Proof. First suppose that $\xi, \eta \in M$. In which case, for each $x, y \in M$,
\[\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}= \dpn{\wh{\xi x \eta}, \wh y}{L^2(M, \tau)}= \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}\]
Now suppose that $\xi \in M$, then for each $\eta \in RC(M)$, by the Cauchy-Schwarz inequality,
\begin{align*}|\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}|&\le \norm{L_\xi}_{B(L^2(M, \tau))}\norm{x}_{M} \norm{y}_{M} \cdot \norm{\eta}_{L^2(M, \tau)}\\ |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}|&\le \norm{x}_{M} \norm{\xi}_{M}\norm{y}_{M} \cdot \norm{\eta}_{L^2(M, \tau)}\end{align*}
so by density of $\wh M$ in $L^{2}(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}= \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in M$ and $\eta \in RC(M)$.
Finally, suppose that $\xi \in LC(M)$ and $\eta \in RC(M)$, then for each $z \in M$,
\begin{align*}\dpn{L_\xi z, \wh y}{L^2(M, \tau)}&= \dpn{\xi z, \wh y}{L^2(M, \tau)}= \dpn{\xi, \wh y z^*}{L^2(M, \tau)}\\&= \dpn{\xi, \wh y J\wh z}{L^2(M, \tau)}= \dpn{y^*\xi, J\wh z}{L^2(M, \tau)}= \dpn{\wh z, Jy^*\xi}{L^2(M, \tau)}\end{align*}
so
\begin{align*}\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}&= \dpn{L_\xi (x \eta), \wh y}{L^2(M, \tau)}= \dpn{x \eta, J y^* \xi}{L^2(M, \tau)}\\ |\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}|&\le \norm{x \eta}_{L^2(M, \tau)}\norm{y}_{M} \cdot \norm{\xi}_{L^2(M, \tau)}\\ |\dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}|&\le \norm{R_\eta}_{B(L^2(M, \tau))}\norm{x}_{M} \norm{y}_{M} \cdot \norm{\xi}_{L^2(M, \tau)}\end{align*}
By density of $\wh M$ in $L^{2}(M, \tau)$, $\dpn{L_\xi R_\eta \wh x, \wh y}{L^2(M, \tau)}= \dpn{R_\eta L_\xi \wh x, \wh y}{L^2(M, \tau)}$ for all $\xi \in LC(M)$ and $\eta \in RC(M)$.$\square$
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