[proofplan] We combine three facts: (i) the [$L^2$ Isometry of the Hilbert Transform](/theorems/3168), giving $\|Hf\|_{L^2} = \|f\|_{L^2}$, which is a strong-type $(2,2)$ bound; (ii) the [Weak-(1,1) Bound for the Hilbert Transform](/theorems/???), giving $\mathcal{L}^1(\{|Hf| > \lambda\}) \le C \lambda^{-1}\|f\|_{L^1}$; and (iii) duality with the adjoint identity $H^* = -H$. The [Marcinkiewicz Interpolation Theorem](/theorems/???) applied with endpoints $(1,1)$ (weak) and $(2,2)$ (strong) yields strong-type $(p,p)$ for every $1 < p \le 2$. The case $p = 2$ is the isometry. For $p > 2$, the adjoint identity $H^* = -H$ converts a bound on $H$ at exponent $p'$ (where $1/p + 1/p' = 1$) into a bound on $H$ at exponent $p$, completing the range $1 < p < \infty$. [/proofplan]

[step:Record the strong-$(2,2)$ and weak-$(1,1)$ endpoint bounds] By the [$L^2$ Isometry of the Hilbert Transform](/theorems/3168), $H: L^2(\mathbb{R}) \to L^2(\mathbb{R})$ is bounded with $\|H\|_{\mathcal{L}(L^2(\mathbb{R}))} = 1$, which is a strong-type $(2,2)$ bound. By the [Weak-(1,1) Bound for the Hilbert Transform](/theorems/???), there is an absolute constant $C_1 > 0$ such that for every $f \in L^1(\mathbb{R})$ and every $\lambda > 0$, \begin{align*} \mathcal{L}^1\!\left(\{x \in \mathbb{R} : |Hf(x)| > \lambda\}\right) \le \frac{C_1}{\lambda}\,\|f\|_{L^1}. \end{align*} Equivalently, $H$ is of weak type $(1,1)$ with constant $C_1$. [/step]

[step:Apply Marcinkiewicz interpolation between $(1,1)$ weak and $(2,2)$ strong to obtain $L^p$ for $1 < p < 2$] Fix $p \in (1, 2)$. We invoke the [Marcinkiewicz Interpolation Theorem](/theorems/???), which states: if a sublinear operator $T$ is of weak type $(p_0, p_0)$ with constant $A_0$ and of weak type $(p_1, p_1)$ with constant $A_1$ for some $1 \le p_0 < p_1 \le \infty$, then for every $\theta \in (0,1)$ and $p \in (p_0, p_1)$ defined by $\frac{1}{p} = \frac{1-\theta}{p_0} + \frac{\theta}{p_1}$, the operator $T$ extends to a bounded operator on $L^p$ with \begin{align*} \|Tf\|_{L^p} \le C(p_0, p_1, p)\,A_0^{1-\theta}\,A_1^{\theta}\,\|f\|_{L^p}. \end{align*} We verify the hypotheses with $T = H$ on the diagonal endpoints $(p_0, q_0) = (1, 1)$ and $(p_1, q_1) = (2, 2)$: - $H$ is sublinear (in fact linear) by construction. - $H$ is of weak type $(1,1)$ with constant $A_0 = C_1$ (Step 1). - $H$ is of strong type $(2,2)$ with constant $A_1 = 1$ (Step 1); in particular, of weak type $(2,2)$. By Marcinkiewicz, $H: L^p(\mathbb{R}) \to L^p(\mathbb{R})$ is bounded for every $p \in (1, 2)$, with norm $C_p$ depending only on $p$. [/step]

[step:Cover the case $p = 2$ by the isometry] For $p = 2$, the [$L^2$ Isometry of the Hilbert Transform](/theorems/3168) directly gives $\|Hf\|_{L^2} = \|f\|_{L^2}$ for all $f \in L^2(\mathbb{R})$, so $C_2 = 1$ works. [/step]

[step:Cover the range $p > 2$ by duality with the adjoint identity $H^* = -H$] Let $p \in (2, \infty)$ and let $p' = p/(p-1) \in (1, 2)$ be the conjugate exponent (so $1/p + 1/p' = 1$). We first identify the adjoint of $H$ with respect to the $L^2$ pairing. For $f, g \in \mathcal{S}(\mathbb{R})$, by Parseval and the multiplier identity $\widehat{Hf}(\xi) = -i\,\mathrm{sgn}(\xi)\,\hat{f}(\xi)$, \begin{align*} (Hf, g)_{L^2} = \frac{1}{2\pi}\,(\widehat{Hf}, \hat{g})_{L^2} = \frac{1}{2\pi}\int_{\mathbb{R}} -i\,\mathrm{sgn}(\xi)\,\hat{f}(\xi)\,\overline{\hat{g}(\xi)}\,d\mathcal{L}^1(\xi). \end{align*} Pulling the multiplier onto the second factor (with conjugation) gives \begin{align*} (Hf, g)_{L^2} = \frac{1}{2\pi}\int_{\mathbb{R}} \hat{f}(\xi)\,\overline{\,i\,\mathrm{sgn}(\xi)\,\hat{g}(\xi)}\,d\mathcal{L}^1(\xi) = (f, \mathcal{F}^{-1}(i\,\mathrm{sgn}\cdot \hat{g}))_{L^2}. \end{align*} The multiplier $i\,\mathrm{sgn}(\xi) = -(-i\,\mathrm{sgn}(\xi))$ corresponds to $-H$, so $\mathcal{F}^{-1}(i\,\mathrm{sgn}\cdot\hat{g}) = -Hg$. Therefore $(Hf, g)_{L^2} = -(f, Hg)_{L^2}$ for all $f, g \in \mathcal{S}(\mathbb{R})$, and by density of $\mathcal{S}(\mathbb{R})$ in $L^2(\mathbb{R})$ this extends to all $f, g \in L^2(\mathbb{R})$. We conclude that the $L^2$-adjoint $H^*$ satisfies $H^* = -H$. Now apply Step 2 at the exponent $p' \in (1, 2)$: $H: L^{p'}(\mathbb{R}) \to L^{p'}(\mathbb{R})$ is bounded with norm $C_{p'}$. The dual operator $H': (L^{p'}(\mathbb{R}))^* \to (L^{p'}(\mathbb{R}))^*$, identified via $(L^{p'})^* \cong L^p$ (by the [Riesz Representation Theorem for $L^p$](/theorems/???), which requires $1 \le p' < \infty$, satisfied since $p' \in (1,2)$), is bounded with the same norm $C_{p'}$. On the dense subspace $\mathcal{S}(\mathbb{R}) \subset L^{p'}(\mathbb{R}) \cap L^p(\mathbb{R})$, the dual $H'$ acts by $H' = (H|_{L^{p'}})^*$, and by the computation above this coincides with the $L^2$-adjoint $H^* = -H$. Therefore $-H: L^p(\mathbb{R}) \to L^p(\mathbb{R})$ is bounded with norm $C_{p'}$, hence so is $H$ with the same norm: \begin{align*} \|Hf\|_{L^p} \le C_{p'}\,\|f\|_{L^p} \quad \text{for all } f \in L^p(\mathbb{R}). \end{align*} Setting $C_p := C_{p'}$ gives the desired bound for $p > 2$. [/step]

[step:Combine the three regimes to conclude $L^p$ boundedness for all $p \in (1, \infty)$] Steps 2, 3, and 4 cover the regimes $1 < p < 2$, $p = 2$, and $2 < p < \infty$, respectively, with a constant $C_p > 0$ depending only on $p$. Since these three regimes are disjoint and exhaust $(1, \infty)$, we conclude that for every $p \in (1, \infty)$ the Hilbert transform extends to a bounded linear operator $H: L^p(\mathbb{R}) \to L^p(\mathbb{R})$ with \begin{align*} \|Hf\|_{L^p} \le C_p\,\|f\|_{L^p} \quad \text{for all } f \in L^p(\mathbb{R}), \end{align*} which is the assertion of the theorem. [/step]