[proofplan]
We show that the map $\Phi_s: u \mapsto (1+|\xi|^2)^{s/2}\hat{u}$ is an isometric bijection from $H^s(\mathbb{R}^n)$ onto $L^2(\mathbb{R}^n)$. Well-definedness of $\Phi_s$ on $\mathcal{S}'$ uses the [Fourier automorphism on $\mathcal{S}'$](/theorems/230) and the fact that $(1+|\xi|^2)^{s/2} \in \mathcal{O}_M$. Uniqueness of the $L^2$ representative follows from the [density of $\mathcal{S}$ in $L^2$](/theorems/229). Surjectivity is constructed by dividing by the weight and taking the inverse Fourier transform. Completeness and separability transfer from $L^2$ via the isometry.
[/proofplan]
Throughout, for $g \in L^1_\mathrm{loc}(\mathbb{R}^n)$, we write $T_g \in \mathcal{S}'(\mathbb{R}^n)$ for the [regular distribution](/page/Distribution) defined by
\begin{align*}
T_g: \mathcal{S}(\mathbb{R}^n) &\to \mathbb{R} \\
\varphi &\mapsto \int_{\mathbb{R}^n} g(x)\,\varphi(x) \, d\mathcal{L}^n(x).
\end{align*}
[step:Show that $\Phi_s$ is well-defined on $\mathcal{S}'(\mathbb{R}^n)$]
The [Fourier transform on $\mathcal{S}'$](/theorems/230) is a topological automorphism, so $\hat{u} \in \mathcal{S}'(\mathbb{R}^n)$ for every $u \in \mathcal{S}'(\mathbb{R}^n)$.
The function $(1+|\xi|^2)^{s/2}$ belongs to $\mathcal{O}_M(\mathbb{R}^n)$ (the space of slowly increasing smooth functions): for every multi-index $\beta$, the derivative $\partial^\beta[(1+|\xi|^2)^{s/2}]$ is a sum of terms each bounded by a constant times $(1+|\xi|^2)^{s/2 - |\beta|/2}$.
More precisely,
\begin{align*}
|\partial^\beta[(1+|\xi|^2)^{s/2}]| &\le C_\beta(1+|\xi|^2)^{s/2 - |\beta|/2} \le C_\beta(1+|\xi|)^{|s|}
\end{align*}
for a constant $C_\beta > 0$ depending only on $\beta$ and $s$.
Since multiplication by an $\mathcal{O}_M$ function is a continuous endomorphism of $\mathcal{S}'(\mathbb{R}^n)$, the product $(1+|\xi|^2)^{s/2}\hat{u}$ is a well-defined tempered distribution.
[/step]
[step:Establish uniqueness of the $L^2$ representative]
Suppose $T_{g_1} = T_{g_2}$ in $\mathcal{S}'(\mathbb{R}^n)$ for $g_1, g_2 \in L^2(\mathbb{R}^n)$.
Then
\begin{align*}
\int_{\mathbb{R}^n} (g_1 - g_2)\,\varphi \, d\mathcal{L}^n &= 0
\end{align*}
for every $\varphi \in \mathcal{S}(\mathbb{R}^n)$.
Since $\mathcal{S}(\mathbb{R}^n)$ is [dense in $L^2(\mathbb{R}^n)$](/theorems/229) and $g_1 - g_2 \in L^2(\mathbb{R}^n)$, choose a sequence $(\varphi_k)_{k \in \mathbb{N}} \subseteq \mathcal{S}(\mathbb{R}^n)$ with $\varphi_k \to g_1 - g_2$ in $L^2(\mathbb{R}^n)$.
Then
\begin{align*}
\|g_1 - g_2\|_{L^2}^2 &= \int_{\mathbb{R}^n} (g_1 - g_2)(g_1 - g_2) \, d\mathcal{L}^n = \lim_{k \to \infty} \int_{\mathbb{R}^n} (g_1 - g_2)\,\varphi_k \, d\mathcal{L}^n = 0,
\end{align*}
where the second equality uses continuity of the $L^2$ inner product.
Hence $g_1 = g_2$ $\mathcal{L}^n$-a.e., establishing part (1): for $u \in H^s(\mathbb{R}^n)$, the function $g \in L^2(\mathbb{R}^n)$ with $T_g = (1+|\xi|^2)^{s/2}\hat{u}$ is unique $\mathcal{L}^n$-a.e., so $\|u\|_{H^s} = \|g\|_{L^2}$ is well-defined.
[guided]
Why do we need density of $\mathcal{S}$ in $L^2$?
The condition $T_{g_1} = T_{g_2}$ means the two functions agree when tested against every Schwartz function.
But to conclude $g_1 = g_2$ a.e., we need the testing class to be large enough to separate $L^2$ functions.
If $\mathcal{S}$ were not dense in $L^2$, there could exist a nonzero $L^2$ function orthogonal to every Schwartz function.
By the [density of $\mathcal{S}$ in $L^2$](/theorems/229), this cannot happen: the Schwartz functions are dense, so a function orthogonal to all of them must be zero a.e.
Concretely: $g_1 - g_2 \in L^2$ and $\int (g_1 - g_2)\varphi \, d\mathcal{L}^n = 0$ for all $\varphi \in \mathcal{S}$.
Choosing $\varphi_k \to g_1 - g_2$ in $L^2$ (possible by density), continuity of the inner product gives
\begin{align*}
\|g_1 - g_2\|_{L^2}^2 &= \lim_{k \to \infty} \int_{\mathbb{R}^n} (g_1 - g_2)\,\varphi_k \, d\mathcal{L}^n = \lim_{k \to \infty} 0 = 0.
\end{align*}
[/guided]
[/step]
[step:Prove that $\Phi_s$ is an isometric bijection from $H^s$ to $L^2$]
Define the map
\begin{align*}
\Phi_s: H^s(\mathbb{R}^n) &\to L^2(\mathbb{R}^n) \\
u &\mapsto g,
\end{align*}
where $g \in L^2(\mathbb{R}^n)$ is the unique (by the previous step) function satisfying $T_g = (1+|\xi|^2)^{s/2}\hat{u}$ in $\mathcal{S}'(\mathbb{R}^n)$.
By definition, $\|\Phi_s(u)\|_{L^2} = \|g\|_{L^2} = \|u\|_{H^s}$, so $\Phi_s$ is an isometry.
Injectivity follows immediately: if $\Phi_s(u_1) = \Phi_s(u_2)$ $\mathcal{L}^n$-a.e., then $\|u_1 - u_2\|_{H^s} = \|\Phi_s(u_1) - \Phi_s(u_2)\|_{L^2} = 0$.
For surjectivity, let $g \in L^2(\mathbb{R}^n)$.
The [regular distribution](/page/Distribution) $T_g \in \mathcal{S}'(\mathbb{R}^n)$ can be divided by the weight: since $(1+|\xi|^2)^{-s/2} \in \mathcal{O}_M(\mathbb{R}^n)$ (verified by the same derivative bounds as above with $s$ replaced by $-s$), the product
\begin{align*}
v &:= (1+|\xi|^2)^{-s/2}\,T_g
\end{align*}
is a well-defined tempered distribution.
Setting $u := \mathcal{F}^{-1}(v)$, where $\mathcal{F}^{-1}$ is the inverse [Fourier transform on $\mathcal{S}'$](/theorems/230), we have $\hat{u} = v$ and
\begin{align*}
(1+|\xi|^2)^{s/2}\hat{u} &= (1+|\xi|^2)^{s/2} \cdot (1+|\xi|^2)^{-s/2}\,T_g = T_g.
\end{align*}
Thus $u \in H^s(\mathbb{R}^n)$ (by definition of the [inhomogeneous Sobolev space](/page/Inhomogeneous%20Sobolev%20Space)) and $\Phi_s(u) = g$.
[/step]
[step:Transfer the inner product structure from $L^2$ to $H^s$]
Linearity of $\Phi_s$ follows from the linearity of the [Fourier transform on $\mathcal{S}'$](/theorems/230) and of multiplication by $(1+|\xi|^2)^{s/2}$.
Since $\Phi_s$ is a linear isometry onto $L^2$, the pullback of the $L^2$ inner product
\begin{align*}
\langle u, v \rangle_{H^s} &:= \langle \Phi_s(u), \Phi_s(v) \rangle_{L^2} = \langle g_u, g_v \rangle_{L^2}
\end{align*}
satisfies all inner product axioms (conjugate symmetry, linearity, and positive definiteness), inherited from $L^2(\mathbb{R}^n)$.
The induced norm satisfies $\langle u, u \rangle_{H^s}^{1/2} = \|g_u\|_{L^2} = \|u\|_{H^s}$.
This establishes part (2).
[/step]
[step:Conclude completeness and separability via the isometric isomorphism]
Since $\Phi_s: H^s \to L^2$ is an isometric bijection and $L^2(\mathbb{R}^n)$ is a separable [Hilbert space](/page/Hilbert%20Space), both completeness and separability transfer to $H^s$ via $\Phi_s^{-1}$.
For completeness: if $(u_k)_{k \in \mathbb{N}}$ is a [Cauchy sequence](/page/Cauchy%20Sequence) in $H^s$, then $(\Phi_s(u_k))_{k \in \mathbb{N}}$ is Cauchy in $L^2$ (since $\Phi_s$ is an isometry), hence converges to some $g \in L^2$, and $u := \Phi_s^{-1}(g) \in H^s$ satisfies $\|u_k - u\|_{H^s} = \|\Phi_s(u_k) - g\|_{L^2} \to 0$.
For separability: if $\{g_k : k \in \mathbb{N}\}$ is a countable dense subset of $L^2$, then $\{\Phi_s^{-1}(g_k) : k \in \mathbb{N}\}$ is a countable dense subset of $H^s$.
This establishes part (3).
[/step]
