$$u(t)=M_{\infty, 1}(t,\cdot),\quad v(t)=m_{1}(t,\cdot).$$ Then both functions, $u(t)$ and $v(t)$, treated as functions in $l^{\infty}(\mathbb{Z}^{d})$ satisfy the same initial condition $u(0)=v(0)=1$ or, what is the same, $u(0,\cdot)=v(0,\cdot)\equiv1$.

According to [@YarBRW:r Thm. 1.3.1] the function $v(t)$ is a solution of the following linear differential equation in $l^{\infty}(\mathbb{Z}^{d})$: $$\label{E188}
\frac{dx}{dt}=\mathscr{A}x +\Delta_{0}x=\mathscr{H}x,$$ whereas according to Theorem \[M:osn\] the function $u(t)$ is a solution of the following linear differential equation in $l^{\infty}(\mathbb{Z}^{d})$: $$\frac{dx}{dt}=\mathscr{A}^{*}x +\Delta_{0}x=\mathscr{H}^{*}x.$$

Since by assumption of this paper the linear operator $\mathscr{A}$ is symmetric, that is $\mathscr{A}=\mathscr{A}^{*}$, then the both functions, $u(t)$ and $v(t)$, satisfy the same linear differential equation  with bounded linear operator in the right-hand side, and the same initial conditions. Due to unique dependence of solutions of the differential equations on the initial condition, in this case $v(t)\equiv u(t)$ for all $t\ge0$ or, what is the same, $$M_{\infty, 1}(t,\cdot) \equiv u(t,\cdot)\equiv v(t,\cdot)\equiv m_{1}(t,\cdot),$$ which finalize the proof of .

\[RforwardKolmEq\]One of possible ways to prove the identity  is to obtain the forward Kolmogorov equation for $F_{\infty}(z;t,y)$, and then by standard arguments to derive the equation for the first moment. However, this way is technically quite cumbersome, so we preferred to give in Theorem \[M:osn\] a direct proof of the forward 