$ (since the number of summands is finite and each summand tends to zero when $ h \to 0 $ due to the relation ).

Similarly we prove that the second sum in equation  tends to zero. As a result, it is proved that $ F_{1}(z; t, x, y) $ is differentiable with respect to $t$ (and also continuous) for each $z, x, y$ and satisfies the differential equation .

Now the inequalities $ 0\leqslant F_{1}(z; t, x, y) \leqslant 1 $, equation  and the boundedness of the operators $\mathscr{A}$ and $ \delta_0 $ in the space $l^{\infty}(\mathbb{Z}^d)$ (by the Schur’s lemma) lead [@Hartman82] to the uniform boundedness with respect to $z, x, y $ and continuity of the derivative with respect to $t$ of the function $F_{1}(z; t, x, y) $. And this results in uniform with respect to $z, x, y$ continuous differentiability of the function $ F_{1}(z; t, x, y)$ with respect to $t$.

\[Rem1\]Owing to the uniform with respect to $ y\in \mathbb{Z}^d $ continuous differentiability of the function $ F_{1}(z; t, x) $ with respect to $t$ (for each $ 0\leqslant z \leqslant \infty $), provided by Lemma \[L1\], equation  can be treated as the 