# Strong maximum principle in entire space

Let $$n\geq 3$$, $$K$$ is a bounded function in $$\mathbb{R}^n$$. If $$u$$ is a nonegative solution but not equal to 0 of $$-\Delta u=Ku^{\frac{n+2}{n-2}}\quad \text{in }\,\mathbb{R}^n.$$ Can we use the strong maximum principle to deduce $$u$$ is positive in $$\mathbb{R}^n$$? I saw the strong maximum principle in the bounded domain, what about the entire space like $$\mathbb{R}^n$$?

• Isn't u=0 a nonnegative solution? Commented Aug 10, 2023 at 2:57
• sorry, here $u$ not equal to 0. Commented Aug 10, 2023 at 4:31
• Yes, the usual proof for subharmonic/superharmonic functions works in any open connected set and gives $u=0$ if u vanishes somewhere. Or else you can use the strong minimun principle in any ball. Commented Aug 10, 2023 at 6:54
• Can you give some details about the proof, thanks. Commented Aug 10, 2023 at 14:32
• I just use that a superharmonic function with a global miniminum is constant in a open connected set. No boundedness assumption on the set is needed. So, if your u is zero somewhere, since it is nonegative, it has a minimum and it is zero everywhere. Commented Aug 10, 2023 at 21:50

Yes, $$u$$ is strictly positive. Assume that $$F=\{x:u(x)=0\}$$ is non-empty. $$F$$ is clearly closed and I show that is open. Let $$x_0 \in F$$ and $$r>0$$ such that $$u-\Delta u=u(1+Ku^{\frac{4}{n-2}}) \geq 0$$ in $$B(x_0,r)$$. Such $$r$$ exists since $$K$$ is bounded. Applying the strong minimum principle in $$B(x_0,r)$$ to the operator $$I-\Delta$$, we see that $$u\equiv 0$$ in $$B(x_0,r)$$.
• how to guarantee that $(1+K u^{\frac{4}{n-2}}) \geq 0$ in $B\left(x_0, r\right)$? since $K$ may be always negative. Commented Aug 14, 2023 at 2:53
• Just use that $u(x) \to 0$ as $x \to x_0$. Commented Aug 14, 2023 at 7:54