= Solution
We place the definitions and the unheaded preliminary requests here before addressing the first labelled property. The <Schwartz space> consists of <smooth functions> with finite <seminorms>
$$
q_{\alpha,\beta}(f)=\sup_{x\in\mathbb R^n}|x^\alpha\partial^\beta f(x)|
$$
for every pair of <multi-indices>. Convergence means convergence in each <seminorm>. The <tempered distribution> space $\mathcal S'(\mathbb R^n)$ is its <continuous dual space>, with weak convergence tested against every <Schwartz function>. A continuous functional satisfies a bound by finitely many of these <seminorms>, equivalently by one sufficiently large weighted <derivative> <seminorm>.
Use the <Fourier transform> convention
$$
\widehat f(\xi)=\int e^{-ix\cdot\xi}f(x)\,dx,\qquad f(x)=\frac1{(2\pi)^n}\int e^{ix\cdot\xi}\widehat f(\xi)\,d\xi.
$$
Differentiation under the integral and <integration by parts> express $\xi^\alpha\partial_\xi^\beta\widehat f$ as a constant of modulus one times the <Fourier transform> of $\partial_x^\alpha(x^\beta f)$. Its supremum is bounded by the $L^1$ norm of that function. For $s>n$, this norm is at most a constant times finitely many Schwartz <seminorms>, using the integrable weight $(1+|x|)^{-s}$. Thus $\mathcal F:\mathcal S\to\mathcal S$ is continuous.
For completeness, <Fourier inversion> follows by inserting $e^{-\varepsilon|\xi|^2}$ in the inverse integral and using <Fubini's theorem>. The result is <convolution> with the <Gaussian approximate identity>
$$
(4\pi\varepsilon)^{-n/2}e^{-|x|^2/(4\varepsilon)}.
$$
It tends to $f$, while integrability of $\widehat f$ allows the damping factor to be removed by <dominated convergence theorem>. Consequently $\mathcal F^2f=(2\pi)^nf(-\cdot)$. Reflection preserves every Schwartz <seminorm>, so the inverse transform is continuous as well. This proves \b[the <Fourier transform isomorphism of the Schwartz space>].
Define the <Fourier transform of a tempered distribution> by transposition,
$$
\langle\widehat u,\psi\rangle=\langle u,\widehat\psi\rangle.
$$
The Schwartz-space continuity just proved makes this a <tempered distribution>. Its inverse is $(2\pi)^{-n}\mathcal R\mathcal F$, where $\langle\mathcal Ru,\psi\rangle=\langle u,\psi(-\cdot)\rangle$. These maps are continuous for weak convergence, since each pairing is a pairing with a fixed transformed test. They are also continuous for the <strong dual topology>, because the Schwartz-space maps take bounded sets to bounded sets.
The <convolution of a tempered distribution with a Schwartz function> is
$$
(u*\varphi)(x)=\langle u_y,\varphi(x-y)\rangle.
$$
Smooth dependence of translated Schwartz functions gives $\partial^\alpha(u*\varphi)(x)=\langle u,\partial^\alpha\varphi(x-\cdot)\rangle$. The finite-<seminorm> estimate and $1+|y|\leq(1+|x|)(1+|x-y|)$ show that each <derivative> has at most <polynomial growth>. In particular, $u*\varphi$ is a <smooth function> defining a <tempered distribution>. It need not itself be a <Schwartz function>; for example $1*\varphi=\int\varphi$.
Writing $\check\varphi(y)=\varphi(-y)$, its distributional pairing is $\langle u*\varphi,\psi\rangle=\langle u,\check\varphi*\psi\rangle$. The inner <convolution> is a <Schwartz function>, and this identity follows by integration in the Schwartz topology, justified by the weighted <seminorm> estimates. A direct <Fubini's theorem> calculation gives $\check\varphi*\widehat\psi=\mathcal F(\widehat\varphi\psi)$. Hence
$$
\langle\widehat{u*\varphi},\psi\rangle=\langle u,\mathcal F(\widehat\varphi\psi)\rangle=\langle\widehat u,\widehat\varphi\psi\rangle,
\qquad\boxed{\widehat{u*\varphi}=\widehat u\,\widehat\varphi.}
$$
Multiplication is well defined because multiplication by $\widehat\varphi$ acts continuously on $\mathcal S$.
Now write the <Hilbert transform> as <convolution> with $K=\pi^{-1}\operatorname{pv}(1/x)$, the <principal-value reciprocal distribution>. The given <Heaviside function> transform, together with $\mathcal F^2H=2\pi H(-\cdot)$, yields
$$
\pi-i\mathcal F\big(\operatorname{pv}(1/x)\big)=2\pi H(-\xi),\qquad\widehat K(\xi)=-i\operatorname{sgn}\xi.
$$
The value of the sign function at zero is irrelevant to its regular <distribution>. Thus the <Hilbert-transform Fourier multiplier> is
$$
\widehat{\mathcal H\varphi}(\xi)=-i\operatorname{sgn}(\xi)\widehat\varphi(\xi).
$$
Applying <Plancherel theorem>, whose normalization here is $\|f\|_2^2=(2\pi)^{-1}\|\widehat f\|_2^2$, proves \b[the isometry]:
$$
\boxed{\|\mathcal H\varphi\|_{L^2}=\|\varphi\|_{L^2}.}
$$
The principal-value integral agrees with this <convolution>: near its singular point subtract $\varphi(x)$, and use odd cancellation; at infinity the Schwartz decay gives convergence. The <Hilbert transform> has domain $\mathcal S$, but generally does not take values in $\mathcal S$, as the tail in part (c) demonstrates.
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