The Hilbert transform is the tempered convolution with pv(1/(pi x)) and has signum Fourier multiplier

For $\phi \in \mathcal{S}(\mathbb{R})$ both integrals in the definition of $W$ converge absolutely: on $\mathrm{\mid }x\mathrm{\mid }>1$ the bound $\mathrm{\mid }\phi (x)\mathrm{/}x\mathrm{\mid }\le p_{20}(\phi )\mathrm{\mid }x{\mathrm{\mid }}^{-3}$ is integrable, and on $\mathrm{\mid }x\mathrm{\mid }<1$ the bound $\mathrm{\mid }\phi (x)-\phi (0)\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }\le p_{01}(\phi )$ from [F10] is integrable on a set of length two. Hence $\mathrm{\mid }⟨W,\phi ⟩\mathrm{\mid }\le \frac{2}{\pi }(p_{20}(\phi )+p_{01}(\phi ))$ and $W$ is a tempered distribution by [F6]. Moreover ${\int }_{\epsilon <\mathrm{\mid }x\mathrm{\mid }<1}\phi (0)\mathrm{/}xdx=0$ by oddness of $1\mathrm{/}x$, so for $0<\epsilon <1$ the truncated pairing $(1\mathrm{/}\pi ){\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }\phi (x)\mathrm{/}xdx$ equals the defining two-piece pairing with the local piece integrated over $\epsilon <\mathrm{\mid }x\mathrm{\mid }<1$; consequently $⟨W,\phi ⟩={\lim }_{\epsilon \downarrow 0}\frac{1}{\pi }{\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }\phi (x)\mathrm{/}xdx$.

Fix $\phi \in \mathcal{S}(\mathbb{R})$ and $0<\epsilon <R$. By [F11] applied on the product of the finite-measure annulus $\{\epsilon <\mathrm{\mid }x\mathrm{\mid }<R\}$ with $\mathbb{R}$, using the integrable majorant $\mathrm{\mid }\phi (\xi )\mathrm{\mid }\mathrm{/}\mathrm{\mid }x\mathrm{\mid }$ from [F8], $I_{\epsilon ,R}:=\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }x\mathrm{\mid }<R}\frac{\widehat{\phi }(x)}{x}dx=\frac{1}{\pi }{\int }_{\mathbb{R}}\phi (\xi ){\mathrm{\Lambda }}_{\epsilon ,R}(\xi )d\xi$ with ${\mathrm{\Lambda }}_{\epsilon ,R}(\xi ):={\int }_{\epsilon <\mathrm{\mid }x\mathrm{\mid }<R}{e}^{-2\pi ix\xi }{x}^{-1}dx$. Writing the exponential in cosine and sine, the cosine term is odd and integrates to zero, while [F13] with $u=2\pi x\xi$ gives ${\mathrm{\Lambda }}_{\epsilon ,R}(\xi )=-2i{\int }_{\epsilon }^R\frac{\sin (2\pi x\xi )}{x}dx=-2i\mathrm{sgn}(\xi )(S(2\pi R\mathrm{\mid }\xi \mathrm{\mid })-S(2\pi \epsilon \mathrm{\mid }\xi \mathrm{\mid }))$ for the partial sine integral $S$ of [F2]; in particular $\mathrm{\mid }{\mathrm{\Lambda }}_{\epsilon ,R}(\xi )\mathrm{\mid }\le 12$.

Fix $f\in \mathcal{S}(\mathbb{R})$ and $x\in \mathbb{R}$. For $0<\epsilon <1$, [F1] gives $H_{\epsilon }f(x)=\frac{1}{\pi }{\int }_{\mathrm{\mid }t\mathrm{\mid }>\epsilon }\frac{f(x-t)}{t}dt$; since ${\int }_{\epsilon <\mathrm{\mid }t\mathrm{\mid }<1}f(x)\mathrm{/}tdt=0$, this equals $\frac{1}{\pi }{\int }_{\epsilon <\mathrm{\mid }t\mathrm{\mid }<1}\frac{f(x-t)-f(x)}{t}dt+\frac{1}{\pi }{\int }_{\mathrm{\mid }t\mathrm{\mid }>1}\frac{f(x-t)}{t}dt$. The tail is absolutely convergent by [F8], and the first integral converges as $\epsilon \downarrow 0$ by [F12], the integrand tending pointwise to $\frac{f(x-t)-f(x)}{t}$ and being dominated on $(-1,1)$ by $p_{01}(f)=\mathrm{\parallel }{f}^{\mathrm{\prime }}{\mathrm{\parallel }}_{\mathrm{\infty }}$ thanks to [F10]. The resulting limit is exactly $⟨W_y,f(x-y)⟩=(W\ast f)(x)$ by the defining formula of $W$ in 1.1 and the convolution definition [F5]. Hence ${\lim }_{\epsilon \downarrow 0}{H}_{\epsilon }f(x)$ exists at every $x$ and equals $(W\ast f)(x)$.

By 1.1 and [F3], $⟨\mathcal{F}W,\phi ⟩=⟨W,\widehat{\phi }⟩={\lim }_{\epsilon \downarrow 0}\frac{1}{\pi }{\int }_{\mathrm{\mid }x\mathrm{\mid }>\epsilon }\widehat{\phi }(x)\mathrm{/}xdx$. Holding $\epsilon$ fixed, [F2] gives ${\mathrm{\Lambda }}_{\epsilon ,R}(\xi )\to {\rho }_{\epsilon }(\xi ):=-2i\mathrm{sgn}(\xi )(\frac{\pi }{2}-S(2\pi \epsilon \mathrm{\mid }\xi \mathrm{\mid }))$ as $R\to \mathrm{\infty }$, with $\mathrm{\mid }{\rho }_{\epsilon }(\xi )\mathrm{\mid }\le \pi +6$. Thus [F12] against $\mathrm{\mid }\phi \mathrm{\mid }$ yields $I_{\epsilon ,R}\to \frac{1}{\pi }\int \phi {\rho }_{\epsilon }$ as $R\to \mathrm{\infty }$. Then $S(2\pi \epsilon \mathrm{\mid }\xi \mathrm{\mid })\le 2\pi \epsilon \mathrm{\mid }\xi \mathrm{\mid }$ for $2\pi \epsilon \mathrm{\mid }\xi \mathrm{\mid }\le 1$ by [F2] makes ${\rho }_{\epsilon }(\xi )\to -i\pi \mathrm{sgn}(\xi )$ pointwise as $\epsilon \downarrow 0$, and a second application of [F12] gives $\frac{1}{\pi }\int \phi {\rho }_{\epsilon }\to \int (-i\mathrm{sgn}\xi )\phi (\xi )d\xi$. Combining the two limits with the pairing identity gives $⟨\mathcal{F}W,\phi ⟩=\int (-i\mathrm{sgn}\xi )\phi (\xi )d\xi$ for every $\phi \in \mathcal{S}$, that is, $\mathcal{F}W=-i\mathrm{sgn}(\xi )$ as tempered distributions.

By 3.1 and [F4] applied to the tempered distribution $W$ and the Schwartz function $f$, $\mathcal{F}(W\ast f)=(\mathcal{F}W)(\mathcal{F}f)=-i\mathrm{sgn}(\xi )\widehat{f}(\xi )$, the product being that of the distribution $-i\mathrm{sgn}$ with the Schwartz function $\widehat{f}\in \mathcal{S}$ supplied by [F9] and [F3]. By 2.2 the same $Hf=W\ast f$ is the pointwise principal-value transform of $f$; thus the principal value defines the tempered convolution with $\mathrm{pv}\frac{1}{\pi x}$ and has the signum Fourier multiplier, as claimed.