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](https://www.ams.org/journals/tran/1984-284-01/S0002-9947-1984-0742415-3/S0002-9947-1984-0742415-3.pdf)


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 question about positivity preserving property of semigroup of Laplacian
The first equality is known as the [Trotter product formula](https://en.wikipedia.org/wiki/Lie_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 $m \to \infty$.


