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Let $\delta > 0$ and define $$ H_\delta(x) = \prod_{j=1}^{N} \cosh(\delta x_j), \quad \forall x \in \mathbb{R}^N. $$ By straightforward calculations we get $\Delta H_{\delta} (x) = \delta^2 H_\delta(x)$. So, $$ \Delta H(x) + H(x) = (-\delta^2 + 1) H(x) \geq a_1H, $$ where $a_1 < 1$ is fixed and $0 \leq \delta \leq \delta_0 := \frac{(1 - a_1)^{1/2}}{2}.$

A paper I'm reading says that, using the function $H$, is easy to prove:

Theorem: Suppose that $u \in C^2(\mathbb{R}^N)$ and $u(x) \leq Ce^{\delta \sum_{i=1}^{N} |x_i|}$ on $\mathbb{R}^N$ for some constants $C > 0$ and $0 \leq \delta \leq \delta_0$ and $$ -\Delta u + u \leq 0 \text{ in }\mathbb{R}^N. $$ Then $u(x) \leq 0$ on $\mathbb{R}^N$.

My attempt: Notice that $a_1 H(x) > 0, $ for all $x \in \mathbb{R}^N$, since $\cosh(t) \geq 1, $ for all $t \in \mathbb{R}$. Then, letting $L u = -\Delta u + u,$ we get $$ L(u) \leq 0 < a_1 H \leq L(H), \text{ in }\mathbb{R}^N. $$ If we suppose that $u(x_0) > 0$ for some $x_0$, we get $u(x) > 0$ for all $x \in B_r(x_0)$ for some $r > 0$. At this point I dont't know what to do. Any help even with another kind of meaximum principle in $\mathbb{R}^N$ is welcome.

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  • $\begingroup$ Can't we just do the following: if $-\Delta u+u\le 0$ pointwise and $u>0$ on $B$, then also $\langle u, -\Delta u\rangle \le - \langle u, u\rangle$ on $L^2(B)$, but $u$ is a test function for the Neumann Laplacian on $L^2(B)$, which doesn't have negative spectrum. $\endgroup$
    – Christian Remling
    Commented Feb 14, 2023 at 17:58
  • $\begingroup$ The original problem is given with the operator $L(u) = - Delta u + a(x) u$, where $a$ is bounded, positive, locally Holder continuous and $\inf_{x \in \mathbb{R}^N} a(x) > 0$ With this hypothesis still works ? $\endgroup$
    – Lucas Linhares
    Commented Feb 14, 2023 at 18:11
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    $\begingroup$ Yes, because $-\Delta+a\ge -\Delta$ if $a\ge 0$. $\endgroup$
    – Christian Remling
    Commented Feb 14, 2023 at 19:18
  • $\begingroup$ I still didn't understand how the result comes from whats you said. What is a test function for the Neumann Laplacian on $L^2(B)$? $\endgroup$
    – Lucas Linhares
    Commented Feb 14, 2023 at 19:56
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    $\begingroup$ The standard proof uses the maximum principle applied to the function $u_\epsilon=u-\epsilon H$. Indeed $-\Delta u_\epsilon +u_\epsilon <0$ and $u_\epsilon (x) \to -\infty$ when $|x| \to \infty$ by the growth assumption on $u$ (maybe we need to adjust the $\delta$). Then $u_\epsilon$ has a maximum at a certain $x_0$ and $u_\epsilon (x_0)>0$ contradicts the equation, since $\Delta u_\epsilon (x_0) \leq 0$. Then $u_\epsilon \leq 0$ and one lets $\epsilon \to 0$. $\endgroup$
    – Giorgio Metafune
    Commented Feb 14, 2023 at 20:03

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