[proofplan] We prove equivalence by routing all three definitions through the Fourier transform. First we compute the distributional Fourier transform of the principal-value kernel $\operatorname{p.v.}(1/(\pi x))$ via a regularised Dirichlet integral and obtain $-i\operatorname{sgn}(\xi)$. Combined with the convolution theorem, this identifies the principal-value definition with the multiplier definition. Next we compute $\widehat{Q_t}(\xi) = -i\operatorname{sgn}(\xi)e^{-t|\xi|}$ for the conjugate Poisson kernel and pass to the limit $t \to 0^+$ by dominated convergence on the Fourier side, identifying the conjugate-Poisson boundary value with the multiplier definition. Finally we check that the multiplier $m(\xi) = -i\operatorname{sgn}(\xi)$ preserves $\mathcal{S}(\mathbb{R})$. [/proofplan] [step:Compute the distributional Fourier transform of the principal-value kernel] Let $K \in \mathcal{S}'(\mathbb{R})$ denote the tempered distribution \begin{align*} K(\varphi) := \frac{1}{\pi}\,\mathrm{p.v.}\!\int_{\mathbb{R}} \frac{\varphi(x)}{x}\, d\mathcal{L}^1(x) = \frac{1}{\pi}\lim_{\varepsilon \to 0^+}\int_{|x| > \varepsilon} \frac{\varphi(x)}{x}\, d\mathcal{L}^1(x), \qquad \varphi \in \mathcal{S}(\mathbb{R}). \end{align*} We claim that the distributional Fourier transform of $K$ is the bounded function \begin{align*} \hat{K}: \mathbb{R} &\to \mathbb{C} \\ \xi &\mapsto -i\operatorname{sgn}(\xi). \end{align*} Let $\varphi \in \mathcal{S}(\mathbb{R})$. By the definition of the Fourier transform of a tempered distribution, $\hat{K}(\varphi) = K(\hat{\varphi})$, so we must compute \begin{align*} \hat{K}(\varphi) = \frac{1}{\pi}\lim_{\varepsilon \to 0^+}\int_{|x|>\varepsilon} \frac{\hat\varphi(x)}{x}\, d\mathcal{L}^1(x). \end{align*} By Fubini's theorem (justified because $\varphi \in L^1(\mathbb{R})$ and $\mathbb{1}_{\{|x|>\varepsilon\}}/x$ is bounded for each fixed $\varepsilon > 0$), we substitute $\hat\varphi(x) = \int_{\mathbb{R}} \varphi(\xi) e^{-i\xi x}\, d\mathcal{L}^1(\xi)$ and exchange the order of integration: \begin{align*} \int_{|x|>\varepsilon} \frac{\hat\varphi(x)}{x}\, d\mathcal{L}^1(x) = \int_{\mathbb{R}} \varphi(\xi) \left(\int_{|x|>\varepsilon} \frac{e^{-i\xi x}}{x}\, d\mathcal{L}^1(x)\right) d\mathcal{L}^1(\xi). \end{align*} Since the kernel $1/x$ is odd, only the imaginary part of $e^{-i\xi x} = \cos(\xi x) - i\sin(\xi x)$ contributes: \begin{align*} \int_{|x|>\varepsilon} \frac{e^{-i\xi x}}{x}\, d\mathcal{L}^1(x) = -2i\int_\varepsilon^\infty \frac{\sin(\xi x)}{x}\, d\mathcal{L}^1(x). \end{align*} For $\xi \neq 0$, the substitution $u = \xi x$, $d\mathcal{L}^1(u) = |\xi|\, d\mathcal{L}^1(x)$ when $\xi > 0$ (with the analogous orientation reversal when $\xi < 0$), gives \begin{align*} \int_\varepsilon^\infty \frac{\sin(\xi x)}{x}\, d\mathcal{L}^1(x) = \operatorname{sgn}(\xi)\int_{\varepsilon|\xi|}^\infty \frac{\sin u}{u}\, d\mathcal{L}^1(u). \end{align*} The Dirichlet integral $\int_0^\infty \frac{\sin u}{u}\, d\mathcal{L}^1(u) = \pi/2$ gives \begin{align*} \lim_{\varepsilon \to 0^+}\int_{|x|>\varepsilon} \frac{e^{-i\xi x}}{x}\, d\mathcal{L}^1(x) = -i\pi\operatorname{sgn}(\xi), \qquad \xi \neq 0. \end{align*} The integrand $\xi \mapsto \int_{|x|>\varepsilon} e^{-i\xi x}/x\, d\mathcal{L}^1(x)$ is uniformly bounded in $\varepsilon, \xi$ (one checks via integration by parts that $|2\int_\varepsilon^\infty \sin(\xi x)/x\, d\mathcal{L}^1(x)| \le 2\pi$), so dominated convergence applied with dominator $2\pi |\varphi(\xi)|$ permits passing the limit through the outer integral: \begin{align*} \hat{K}(\varphi) = \frac{1}{\pi}\int_{\mathbb{R}} \varphi(\xi) \cdot (-i\pi\operatorname{sgn}(\xi))\, d\mathcal{L}^1(\xi) = \int_{\mathbb{R}} \varphi(\xi) \cdot (-i\operatorname{sgn}(\xi))\, d\mathcal{L}^1(\xi). \end{align*} Hence $\hat{K} = -i\operatorname{sgn}(\xi)$ as a regular tempered distribution. [/step] [step:Identify the principal-value definition with the multiplier definition via convolution] Let $f \in \mathcal{S}(\mathbb{R})$. The principal-value definition reads \begin{align*} H_{\mathrm{p.v.}} f := \frac{1}{\pi}\mathrm{p.v.}\!\int_{\mathbb{R}} \frac{f(\cdot - y)}{y}\, d\mathcal{L}^1(y) = (K * f), \end{align*} the convolution of the tempered distribution $K$ with $f$. By the convolution theorem for the Fourier transform of $\mathcal{S}' \times \mathcal{S}$, \begin{align*} \widehat{K * f}(\xi) = \hat{K}(\xi) \cdot \hat{f}(\xi) = -i\operatorname{sgn}(\xi)\hat{f}(\xi). \end{align*} This shows $\widehat{H_{\mathrm{p.v.}} f}(\xi) = -i\operatorname{sgn}(\xi)\hat{f}(\xi)$, which is the multiplier definition $H_m f$ on the Fourier side. Since the Fourier transform is injective on $\mathcal{S}'(\mathbb{R})$, $H_{\mathrm{p.v.}}f = H_m f$ as elements of $\mathcal{S}(\mathbb{R})$ (provided we verify that the result is Schwartz, which Step 4 does). [/step] [step:Compute the Fourier transform of the conjugate Poisson kernel and pass to the limit] For $t > 0$, define the conjugate Poisson kernel \begin{align*} Q_t: \mathbb{R} &\to \mathbb{R} \\ x &\mapsto \frac{x}{\pi(x^2 + t^2)}. \end{align*} We claim that \begin{align*} \widehat{Q_t}(\xi) = -i\operatorname{sgn}(\xi)e^{-t|\xi|}, \qquad \xi \in \mathbb{R}. \end{align*} This follows from the analogous Poisson identity $\widehat{P_t}(\xi) = e^{-t|\xi|}$ for $P_t(x) = t/(\pi(x^2+t^2))$ via the holomorphic extension: writing $P_t + iQ_t = \frac{1}{\pi i(x - it)} \cdot i = \frac{1}{\pi(it - x)}$ (or equivalently the boundary value of the Cauchy kernel on the upper half-plane), and Fourier-transforming the Cauchy kernel gives $\widehat{P_t} + i\widehat{Q_t} = e^{-t|\xi|}(1 + \operatorname{sgn}(\xi))/2 \cdot 2 \cdot \mathbb{1}_{\{\xi>0\}}$ — concretely, evaluating directly, \begin{align*} \widehat{Q_t}(\xi) = \int_{\mathbb{R}} \frac{x}{\pi(x^2 + t^2)} e^{-i\xi x}\, d\mathcal{L}^1(x) = -i\operatorname{sgn}(\xi)e^{-t|\xi|}, \end{align*} where the integral is computed by residues (closing in the upper or lower half-plane according to $\operatorname{sgn}(\xi)$, with the simple pole at $x = \mp it$). For $f \in \mathcal{S}(\mathbb{R})$, since $Q_t \in L^2(\mathbb{R})$ and $f \in L^2(\mathbb{R})$, we have $Q_t * f \in L^2(\mathbb{R})$ with $\widehat{Q_t * f}(\xi) = \widehat{Q_t}(\xi)\hat{f}(\xi) = -i\operatorname{sgn}(\xi)e^{-t|\xi|}\hat{f}(\xi)$. We claim $(Q_t * f)(x) \to H_m f(x)$ pointwise as $t \to 0^+$. By Fourier inversion, \begin{align*} (Q_t * f)(x) = \frac{1}{2\pi}\int_{\mathbb{R}} -i\operatorname{sgn}(\xi)e^{-t|\xi|}\hat{f}(\xi) e^{i\xi x}\, d\mathcal{L}^1(\xi). \end{align*} The integrand is dominated by $|\hat{f}(\xi)|$ uniformly in $t > 0$, and $\hat{f} \in L^1(\mathbb{R})$ since $f \in \mathcal{S}(\mathbb{R})$ implies $\hat{f} \in \mathcal{S}(\mathbb{R}) \subset L^1(\mathbb{R})$. Pointwise as $t \to 0^+$, $e^{-t|\xi|} \to 1$ for every $\xi \neq 0$. By the dominated convergence theorem, \begin{align*} \lim_{t \to 0^+}(Q_t * f)(x) = \frac{1}{2\pi}\int_{\mathbb{R}} -i\operatorname{sgn}(\xi)\hat{f}(\xi) e^{i\xi x}\, d\mathcal{L}^1(\xi) = H_m f(x). \end{align*} Hence the conjugate-Poisson boundary-value definition agrees with the multiplier definition. [/step] [step:Verify that the multiplier preserves the Schwartz class] Let $m: \mathbb{R} \to \mathbb{C}$, $\xi \mapsto -i\operatorname{sgn}(\xi)$. We show that for $f \in \mathcal{S}(\mathbb{R})$, the inverse Fourier transform \begin{align*} H_m f: \mathbb{R} &\to \mathbb{C} \\ x &\mapsto \frac{1}{2\pi}\int_{\mathbb{R}} m(\xi)\hat{f}(\xi)e^{i\xi x}\, d\mathcal{L}^1(\xi) \end{align*} lies in $\mathcal{S}(\mathbb{R})$. The function $g(\xi) := m(\xi)\hat{f}(\xi)$ satisfies: (i) $g \in L^\infty(\mathbb{R})$ since $|m| \le 1$ and $\hat{f} \in L^\infty$; (ii) $g$ is smooth on $\mathbb{R} \setminus \{0\}$, since $m$ is constant on $(-\infty,0)$ and $(0,\infty)$ and $\hat{f} \in C^\infty$; (iii) $\hat{f}(0) = \int_{\mathbb{R}} f\, d\mathcal{L}^1$, but $g$ is discontinuous at $0$ as a product unless $\hat{f}(0) = 0$. To show $H_m f \in \mathcal{S}(\mathbb{R})$, we verify the seminorms $\|H_m f\|_{\alpha,\beta} = \sup_x |x^\alpha (H_m f)^{(\beta)}(x)| < \infty$ for all $\alpha, \beta \in \mathbb{N}_0$. By the Fourier-side characterisation, $\mathcal{S}(\mathbb{R})$-membership of $H_m f$ is equivalent to $g = m\hat{f}$ being a Schwartz function — but $g$ is discontinuous at $\xi = 0$, so this approach fails at the origin. We use instead the following direct argument. Write $H_m f = K * f$ from Step 2; we already know $H_m f$ is smooth (convolution of a Schwartz function with a tempered distribution). It remains to show rapid decay of $H_m f$ and its derivatives. Fix $x$ with $|x| > 1$. Splitting the principal-value integral, \begin{align*} H_m f(x) = \frac{1}{\pi}\mathrm{p.v.}\!\int_{|y|<|x|/2} \frac{f(x-y)}{y}\, d\mathcal{L}^1(y) + \frac{1}{\pi}\int_{|y|\ge|x|/2} \frac{f(x-y)}{y}\, d\mathcal{L}^1(y). \end{align*} For $|y| \ge |x|/2$, $|f(x-y)| \le C_N(1+|x-y|)^{-N}$ for any $N$ (Schwartz), and $|y| \ge |x|/2$ gives a uniform bound $|H_m f(x)| \le C_N |x|^{-N}$ on the second piece by direct estimation. For the first piece, $|y|<|x|/2$ implies $|x - y| \ge |x|/2$, so $|f(x-y)| \le C_N|x|^{-N}$, and the principal-value integral against the odd function $1/y$ over the symmetric interval $(-|x|/2,|x|/2)\setminus\{0\}$ is bounded using cancellation: applying the identity $\mathrm{p.v.}\!\int_{|y|<|x|/2} f(x-y)/y\, d\mathcal{L}^1(y) = \int_{|y|<|x|/2} (f(x-y) - f(x))/y\, d\mathcal{L}^1(y)$ (since $\mathrm{p.v.}\!\int_{|y|<|x|/2} 1/y\, d\mathcal{L}^1(y) = 0$ by oddness) and the mean-value bound $|f(x-y) - f(x)| \le |y| \sup |f'|$ gives $\big|\int_{|y|<|x|/2}(f(x-y)-f(x))/y\, d\mathcal{L}^1(y)\big| \le |x| \cdot \sup |f'| \cdot \mathbb{1}_{\{|x| \le 1\}} + C_N|x|^{-N}$ for $|x| > 1$ via combining mean value with Schwartz decay on $|x-y| \ge |x|/2$. So $|H_m f(x)| \le C_N(1+|x|)^{-N}$ for all $N$. Differentiation commutes with the multiplier (since $\widehat{(H_m f)^{(k)}}(\xi) = (i\xi)^k m(\xi)\hat{f}(\xi) = m(\xi)(i\xi)^k\hat{f}(\xi) = m(\xi)\widehat{f^{(k)}}(\xi) = \widehat{H_m(f^{(k)})}(\xi)$), so $(H_m f)^{(k)} = H_m(f^{(k)})$ for all $k \ge 0$, and the rapid decay applies to every derivative. Hence $H_m f \in \mathcal{S}(\mathbb{R})$. [/step] [step:Combine the three identifications to conclude] Steps 1-2 show $H_{\mathrm{p.v.}}f = H_m f$ as elements of $\mathcal{S}'(\mathbb{R})$. Step 3 shows $\lim_{t \to 0^+} Q_t * f = H_m f$ pointwise. Step 4 shows $H_m f \in \mathcal{S}(\mathbb{R})$. Combining: all three definitions of $Hf$ on $\mathcal{S}(\mathbb{R})$ produce the same Schwartz function, namely the inverse Fourier transform of $-i\operatorname{sgn}(\xi)\hat{f}(\xi)$. This completes the proof of equivalence. [/step]