The following is a problem in a text on Critical Point Theory (in Portuguese) I am reading. I would like to know if my reasoning is correct.
One observation: the Ekeland Variational Principle cited in the statement of the problem is only used to ascertain that a Palais-Smale sequence exists.
The problem:
Use the Ekeland Variational Principle to solve the following sublinear problem: $$ (P) \quad \begin{cases} - \Delta u + u = h(x)|u|^{q - 2}u \quad \text{ in } \Bbb{R}^N, \\ u \in H^1(\Bbb{R}^N), \end{cases} $$ where $1 < q < 2$, $h \in L^{\frac{2^*}{2^* - q}}(\Bbb{R}^N) \cap L^\infty(\Bbb{R}^N)$, $h \geq 0$, $h \neq 0$.
My solution: The weak solutions for the problem $(P)$ are the critical points of the functional $$ I(u) = \frac12 ||u||^2 - \frac{1}{q} \int_{\Bbb{R}^N} h(x) |u|^q \ dx, $$ where $||u|| = (||\nabla u||_2^2 + ||u||_2^2)^{1/2}$ is the norm in $H^1 (\Bbb{R}^N)$. Also, $I$ is coercive and weakly lower semicontinuous, hence bounded from below.
Consider a Palais-Smale sequence at the level of the infimum of the functional, that is, $(u_n) \subset H^1 (\Bbb{R}^N)$ such that $$ I(u_n) \to c = \inf_{H^1(\Bbb{R}^N)}I, \quad I'(u_n) \to 0. $$ Such a sequence exists by the Ekeland Variational Principle. Then $(u_n)$ is bounded, up to a subsequence. Otherwise we would have $I(u_n) \to \infty$, a contradiction. Therefore, since $H^1 (\Bbb{R}^N)$ is reflexive, there exists $u \in H^1 (\Bbb{R}^N)$ such that $u_n \rightharpoonup u$, up to a subsequence. Let us show that $u$ is a critical point.
Let $\phi \in C_c^\infty (\Bbb{R}^N)$. Then, by the compact Sobolev embeddings, $$ u_n|_{\text{supp} \phi} \to u|_{\text{supp} \phi} \quad \text{ in } L^s(\Bbb{R}^N), $$ for $s \in [1, 2^*)$, up to a subsequence. By Vainberg's Theorem there is a subsequence, again denoted $(u_n)$, and $g \in L^s(\text{supp} \phi)$ such that $$ u_n(x) \to u(x) \quad \text{ a.e. in } \text{supp} \phi $$ and $$ |u_n(x)| \leq g(x) \quad \text{ a.e. in } \text{supp} \phi. $$ Therefore, $$ h(x)|u_n(x)|^{q - 2}u_n(x) \phi(x) \to h(x)|u(x)|^{q - 2}u(x) \phi(x) \quad \text{ a.e. in } \Bbb{R}^N $$ and $$ h(x) |u_n(x)|^{q - 2}u_n(x) \phi(x) \leq h(x) g^{q - 1}(x) \phi(x) \quad \text{ a.e. in } \Bbb{R}^N. $$ Note that $hg^{q - 1}\phi \in L^1 (\Bbb{R}^N)$. By the Dominated Convergence Theorem, $$ \int_{\Bbb{R}^N} h(x) |u_n|^{q - 2}u_n \phi \ dx \to \int_{\Bbb{R}^N} h(x) |u|^{q - 2}u \phi \ dx. $$ Also, note that $$ v \mapsto \int_{\Bbb{R}^N} \nabla v \cdot \nabla \phi \ dx + \int_{\Bbb{R}^N} v \phi \ dx $$ is a continuous linear functional on $W^{1, p}(\Bbb{R}^N)$. Therefore, by the weak convergence, $$ \int_{\Bbb{R}^N} \nabla u_n \cdot \nabla \phi \ dx + \int_{\Bbb{R}^N} u_n \phi \ dx \to \int_{\Bbb{R}^N} \nabla u \cdot \nabla \phi \ dx + \int_{\Bbb{R}^N} u \phi \ dx. $$
Since $I'(u_n) \to 0$, it follows that $$ 0 = \lim I(u_n) \phi = I'(u)\phi. $$This holds for all $\phi \in C_c^\infty (\Bbb{R}^N)$. By density, $$ I'(u)v = 0 \quad \forall v \in H^1(\Bbb{R}^N), $$ and therefore $u$ is a weak solution of $(P)$.
Let $\phi \in C_c^\infty (\Bbb{R}^N)$, $\phi > 0$. Then $$ I(t \psi) = \frac{t^2}{2}||\psi||^2 - \frac{t^q}{q} \int_{\Bbb{R}^N} h(x)|\psi|^q \ dx \leq 0 $$ it $t$ is sufficiently small. Since $I(u) \leq I(t \psi)$, $u$ is not the trivial solution.
My question:
Is the above correct?
Thanks in advance and kind regards.