[proofplan]
The proof is a direct computation with Fourier multipliers on the dense subspace $\mathcal{S}(\mathbb{R}^n)$. Since the semiclassical Sobolev norm is defined by applying the multiplier $\langle hD\rangle^r$ and taking the $L^2$ norm, multiplication of the symbols gives an exact norm identity. The identity implies that $\Lambda_h^s$ is an isometry from $\mathcal{S}(\mathbb{R}^n)$ with the $H_h^r$ norm into $\mathcal{S}(\mathbb{R}^n)$ with the $H_h^{r-s}$ norm, so it extends uniquely by completion. Applying the same argument to $-s$ gives the inverse map, and the multiplier identities show that the two extensions compose to the identity.
[/proofplan]
[step:Compute the Sobolev norm after applying the order reduction multiplier]
Let
\begin{align*}
\Lambda_h^s: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)
\end{align*}
denote the Fourier multiplier defined by
\begin{align*}
\widehat{\Lambda_h^s u}(\xi) = \langle h\xi\rangle^s \hat{u}(\xi)
\end{align*}
for every $u \in \mathcal{S}(\mathbb{R}^n)$ and every $\xi \in \mathbb{R}^n$. By the definition of the semiclassical Sobolev norm,
\begin{align*}
\|\Lambda_h^s u\|_{H_h^{r-s}(\mathbb{R}^n)} = \|\langle hD\rangle^{r-s}\Lambda_h^s u\|_{L^2(\mathbb{R}^n)}.
\end{align*}
The Fourier symbol of $\langle hD\rangle^{r-s}\Lambda_h^s$ is
\begin{align*}
\langle h\xi\rangle^{r-s}\langle h\xi\rangle^s = \langle h\xi\rangle^r.
\end{align*}
Therefore
\begin{align*}
\|\Lambda_h^s u\|_{H_h^{r-s}(\mathbb{R}^n)} = \|\langle hD\rangle^r u\|_{L^2(\mathbb{R}^n)} = \|u\|_{H_h^r(\mathbb{R}^n)}.
\end{align*}
[guided]
We first prove the desired mapping property on the test space where all Fourier multipliers are defined pointwise without completion issues. Define
\begin{align*}
\Lambda_h^s: \mathcal{S}(\mathbb{R}^n) \to \mathcal{S}(\mathbb{R}^n)
\end{align*}
by the Fourier-side formula
\begin{align*}
\widehat{\Lambda_h^s u}(\xi) = \langle h\xi\rangle^s \hat{u}(\xi),
\end{align*}
where $u \in \mathcal{S}(\mathbb{R}^n)$ and $\xi \in \mathbb{R}^n$. The key point is that the Sobolev norm itself is defined by another Fourier multiplier. Namely,
\begin{align*}
\|v\|_{H_h^{r-s}(\mathbb{R}^n)} = \|\langle hD\rangle^{r-s}v\|_{L^2(\mathbb{R}^n)}
\end{align*}
for $v \in \mathcal{S}(\mathbb{R}^n)$. Taking $v = \Lambda_h^s u$, we get
\begin{align*}
\|\Lambda_h^s u\|_{H_h^{r-s}(\mathbb{R}^n)} = \|\langle hD\rangle^{r-s}\Lambda_h^s u\|_{L^2(\mathbb{R}^n)}.
\end{align*}
Why does this simplify exactly rather than merely up to a constant? Both operators are Fourier multipliers, so their composition has symbol equal to the product of the symbols. Thus the symbol of $\langle hD\rangle^{r-s}\Lambda_h^s$ is
\begin{align*}
\langle h\xi\rangle^{r-s}\langle h\xi\rangle^s = \langle h\xi\rangle^r.
\end{align*}
Hence
\begin{align*}
\langle hD\rangle^{r-s}\Lambda_h^s u = \langle hD\rangle^r u
\end{align*}
as tempered distributions, and in fact as Schwartz functions. Taking the $L^2(\mathbb{R}^n)$ norm gives
\begin{align*}
\|\Lambda_h^s u\|_{H_h^{r-s}(\mathbb{R}^n)} = \|\langle hD\rangle^r u\|_{L^2(\mathbb{R}^n)} = \|u\|_{H_h^r(\mathbb{R}^n)}.
\end{align*}
This is the whole order reduction mechanism: applying $\Lambda_h^s$ lowers the Sobolev order by exactly $s$, and the norm is preserved.
[/guided]
[/step]
[step:Extend the multiplier continuously from $\mathcal{S}(\mathbb{R}^n)$ to $H_h^r(\mathbb{R}^n)$]
The identity
\begin{align*}
\|\Lambda_h^s u\|_{H_h^{r-s}(\mathbb{R}^n)} = \|u\|_{H_h^r(\mathbb{R}^n)}
\end{align*}
for $u \in \mathcal{S}(\mathbb{R}^n)$ shows that $\Lambda_h^s$ is uniformly continuous from $\mathcal{S}(\mathbb{R}^n)$ equipped with the $H_h^r(\mathbb{R}^n)$ norm to $\mathcal{S}(\mathbb{R}^n)$ equipped with the $H_h^{r-s}(\mathbb{R}^n)$ norm. Since $H_h^r(\mathbb{R}^n)$ is the completion of $\mathcal{S}(\mathbb{R}^n)$ in the $H_h^r$ norm, there is a unique continuous extension
\begin{align*}
\Lambda_h^s: H_h^r(\mathbb{R}^n) \to H_h^{r-s}(\mathbb{R}^n).
\end{align*}
The same norm identity passes to the extension by continuity, so the extended map is an isometry.
[/step]
[step:Identify the inverse as the multiplier of opposite order]
Applying the preceding argument with $s$ replaced by $-s$ and $r$ replaced by $r-s$ gives a continuous isometry
\begin{align*}
\Lambda_h^{-s}: H_h^{r-s}(\mathbb{R}^n) \to H_h^r(\mathbb{R}^n).
\end{align*}
On $\mathcal{S}(\mathbb{R}^n)$, the Fourier symbols multiply as
\begin{align*}
\langle h\xi\rangle^s\langle h\xi\rangle^{-s} = 1.
\end{align*}
Thus
\begin{align*}
\Lambda_h^s\Lambda_h^{-s}u = u
\end{align*}
and
\begin{align*}
\Lambda_h^{-s}\Lambda_h^s u = u
\end{align*}
for every $u \in \mathcal{S}(\mathbb{R}^n)$. Since $\mathcal{S}(\mathbb{R}^n)$ is dense in the relevant semiclassical Sobolev spaces and both compositions are continuous, the identities extend to all of $H_h^{r-s}(\mathbb{R}^n)$ and $H_h^r(\mathbb{R}^n)$ respectively. Therefore $\Lambda_h^{-s}$ is the inverse of $\Lambda_h^s$ on the corresponding semiclassical Sobolev spaces.
[/step]
