[proofplan] The proof rests on two pillars: a Littlewood--Paley decomposition of $\mathbb{R}\setminus\{0\}$ into the dyadic intervals $I_k$, $I_k^-$, and the [Littlewood--Paley Square Function Characterisation of $L^p$](/theorems/???), which converts an $L^p$ norm into the $L^p$ norm of a vector-valued square function. The strategy is to apply a sharp dyadic Littlewood--Paley projection $S_k$ adapted to $I_k \cup I_k^-$, write $T_m f = \sum_k T_m(S_k f)$, and integrate by parts on each dyadic piece against the Stieltjes measure $dm|_{I_k}$ to express $T_m S_k f$ as a *single-scale* convolution operator whose symbol has $L^\infty$-norm controlled by the variation of $m$ on $I_k$. The vector-valued Marcinkiewicz interpolation argument then bundles these single-scale operators into a $\ell^2$-valued [Calderón--Zygmund operator](/theorems/???), whose $L^p$ boundedness for $1 < p < \infty$ follows from the vector-valued [Calderón--Zygmund Theorem](/theorems/???). Re-applying the Littlewood--Paley square function inverse closes the loop. [/proofplan]

[step:Set up a dyadic Littlewood--Paley resolution adapted to the intervals $I_k \cup I_k^-$] For $k \in \mathbb{Z}$, let $I_k := [2^k, 2^{k+1})$ and $I_k^- := (-2^{k+1}, -2^k]$, and set $J_k := I_k \cup I_k^-$. Choose $\hat\psi \in C^\infty_c(\mathbb{R})$ supported in $\{1/2 \le |\xi| \le 4\}$ with $\hat\psi \equiv 1$ on $\{1 \le |\xi| \le 2\}$ and $0 \le \hat\psi \le 1$. Define \begin{align*} \hat\psi_k : \mathbb{R} &\to \mathbb{R} \\ \xi &\mapsto \hat\psi(2^{-k}\xi), \end{align*} so that $\hat\psi_k$ is supported in $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$ and equals $1$ on $J_k$. The Littlewood--Paley projector is \begin{align*} \Delta_k : \mathcal{S}'(\mathbb{R}) &\to \mathcal{S}'(\mathbb{R}) \\ f &\mapsto \mathcal{F}^{-1}\bigl(\hat\psi_k\,\hat f\,\bigr). \end{align*} Observe that the family $(\hat\psi_k)_{k \in \mathbb{Z}}$ has finite overlap: any $\xi \ne 0$ lies in $\operatorname{supp}\hat\psi_k$ for at most $4$ values of $k$. This finite-overlap property will let us pass between $f$ and $\sum_k \Delta_k f$ using the [Littlewood--Paley Square Function Characterisation of $L^p$](/theorems/???). For the remainder of the argument fix $f \in \mathcal{S}(\mathbb{R})$; the conclusion for $L^p$ then follows by density. [/step]

[step:Express each piece $T_m \Delta_k f$ as a single-scale operator using the Stieltjes representation of $m$ on $J_k$] Since $\hat\psi_k$ is supported on $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$, the function $m\,\hat\psi_k$ depends only on $m\big|_{J'_k}$ where $J'_k := \{\xi : 2^{k-1} \le |\xi| \le 2^{k+2}\}$. The set $J'_k$ is the union of the four consecutive dyadic intervals $I_{k-1}, I_k, I_{k+1}, I_{k+1}^{-}$ on the positive side and the symmetric ones on the negative side; hence the total variation of $m$ on $J'_k$ is bounded by $8A$ (eight dyadic-interval contributions of $\le A$ each). Fix $k$ and define \begin{align*} m_k : \mathbb{R} &\to \mathbb{R} \\ \xi &\mapsto m(\xi)\,\hat\psi_k(\xi). \end{align*} The pointwise representation $\widehat{T_m \Delta_k f}(\xi) = m(\xi)\,\hat\psi_k(\xi)\,\hat f(\xi) = m_k(\xi)\,\hat f(\xi)$ shows that $T_m \Delta_k f = T_{m_k} f$. Since $\hat\psi_k \in C^\infty_c$ and $m \in L^\infty$ with bounded variation on $J'_k$, $m_k$ has bounded variation on $\mathbb{R}$ with $\operatorname{Var}(m_k) \le \operatorname{Var}(m;J'_k)\,\|\hat\psi_k\|_{L^\infty} + \|m\|_{L^\infty}\,\operatorname{Var}(\hat\psi_k) \le 8A + A\,V_\psi$, where $V_\psi := \operatorname{Var}(\hat\psi)$ depends only on $\hat\psi$. In particular $\|m_k\|_{L^\infty} \le \|m\|_{L^\infty} \le A$ and $\operatorname{Var}(m_k) \le C_\psi\,A$ where $C_\psi := 8 + V_\psi$. [/step]

[step:Reduce $L^p$ boundedness of $T_m$ to a vector-valued $\ell^2$ estimate via the Littlewood--Paley square function] Apply the [Littlewood--Paley Square Function Characterisation of $L^p$](/theorems/???) to $T_m f$. The hypothesis "$1 < p < \infty$" is given, and the resolution $(\hat\psi_k)_{k \in \mathbb{Z}}$ has been verified to have finite overlap and to form a Littlewood--Paley resolution after appropriate normalisation. The theorem yields constants $0 < a_p \le b_p < \infty$ depending only on $p$ and on $\hat\psi$ such that \begin{align*} a_p\,\|T_m f\|_{L^p(\mathbb{R})} \le \Bigl\|\Bigl(\sum_{k \in \mathbb{Z}} |\Delta_k T_m f|^2\Bigr)^{1/2}\Bigr\|_{L^p(\mathbb{R})} \le b_p\,\|T_m f\|_{L^p(\mathbb{R})}. \end{align*} The Fourier multipliers $T_m$ and $\Delta_k$ commute (both diagonal in frequency), so $\Delta_k T_m f = T_m \Delta_k f = T_{m_k} f$. Substituting and applying the same characterisation to $f$ in reverse, \begin{align*} \|T_m f\|_{L^p(\mathbb{R})} \le a_p^{-1}\,\Bigl\|\Bigl(\sum_{k \in \mathbb{Z}} |T_{m_k} f|^2\Bigr)^{1/2}\Bigr\|_{L^p(\mathbb{R})}. \end{align*} Therefore the proof reduces to the vector-valued estimate \begin{align*} \Bigl\|\Bigl(\sum_{k \in \mathbb{Z}} |T_{m_k} f|^2\Bigr)^{1/2}\Bigr\|_{L^p(\mathbb{R})} \le C_p\,A\,\|f\|_{L^p(\mathbb{R})}. \end{align*} [/step]

[step:Establish a kernel estimate for each single-scale multiplier $T_{m_k}$ via integration by parts] Let \begin{align*} K_k : \mathbb{R} &\to \mathbb{C} \\ x &\mapsto \mathcal{F}^{-1}(m_k)(x) = \frac{1}{2\pi}\int_\mathbb{R} m_k(\xi)\,e^{i\xi x}\,d\mathcal{L}^1(\xi) \end{align*} denote the convolution kernel of $T_{m_k}$, viewed initially as a tempered distribution and shown below to be a bounded measurable function for $x \ne 0$. \textbf{Pointwise size bound: $|K_k(x)| \le C\,A\,\min(2^k,\,|x|^{-1})$.} \textbf{(i) Direct bound at low frequency relative to $|x|$.} Since $m_k$ is supported in $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$ and $\|m_k\|_{L^\infty(\mathbb{R})} \le A$, \begin{align*} |K_k(x)| \le \frac{1}{2\pi}\int_{\operatorname{supp}m_k} |m_k(\xi)|\,d\mathcal{L}^1(\xi) \le \frac{A}{2\pi}\,\mathcal{L}^1(\operatorname{supp}m_k) \le C\,A\,2^k. \end{align*} \textbf{(ii) Sharp estimate at high frequency relative to $|x|$ via integration by parts.} Fix $x \ne 0$. Apply integration by parts in the Fourier integral once (the boundary terms vanish since $m_k$ has compact support). With the distributional derivative $dm_k$ on $\mathbb{R}$ — which is a finite signed Radon measure on $\mathbb{R}$ with total variation $\operatorname{Var}(m_k) \le C_\psi A$ established in Step 2 — \begin{align*} ix\,K_k(x) = \frac{1}{2\pi}\int_\mathbb{R} ix\,m_k(\xi)\,e^{i\xi x}\,d\mathcal{L}^1(\xi) = \frac{1}{2\pi}\int_\mathbb{R} \bigl(\partial_\xi e^{i\xi x}\bigr)\,m_k(\xi)\,d\mathcal{L}^1(\xi) = -\frac{1}{2\pi}\int_\mathbb{R} e^{i\xi x}\,dm_k(\xi), \end{align*} where the last equality uses integration by parts in the Stieltjes sense. Taking absolute values, \begin{align*} |x|\,|K_k(x)| \le \frac{1}{2\pi}\,|m_k|(\mathbb{R}) = \frac{\operatorname{Var}(m_k)}{2\pi} \le \frac{C_\psi A}{2\pi}. \end{align*} Hence $|K_k(x)| \le C_\psi A\,|x|^{-1}$ for all $x \ne 0$. Combining (i) and (ii), \begin{align*} |K_k(x)| \le C_\psi A \cdot \min\bigl(2^k,\,|x|^{-1}\bigr) \qquad \text{for } x \ne 0. \end{align*} \textbf{Sharper size bound via two integrations by parts: $|K_k(x)| \le C\,A\,2^{-k}\,|x|^{-2}$.} The single integration by parts in (ii) used only the BV-regularity of $m$. To obtain the second power of $|x|^{-1}$ we exploit the *smoothness* of the cutoff $\hat\psi$. Write $m_k(\xi) = \hat\psi(2^{-k}\xi)\,m(\xi)$ and split the IBP identity from (ii): \begin{align*} ix\,K_k(x) = -\frac{1}{2\pi}\int_\mathbb{R} e^{i\xi x}\,dm_k(\xi) = -\frac{1}{2\pi}\Bigl[\underbrace{\int_\mathbb{R} e^{i\xi x}\,2^{-k}\hat\psi'(2^{-k}\xi)\,m(\xi)\,d\mathcal{L}^1(\xi)}_{=: I_1(x)} + \underbrace{\int_\mathbb{R} e^{i\xi x}\,\hat\psi(2^{-k}\xi)\,dm(\xi)}_{=: I_2(x)}\Bigr], \end{align*} where we used the Leibniz product rule for the Stieltjes derivative: $dm_k = 2^{-k}\hat\psi'(2^{-k}\xi)\,m(\xi)\,d\mathcal{L}^1(\xi) + \hat\psi(2^{-k}\xi)\,dm(\xi)$. Both $I_1$ and $I_2$ are integrals against $e^{i\xi x}$ of a function with bounded $L^1(d\mathcal{L}^1)$- or BV-norm, of size $\le C_\psi A$ uniformly in $k$ (using $|m| \le A$, $|\hat\psi'| \le C$, $\hat\psi$ bounded, and $|m'|(J'_k) \le 8A$). \textbf{Sub-claim.} $|I_1(x)| \le C_\psi A\,|x|^{-1}$ and $|I_2(x)| \le C_\psi A\,|x|^{-1}$ for $x \ne 0$. \textit{Estimate of $I_1$.} The integrand $g_1(\xi) := 2^{-k}\hat\psi'(2^{-k}\xi)\,m(\xi)$ is supported on $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$ (where $\hat\psi'(2^{-k}\xi) \ne 0$). Integrate by parts once more in $\xi$, using $\partial_\xi e^{i\xi x} = ix\,e^{i\xi x}$: \begin{align*} ix\,I_1(x) = \int_\mathbb{R}\bigl(\partial_\xi e^{i\xi x}\bigr)\,g_1(\xi)\,d\mathcal{L}^1(\xi) = -\int_\mathbb{R} e^{i\xi x}\,dg_1(\xi), \end{align*} where $dg_1$ is the Stieltjes derivative of the BV-function $g_1$. By Leibniz, $dg_1 = 2^{-k}\hat\psi'(2^{-k}\xi)\,dm + 2^{-2k}\hat\psi''(2^{-k}\xi)\,m\,d\mathcal{L}^1$. The total variation of $dg_1$ is bounded by \begin{align*} |dg_1|(\mathbb{R}) \le 2^{-k}\,\|\hat\psi'\|_{L^\infty}\cdot|m'|(J'_k) + 2^{-2k}\,\|\hat\psi''\|_{L^\infty}\cdot\|m\|_{L^\infty(J'_k)}\cdot\mathcal{L}^1(\operatorname{supp}\hat\psi'(2^{-k}\cdot)) \le C_\psi\,A\,2^{-k}, \end{align*} since $|m'|(J'_k) \le 8A$ and $\mathcal{L}^1(\operatorname{supp}\hat\psi'(2^{-k}\cdot)) \asymp 2^k$. Therefore \begin{align*} |x|\,|I_1(x)| \le |dg_1|(\mathbb{R}) \le C_\psi\,A\,2^{-k}, \quad\text{i.e.}\quad |I_1(x)| \le C_\psi A\,2^{-k}\,|x|^{-1}. \end{align*} \textit{Estimate of $I_2$.} The integrand pairs $\hat\psi(2^{-k}\xi)$ (smooth, supported on $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$) against $dm$. We integrate by parts in the *smooth* factor: since $\hat\psi(2^{-k}\xi)$ vanishes at the endpoints of any interval containing the support, and $dm$ is a Radon measure, pair-by-parts gives \begin{align*} ix\,I_2(x) = \int_\mathbb{R}\bigl(\partial_\xi e^{i\xi x}\bigr)\hat\psi(2^{-k}\xi)\,dm(\xi) = -\int_\mathbb{R}e^{i\xi x}\,d\bigl[\hat\psi(2^{-k}\xi)\,dm(\xi)\bigr] - \int_\mathbb{R}e^{i\xi x}\,2^{-k}\hat\psi'(2^{-k}\xi)\,dm(\xi), \end{align*} where the first term on the right uses the BV-regularity of $m$ (giving an absolutely continuous distribution) and the second comes from differentiating the smooth factor. The first term vanishes because the relevant Stieltjes IBP for two BV functions gives no contribution at infinity (both factors compactly supported). What remains is \begin{align*} |x|\,|I_2(x)| \le \int_\mathbb{R} 2^{-k}|\hat\psi'(2^{-k}\xi)|\,d|m|(\xi) \le 2^{-k}\,\|\hat\psi'\|_{L^\infty}\,|m'|(J'_k) \le C_\psi\,A\,2^{-k}. \end{align*} Hence $|I_2(x)| \le C_\psi A\,2^{-k}\,|x|^{-1}$, proving the sub-claim. From the identity $ix\,K_k(x) = -(I_1 + I_2)/(2\pi)$ and the sub-claim, \begin{align*} |x|\,|K_k(x)| \le \frac{|I_1(x)| + |I_2(x)|}{2\pi} \le C_\psi A\,2^{-k}\,|x|^{-1}, \end{align*} which rearranges to \begin{align*} |K_k(x)| \le C_\psi A\,2^{-k}\,|x|^{-2} \qquad \text{for } x \ne 0. \end{align*} \textbf{Pointwise smoothness bound: $|\partial_x K_k(x)| \le C\,A\,\min(2^{2k},\,2^{-k}|x|^{-3})$.} The Fourier multiplier of $\partial_x K_k$ is $i\xi\,m_k$. The function $\xi\,m_k$ is BV with variation $\le 2^{k+2}\,\operatorname{Var}(m_k) + \|m_k\|_{L^\infty}\operatorname{Var}(\xi\,\mathbb{1}_{\operatorname{supp}m_k}) \le C_\psi A\,2^k$, using $|\xi| \le 2^{k+2}$ on the support. \textbf{(i$'$) Direct bound.} As in (i), $|\partial_x K_k(x)| \le \frac{1}{2\pi}\int |\xi\,m_k(\xi)|\,d\mathcal{L}^1(\xi) \le C\,A\,2^{2k}$. \textbf{(ii$'$) Sharp estimate by repeating the size argument.} Apply the same two-step argument used for $|K_k|$, but to $\partial_x K_k$ — i.e. with multiplier $i\xi\,m_k(\xi) = i\xi\,\hat\psi(2^{-k}\xi)\,m(\xi)$ in place of $m_k$. Write $i\xi\,m_k(\xi) = \tilde\psi_k(\xi)\,m(\xi)$ where $\tilde\psi_k(\xi) := i\xi\,\hat\psi(2^{-k}\xi)$. The function $\tilde\psi_k$ is smooth, supported on $\{2^{k-1} \le |\xi| \le 2^{k+2}\}$, and satisfies $\|\tilde\psi_k\|_{L^\infty} \le C\,2^k$, $\|\tilde\psi_k'\|_{L^\infty} \le C$, $\|\tilde\psi_k''\|_{L^\infty} \le C\,2^{-k}$ (chain rule, with $\hat\psi$ Schwartz). Repeating the IBP-twice argument verbatim with $\tilde\psi_k$ in place of $\hat\psi(2^{-k}\cdot)$: the analogue of $|I_1|$ becomes \begin{align*} |x|^2\,|\partial_x K_k(x)| \le \frac{1}{2\pi}\Bigl(\|\tilde\psi_k''\|_{L^\infty}\,\|m\|_{L^\infty}\,\mathcal{L}^1(\operatorname{supp}\tilde\psi_k) + \|\tilde\psi_k'\|_{L^\infty}\,|m'|(J'_k)\Bigr) \le C_\psi A, \end{align*} using $\|\tilde\psi_k''\|_{L^\infty} \cdot \mathcal{L}^1(\operatorname{supp}) \le C\,2^{-k}\cdot 2^k = C$ and $\|\tilde\psi_k'\|_{L^\infty}\cdot|m'|(J'_k) \le C\cdot 8A = C\,A$. Hence $|\partial_x K_k(x)| \le C_\psi A\,|x|^{-2}$. To gain the additional $2^{-k}|x|^{-1}$ factor we apply *one more* integration by parts on the smooth piece (a third $\partial_\xi$), using that $\tilde\psi_k$ has bounded third derivative $\|\tilde\psi_k'''\|_{L^\infty} \le C\,2^{-2k}$. This produces \begin{align*} |x|^3\,|\partial_x K_k(x)| \le C\Bigl(\|\tilde\psi_k'''\|_{L^\infty}\,\|m\|_{L^\infty}\,\mathcal{L}^1(\operatorname{supp}) + \|\tilde\psi_k''\|_{L^\infty}\,|m'|(J'_k)\Bigr) \le C\,A\,(2^{-2k}\cdot 2^k + 2^{-k}\cdot A) \le C_\psi A\,2^{-k}, \end{align*} giving $|\partial_x K_k(x)| \le C_\psi A\,2^{-k}\,|x|^{-3}$ for $x \ne 0$. Combined with (i$'$), \begin{align*} |\partial_x K_k(x)| \le C_\psi A \cdot \min\bigl(2^{2k},\,2^{-k}|x|^{-3}\bigr). \end{align*} \textbf{Vector-valued $\ell^2$ kernel size estimate.} Define $K: \mathbb{R} \setminus\{0\} \to \ell^2(\mathbb{Z})$ by $K(x) := (K_k(x))_{k \in \mathbb{Z}}$. For $x \ne 0$, fix the threshold $k_0 \in \mathbb{Z}$ with $2^{k_0} \asymp |x|^{-1}$. Use the regime-appropriate bound: - For $k \le k_0$ (low frequency relative to $|x|^{-1}$), use $|K_k(x)| \le C_\psi A\,2^k$ from (i). - For $k > k_0$ (high frequency), use $|K_k(x)| \le C_\psi A\,2^{-k}|x|^{-2}$ from the sharper estimate above. Squaring and summing, \begin{align*} \|K(x)\|_{\ell^2}^2 = \sum_{k \in \mathbb{Z}}|K_k(x)|^2 \le \sum_{k \le k_0}(C_\psi A\,2^k)^2 + \sum_{k > k_0}(C_\psi A\,2^{-k}|x|^{-2})^2. \end{align*} The first sum is geometric with ratio $4$ in $k$: \begin{align*} \sum_{k \le k_0} 2^{2k} = \frac{2^{2(k_0 + 1)}}{2^2 - 1} \cdot \frac{1}{4/3} \asymp 2^{2k_0} \asymp |x|^{-2}. \end{align*} The second sum is geometric with ratio $1/4$: \begin{align*} |x|^{-4}\sum_{k > k_0} 2^{-2k} = |x|^{-4} \cdot \frac{2^{-2(k_0 + 1)}}{1 - 1/4} \asymp |x|^{-4}\cdot 2^{-2k_0} \asymp |x|^{-4}\cdot |x|^2 = |x|^{-2}. \end{align*} Adding the two contributions, \begin{align*} \|K(x)\|_{\ell^2}^2 \le C_\psi^2 A^2\,|x|^{-2}, \qquad \text{equivalently,} \qquad \|K(x)\|_{\ell^2} \le C_\psi A\,|x|^{-1} \qquad \text{for } x \ne 0. \end{align*} \textbf{Hörmander smoothness condition.} For $|x| > 2|y|$, by the [mean value theorem](/theorems/???) along the segment from $x - y$ to $x$, \begin{align*} |K_k(x - y) - K_k(x)| \le |y|\,\sup_{0 \le t \le 1}|\partial_x K_k(x - ty)|, \end{align*} and $|x - ty| \ge |x|/2$ for $t \in [0,1]$ when $|x| > 2|y|$. Substituting the smoothness bound $|\partial_x K_k| \le C_\psi A\,\min(2^{2k},\,2^{-k}|x|^{-3})$ (with $|x|/2$ in place of $|x|$, which only changes the constant), \begin{align*} |K_k(x - y) - K_k(x)| \le C_\psi A\,|y|\,\min\bigl(2^{2k},\,2^{-k}|x|^{-3}\bigr). \end{align*} Squaring and summing over $k$ via the same threshold split with $2^{k_1} \asymp |x|^{-1}$: - For $k \le k_1$, use $|K_k(x-y) - K_k(x)| \le C_\psi A\,|y|\,2^{2k}$, contributing $C_\psi^2 A^2|y|^2\sum_{k \le k_1} 2^{4k} \asymp C_\psi^2 A^2|y|^2\,2^{4k_1} \asymp C_\psi^2 A^2|y|^2|x|^{-4}$. - For $k > k_1$, use $|K_k(x-y) - K_k(x)| \le C_\psi A\,|y|\,2^{-k}|x|^{-3}$, contributing $C_\psi^2 A^2|y|^2|x|^{-6}\sum_{k > k_1} 2^{-2k} \asymp C_\psi^2 A^2|y|^2|x|^{-6}\,2^{-2k_1} \asymp C_\psi^2 A^2|y|^2|x|^{-6}\cdot|x|^2 = C_\psi^2 A^2|y|^2|x|^{-4}$. Hence \begin{align*} \|K(x - y) - K(x)\|_{\ell^2} \le C_\psi A\,|y|\,|x|^{-2}. \end{align*} Integrating against $|x|^{-2}$ over $\{|x| > 2|y|\}$, \begin{align*} \int_{|x| > 2|y|}\|K(x - y) - K(x)\|_{\ell^2}\,d\mathcal{L}^1(x) \le C_\psi A\,|y|\int_{|x| > 2|y|} |x|^{-2}\,d\mathcal{L}^1(x) = C_\psi A\,|y|\cdot 2/(2|y|) = C_\psi A, \end{align*} uniformly in $y \in \mathbb{R}\setminus\{0\}$. [/step]

[step:Verify $L^2$ boundedness of the vector-valued operator and apply the vector-valued Calderón--Zygmund theorem] Define the $\ell^2$-valued operator \begin{align*} T : \mathcal{S}(\mathbb{R}) &\to L^p(\mathbb{R};\ell^2(\mathbb{Z})) \\ f &\mapsto (T_{m_k} f)_{k \in \mathbb{Z}}. \end{align*} By Plancherel, \begin{align*} \|T f\|_{L^2(\mathbb{R};\ell^2)}^2 = \int_\mathbb{R} \sum_{k} |T_{m_k}f(x)|^2\,d\mathcal{L}^1(x) = \int_\mathbb{R}\sum_{k}|m_k(\xi)|^2\,|\hat f(\xi)|^2\,d\mathcal{L}^1(\xi). \end{align*} Using $|m_k(\xi)| \le A\,\hat\psi_k(\xi)$ and the finite-overlap property $\sum_k \hat\psi_k(\xi)^2 \le 4$ for every $\xi \ne 0$, \begin{align*} \|Tf\|_{L^2(\mathbb{R};\ell^2)}^2 \le 4 A^2 \,\|\hat f\|_{L^2(\mathbb{R})}^2 = 4 A^2\,(2\pi)\,\|f\|_{L^2(\mathbb{R})}^2, \end{align*} so $T: L^2(\mathbb{R}) \to L^2(\mathbb{R};\ell^2)$ boundedly with norm $\le C\,A$. Combined with the Hörmander kernel estimate from the previous step, $T$ is a vector-valued Calderón--Zygmund operator from $L^2(\mathbb{R})$ to $L^2(\mathbb{R};\ell^2(\mathbb{Z}))$ with kernel $K$. By the vector-valued [Calderón--Zygmund Theorem](/theorems/???), whose hypotheses ($L^2$ boundedness with norm $\le M$ and Hörmander smoothness with constant $\le M$) we have just verified with $M = C_\psi\,A$, $T$ extends to a bounded operator $L^p(\mathbb{R}) \to L^p(\mathbb{R};\ell^2(\mathbb{Z}))$ for every $1 < p < \infty$, with \begin{align*} \|Tf\|_{L^p(\mathbb{R};\ell^2)} = \Bigl\|\Bigl(\sum_k |T_{m_k}f|^2\Bigr)^{1/2}\Bigr\|_{L^p(\mathbb{R})} \le C_p\,A\,\|f\|_{L^p(\mathbb{R})}. \end{align*} [/step]

[step:Combine the vector-valued estimate with the Littlewood--Paley equivalence to conclude] From the reduction in Step 3 and the bound just established in Step 5, \begin{align*} \|T_m f\|_{L^p(\mathbb{R})} \le a_p^{-1}\,\Bigl\|\Bigl(\sum_{k \in \mathbb{Z}} |T_{m_k} f|^2\Bigr)^{1/2}\Bigr\|_{L^p(\mathbb{R})} \le a_p^{-1}\,C_p\,A\,\|f\|_{L^p(\mathbb{R})}. \end{align*} Setting $C_p' := a_p^{-1}\,C_p$, which depends only on $p$, gives $\|T_m f\|_{L^p} \le C_p'\,A\,\|f\|_{L^p}$ for all $f \in \mathcal{S}(\mathbb{R})$. By density of $\mathcal{S}(\mathbb{R})$ in $L^p(\mathbb{R})$ (valid because $1 < p < \infty$), $T_m$ extends to a bounded operator on $L^p(\mathbb{R})$ with $\|T_m\|_{\mathcal{L}(L^p(\mathbb{R}))} \le C_p'\,A$, completing the proof. [/step]