A question about positivity preserving property of semigroup of Laplacian Sry this the second question from the following article, I am asking in this week.
At page 6 (126), 3th line, of the following article.
THE HEAT EQUATION WITH A SINGULAR POTENTIAL
the authors say by positivity preserving property of the semigroup $\{e^{t \Delta}\}$, we have:
$$ e^{\delta (\Delta+ V_n)}v_0 = \lim_{m \to \infty} (e^{\delta \Delta/m } e^{(\delta/m) V_n})^m v_0 \leq e^{\delta \lambda} e^{\delta \Delta} v_0$$
Here $\delta >0$, $V_n$ is a non-negative bounded function, $\| V_n\|_{\infty}\leq \lambda$ and $v_0$ is an initial value.
I can't understand this inequality. Why this limiting procedure is necessary and where the positivity preserving property is used?
 A: The first equality is known as the Trotter product formula.  There are some hypotheses to check, and one should pay attention to the mode of convergence, but at a formal level, it's what you'd expect.  I believe you can find details in Reed and Simon.
For the second, you may show that for any $t \ge 0$ and any integer $k \ge 0$, we have
$$(e^{t\Delta} e^{tV_n})^k v_0 \le e^{k t \lambda} e^{k t \Delta} v_0 \tag{*}.$$
Let $v_k$ denote the function on the left side.
First, note that $v_0$ is assumed to be nonnegative, and you can then check (using the positivity preserving of $e^{t\Delta}$) that each $v_k$ is also nonnegative.
Now you can prove (*) by induction.  The base case $k=0$ is trivial.  Suppose that (*) holds for $k$.  Since $v_k \ge 0$, we have $$e^{t V_n} v_k \le e^{t \lambda} v_k \le e^{(k+1)t \lambda} e^{kt\Delta} v_0$$ by the induction hypothesis.  So, by the positivity preserving of $e^{t \Delta}$, we have
$$\begin{align*}v_{k+1} &= e^{t \Delta} (e^{t V_n} v_k)\\ &\le e^{t\Delta} (e^{(k+1)t \lambda} e^{kt\Delta} v_0) \\ &= e^{(k+1)t \lambda} e^{t \Delta} e^{k t \Delta} v_0 \\ &= e^{(k+1)t \lambda} e^{(k+1)t \Delta} v_0 \end{align*}$$
as desired.
Now applying (*) with $k=m$ and $t = \delta/m$, we have
$$ (e^{\delta \Delta/m } e^{(\delta/m) V_n})^m v_0 \leq e^{\delta \lambda} e^{\delta \Delta} v_0$$
and now pass to the limit as $m \to \infty$.
