[proofplan]
The proof is a pointwise comparison of the Fourier weights $\langle h\xi\rangle^s$ and $\langle \xi\rangle^s$. First we bound the positive ratio $\langle h\xi\rangle/\langle \xi\rangle$ above and below by constants depending only on $h$. Raising this ratio to the real power $s$ gives the correct constants, including when $s < 0$ because the inequalities reverse under negative powers. Multiplying these pointwise bounds by $|\langle \xi\rangle^s\hat u(\xi)|$ and taking the $L^2$ norm gives the desired equivalence.
[/proofplan]
[step:Compare the Japanese brackets pointwise]
For $\xi \in \mathbb{R}^n$, define the positive scalar ratio $r_h: \mathbb{R}^n \to (0,\infty)$ by
\begin{align*}
r_h(\xi) := \frac{\langle h\xi\rangle}{\langle \xi\rangle}.
\end{align*}
Then
\begin{align*}
r_h(\xi)^2
=
\frac{1 + h^2|\xi|^2}{1 + |\xi|^2}.
\end{align*}
Since $|\xi|^2 \ge 0$, this quotient lies between $\min\{1,h^2\}$ and $\max\{1,h^2\}$. Taking positive square roots gives
\begin{align*}
\min\{1,h\}
\le
r_h(\xi)
\le
\max\{1,h\}.
\end{align*}
[guided]
The only quantity that matters is the ratio between the two Fourier weights. We therefore define
\begin{align*}
r_h: \mathbb{R}^n \to (0,\infty), \qquad
\xi \mapsto \frac{\langle h\xi\rangle}{\langle \xi\rangle}.
\end{align*}
Using the definition $\langle \xi\rangle = (1 + |\xi|^2)^{1/2}$, we compute
\begin{align*}
r_h(\xi)^2
=
\frac{\langle h\xi\rangle^2}{\langle \xi\rangle^2}
=
\frac{1 + h^2|\xi|^2}{1 + |\xi|^2}.
\end{align*}
This is a weighted average of $1$ and $h^2$ with non-negative weights $1$ and $|\xi|^2$. Hence it cannot be smaller than $\min\{1,h^2\}$ and cannot be larger than $\max\{1,h^2\}$. Because all quantities are positive, taking square roots preserves the inequalities:
\begin{align*}
\min\{1,h\}
\le
\frac{\langle h\xi\rangle}{\langle \xi\rangle}
\le
\max\{1,h\}.
\end{align*}
This is the whole mechanism behind the norm equivalence: for fixed $h > 0$, the two weights differ only by a bounded positive multiplier.
[/guided]
[/step]
[step:Raise the multiplier ratio to the real order $s$]
For every $\xi \in \mathbb{R}^n$,
\begin{align*}
\frac{\langle h\xi\rangle^s}{\langle \xi\rangle^s}
=
r_h(\xi)^s.
\end{align*}
From the preceding step and the monotonicity of $t \mapsto t^s$ on $(0,\infty)$ when $s \ge 0$, and its order-reversing monotonicity when $s < 0$, we obtain the uniform pointwise bound
\begin{align*}
\min\{1,h^s\}
\le
\frac{\langle h\xi\rangle^s}{\langle \xi\rangle^s}
\le
\max\{1,h^s\}.
\end{align*}
[/step]
[step:Transfer the pointwise bounds to the Sobolev norms]
Let $u \in H^s(\mathbb{R}^n)$. By the definition of $H^s(\mathbb{R}^n)$, the function
\begin{align*}
\xi \mapsto \langle \xi\rangle^s \hat u(\xi)
\end{align*}
belongs to $L^2(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n), \mathcal{L}^n)$, where $\mathcal{L}^n$ is the $n$-dimensional Lebesgue measure specified in the theorem statement. Multiplying the pointwise estimate from the previous step by $|\langle \xi\rangle^s \hat u(\xi)|$ gives, for $\mathcal{L}^n$-a.e. $\xi \in \mathbb{R}^n$,
\begin{align*}
\min\{1,h^s\}\,|\langle \xi\rangle^s \hat u(\xi)|
\le
|\langle h\xi\rangle^s \hat u(\xi)|
\le
\max\{1,h^s\}\,|\langle \xi\rangle^s \hat u(\xi)|.
\end{align*}
Taking the $L^2(\mathbb{R}^n)$ norm with respect to $\mathcal{L}^n$ gives
\begin{align*}
\min\{1,h^s\}\,\|\langle \xi\rangle^s \hat u(\xi)\|_{L^2(\mathbb{R}^n)}
\le
\|\langle h\xi\rangle^s \hat u(\xi)\|_{L^2(\mathbb{R}^n)}
\le
\max\{1,h^s\}\,\|\langle \xi\rangle^s \hat u(\xi)\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Using the definitions of the two norms, this is exactly
\begin{align*}
\min\{1,h^s\}\,\|u\|_{H^s(\mathbb{R}^n)} \le \|u\|_{H^s_h(\mathbb{R}^n)} \le \max\{1,h^s\}\,\|u\|_{H^s(\mathbb{R}^n)}.
\end{align*}
Thus $\|\cdot\|_{H^s_h(\mathbb{R}^n)}$ and $\|\cdot\|_{H^s(\mathbb{R}^n)}$ are equivalent norms on $H^s(\mathbb{R}^n)$.
[/step]
