Step 1: The map $\iota$ is well-defined. Let $f \in W^{k,2}(\mathbb{R}^n) \subseteq L^2(\mathbb{R}^n)$. Since $L^2$ functions define tempered distributions via $T_f(\phi) = \int f\phi\, d\mathcal{L}^n$, we have $T_f \in \mathcal{S}'(\mathbb{R}^n)$. We must check that $T_f \in H^k(\mathbb{R}^n)$, i.e. that $(1 + |\xi|^2)^{k/2}\widehat{T_f}$ is represented by an $L^2$ function.

By the Plancherel theorem, $\widehat{T_f} = T_{\hat{f}}$ where $\hat{f} \in L^2(\mathbb{R}^n)$ is the $L^2$ Fourier transform. Therefore $(1 + |\xi|^2)^{k/2}\widehat{T_f} = T_{(1+|\xi|^2)^{k/2}\hat{f}}$, and we need $(1 + |\xi|^2)^{k/2}\hat{f} \in L^2(\mathbb{R}^n)$.

Step 2: Norm equivalence. The key identity is the comparison of polynomial weights. For any $\xi \in \mathbb{R}^n$:
\begin{align*} \sum_{|\alpha| \le k} |\xi^\alpha|^2 \le (1 + |\xi|^2)^k \le C_{n,k} \sum_{|\alpha| \le k} |\xi^\alpha|^2, \end{align*}
where the left inequality follows from the binomial expansion (each $|\xi^\alpha|^2$ with $|\alpha| \le k$ is a term in the expansion of $(1 + |\xi|^2)^k$), and the right inequality holds with a constant $C_{n,k}$ depending on the number of multi-indices.

By Plancherel, $\|\partial^\alpha f\|_{L^2}^2 = \|\xi^\alpha \hat{f}\|_{L^2}^2$ for each multi-index $\alpha$. Therefore:
\begin{align*} \|f\|_{W^{k,2}}^2 = \sum_{|\alpha| \le k} \|\partial^\alpha f\|_{L^2}^2 = \sum_{|\alpha| \le k} \|\xi^\alpha \hat{f}\|_{L^2}^2 = \int_{\mathbb{R}^n} \sum_{|\alpha| \le k} |\xi^\alpha|^2 |\hat{f}(\xi)|^2\, d\mathcal{L}^n(\xi). \end{align*}
Combining with the polynomial comparison:
\begin{align*} \|f\|_{W^{k,2}}^2 \le \int_{\mathbb{R}^n} (1 + |\xi|^2)^k |\hat{f}(\xi)|^2\, d\mathcal{L}^n(\xi) = \|\iota(f)\|_{H^k}^2 \le C_{n,k}\,\|f\|_{W^{k,2}}^2. \end{align*}

Step 3: Bijectivity. Injectivity: if $\iota(f) = T_f = 0$ in $\mathcal{S}'$, then $f = 0$ a.e. Surjectivity: if $u \in H^k(\mathbb{R}^n)$, then $\hat{u} = T_g$ for some $g \in L^1_{\mathrm{loc}}$ with $(1 + |\xi|^2)^{k/2}g \in L^2$. In particular $g \in L^2$, so $u = T_f$ where $f = \mathcal{F}^{-1}(g) \in L^2$. The norm equivalence from Step 2 then shows $\xi^\alpha g \in L^2$ for all $|\alpha| \le k$, giving $\partial^\alpha f \in L^2$ by Plancherel, hence $f \in W^{k,2}$.

Step 4: The case $k = 0$. When $k = 0$, the weight is $(1 + |\xi|^2)^0 = 1$ and $\sum_{|\alpha| \le 0} |\xi^\alpha|^2 = 1$. Both sides of the comparison are identically $1$, so the norms are equal: $\|\iota(f)\|_{H^0} = \|\hat{f}\|_{L^2} = \|f\|_{L^2}$.