Let $\Omega$ be a non-empty open bounded subset of $\mathbb{R}^N$ with $N\ge 3$. Let $f:\Omega\times\mathbb{R}\rightarrow\mathbb{R}$ be a measurable function such that
- $\exists a,b>0, \exists\alpha\in(0,\frac{N+2}{N-2}), \forall x \in\Omega, \forall t\ge0, f(x,t)\le a+bt^\alpha;$
- $\forall x\in\Omega, \forall t<0, f(x,t)=0.$
Let $\lambda<\lambda_1(\Omega)$, where $\lambda_1(\Omega)$ is the first eigenvalue of $-\Delta$ on $H^1_0(\Omega)$.
Suppose that $u\in H^1_0(\Omega)$ is such that: $$\forall\varphi\in H^1_0(\Omega), \int_\Omega\nabla u(x)\cdot \nabla \varphi(x)\operatorname{d}x=\lambda \int_\Omega u(x) \varphi(x)\operatorname{d}x+\int_\Omega f(x,u(x))\varphi(x)\operatorname{d}x$$
Then, by a bootstrap argument (see Nonlinear Analysis and Semilinear Elliptic Problem by Antonio Ambrosetti and Andrea Malchiodi, page 11, Theorem 1.16), we deduce that $u\in C^2(\Omega)$ and so also for almost every $x\in\Omega$ we have that: $$-\Delta u(x)=\lambda u(x)+f(x,u(x)).$$
I want to prove that $u\ge0$.
What i tried so far is the following.
First, $u\in C(\Omega)$, and so $D:=\{x\in\Omega\ |\ u(x)<0\}$ is an open subset of $\mathbb{R}^N$.
Now, suppose that $D\neq\emptyset$ to get a contradition. Then $\partial D\subset \{x\in\Omega\ |\ u(x)=0 \}\cup\partial\Omega$ and so I've thought that $v:=u|_D\in H^1_0(D)$. If so, then: $$\forall\varphi\in H^1_0(D), \int_\Omega\nabla v(x)\cdot \nabla \varphi(x)\operatorname{d}x=\lambda \int_\Omega v(x) \varphi(x)\operatorname{d}x+\int_\Omega f(x,v(x))\varphi(x)\operatorname{d}x \\ = \lambda \int_\Omega v(x) \varphi(x)\operatorname{d}x$$ and so $v$ is an eigenfunction relative to the eigenvalue $\lambda$ of the operator $-\Delta$ defined on $H^1_0(D)$. But by the characterization by the Rayleigh quotient of the first eigenvalue of $-\Delta$ on an open subset of $\mathbb{R}^N$, we have that $\lambda_1(\Omega)\le\lambda_1(D)$. However: $\lambda<\lambda_1(\Omega)\le\lambda_1(D)\le\lambda$, absurd. Then we have $D=\emptyset$ and the conclusion follows.
Problem: I've no idea how to prove that $v\in H^1_0(D)$.
In fact, it seems that this could be not the case, as we can guess by the (only, so far) answer to this related question.
Can we somehow fix the previous argument to get the conclusion? Or maybe can anyone provide another argument?
EDIT: the answer given by ktoi to the related question show that actually $v\in H^1_0(D)$ and then completes the argument given above.
I don't know how to prove $v\in H^1_0(\Omega$, but your hypothesis $\lambda<\lambda_1(\Omega)$ lends a solution in any case. Choosing the $H^1_0(\Omega)$ function $\varphi=\min(u,0)$ gives $\int_\Omega |D\varphi|^2=\lambda \int_\Omega \varphi^2$. From the variational principle, this forces either $\lambda=\lambda_1(\Omega)$ or $\varphi\equiv 0$.