[proofplan] The proof uses only the definition of residual membership and the elementary multiplicative stability of symbol estimates. For each prescribed power $h^N$, write the residual symbol $r$ as $h^N r_N$ with $r_N$ still in the same symbol class. Leibniz' rule shows that multiplying $r_N$ by $a$ produces a symbol controlled by the product order function $m_1m_2$, so $ar$ has the same arbitrary $h^N$ gain. Since $N$ is arbitrary, $ar$ is residual in the product class. [/proofplan]

[step:Convert residual decay of $r$ into an arbitrary fixed power of $h$]Fix $N \in \mathbb{N}_0$. Since $r \in h^\infty S(m_1)$, by definition there exists a symbol \begin{align*} r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} with $r_N \in S(m_1)$ such that \begin{align*} r = h^N r_N. \end{align*} Therefore \begin{align*} ar = h^N a r_N. \end{align*} It remains to verify that $a r_N \in S(m_1m_2)$.[/step]

[guided]The phrase $r \in h^\infty S(m_1)$ means that $r$ gains every power of the semiclassical parameter $h$ while retaining the $S(m_1)$ symbol estimates after removing that power. Thus, once we choose a specific integer $N \in \mathbb{N}_0$, the definition gives a symbol \begin{align*} r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} such that $r_N \in S(m_1)$ and \begin{align*} r = h^N r_N. \end{align*} Multiplying by $a$ gives \begin{align*} ar = a h^N r_N = h^N a r_N. \end{align*} So the whole problem has been reduced to one ordinary symbol-class question: if $a \in S(m_2)$ and $r_N \in S(m_1)$, then their product should lie in $S(m_1m_2)$. Once that is proved, the displayed identity shows that $ar$ lies in $h^N S(m_1m_2)$ for the chosen $N$.[/guided]

[step:Use Leibniz' rule to place the product in $S(m_1m_2)$]Let $\alpha,\beta \in \mathbb{N}_0^n$ be multiindices. Since $a \in S(m_2)$ and $r_N \in S(m_1)$, for every pair of multiindices $\gamma,\eta \in \mathbb{N}_0^n$ there exist constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$, \begin{align*} |\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} m_2(x,\xi) \end{align*} and \begin{align*} |\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} m_1(x,\xi). \end{align*} By Leibniz' rule, \begin{align*} \partial_x^\alpha \partial_\xi^\beta(a r_N) = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} (\partial_x^\gamma \partial_\xi^\eta a)(\partial_x^{\alpha-\gamma} \partial_\xi^{\beta-\eta} r_N) \end{align*} Define the finite Leibniz constant \begin{align*} C_{\alpha,\beta,N} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} A_{\gamma,\eta} R_{\alpha-\gamma,\beta-\eta}. \end{align*} Then, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$, \begin{align*} |\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N} m_1(x,\xi)m_2(x,\xi). \end{align*} Thus $a r_N \in S(m_1m_2)$.[/step]

[guided]We now prove the ordinary product estimate that was isolated in the previous step. Fix multiindices $\alpha,\beta \in \mathbb{N}_0^n$. The definition of $a \in S(m_2)$ gives constants $A_{\gamma,\eta}>0$ for every $\gamma,\eta \in \mathbb{N}_0^n$ such that \begin{align*} |\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} m_2(x,\xi) \end{align*} for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$. Likewise, since $r_N \in S(m_1)$, there are constants $R_{\gamma,\eta}>0$ such that \begin{align*} |\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} m_1(x,\xi). \end{align*} Leibniz' rule distributes the $\alpha$ derivatives in $x$ and the $\beta$ derivatives in $\xi$ between the two factors: \begin{align*} \partial_x^\alpha \partial_\xi^\beta(a r_N) = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} (\partial_x^\gamma \partial_\xi^\eta a)(\partial_x^{\alpha-\gamma} \partial_\xi^{\beta-\eta} r_N). \end{align*} Define the finite Leibniz constant by \begin{align*} C_{\alpha,\beta,N} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma} \binom{\beta}{\eta} A_{\gamma,\eta} R_{\alpha-\gamma,\beta-\eta}. \end{align*} Taking absolute values and applying the two symbol estimates to each summand gives, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$, \begin{align*} |\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N} m_1(x,\xi)m_2(x,\xi). \end{align*} Because this estimate holds for every pair $\alpha,\beta$, it is exactly the defining estimate for $a r_N \in S(m_1m_2)$.[/guided]

[step:Conclude residual membership in the product class] For the fixed $N \in \mathbb{N}_0$, we have shown that \begin{align*} ar = h^N c_N \end{align*} with \begin{align*} c_N := a r_N \in S(m_1m_2). \end{align*} Therefore $ar \in h^N S(m_1m_2)$. Since $N \in \mathbb{N}_0$ was arbitrary, \begin{align*} ar \in \bigcap_{N=0}^{\infty} h^N S(m_1m_2) = h^\infty S(m_1m_2). \end{align*} [/step]

[step:Repeat the argument with the $S_\delta$ symbol estimates]Fix $\delta$ satisfying \begin{align*} 0 \leq \delta < \frac{1}{2}, \end{align*} and assume $r \in h^\infty S_\delta(m_1)$ and $a \in S_\delta(m_2)$. For each $N \in \mathbb{N}_0$, choose \begin{align*} r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} with $r_N \in S_\delta(m_1)$ and $r = h^N r_N$. The $S_\delta$ estimates give constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that \begin{align*} |\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)} m_2(x,\xi) \end{align*} and \begin{align*} |\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)} m_1(x,\xi). \end{align*} Applying Leibniz' rule to $\partial_x^\alpha \partial_\xi^\beta(a r_N)$, each summand has total derivative order $|\alpha|+|\beta|$, so the powers of $h$ combine to give exactly the factor \begin{align*} h^{-\delta(|\alpha|+|\beta|)}. \end{align*} Define \begin{align*} C_{\alpha,\beta,N,\delta} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma}\binom{\beta}{\eta} A_{\gamma,\eta}R_{\alpha-\gamma,\beta-\eta}. \end{align*} Therefore, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$, \begin{align*} |\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N,\delta} h^{-\delta(|\alpha|+|\beta|)} m_1(x,\xi)m_2(x,\xi). \end{align*} Hence $a r_N \in S_\delta(m_1m_2)$, and so \begin{align*} ar \in h^N S_\delta(m_1m_2). \end{align*} Since $N$ was arbitrary, \begin{align*} ar \in h^\infty S_\delta(m_1m_2). \end{align*} This proves both assertions.[/step]

[guided]The $S_\delta$ case follows the same structure, but we must track the derivative loss in powers of $h$. Fix $\delta$ with \begin{align*} 0 \leq \delta < \frac{1}{2}, \end{align*} and assume $r \in h^\infty S_\delta(m_1)$ and $a \in S_\delta(m_2)$. For a chosen $N \in \mathbb{N}_0$, residual membership gives \begin{align*} r_N:T^*\mathbb{R}^n \times (0,h_0] \to \mathbb{C} \end{align*} with $r_N \in S_\delta(m_1)$ and $r=h^N r_N$. For each pair of multiindices $\gamma,\eta \in \mathbb{N}_0^n$, the $S_\delta$ estimates provide constants $A_{\gamma,\eta}>0$ and $R_{\gamma,\eta}>0$ such that \begin{align*} |\partial_x^\gamma \partial_\xi^\eta a(x,\xi;h)| \leq A_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)}m_2(x,\xi) \end{align*} and \begin{align*} |\partial_x^\gamma \partial_\xi^\eta r_N(x,\xi;h)| \leq R_{\gamma,\eta} h^{-\delta(|\gamma|+|\eta|)}m_1(x,\xi). \end{align*} Apply Leibniz' rule to $\partial_x^\alpha\partial_\xi^\beta(a r_N)$. In the summand indexed by $\gamma \leq \alpha$ and $\eta \leq \beta$, the two derivative orders add as \begin{align*} |\gamma|+|\eta|+|\alpha-\gamma|+|\beta-\eta|=|\alpha|+|\beta|. \end{align*} Thus the two $h$-losses multiply to exactly $h^{-\delta(|\alpha|+|\beta|)}$, not to a worse power. Define the finite Leibniz constant \begin{align*} C_{\alpha,\beta,N,\delta} = \sum_{\gamma \leq \alpha} \sum_{\eta \leq \beta} \binom{\alpha}{\gamma}\binom{\beta}{\eta} A_{\gamma,\eta}R_{\alpha-\gamma,\beta-\eta}. \end{align*} Therefore, for all $(x,\xi,h) \in T^*\mathbb{R}^n \times (0,h_0]$, \begin{align*} |\partial_x^\alpha \partial_\xi^\beta(a r_N)(x,\xi;h)| \leq C_{\alpha,\beta,N,\delta} h^{-\delta(|\alpha|+|\beta|)}m_1(x,\xi)m_2(x,\xi). \end{align*} Hence $a r_N \in S_\delta(m_1m_2)$, so $ar=h^N a r_N \in h^N S_\delta(m_1m_2)$. Since $N$ was arbitrary, $ar \in h^\infty S_\delta(m_1m_2)$.[/guided]