Strong positivity of Neumann Laplacian

Question

There are many places in the literature where the positivity of some semigroups is treated. However I did not know anyone which states and proves the strong positivity even for the basic semigroups like the Neumann laplacian semigroup.

Here is a simplified mathematical problem:

$$ \begin{cases}\dfrac{\partial u}{\partial t}(t,x)-\Delta u(t,x)+u(t,x)=0, & (t,x)\in (0,T)\times\Omega \\ \dfrac{\partial u}{\partial\nu}(t,x)=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega \end{cases}, $$

where $\Omega\subset\mathbb{R}^N$ is an open, bounded set with smooth boundary, and $f\in L^{\infty}(\Omega)^+=\{g\in L^{\infty}(\Omega)\ |\ g(x)\geq 0,\ \text{a.e. on}\ \Omega\}$.

If we denote by $S(t)$ the semigroup generated by $-\Delta+I$ with Neumann b.c. on $L^2(\Omega)$ then $S(t)$ is a positive semigroup, i.e. $u(t,\cdot)=S(t)f\in L^{\infty}(\Omega)^+$ for any $t\in [0,T]$. See for example W. Arendt - Heat Kernels (Theorem 3.3.1) https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.020/arendt/downloads/internetseminar.pdf.

In some articles I found, without proof or references that if $f\in L^{\infty}(\Omega)^+$ and $f\neq 0$ then $u(t,\cdot)\in \text{int}(L^{\infty}(\Omega)^+)$, i.e. there is some constant $c(t,u_0)>0$ such that $u(t,x)>c(t,u_0)$ a.e. on $\Omega$. How can we prove that?

It looks like a parabolic Harnack-type inequality is needed here...

Answer 1

As other users have indicated in the comments, for sufficiently smooth domains one can get it by combining, for instance, elliptic regularity with Hopf's boundary point lemma (and then go from the elliptic to the parabolic case by, for instance, a semigroup argument).

However, the same result remains true in much more general situations; for the Neumann Laplace, for instance, it suffices if the domain is connected and has the extension property for the Sobolev space $H^1$. Here is general operator theoretic argument that can be used in such situations:

Let $X$ and $Y$ be an $L^p$- and an $L^q$-space or, more generally, Banach lattices and let $T: X \to Y$ be a bounded linear operator which is positive in the sense that $Tf \ge 0$ whenever $f \ge 0$.

1. A vector $0 \le f \in X$ is called a quasi-interior point of $X_+$ if for every non-zero functional $0 \le \varphi \in X'$ one has $\langle \varphi, f \rangle > 0$.

2. Example: If $0 \le f \in X = L^p(\Omega,\mu)$ for a $\sigma$-finite measure space $(\Omega,\mu)$ then the following holds: (a) if $1 \le p < \infty$, then $f$ is quasi-interior point if and only if $f(\omega) > 0$ for almost all $\omega \in \Omega$. (b) If $p=\infty$, then $f$ is a quasi-interior point if and only if $f \ge \varepsilon 1$ for some $\varepsilon$. This distinction between the cases $p=\infty$ and $p < \infty$ will be very important at the end of our argument.

3. Proposition. Assume that there exists $0 \le g \in X$ such that $Tg$ is a quasi interior point of $Y_+$. Then for every quasi-interior point $f$ of $X_+$ the vector $Tf$ is a quasi-interior point of $Y_+$.

4. Proof of the proposition. This follows easily from a duality argument, since the assumption on $Tg$ implies that the dual operator $T'$ does not have any non-zero positive functionals in its kernel. For more details see for instance (warning: self-promotion ahead!) Proposition 2.21 in this article by Martin Weber and myself (there the notion almost interior point is used; in Banach lattices this coincides with the notion quasi-interior point).

5. How to apply it to the heat equation with Neumann boundary conditions? The solution semigroup $S$ is irreducible and analytic on $L^2$. Hence, it even has the property that $S(t)f$ is a quasi-interior point of $L^2$ for every $t > 0$ and every non-zero function $f \ge 0$; see for instance Theorem C-III-3.2(b) in this classical book on positive semigroups. Moreover, as we assumed $\Omega$ to have the extension property, the Neumann semigroup $S$ is ultracontractive, so $S(t)$ maps $L^2$ into $L^\infty$ for each $t > 0$. Moreover, the constant function $1$, which is a quasi-interior point in the positive cone of $L^\infty$, is a fixed point of the semigroup, i.e., $S(t) 1 = 1$ for all times $t$. Hence, we can now apply the proposition: for every $t > 0$ and every non-zero $0 \le f \in L^2$ the vector $S(t/2)f$ is a quasi-interior point of $L^2$, so by the proposition $S(t)f = S(t/2)S(t/2)f$ is a quasi-initerior point in $L^\infty$.

6. Generalizations: A similar type of argument can be used (if one combines it with some additional tools) for more general second order differential operators with rough coefficients on rough domains, and e.g. also with Robin boundary conditions - see (warning: self-promotion, again!) this article by Wolfgang Arendt, Tom ter Elst and myself.