[proofplan]
We characterise $\dot{H}^k$ membership in terms of top-order distributional derivatives. First, we verify the derivative condition is well-defined on equivalence classes (the only polynomial in $L^2$ is zero, so if $\partial^\alpha f = T_{g_\alpha}$ for one representative, the same holds for all). Second, we establish a two-sided pointwise bound between $|\xi|^{2k}$ and $\sum_{|\alpha|=k}|\xi^\alpha|^2$ via the multinomial theorem and compactness of $\mathbb{S}^{n-1}$. Third, we use the Fourier exchange identity $\widehat{\partial^\alpha f} = (i\xi)^\alpha\hat{f}$ and the [Plancherel Theorem](/theorems/247) to translate between the derivative condition and the $\dot{H}^k$ norm.
[/proofplan]
[step:Verify the derivative condition is well-defined on equivalence classes]
If $f_1 - f_2 = T_p \in T_{\mathcal{P}}$ for some polynomial $p$, then $\partial^\alpha f_1 - \partial^\alpha f_2 = T_{\partial^\alpha p}$ for every multi-index $\alpha$.
Suppose $\partial^\alpha f_1 = T_{g_\alpha}$ for some $g_\alpha \in L^2(\mathbb{R}^n)$ and $|\alpha| = k$.
Then $\partial^\alpha f_2 = T_{g_\alpha - \partial^\alpha p}$.
For $g_\alpha - \partial^\alpha p$ to belong to $L^2(\mathbb{R}^n)$, we need $\partial^\alpha p \in L^2(\mathbb{R}^n)$.
But the only polynomial in $L^2(\mathbb{R}^n)$ is the zero polynomial (by the argument in the [Canonical Isomorphism $\dot{H}^0 \cong L^2$](/theorems/471)).
Therefore $\partial^\alpha p = 0$ and $\partial^\alpha f_2 = T_{g_\alpha}$.
This shows that the condition and the function $g_\alpha$ depend only on the equivalence class $[f]$.
[/step]
[step:Establish a two-sided bound between $|\xi|^{2k}$ and $\sum_{|\alpha|=k}|\xi^\alpha|^2$]
For $\xi \in \mathbb{R}^n$ and $k \in \mathbb{N}$, the multinomial theorem gives
\begin{align*}
|\xi|^{2k} &= (|\xi_1|^2 + \cdots + |\xi_n|^2)^k = \sum_{|\alpha|=k} \binom{k}{\alpha}\,|\xi^\alpha|^2,
\end{align*}
where $\binom{k}{\alpha} = k!/(\alpha_1!\cdots\alpha_n!)$ and $\xi^\alpha = \xi_1^{\alpha_1}\cdots\xi_n^{\alpha_n}$.
For the upper bound: each $|\xi^\alpha| = |\xi_1|^{\alpha_1}\cdots|\xi_n|^{\alpha_n} \le |\xi|^{|\alpha|} = |\xi|^k$ (since $|\xi_j| \le |\xi|$), so $\sum_{|\alpha|=k}|\xi^\alpha|^2 \le N_{n,k}\,|\xi|^{2k}$ where $N_{n,k} = \#\{\alpha \in \mathbb{N}_0^n : |\alpha| = k\}$.
For the lower bound: both $|\xi|^{2k}$ and $\sum_{|\alpha|=k}|\xi^\alpha|^2$ are continuous, homogeneous of degree $2k$.
On the unit sphere $\mathbb{S}^{n-1}$, the function $\sum_{|\alpha|=k}|\xi^\alpha|^2$ is strictly positive (if it vanished at some $\xi_0 \in \mathbb{S}^{n-1}$, then $\xi_0^\alpha = 0$ for all $|\alpha| = k$, forcing $\xi_0 = 0$, contradicting $|\xi_0| = 1$).
By compactness of $\mathbb{S}^{n-1}$, the ratio attains a positive minimum $c_{n,k} > 0$.
Homogeneity extends the bound to all of $\mathbb{R}^n_0$, and both sides vanish at $\xi = 0$.
Therefore there exist $0 < c_{n,k} \le C_{n,k} < \infty$ with
\begin{align*}
c_{n,k}\sum_{|\alpha|=k}|\xi^\alpha|^2 &\le |\xi|^{2k} \le C_{n,k}\sum_{|\alpha|=k}|\xi^\alpha|^2 \quad \text{for all } \xi \in \mathbb{R}^n.
\end{align*}
[/step]
[step:Translate between the derivative condition and the $\dot{H}^k$ norm via Fourier]
The distributional derivative satisfies $\widehat{\partial^\alpha f} = (i\xi)^\alpha\hat{f}$ in $\mathcal{S}'(\mathbb{R}^n)$.
On $\mathbb{R}^n_0$, working with $T$-representatives:
\begin{align*}
(\widehat{\partial^\alpha f})_{T\text{-rep}}(\xi) &= i^{|\alpha|}\,\xi^\alpha\,\hat{f}_{T\text{-rep}}(\xi).
\end{align*}
If $\partial^\alpha f = T_{g_\alpha}$ for $g_\alpha \in L^2(\mathbb{R}^n)$, then by the [Compatibility of the Distributional and Classical Fourier Transforms](/theorems/718) (extended to $L^2$ via density), $\widehat{T_{g_\alpha}} = T_{\hat{g}_\alpha}$, so $\hat{g}_\alpha(\xi) = i^{|\alpha|}\xi^\alpha\hat{f}_{T\text{-rep}}(\xi)$ $\mathcal{L}^n$-a.e.
By the [Plancherel Theorem](/theorems/247):
\begin{align*}
\|g_\alpha\|_{L^2}^2 &= (2\pi)^{-n}\|\hat{g}_\alpha\|_{L^2}^2 = (2\pi)^{-n}\int_{\mathbb{R}^n} |\xi^\alpha|^2\,|\hat{f}_{T\text{-rep}}(\xi)|^2 \, d\mathcal{L}^n(\xi),
\end{align*}
since $|i^{|\alpha|}|^2 = 1$.
Conversely, if $|\xi|^k\hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$, then $\xi^\alpha\hat{f}_{T\text{-rep}} \in L^2(\mathbb{R}^n)$ for each $|\alpha| = k$ (by the upper bound), so $g_\alpha := (2\pi)^{-n}\mathcal{F}^{-1}(i^{|\alpha|}\xi^\alpha\hat{f}_{T\text{-rep}}) \in L^2(\mathbb{R}^n)$ and $\partial^\alpha f = T_{g_\alpha}$.
[/step]
[step:Integrate the two-sided bound to obtain the norm equivalence]
Applying the two-sided bound and integrating against $|\hat{f}_{T\text{-rep}}(\xi)|^2$:
\begin{align*}
c_{n,k}\sum_{|\alpha|=k}\int_{\mathbb{R}^n}|\xi^\alpha|^2\,|\hat{f}_{T\text{-rep}}|^2 \, d\mathcal{L}^n &\le \int_{\mathbb{R}^n}|\xi|^{2k}\,|\hat{f}_{T\text{-rep}}|^2 \, d\mathcal{L}^n \le C_{n,k}\sum_{|\alpha|=k}\int_{\mathbb{R}^n}|\xi^\alpha|^2\,|\hat{f}_{T\text{-rep}}|^2 \, d\mathcal{L}^n.
\end{align*}
The middle term is $\|[f]\|_{\dot{H}^k}^2$ by definition.
Each integral on the left and right equals $(2\pi)^n\|g_\alpha\|_{L^2}^2$ by the Fourier-side translation.
Absorbing $(2\pi)^n$ into the constants gives
\begin{align*}
c_{n,k}\sum_{|\alpha|=k}\|g_\alpha\|_{L^2}^2 &\le \|[f]\|_{\dot{H}^k}^2 \le C_{n,k}\sum_{|\alpha|=k}\|g_\alpha\|_{L^2}^2.
\end{align*}
[/step]
