[proofplan]
The first estimate is a direct comparison between the semiclassical derivative $(hD_x)^\beta$ and the ordinary derivative $D_x^\beta$: the factor $h^{|\beta|}$ is uniformly bounded for $0<h\leq h_0$. The Fourier-side estimate follows from the scaling identity $\mathcal{F}_h a(\xi)=h^{-n/2}\mathcal{F}a(\xi/h)$ and the chain rule, since each $\xi$-derivative produces one factor of $h^{-1}$. After multiplying by $h^{n/2+|\beta|}$ and setting $\eta=\xi/h$, the estimate becomes exactly a Schwartz seminorm bound for the ordinary Fourier transform $\mathcal{F}a$.
[/proofplan]
[step:Compare semiclassical derivatives with ordinary Schwartz seminorms]
Fix multi-indices $\alpha,\beta\in\mathbb{N}_0^n$. Since $D_{x_j}=-i\partial_{x_j}$, the operators $D_x^\beta$ and $\partial_x^\beta$ differ only by a complex constant of modulus $1$. By definition of $(hD_x)^\beta$,
\begin{align*}
(hD_x)^\beta u_h=h^{|\beta|}D_x^\beta u_h.
\end{align*}
Therefore, for every $0<h\leq h_0$ and every $x\in\mathbb{R}^n$,
\begin{align*}
|x^\alpha(hD_x)^\beta u_h(x)|=h^{|\beta|}|x^\alpha D_x^\beta u_h(x)|.
\end{align*}
Since $0<h\leq h_0$, we have $h^{|\beta|}\leq h_0^{|\beta|}$. The uniformly Schwartz hypothesis applied to the pair $(\alpha,\beta)$ gives a finite constant
\begin{align*}
M_{\alpha\beta}:=\sup_{0<h\leq h_0}\sup_{x\in\mathbb{R}^n}|x^\alpha D_x^\beta u_h(x)|.
\end{align*}
Thus
\begin{align*}
\sup_{0<h\leq h_0}\sup_{x\in\mathbb{R}^n}|x^\alpha(hD_x)^\beta u_h(x)|\leq h_0^{|\beta|}M_{\alpha\beta}.
\end{align*}
Taking $C_{\alpha\beta}:=1+h_0^{|\beta|}M_{\alpha\beta}$ gives $C_{\alpha\beta}>0$ and proves the first assertion.
[guided]
Fix multi-indices $\alpha,\beta\in\mathbb{N}_0^n$. The only new feature in the expression $x^\alpha(hD_x)^\beta u_h(x)$ is the power of $h$ attached to the derivative. Since $D_{x_j}=-i\partial_{x_j}$, the operator $D_x^\beta$ differs from the usual partial derivative $\partial_x^\beta$ only by a complex phase of modulus $1$, so it has the same absolute-value estimates. By the definition of semiclassical differentiation,
\begin{align*}
(hD_x)^\beta u_h=h^{|\beta|}D_x^\beta u_h.
\end{align*}
Hence, for every $x\in\mathbb{R}^n$ and every $0<h\leq h_0$,
\begin{align*}
|x^\alpha(hD_x)^\beta u_h(x)|=h^{|\beta|}|x^\alpha D_x^\beta u_h(x)|.
\end{align*}
The point of the uniformly Schwartz assumption is that the ordinary seminorm on the right is bounded uniformly in $h$. Define
\begin{align*}
M_{\alpha\beta}:=\sup_{0<h\leq h_0}\sup_{x\in\mathbb{R}^n}|x^\alpha D_x^\beta u_h(x)|.
\end{align*}
This number is finite by hypothesis. Also $h^{|\beta|}\leq h_0^{|\beta|}$ because $0<h\leq h_0$. Combining these two bounds gives
\begin{align*}
\sup_{0<h\leq h_0}\sup_{x\in\mathbb{R}^n}|x^\alpha(hD_x)^\beta u_h(x)|\leq h_0^{|\beta|}M_{\alpha\beta}.
\end{align*}
Thus the desired estimate holds with the strictly positive constant $C_{\alpha\beta}:=1+h_0^{|\beta|}M_{\alpha\beta}$.
[/guided]
[/step]
[step:Differentiate the scaled Fourier transform]
Fix an integer $N\geq 0$ and a multi-index $\beta\in\mathbb{N}_0^n$. Define the scaling map $S_h:\mathbb{R}^n\to\mathbb{R}^n$ by
\begin{align*}
S_h(\xi)=\xi/h.
\end{align*}
For the ordinary Fourier variable $\eta\in\mathbb{R}^n$, write $D_{\eta_j}:=-i\partial_{\eta_j}$ and $D_\eta^\beta:=D_{\eta_1}^{\beta_1}\cdots D_{\eta_n}^{\beta_n}$. For the semiclassical Fourier variable $\xi\in\mathbb{R}^n$, write $D_{\xi_j}:=-i\partial_{\xi_j}$ and $D_\xi^\beta:=D_{\xi_1}^{\beta_1}\cdots D_{\xi_n}^{\beta_n}$.
The definition of $\mathcal{F}_h a$ gives
\begin{align*}
\mathcal{F}_h a(\xi)=h^{-n/2}(\mathcal{F}a)(S_h(\xi)).
\end{align*}
Applying the chain rule once for each derivative in the multi-index $\beta$ gives
\begin{align*}
D_\xi^\beta\mathcal{F}_h a(\xi)=h^{-n/2-|\beta|}(D_\eta^\beta\mathcal{F}a)(\xi/h).
\end{align*}
The constant powers of $-i$ in $D_\xi^\beta$ and $D_\eta^\beta$ agree on both sides, so no additional absolute-value factor appears.
[/step]
[step:Reduce the Fourier estimate to one ordinary Schwartz seminorm]
Using the identity from the previous step, for every $0<h\leq h_0$ and every $\xi\in\mathbb{R}^n$,
\begin{align*}
h^{n/2+|\beta|}\left(1+\frac{|\xi|}{h}\right)^N|D_\xi^\beta\mathcal{F}_h a(\xi)|=\left(1+\frac{|\xi|}{h}\right)^N|(D_\eta^\beta\mathcal{F}a)(\xi/h)|.
\end{align*}
Set $\eta:=\xi/h\in\mathbb{R}^n$. Since $h>0$, as $\xi$ ranges over $\mathbb{R}^n$, the variable $\eta=\xi/h$ also ranges over all of $\mathbb{R}^n$. Therefore
\begin{align*}
\sup_{0<h\leq h_0}\sup_{\xi\in\mathbb{R}^n}h^{n/2+|\beta|}\left(1+\frac{|\xi|}{h}\right)^N|D_\xi^\beta\mathcal{F}_h a(\xi)|=\sup_{\eta\in\mathbb{R}^n}(1+|\eta|)^N|(D_\eta^\beta\mathcal{F}a)(\eta)|.
\end{align*}
Because $a\in\mathcal{S}(\mathbb{R}^n)$, the standard Fourier transform automorphism theorem for Schwartz functions applies to $a$ and gives $\mathcal{F}a\in\mathcal{S}(\mathbb{R}^n)$. Hence the seminorm
\begin{align*}
M_{N\beta}:=\sup_{\eta\in\mathbb{R}^n}(1+|\eta|)^N|(D_\eta^\beta\mathcal{F}a)(\eta)|
\end{align*}
is finite. Define the strictly positive constant $C_{N\beta}:=1+M_{N\beta}$. Since $M_{N\beta}\leq C_{N\beta}$, this gives the required Fourier-side estimate.
[/step]
[step:Conclude both uniform estimates]
The first step proves the uniform estimate for $x^\alpha(hD_x)^\beta u_h$ with $C_{\alpha\beta}=1+h_0^{|\beta|}M_{\alpha\beta}$. The final step proves the Fourier-side estimate for the fixed Schwartz function $a$ with
\begin{align*}
C_{N\beta}=1+\sup_{\eta\in\mathbb{R}^n}(1+|\eta|)^N|(D_\eta^\beta\mathcal{F}a)(\eta)|.
\end{align*}
Both constants are finite, strictly positive, and independent of $h$, so the theorem follows.
[/step]
