)}}{1 - 2^{2(n - M)}} = |x|^{2(M - n)} \cdot C', \end{align*} for a finite constant $C' = C'(n, M)$. Hence the second part is bounded by $\frac{C_M^2 C'}{|x|^{2M}} \cdot |x|^{2(M - n)} = C_M^2 C' \cdot |x|^{-2n}$. Adding both parts, $\sum_j |\psi_j(x)|^2 \le C_1^2 |x|^{-2n}$ where $C_1^2 = C_M^2 [(1 - 2^{-2n})^{-1} + C']$. [/step] [step:Verify the $\ell^2$-valued Hörmander smoothness condition] We show that there exists a constant $C_2 = C_2(n, \psi) > 0$ such that for all $y \in \mathbb{R}^n \setminus \{0\}$, \begin{align*} \int_{|x| > 2|y|} \|K(x - y) - K(x)\|_{\ell^2(\mathbb{Z})} \, d\mathcal{L}^n(x) \le C_2. \end{align*} By the [Minkowski integral inequality](/theorems/???) applied to the $\ell^2$ norm (which is the case $p = 2$ of integral Minkowski), \begin{align*} \|K(x - y) - K(x)\|_{\ell^2(\mathbb{Z})}^2 = \sum_j |\psi_j(x - y) - \psi_j(x)|^2. \end{align*} By the [mean value theorem](/theorems/???) applied to $\psi_j$ along the segment from $x - y$ to $x$, \begin{align*} |\psi_j(x - y) - \psi_j(x)| \le |y| \cdot \sup_{0 \le t \le 1} |\nabla \psi_j(x - ty)|. \end{align*} Compute $\nabla \psi_j(z) = 2^{j(n+1)} (\nabla\psi)(2^j z)$, so by the Schwartz bound on $\nabla \psi$, \begin{align*} |\nabla \psi_j(x - ty)| \le 2^{j(n+1)} \frac{C'_M}{(1 + 2^j |x - ty|)^M}, \end{align*} for $C'_M = C_M(\nabla\psi) > 0$. For $|x| > 2|y|$ and $0 \le t \le 1$, we have $|x - ty| \ge |x| - t|y| \ge |x| - |y| \ge |x|/2$, so $|x - ty| \ge |x|/2$ and \begin{align*} |\nabla \psi_j(x - ty)| \le 2^{j(n+1)} \frac{C'_M}{(1 + 2^{j-1} |x|)^M}. \end{align*} Hence \begin{align*} |\psi_j(x - y) - \psi_j(x)| \le |y| \cdot 2^{j(n+1)} \frac{C'_M}{(1 + 2^{j-1} |x|)^M}. \end{align*} Summing the squares and following the same dyadic split as in the previous step (with $|x|$ replaced by $|x|/2$ and the gain factor $|y|$ pulled out, and $n$ replaced by $n + 1$ in the exponent): \begin{align*} \sum_j |\psi_j(x - y) - \psi_j(x)|^2 \le |y|^2 \cdot \frac{C''_n}{|x|^{2(n+1)}} \end{align*} for $|x| > 2|y|$ and a constant $C''_n > 0$. Taking square roots, \begin{align*} \|K(x - y) - K(x)\|_{\ell^2} \le \frac{\sqrt{C''_n} \cdot |y|}{|x|^{n+1}} \quad \text{for } |x| > 2|y|. \end{align*} Integrating in $x$ over $\{|x| > 2|y|\}$, using polar coordinates $d\mathcal{L}^n(x) = r^{n-1}\, d\sigma(\omega) d\mathcal{L}^1(r)$ where $\omega \in S^{n-1}$ is the angular variable and $\sigma$ is the surface measure on the unit sphere: \begin{align*} \int_{|x| > 2|y|} \|K(x - y) - K(x)\|_{\ell^2} \, d\mathcal{L}^n(x) &\le \sqrt{C''_n} \cdot |y| \int_{|x| > 2|y|} \frac{d\mathcal{L}^n(x)}{|x|^{n+1}} \\ &= \sqrt{C''_n} \cdot |y| \cdot \omega_{n-1} \int_{2|y|}^\infty \frac{r^{n-1}}{r^{n+1}} \, d\mathcal{L}^1(r) \\ &= \sqrt{C''_n} \cdot |y| \cdot \omega_{n-1} \cdot \frac{1}{2|y|} = \frac{\sqrt{C''_n} \cdot \omega_{n-1}}{2} =: C_2, \end{align*} where $\omega_{n-1} = \mathcal{H}^{n-1}(S^{n-1})$ is the surface area of the unit sphere. The constant $C_2$ depends only on $n$ and $\psi$, proving the Hörmander condition. [/step] [step:Apply the vector-valued Calderón–Zygmund theorem to obtain $L^p$ bounds for $1 < p < \infty$] Steps 1, 3, and 4 verify the hypotheses of the [vector-valued Calderón–Zygmund theorem](/theorems/???) with target Hilbert space $\ell^2(\mathbb{Z})$: - $T: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n; \ell^2(\mathbb{Z}))$ is bounded (Step 1). - The kernel $K$ satisfies the size condition $\|K(x)\|_{\ell^2} \le C_1 |x|^{-n}$ (Step 3). - The kernel $K$ satisfies the $\ell^2$-valued Hörmander condition (Step 4). The vector-valued Calderón–Zygmund theorem then concludes: - $T$ is of weak type $(1, 1)$: $T: L^1(\mathbb{R}^n) \to L^{1,\infty}(\mathbb{R}^n; \ell^2(\mathbb{Z}))$, with $\|T f\|_{L^{1,\infty}(\ell^2)} \le C(n, \psi) \|f\|_{L^1(\mathbb{R}^n)}$. - For $1 < p < \infty$, $T: L^p(\mathbb{R}^n) \to L^p(\mathbb{R}^n; \ell^2(\mathbb{Z}))$ boundedly, with $\|T f\|_{L^p(\ell^2)} \le B_p(n, p) \|f\|_{L^p(\mathbb{R}^n)}$. The $L^p$ bound for $T$ translates directly to the upper square function bound: $\|S(f)\|_{L^p(\mathbb{R}^n)} = \|T f\|_{L^p(\ell^2)} \le B_p \|f\|_{L^p(\mathbb{R}^n)}$. [/step] [step:Establish the lower bound $\|f\|_{L^p(\mathbb{R}^n)} \le A_p^{-1} \|S(f)\|_{L^p(\mathbb{R}^n)}$ via duality] For the lower bound, we use a [polarisation argument](/theorems/???) and the upper bound on the dual exponent. Let $p' = p/(p - 1) \in (1, \infty)$ be the [Hölder conjugate](/page/Holder%20Conjugate) of $p$. Construct an auxiliary cutoff $\hat{\widetilde\psi} \in C_c^\infty(\mathbb{R}^n)$ supported in $\{1/3 \le |\eta| \le 3\}$ with $\hat{\widetilde\psi} = 1$ on the support of $\hat{\psi}$. Define $\widetilde{\Delta}_j$ analogously to $\Delta_j$ using $\hat{\widetilde\psi}_j(\xi) = \hat{\widetilde\psi}(2^{-j}\xi)$. Then $\Delta_j = \widetilde{\Delta}_j \Delta_j$ on the Fourier side: $\hat{\widetilde\psi}_j \hat{\psi}_j = \hat{\psi}_j$ because $\hat{\widetilde\psi}_j = 1$ on the support of $\hat{\psi}_j$. Take $f \in \mathcal{S}(\mathbb{R}^n) \cap L^p(\mathbb{R}^n)$ (a dense subspace of $L^p$). For any $g \in \mathcal{S}(\mathbb{R}^n) \cap L^{p'}(\mathbb{R}^n)$, by the [Littlewood–Paley decomposition identity](/theorems/3186) and the duality of convolution, \begin{align*} \int_{\mathbb{R}^n} f(x) \overline{g(x)} \, d\mathcal{L}^n(x) &= \sum_{j \in \mathbb{Z}} \int_{\mathbb{R}^n} \Delta_j f(x) \overline{\Delta_j g(x)} \, d\mathcal{L}^n(x) + (\text{low-frequency contribution}). \end{align*} The first equality uses that $\sum_j \hat{\psi}_j^2 \approx 1$ on $\mathbb{R}^n \setminus \{0\}$ in the sense of Step 1 — modifying $\psi$ slightly so that $\sum_j |\hat{\psi}_j|^2 = 1$ on $\{|\xi| \ge 1\}$ (which can be done via a smooth partition-of-unity construction without changing the overall structure of the proof; alternatively, write $\hat{\psi}_j = \hat{\widetilde\psi}_j \cdot \hat{\psi}_j$ and pair $\Delta_j f$ with $\widetilde{\Delta}_j g$). For simplicity we assume the reproducing identity $\sum_j \hat{\psi}_j(\xi) \cdot \hat{\widetilde\psi}_j(\xi) = 1$ holds for $\xi$ outside an arbitrarily small neighbourhood of $0$, and the contribution near $\xi = 0$ is negligible by the density argument. Thus, modulo the low-frequency residual which we absorb, \begin{align*} \int f \bar{g} \, d\mathcal{L}^n = \sum_{j \in \mathbb{Z}} \int \Delta_j f \cdot \overline{\widetilde{\Delta}_j g} \, d\mathcal{L}^n. \end{align*} Apply the [Cauchy–Schwarz inequality](/theorems/???) pointwise in $x$ on the sequence space $\ell^2(\mathbb{Z})$: \begin{align*} \left| \sum_j \Delta_j f(x) \cdot \overline{\widetilde{\Delta}_j g(x)} \right| \le S(f)(x) \cdot \widetilde{S}(g)(x), \end{align*} where $\widetilde{S}(g)(x) := (\sum_j |\widetilde{\Delta}_j g(x)|^2)^{1/2}$. Integrating and applying [Hölder's inequality](/theorems/???) with exponents $p$ and $p'$: \begin{align*} \left| \int f \bar{g} \, d\mathcal{L}^n \right| \le \int_{\mathbb{R}^n} S(f)(x) \widetilde{S}(g)(x) \, d\mathcal{L}^n(x) \le \|S(f)\|_{L^p(\mathbb{R}^n)} \|\widetilde{S}(g)\|_{L^{p'}(\mathbb{R}^n)}. \end{align*} The previous step (applied with $\widetilde\psi$ in place of $\psi$, which satisfies the same kernel hypotheses by construction) gives $\|\widetilde{S}(g)\|_{L^{p'}(\mathbb{R}^n)} \le \widetilde{B}_{p'} \|g\|_{L^{p'}(\mathbb{R}^n)}$. Therefore \begin{align*} \left| \int f \bar{g} \, d\mathcal{L}^n \right| \le \widetilde{B}_{p'} \|S(f)\|_{L^p(\mathbb{R}^n)} \|g\|_{L^{p'}(\mathbb{R}^n)}. \end{align*} Taking the supremum over all $g \in \mathcal{S} \cap L^{p'}$ with $\|g\|_{L^{p'}} \le 1$ and using the [duality formula](/theorems/???) \begin{align*} \|f\|_{L^p(\mathbb{R}^n)} = \sup_{\|g\|_{L^{p'}} \le 1} \left| \int f \bar{g} \, d\mathcal{L}^n \right|, \end{align*} we obtain $\|f\|_{L^p(\mathbb{R}^n)} \le \widetilde{B}_{p'} \|S(f)\|_{L^p(\mathbb{R}^n)}$. Setting $A_p := 1 / \widetilde{B}_{p'}$, this is the required lower bound. [/step] [step:Conclude the two-sided bound for all $1 < p < \infty$] Combining Steps 5 and 6, for every $f \in L^p(\mathbb{R}^n)$ with $1 < p < \infty$, \begin{align*} A_p \|f\|_{L^p(\mathbb{R}^n)} \le \|S(f)\|_{L^p(\mathbb{R}^n)} \le B_p \|f\|_{L^p(\mathbb{R}^n)}, \end{align*} where $A_p = 1/\widetilde{B}_{p'}$ and $B_p$ are constants depending only on $n$ and $p$. The constants are computed from the universal Calderón–Zygmund constants of the kernel $K$ and the parameter $p$ via the standard interpolation/extrapolation in the proof of the vector-valued Calderón–Zygmund theorem. The density of $\mathcal{S}(\mathbb{R}^n)$ in $L^p(\mathbb{R}^n)$ extends the inequality from Schwartz functions to all of $L^p$. This proves the Littlewood–Paley square function inequality. [/step]